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This article is cited in 7 scientific papers (total in 7 papers) Feynman checkers: towards algorithmic quantum theory M. B. Skopenkovabc, A. V. Ustinovad a Faculty of Mathematics, HSE University b Institute for Information Transmission Problems of the Russian Academy of Sciences (Kharkevich Institute) c King Abdullah University of Science and Technology, Thuwal, Saudi Arabia d Khabarovsk Division of the Institute for Applied Mathematics, Far Eastern Branch, Russian Academy of Sciences
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     1. IntroductionWe survey and develop the most elementary model of electron motion introduced by Feynman. In this game, a checker moves on a checkerboard by simple rules, and we count the turns (see Definition 2 and Figure 1). Feynman checkers can be viewed as a one-dimensional quantum walk, or an Ising model, or the count of Young diagrams of certain type.  Figure 1.The probability to find an electron in a small square around a given point (white depicts strong oscillations of the probability). Left: in the basic model from § 2 (cf. [52], Fig. 6). Middle: in the upgrade from § 3 for a smaller square side. Right: in continuum theory. For the latter, the relative probability density is shown. 1.1. MotivationThe simplest way to understand what is the model about is to consider the classical double-slit experiment (see Fig. 2). In this experiment, a (coherent) beam of electrons is directed towards a plate pierced by two parallel slits, and the part of the beam passing through the slits is observed on a screen behind the plate. If one of the slits is closed, then the beam illuminates a spot on the screen. If both slits are open, one would expect a larger spot, but in fact one observes a sequence of bright and dark bands (interferogram). This shows that electrons behave like a wave: waves travel through both slits, and the contributions of the two paths either amplify or cancel each other depending on the final phases. Further, if electrons are sent through the slits one at time, then single dots appear on the screen, as expected. Remarkably, however, the same interferogram with bright and dark bands emerges when the electrons are allowed to build up one by one. One cannot predict where a particular electron hits the screen; all we can do is to compute the probability to find the electron at a given place. The Feynman sum-over-paths (or path integral) method of computing such probabilities is to assign phases to all possible paths and to sum up the resulting waves (see [11], [12]). Feynman checkers (or Feynman checkerboard) is based on a particularly simple combinatorial rule for those phases in the case of an electron freely moving (or better jumping) along a line. 1.2. Background1.2.1. The beginningThe checkers model was invented by Feynman in the 1940s [42] and first published in 1965; see [12]. In Problem 2-6 there, a function (called kernel; see (2)) on a lattice of small step $\varepsilon$ was constructed and the following task was posed: “If the time interval is very long ($t_b-t_a\gg \hbar/mc^2$) and the average velocity is small ($x_b-x_a\ll c(t_b-t_a)$), show that the resulting kernel is approximately the same as that for a free particle (given in [12], (3-3)), except for a factor $\exp[(imc^2/\hbar)(t_b-t_a)]$.” Mathematically, this means that the kernel (divided by $2i\varepsilon\exp[(-imc^2/\hbar)(t_b- t_a)]$) is asymptotically equal to the free-particle kernel (24) (this is equation (3-3) from [12]) in the triple limit as time tends to infinity, whereas the average velocity and the lattice step tend to zero (see Table 1 and Fig. 3). Both scaling by the lattice step and letting it tend to zero were understood, for otherwise the ‘exceptional’ factor mentioned above would be different (see Example 4). We show that this assertion, although incorrect literally, holds under mild assumptions (see Corollary 5). Although the Feynman problem might seem self-evident for theoretical physicists, even the first step of a mathematical solution (disproving the assertion as stated) cannot be found in the literature. As usual, the difficulty is to prove the convergence rather than to guess the limit. Table 1.Expressions for the propagators of a particle freely moving in one space dimension and one time dimension. The meaning of the norm square of a propagator is the relative probability density to find the particle at a particular point, or alternatively, the charge density at the point
In 1972 Narlikar [36] discovered that the above kernel reproduces the spin-$1/2$ retarded propagator in the different limit as the lattice step tends to zero but time stays fixed (see Table 1, Figs. 4 and 1, and Corollary 6). In 1984 Jacobson and Schulman [23] repeated this derivation, applied the stationary phase method among other bright ideas, and found the probability of changing the direction of motion (cf. Theorem 6). The remarkable works of that period contain no mathematical proofs, but only approximate computations without estimates for the error. 1.2.2. Ising modelIn 1981 Gersch noticed that Feynman checkers can be viewed as a one-dimensional Ising model with imaginary temperature or edge weights (see § 2.2 and [18], [23], § 3). Imaginary values of these quantities are usual in physics (for instance, in quantum field theory or in alternating current networks). Due to imaginarity, the contributions of most configurations cancel each other, which makes the model highly non-trivial in spite of being one-dimensional. In particular, the model exhibits a phase transition (see Figs. 1 and 5). Surprisingly, the latter seems to have never been reported before. Phase transitions were studied only in more complicated one-dimensional Ising models (see [25], § III, and [35]), in spite of a known equivalent result, which we are going to discuss now (see Theorem 1, (B), and Corollary 3).  Figure 5.The distribution of the electron position $x$ at time $t=100$ in the natural units for the basic model from § 2 ((a), the dots). Its normalized logarithm ((b), the dots) and cumulative distribution function ((c), the dots). Their (weak) scaling limits as $t\to+\infty$ (curves). Figure (b) is also the graph of (minus the imaginary part of) the limiting free energy density in the Ising model. The non-analyticity of the curves reflects a phase transition. 1.2.3. Quantum walksIn 2001 Ambainis, Bach, Nayak, Vishwanath, and Watrous performed a breakthrough [1]. They studied Feynman checkers under the names of one-dimensional quantum walk and Hadamard walk; although those cleverly defined models were completely equivalent to Feynman’s simple one. They computed the large-time limit of the model (see Theorem 2). They discovered several striking properties which are in sharp contrast with both continuum theory and the classical random walks. First, the most probable average electron velocity in the model equals $1/\sqrt{2}$ of the speed of light, and the probability of exceeding this value is very small (see Fig. 5, (a), Fig. 1 to the left, and Theorem 1, (B)). Second, if an absorbing boundary is put immediately to the left of the starting position, then the probability that the electron is absorbed on the boundary is $2/\pi$. Third, if an additional absorbing boundary is put at $x>0$, the probability that the electron is absorbed to the left increases actually, approaching $1/\sqrt{2}$ in the limit as $x\to+\infty$. Recall that in the classical case both absorption probabilities are $1$. In addition, they found many combinatorial identities and expressed the above kernel through the values of Jacobi polynomials at a particular point (see Remark 3; cf. [48], § 2). Konno studied a biased quantum walk [29], [30], which is also essentially equivalent to Feynman checkers (see Remark 4). He found the distribution of the electron position in the (weak) large-time limit (see Fig. 5 and Theorem 1, (B)). This result was proved mathematically by Grimmett, Janson, and Scudo [20]. In their seminal paper [49] from 2012, Sunada and Tate found and proved a large-time asymptotic formula for the distribution (see Theorems 2–4). But even this result still could not solve the Feynman problem because the error estimate was not uniform in the lattice step. In 2018 Maeda et al. proved an estimate for the maximum of the distribution for large time ([33], Theorem 2.1). In the 2000s quantum walks were generalized to arbitrary graphs and applied to quantum algorithms (see Fig. 6 and [17] for an implementation). We refer to the surveys by Cantero, Grünbaum, Moral, and Velázquez [7], Konno [30], [31], Kempe [27], and Venegas-Andraca [51] for further details in this direction.  Figure 6.Implementation of the basic model from § 2 on a quantum computer using quantum-circuit language (a). The output is a random bit-string coding electron position $x$ at time $t=4$ (cf. Fig. 8). The strings 000, 010, 100, 110 code $x=4,2,0,-2$, respectively. The distribution of $x$ ((b), left) and a histogram for the quantum computer IBM Lima ((b), right) [45; § 19]. See [17] for details. 1.2.4. Lattice quantum field theoriesIn a more general context, this is a direction towards the creation of Minkowskian lattice quantum field theory, with both space and time being discrete [2]. In the 1970s Wegner and Wilson introduced lattice gauge theory as a computational tool for gauge theory describing all known interactions (except gravity); see [34] for a popular-science introduction. This culminated in determining the proton mass theoretically with error less than $2\%$ in a sense (this error bound was based on some additional assumptions). This theory is Euclidean in the sense that it involves imaginary time. Likewise, an asymptotic formula for the propagator for the (massless) Euclidean lattice Dirac equation (see [28], Theorem 4.3) played a crucial role in the continuum limit of the Ising model found by Chelkak and Smirnov [8]. Similarly, asymptotic formulae for the Minkowskian propagator (Theorems 2–5) can be useful for the missing Minkowskian lattice quantum field theory. Several authors argue that Feynman checkers have the advantage of no fermion doubling inherent in Euclidean lattice theories and avoid the Nielsen–Ninomiya no-go theorem [4], [15]. Several upgrades of Feynman checkers have been discussed. For instance, Gaveau and Schulman [16] (1989) and Ord [38] (1997) added an electromagnetic field to the model. That time they achieved neither exact charge conservation nor generalization to non-Abelian gauge fields; this is fixed in Definition 3. Another example was adding mass matrix by Jizba [24]. It was an old dream to incorporate also checker paths turning backwards in time or forming cycles (see [42], pp. 481–483, and [22]); this would mean creation of electron-positron pairs, celebrating a passage from quantum mechanics to quantum field theory. One looked for a combinatorial model reproducing the Feynman propagator, rather than the retarded one, in the continuum limit (see Table 1). Known constructions (such as the hopping expansion) did not lead to the Feynman propagator because of certain lattice artifacts (for example, the title of [39] is misleading: the Feynman propagator is not discussed there). In the massless case, a non-combinatorial construction of Feynman propagator on the lattice was provided by Bender, Mead, Milton, and Sharp in [2], § 9F, and [3], § IV. In Definition 6 the desired combinatorial construction is finally given. Another long-standing open problem is to generalize the model to the $4$-dimensional real world. In his Nobel prize lecture Feynman mentioned his own unsuccessful attempts. There are several recent approaches; for instance, by Foster and Jacobson from 2017 [15]. As written in [15], § 7.1, itself, their generalizations are not yet as simple and beautiful as the original two-dimensional model. 1.2.5. On physical and mathematical worksThe physics literature on the subject is quite extensive [51], and we cannot mention all remarkable works in this brief overview. Surprisingly, in the whole literature we have not found the conspicuous property of concentration of measure for lattice step tending to zero (see Corollary 7). Many papers are well written, insomuch that physical theorems and proofs there could be taken for mathematical ones, although some of the theorems are incorrect as stated (see subsection ‘Underwater rocks’ at the end of § 12). We are aware of just a few mathematical works on the subject, such as [20], [49], and [33]. 1.3. ContributionsWe solve mathematically a problem by Feynman from 1965, which was to prove that his model is consistent with the continuum one, namely, it reproduces the usual quantum-mechanical free-particle kernel for large time, small average velocity, and small lattice step (see Corollary 5). We compute large-time and small-lattice-step asymptotic formulae for the lattice propagator, which are uniform in the parameters of the model (see Theorems 2 and 5). For the first time we observe and prove the concentration of measure in the continuum limit: the average velocity of an electron emitted by a point source is close to the speed of light with high probability (see Corollary 7). These results can be interpreted as asymptotic properties of Young diagrams (see Corollary 2) and Jacobi polynomials (see Remark 3). All these results are proved mathematically for the first time. For their statements just Definition 2 is sufficient. In Definitions 3 and 6 we perform a coupling with lattice gauge theory and the second quantization of the model, promoting Feynman checkers to a full-fledged lattice quantum field theory. 1.4. Organization of the paper and further directionsFirst we give definitions and the precise statements of the results, and in the process we provide a zero-knowledge examples for basic concepts of quantum theory. These are precisely those examples that Feynman presented first in his own books: Feynman checkers (see § 3) is the first specific example in the whole book [12]. The thin-film reflection (see § 7) is the first example in [11]; see Figs. 10 and 11 there. Thus we hope that these examples could be enlightening to readers unfamiliar with quantum theory. We start with the simplest (and rough) particular case of the model and upgrade it step by step in each subsequent section. Before each upgrade, we summarize which physical question it addresses, which simplifying assumptions it resolves or imposes additionally, and which experimental or theoretical results it reproduces. Some upgrades (§§ 7–9) are just announced to be discussed in subsequent publications (for instance, in [46]). Our aim is $(1+1)$-dimensional lattice quantum electrodynamics (‘QED’) but the last step on this way (mentioned in § 10) is still undone. Open problems are collected in § 11. For easier navigation, we present the upgrades-dependence chart: Hopefully this is a promising path to make quantum field theory rigorous and algorithmic. An algorithmic quantum field theory would be one which, given an experimentally observable quantity and a number $\delta>0$, would provide a precise statement of an algorithm predicting a value for this quantity with accuracy $\delta$. (Surely, the predicted value need not agree with the experiment for $\delta$ less than the accuracy of the theory itself.) See Algorithm 1 for a toy example. This would be an extension of constructive quantum field theory (which is currently far from being algorithmic). Application of quantum theory to computer science is in mainstream now, but the opposite direction could provide benefits as well. (Re)thinking algorithmically is a way to make a subject available to non-specialists, as it is happening with, say, algebraic topology. The paper is written at the mathematical level of rigour, in the sense that all the definitions, conventions, and theorems (including corollaries, propositions, lemmas) should be understood literally. Theorems remain true, even if cut out from the text. The proofs of theorems use the statements but not the proofs of other theorems. Most statements are much less technical than the proofs; hence the proofs are kept in a separate section (§ 12) and long computations are kept in [45]. In the process of proof we give a zero-knowledge introduction to the main tools to study the model: combinatorial identities, the Fourier transform, the method of moments, the stationary phase method. Remarks are informal and usually not used elsewhere (hence skippable). Neither is the next outside the definitions, theorems, and proofs used formally. 2. Basic model (Hadamard walk)Question: what is the probability to find an electron in the square $(x,t)$ if it was emitted from $(0,0)$? Assumptions: no self-interaction, no creation of electron-positron pairs, fixed mass and lattice step, point source; the electron moves in a plane by ‘drifting uniformly along the $t$-axis’ or along a line (and then $t$ is time). Results: double-slit experiment (qualitative explanation), charge conservation, large-time limiting distribution. 2.1. Definition and examplesWe first give an informal definition of the model in the spirit of [11] and then a precise one. On an infinite checkerboard a checker moves to the diagonal-neighbouring squares, either upwards-right or upwards-left (see Fig. 7, (a)). To each path $s$ of the checker, assign a vector $a(s)$ as follows (see Fig. 7, (b)). Take a stopwatch that can time the checker as it moves. Initially the stopwatch hand points upwards. While the checker moves straight, the hand does not rotate, but each time when the checker changes the direction, the hand rotates through $90^\circ$ clockwise (independently of the direction the checker turns). The final direction of the hand is the direction of the required vector $a(s)$. The length of the vector is set to be $1/2^{(t-1)/2}$, where $t$ is the total number of moves (this is just a normalization to be interpreted below). For instance, for the path $s$ in Fig. 7 ((b), top) the vector is $a(s)=(0,-1/2)$. Denote by ${a}(x,t):=\sum_s {a}(s)$ the sum over all the checker paths from the square $(0,0)$ to the square $(x,t)$ starting with the upwards-right move. For instance, $a(1,3)=(0,-1/2)+(1/2,0)=(1/2,-1/2)$; see Fig. 7, (c). The square of the length of the vector $a(x,t)$ is called the probability to find an electron in the square $(x,t)$, if it was emitted from $(0,0)$ (see § 2.2 for a discussion of the terminology). The vector $a(x,t)$ itself is called an arrow (see [11], Fig. 6). Let us summarize this construction rigorously. Definition 1. A checker path is a finite sequence of integer points in the plane such that the vector from each point (except the last one) to the next one equals either $(1,1)$ or $(-1,1)$. A turn is a point on the path (not the first nor the last one) such that the vectors from this point to the next and the previous ones are orthogonal. The arrow is the complex number where we sum over all checker paths $s$ from $(0,0)$ to $(x,t)$ with the first step to $(1,1)$, and $\operatorname{turns}(s)$ is the number of turns in $s$. Hereafter, an empty sum is $0$ by definition. Denote Points (or squares) $(x,t)$ with even and odd $x+t$ are called black and white, respectively.  Figure 8.(a) The arrows $a(x,t)$ and the probabilities $P(x,t)$ for small $x$, $t$; the scale depends on the row. (b) The arrows $10a(x,t)$ for $t\leqslant50$. Cf. Fig. 6. Table 2.The arrows $a(x,t)$ for small $x$, $t$
Figure 8 and Table 2 depict the arrows $a(x,t)$ and the probabilities $P(x,t)$ for small $x$, $t$. Figure 9 depicts the graphs of $P(x,1000)$, $a_1(x,1000)$, and $a_2(x,1000)$ as functions in an even number $x$. We see that under variation of the final position $x$ at some fixed large time $t$, right after the peak the probability falls to very small, although still non-zero values. What is particularly interesting, is the unexpected position of the peak, far from $x=t$. In Fig. 1 to the left, the colour of a point $(x,t)$ with even $x+t$ indicates the value of $P(x,t)$. Notice that the sides of the visible angle are not the lines $t=\pm x$, as one could expect, but the lines $t=\pm \sqrt{2}\,x$ (see Theorem 1, (A)). 2.2. Physical interpretationLet us comment on the possible interpretations of the model and ensure that it captures paradoxical behaviour of electrons. There are two different physical interpretations (see Table 3) and a combinatorial one. Table 3.The physical interpretations of Feynman checkers
2.2.1. The standard interpretationHere the $x$- and $t$-coordinates are interpreted as the electron position and time respectively. Sometimes (for instance, in Example 1) we, a bit informally, interpret both as the position, and assume that the electron performs a ‘uniform classical motion’ along the $t$-axis. We work in the natural system of units, where the speed of light and the Plank and Boltzmann constants equal $1$. Thus the lines $x=\pm t$ represent a motion with the speed of light. Any checker path lies above both lines, that is, in the light cone, which means agreement with relativity: the speed of electron cannot exceed the speed of light. To think of $P(x,t)$ as a probability, consider the $t$-coordinate as fixed and the squares $(-t,t), (-t+2,t), \dots, (t,t)$ as all the possible outcomes of an experiment. For instance, the $t$th horizontal can be a screen detecting the electron. We shall see that all numbers $P(x,t)$ on the same horizontal add up to $1$ (Proposition 2) and thus can indeed be considered as probabilities. Notice that the probability to find the electron in a set $X\subset\mathbb{Z}$ is $P(X,t):=\sum_{x\in X}P(x,t)=\sum_{x\in X}|a(x,t)|^2$ rather than $\bigl|\sum_{x\in X}a(x,t)\bigr|^2$ (cf. [11], Fig. 50). In reality, one cannot measure the electron position exactly. A fundamental limitation is the electron reduced Compton wavelength $\lambda=1/m\approx 4\cdot 10^{-13}$ meters, where $m$ is the electron mass. Physically, the basic model approximates the continuum by a lattice of step exactly $\lambda$. But that is still a rough approximation: one needs even smaller step to prevent accumulation of approximation error at larger distances and times. For instance, Figs. 1 and 10 to the left show a finite-lattice-step effect (renormalization of the speed of light): the average velocity $x/t$ cannot exceed $1/\sqrt{2}$ of the speed of light with high probability. (An explanation in physical terms: lattice regularization cuts off distances smaller than the lattice step, hence small wavelengths, hence large momenta, and hence large velocities.) A more precise model is given in § 3: compare the plots in Fig. 1. As we shall see now, the model captures qualitatively the unbelievable behaviour of electrons. (For correct quantitative results like the exact shape of an interferogram, an upgrade involving a coherent source is required; see § 6.) The probability to find an electron in the square $(x,t)$ subject to absorption in a subset $B\subset\mathbb{Z}^2$ is defined analogously to $P(x,t)$, only summation is over the checker paths $s$ that have no common points with $B$ except possibly $(0,0)$ and $(x,t)$. The probability is denoted by $P(x,t, \text{ bypass } B)$. Informally, this means an additional outcome of the experiment: the electron is absorbed and does not reach the screen. In the following example, we view the two black squares $(\pm1,1)$ on the horizontal $t=1$ as two slits in a horizontal plate (cf. Fig. 2). Example 1 (double-slit experiment). Distinct paths cannot be viewed as ‘mutually exclusive’: Absorption can occasionally increase probabilities: The standard interpretation of Feynman checkers is also known as an Hadamard walk, or more generally, a 1-dimensional quantum walk or a quantum lattice gas. All of these are equivalent but lead to generalizations in distinct directions [51], [30], [52]. For instance, a unification of the upgrades from §§ 3–7 gives a general inhomogeneous quantum walk. The striking properties of quantum walks discussed in § 1.2 are stated precisely as follows:
$$
\begin{equation*}
\begin{aligned} \, \sum_{t=1}^{\infty}P\bigl(0,t,\text{ bypass }\{x=0\}\bigr)&= \frac{2}{\pi}\leqslant \sum_{t=1}^{\infty} P\bigl(0,t, \text{ bypass } \{x=0\}\cup \{x=n\}\bigr) \\ &\to \frac{1}{\sqrt{2}} \quad \text{as }n\to+\infty. \end{aligned}
\end{equation*}
\notag
$$
Recently Dmitriev [10] has found $\sum_{t=1}^{\infty}P(n,t, \text{ bypass } \{x=n\})$ for a few $n\ne 0$ (see Problem 6). For example, for $n=3$ we obtain $8/\pi-2$. Similar numbers appear in a simple random walk on $\mathbb{Z}^2$ (see [43], Table 2). 2.2.2. The spin-chain interpretationThere is a very different physical interpretation of the same model: a $1$-dimensional Ising model with imaginary temperature and fixed magnetization. Recall that a configuration in the Ising model is a sequence $\sigma=(\sigma_1,\dots,\sigma_t)$ of $\pm 1$ of fixed length. The magnetization and the energy of the configuration are $\sum_{k=1}^{t}\sigma_k/t$ and $H(\sigma)=\sum_{k=1}^{t-1}(1-\sigma_k\sigma_{k+1})$, respectively. The probability of the configuration is $e^{-\beta H(\sigma)}/Z(\beta)$, where the inverse temperature $\beta=1/T>0$ is a parameter and the partition function $Z(\beta):=\sum_\sigma e^{-\beta H(\sigma)}$ is a normalization factor. Additional restrictions on configurations $\sigma$ are usually imposed. Now, moving the checker along a path $s$, write ‘$+$’ for each upwards-right move, and ‘$-$’ for each upwards-left one; see Fig. 11 to the left. The resulting sequence of signs is a configuration in the Ising model, the number of turns in $s$ is one half of the configuration energy, and the ratio of the final $x$- and $t$-coordinates is the magnetization. Thus $a(x,t)=\sum_s a(s)$ coincides (up to a factor independent of $x$) with the partition function for the Ising model at the imaginary inverse temperature $\beta=i\pi/4$ under the fixed magnetization $x/t$:
$$
\begin{equation*}
a(x,t)=2^{(1-t)/2}i\sum_{\substack{(\sigma_1,\dots,\sigma_t)\in\{+1,-1\}^t: \\ \sigma_1=+1,\,\sum_{k=1}^{t-1}\sigma_k=x}} \exp\biggl(\frac{i\pi}{4}\sum_{k=1}^{t-1}(\sigma_k\sigma_{k+1}-1)\biggr).
\end{equation*}
\notag
$$
Notice a crucial difference of the resulting spin-chain interpretation from both the usual Ising model and the above standard interpretation. In the last two models, the magnetization $x/t$ and the average velocity $x/t$ were random variables; now the magnetization $x/t$ (not to be confused with an external magnetic field) is an external condition. The configuration space in the spin-chain interpretation consists of sequences of ‘$+$’ and ‘$-$’ signs with fixed numbers of pluses and minuses. Summation over the configurations with different $x$ or $t$ would make no sense: for example, recall that $P(X,t)=\sum_{x\in X}|a(x,t)|^2$ rather than $\bigl|\sum_{x\in X}a(x,t)\bigr|^2$. Varying the magnetization $x/t$, viewed as an external condition, we observe a phase transition: (the imaginary part of) the limiting free energy density $-\log a(x,t)/\beta t$ is non-analytic as $x/t$ passes over $\pm 1/\sqrt{2}$ (see Fig. 5, (b), and Corollary 3). A phase transition emerges as $t\to+\infty$. It is interesting to study other order parameters, for instance, the ‘probability’ $i\,a_2(x,t)/a(x,t)$ of equal signs at the endpoints of the spin chain (see Fig. 10 and Problems 4 and 5). These quantities are complex just because the temperature is imaginary. (A comment for experts: the phase transition is not related to the accumulation of zeros of the partition function in the plane of the complex parameter $\beta$ as in [25], § III, and [35]. In our situation $\beta=i\pi/4$ is fixed, the real parameter $x/t$ is variable, and the partition function $a(x,t)$ has no zeros at all; see [37], Theorem 1.) 2.2.3. The Young-diagram interpretationOur results also have a combinatorial interpretation. The number of steps (or inner corners) in a Young diagram with $w$ columns of heights $x_1,\dots,x_w$ is the number of elements in the set $\{x_1,\dots,x_w\}$; see Fig. 11, left. Then the quantity $2^{(h+w-1)/2}a_1(h-w,h+w)$ is the difference between the number of Young diagrams with an odd and an even number of steps that have exactly $w$ columns and $h$ rows. An interesting behaviour starts already for $h=w$ (see Proposition 4). For even $h=w$ the difference vanishes. For odd $h=w=2n+1$ it is $(-1)^n\binom{2n}{n}$. Such $4$-periodicity persists roughly for $h$ close to $w$ (see [44], Theorem 2). For fixed half-perimeter $h+w$, the difference oscillates slowly as $h/w$ increases, attains a peak at $h/w\approx 3+2\sqrt{2}$, and then drops sharply to very small values (see Corollary 2 and Theorems 2–4). Similarly, $2^{(h+w-1)/2}a_2(h-w,h+w)$ is the difference between the numbers of Young diagrams with an odd and an even number of steps that have exactly $w$ columns and fewer than $h$ rows. The behaviour is similar. The upgrade in § 3 is related to Stanley character polynomials (see [48], § 2). 2.2.4. A discussion of the definitionNow compare Definition 1 with the ones in the literature. The notation ‘$a$’ comes from ‘arrow’ and ‘probability amplitude’; other names are ‘wavefunction’, ‘kernel’, ‘the Green function’, ‘propagator’. More traditional notations are ‘$\psi$’, ‘$K$’, ‘$G$’, ‘$\Delta$’, ‘$S$’, depending on the context. We prefer neutral notation, suitable for all contexts. The factor of $i$ and the minus sign in the definition are irrelevant (and absent from the original definition: see [12], Problem 2.6). They come from the ordinary starting direction and rotation direction of the stopwatch hand, and reduce the number of minus signs in what follows. The normalization factor $2^{(1-t)/2}$ can be explained by analogy with the classical random walk. If the checker were performing just a random walk, choosing one of the two possible directions at each step (after the obligatory first upwards-right move), then $|a(s)|^2=2^{1-t}$ would be the probability of the path $s$. This analogy should be taken with a grain of salt: in quantum theory the ‘probability of a path’ has absolutely no sense (recall Example 1). The reason is that the path is not something one can measure: a measurement of the electron position at one moment affects strongly the motion at all later moments. Conceptually, one should also fix the direction of the last move of the path $s$ (see [12], the bottom of p. 35). Luckily, this is not required (and thus not done) in the present paper, but this becomes crucial in further upgrades (see § 4 for an explanation). One could ask where does the definition come from. Following Feynman, we do not try to explain or ‘derive’ it physically. This quantum model cannot be obtained from a classical one using the standard Feynman sum-over-paths approach: there is simply no clear classical analogue of a spin $1/2$ particle (cf. § 4) and no true understanding of spin. ‘Derivations’ in [1], [4], and [36] appeal to much more complicated notions than the model itself. The true motivation for the model is its simplicity, consistency with basic principles (like probability conservation), and agreement with experiment (which means here the correct continuum limit; see Corollary 6). 2.3. Identities and asymptotic formulaeLet us state several well-known basic properties of the model. The proofs are given in § 12.1. First, the arrow coordinates $a_1(x,t)$ and $a_2(x,t)$ satisfy the following recurrence relation. This mimics the $(1+1)$-dimensional Dirac equation in the Weyl basis (see [41], (19.4) and (3.31))
$$
\begin{equation}
\begin{pmatrix} m &\dfrac{\partial}{\partial x}-\dfrac{\partial}{\partial t} \\ \dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial t} & m \end{pmatrix} \begin{pmatrix} a_2(x,t) \\ a_1(x,t) \end{pmatrix}=0,
\end{equation}
\tag{1}
$$
only the derivatives are replaced by finite differences, $m=1$, and the normalization factor $1/\sqrt{2}$ is added. For the upgrade in § 3 this analogy becomes transparent (see Remark 2). The Weyl basis is not unique, thus there are several forms of equation (1); cf. [23], (1). Dirac’s equation implies the conservation of probability. Proposition 2 (probability/charge conservation). For each integer $t\geqslant 1$ we get $\sum_{x\in\mathbb{Z}}P(x,t)=1$. For $a_1(x,t)$ and $a_2(x,t)$, there is an ‘explicit’ formula (more ones are given in § 13). Proposition 3 (‘explicit’ formula). For any integers $|x|<t$ such that $x+t$ is even we have The following proposition is a straightforward corollary of the explicit formula. Proposition 4 (particular values). For each integer $1\leqslant k\leqslant t-1$ the numbers ${a_1(-t+2k,t)}$ and $a_2(-t+2k,t)$ are the coefficients at $z^{t-k-1}$ and $z^{t-k}$ in the expansion of the polynomial $2^{(1-t)/2}(1+z)^{t-k-1}(1-z)^{k-1}$. In particular, In § 3.1 we give more identities. The sequences $2^{(t-1)/2}a_1(x,t)$ and $2^{(t-1)/2}a_2(x,t)$ are present in the on-line encyclopedia of integer sequences (see [47], A098593 and A104967). The following remarkable result was discovered in [1], § 4 (see Figs. 5 and 10), stated precisely and derived heuristically in [29], Theorem 1, and proved mathematically in [20], Theorem 1. See a short exposition of the latter proof in § 12.2 and generalizations in § 3.2. Theorem 1 (large-time limiting distribution; see Fig. 10). (A) For each $v\in\mathbb{R}$ we have (B) We have the following convergence in distribution as $t\to+\infty$:
$$
\begin{equation*}
tP(\lceil vt\rceil,t)\overset{d}\to F'(v)=\begin{cases} \dfrac{1}{\pi(1-v)\sqrt{1-2v^2}} & \textit{if}\ |v|< \dfrac{1}{\sqrt{2}}\,, \\ 0 & \textit{if}\ |v|\geqslant \dfrac{1}{\sqrt{2}}\,. \end{cases}
\end{equation*}
\notag
$$
(C) For each integer $r\geqslant 0$ we have Theorem 1, (B), demonstrates a phase transition in Feynman checkers if interpreted as an Ising model at imaginary temperature and fixed magnetization. Recall that then the magnetization $v$ is an external condition (rather than a random variable) and $P(\lceil vt\rceil,t)$ is the norm square of the partition function (rather than a probability). The distributional limit of $tP(\lceil vt\rceil,t)$ is discontinuous at $v=\pm 1/\sqrt{2}$, reflecting a phase transition (cf. Corollary 3). 3. Mass (biased quantum walk)Question: what is the probability to find an electron of mass $m$ in the square $(x,t)$, if it was emitted from $(0,0)$? Assumptions: the mass and the lattice step are now arbitrary. Results: analytic expressions for the probability for large time or a small lattice step, concentration of measure. 3.1. IdentitiesDefinition 2. Fix $\varepsilon>0$ and $m\geqslant 0$ called the lattice step and particle mass, respectively. Consider the lattice $\varepsilon\mathbb{Z}^2= \{(x,t)\colon x/\varepsilon, t/\varepsilon\in\mathbb{Z}\}$ (see Fig. 3). A checker path $s$ is a finite sequence of lattice points such that the vector from each point (except the last one) to the next one equals either $(\varepsilon,\varepsilon)$ or $(-\varepsilon,\varepsilon)$. Denote by $\operatorname{turns}(s)$ the number of points in $s$ (distinct from the first and last ones) such that the vectors from the point to the next and to the previous ones are orthogonal. For each $(x,t)\in\varepsilon\mathbb{Z}^2$, where $t>0$, denote by the sum over all checker paths $s$ on $\varepsilon\mathbb{Z}^2$ from $(0,0)$ to $(x,t)$ with the first step to $(\varepsilon,\varepsilon)$. Denote and One interprets $P(x,t,m,\varepsilon)$ as the probability to find an electron of mass $m$ in the square $\varepsilon\times\varepsilon$ with centre $(x,t)$ if the electron was emitted from the origin. Notice that the quantity $m\varepsilon$, hence $P(x,t,m,\varepsilon)$, is dimensionless in the natural units, where $\hbar=c=1$. In Fig. 1, the middle, the colour of a point $(x,t)$ shows the value of $P(x,t,1,0.5)$ (if $x+t$ is an integer). Table 4 shows $a(x,t,m,\varepsilon)$ for small $x$ and $t$. Recently I. Novikov proved elegantly that the probability does not vanish inside the light cone: $P(x,t,m,\varepsilon)\ne 0$ for $m>0$, $|x|<t$ and even $(x+t)/\varepsilon$ (see [37], Theorem 1). Table 4.The quantities $a(x,t,m,\varepsilon)$ for small $x$ and $t$
Example 2 (boundary values). We have $a(t,t,m,\varepsilon)=i(1+m^2\varepsilon^2)^{(1-t/\varepsilon)/2}$ and $a(2\varepsilon-t,t,m,\varepsilon) =m\varepsilon(1+m^2\varepsilon^2)^{(1-t/\varepsilon)/2}$ for each $t\in \varepsilon\mathbb{Z}$, $t>0$, and $a(x,t,m,\varepsilon)=0$ for $x>t$ or $x\leqslant -t$. Example 3 (massless and heavy particles). For each point $(x,t)\in \varepsilon\mathbb{Z}^2$, where $t>0$, we have and Let us list the known combinatorial properties of the model [51], [30]; see § 12.1 for simple proofs. Proposition 5 (Dirac’s equation). For each $(x,t)\in \varepsilon\mathbb{Z}^2$, where $t>0$, we have Remark 2. These equations reproduce Dirac’s equation (1) in the continuum limit $\varepsilon\to 0$: for $C^2$-functions $a_1,a_2\colon\mathbb{R}\times (0,+\infty)\to \mathbb{R}$ satisfying (3) and (4) on $\varepsilon\mathbb{Z}^2$, the left-hand side of (1) is ${O}_m\bigl(\varepsilon\cdot(\|a_1\|_{C^2}+\|a_2\|_{C^2})\bigr)$. Proposition 7 (Klein–Gordon equation). For each $(x,t)\in \varepsilon\mathbb{Z}^2$, where $t>\varepsilon$, we have This equation reproduces the Klein–Gordon equation
$$
\begin{equation*}
\frac{\partial^2 a}{\partial t^2}- \frac{\partial^2 a}{\partial x^2}+m^2a=0
\end{equation*}
\notag
$$
in the limit $\varepsilon\to 0$. Proposition 8 (symmetry). For each $(x,t)\in \varepsilon\mathbb{Z}^2$, where $t>0$, we have and Proposition 9 (Huygens’ principle). For all $x,t,t'\in\varepsilon\mathbb{Z}$, where $t>t'>0$, we have Informally, Huygens’ principle means that each black square $(x',t')$ on the horizontal with index $t'/\varepsilon$ acts like an independent point source, with amplitude and phase determined by $a(x',t',m, \varepsilon)$. Proposition 10 (equal-time recurrence relation). For each $(x,t)\in\varepsilon\mathbb{Z}^2$, where $t>0$, we have This enables us to compute $a_{1}(x,t)$ and $a_2(x,t)$ quickly on distant horizontals, starting from $x=2\varepsilon-t$ and $x=t$, respectively (see Example 2). Proposition 11 (‘explicit’ formula). For $(x,t)\in\varepsilon\mathbb{Z}^2$ with $|x|<t$ and $(x+t)/\varepsilon$ even, Remark 3. For each $|x|\geqslant t$ we have $a(x,t,m,\varepsilon)=0$ unless $0<t=x\in\varepsilon\mathbb{Z}$, which gives $a(t,t,m,\varepsilon)=(1+m^2\varepsilon^2)^{(1-t/\varepsilon)/2}i$. Beware that Proposition 11 is not applicable for $|x|\geqslant t$. By the definition of the Gauss hypergeometric function, we can rewrite the formulae as follows:
$$
\begin{equation*}
\begin{aligned} \, a_1(x,t,m,\varepsilon)&=m\varepsilon(1+m^2\varepsilon^2)^{(1-t/\varepsilon)/2} \cdot{}_2F_1\biggl(1-\frac{x+t}{2\varepsilon}\,,1+\frac{x-t}{2\varepsilon}\,;1; -m^2\varepsilon^2\biggr), \\ a_2(x,t,m,\varepsilon)&= m^2\varepsilon^2(1+m^2\varepsilon^2)^{(1-t/\varepsilon)/2}\cdot{}_2 F_1\biggl(2-\frac{x+t}{2\varepsilon}\,,1+\frac{x-t}{2\varepsilon}\,;2; -m^2\varepsilon^2\biggr) \\ &\qquad\times\biggl(1-\frac{x+t}{2\varepsilon}\biggr). \end{aligned}
\end{equation*}
\notag
$$
This gives a lot of identities. For instance, Gauss’s contiguous relations (see [19], 9.137) connect the values of $a(x,t,m,\varepsilon)$ at any three neighbouring lattice points; cf. Propositions 5 and 10. In terms of the Jacobi polynomials,
$$
\begin{equation*}
\begin{aligned} \, a_1(x,t,m,\varepsilon)&=m\varepsilon(1+m^2\varepsilon^2)^{(x/\varepsilon-1)/2} P_{(x+t)/(2\varepsilon)-1}^{(0,-x/\varepsilon)} \biggl(\frac{1-m^2\varepsilon^2}{1+m^2\varepsilon^2}\biggr), \\ a_2(x,t,m,\varepsilon)&= -m^2\varepsilon^2(1+m^2\varepsilon^2)^{(x/\varepsilon-3)/2} P_{(x+t)/(2\varepsilon)-2}^{(1,1-x/\varepsilon)} \biggl(\frac{1-m^2\varepsilon^2}{1+m^2\varepsilon^2}\biggr). \end{aligned}
\end{equation*}
\notag
$$
There is a similar expression through Kravchuk polynomials (cf. Proposition 4). In terms of Stanley character polynomials (defined in [48], § 2), Proposition 12 (Fourier integral). Set $\omega_p:=\dfrac{1}{\varepsilon} \arccos\dfrac{\cos(p\varepsilon)}{\sqrt{1+m^2\varepsilon^2}}$ . Then for each $m>0$ and $(x,t)\in \varepsilon\mathbb{Z}^2$ such that $t>0$ and $(x+t)/\varepsilon$ is even we have The Fourier integral represents a wave emitted by a point source as a superposition of waves of wavelength $2\pi/p$ and frequency $\omega_p$. Proposition 13 (full space-time Fourier transform). Denote $\delta_{x\varepsilon}:=1$ if $x=\varepsilon$, and $\delta_{x\varepsilon}:=0$ if $x\ne\varepsilon$. For each $m>0$ and $(x,t)\in \varepsilon\mathbb{Z}^2$ such that $t>0$ and $(x+t)/\varepsilon$ is even we get 3.2. Asymptotic formulae3.2.1. The large-time limitFor fixed $m$, $\varepsilon$, and large time $t$, the function $a(x,t,m,\varepsilon)$
 Figure 12.The graphs of $a_1(x,50,4,0.5)$ (left, the dots), its analytic approximation (left, the solid curves) given by Theorems 2 (a), 3 (b), and 4 (c), and their ratio (the plots on the right). The graph (c), left, shows the absolute value of this function on the logarithmic scale. We start by discussing regime (A). Let us state our main theorem: an analytic approximation of $a(x,t,m,\varepsilon)$, which is very accurate for $x/t$ between the peaks but not too close to them (see Fig. 12, (a)). Theorem 2 (large-time asymptotic formula between the peaks). For each $\delta>0$ there is $C_\delta>0$ such that for any $m,\varepsilon>0$ and $(x,t)\in\varepsilon\mathbb{Z}^2$ satisfying Hereafter the notation $f(x,t,m,\varepsilon)=g(x,t,m,\varepsilon)+ O_\delta\bigl(h(x,t,m,\varepsilon)\bigr)$ means that there is a constant $C(\delta)$ (depending on $\delta$ but not on $x$, $t$, $m$, $\varepsilon$) such that for any $x$, $t$, $m$, $\varepsilon$, and $\delta$ satisfying the assumptions of the theorem we have $|f(x,t,m,\varepsilon)-g(x,t,m,\varepsilon)|\leqslant C(\delta)h(x,t,m,\varepsilon)$. The main terms in Theorem 2 were computed originally in [1], Theorem 2, in the particular case when $m=\varepsilon=1$. The error terms were estimated in [49], Proposition 2.2; that estimate had the same order in $t$ but was not uniform in $m$ and $\varepsilon$ (and could not be uniform as it stands, otherwise one would get a contradiction with Corollary 6 as $\varepsilon\to 0$). Getting the uniform estimates (9) and (10) has required the most accurate version of the stationary phase method and additional Steps 3 and 4 of the proof (see § 12.3). Different approaches (Darboux’s asymptotic formula for Legendre polynomials and the Hardy–Littlewood circle method) were used in [6], Theorem 3, and [44], Theorem 2, to get (9) in the particular case when $x=0$ and get a rougher approximation for $x$ close to $0$, respectively. Theorem 2 has several interesting corollaries. First, it allows one to pass to the large-time distributional limit (see Fig. 10) and generalize Theorem 1. Corollary 1 (large-time limiting distribution). For any $m>0$ and $\varepsilon\leqslant 1/m$ we have Recall that all our results can be stated in terms of Young diagrams or Jacobi polynomials. Corollary 2 (steps of Young diagrams; see Fig. 11). Denote by $n_+(h\times w)$ and $n_-(h\times w)$ the numbers of Young diagrams with exactly $h$ rows and $w$ columns that have an even and an odd number of steps (defined in § 2.2.3), respectively. Then for almost every $r>1$ we have For the regimes (B) and (C) around and outside the peaks, we state two theorems by Sunada–Tate without discussing the proofs in detail. (To be precise, the main terms in the literature (see [1], Theorem 2, and [49], Propositions 2.2, 3.1, and 4.1) are slightly different from (9), (13), (17); but this difference is within the error term. In practice, the last three approximations are better by several orders of magnitude.) Regime (B) around the peaks is described by the Airy function (see Fig. 12, (b))
$$
\begin{equation*}
\operatorname{Ai}(\lambda):=\frac{1}{\pi}\int_{0}^{+\infty} \cos\biggl(\lambda p+\frac{p^{3}}{3}\biggr)\,dp.
\end{equation*}
\notag
$$
Theorem 3 (large-time asymptotic formula around the peaks; see [49], Proposition 3.1). For each $m,\varepsilon,\Delta>0$ and each sequence $(x_n,t_n)\in\varepsilon\mathbb{Z}^2$ satisfying We write $f(x_n,t_n,m,\varepsilon)=O_{m,\varepsilon,\{x_n\}}\bigl(g(n)\bigr)$ if there is a constant $C(m,\varepsilon,\{x_n\})$ (depending on $m$, $\varepsilon$, and the whole sequence $\{x_n\}_{n=1}^\infty$ but not on $n$) such that for each $n$ we have
$$
\begin{equation*}
|f(x_n,t_n,m,\varepsilon)|\leqslant C(m,\varepsilon,\{x_n\})g(n).
\end{equation*}
\notag
$$
Recently Zakorko has extended (13) and (14) to a larger neighbourhood of the peak (see [53]). Exponential decay outside the peaks was stated without proof in [1], Theorem 1. A proof appeared only a decade later, when the following asymptotic formula was established (see Fig. 12, (c)). Theorem 4 (large-time asymptotic formula outside the peaks; see [49], Proposition 4.1). For any $m,\varepsilon,\Delta>0$, $v\in (-1;1)$ and each sequence $(x_n,t_n)\in\varepsilon\mathbb{Z}^2$ satisfying Function (19) is positive and convex in $v$ in the region $1/\sqrt{1+m^2\varepsilon^2}<|v|<1$ (see [49], Theorem 1.4). Corollary 3 (limiting free energy density; see Fig. 5, (b)). For any $m,\varepsilon>0$, $v\in(-1;1)$, and $H(v,m,\varepsilon)$ given by (19) we have This means a phase transition in the Ising model in a strong sense: the limiting free energy density (with the imaginary part proportional to the left side) is not analytic in $v$. 3.2.2. The Feynman triple limitTheorem 2 allows us to pass to the limit as $(1/t,x/t,\varepsilon)\to 0$ as follows. Corollary 4 (simpler and rougher asymptotic formula). Under the assumptions of Theorem 2 we have Corollary 5 (Feynman triple limit; see Fig. 3). For each $m\geqslant 0$ and each sequence $(x_n,t_n,\varepsilon_n)$ such that $(x_n,t_n)\in \varepsilon_n\mathbb{Z}^2$, $(x_n+t_n)/\varepsilon_n$ is even, and we have the equivalence For equivalence (22), assumptions (21) are essential and sharp, as the next example shows. Example 4. Equivalence (22) holds for neither $(x_n,t_n,\varepsilon_n)=(n^3,n^4,1/n^4)$, nor $(0,4n^2,1/{2n})$, nor $(0,2n\varepsilon,\varepsilon)$, where $\varepsilon=\mathrm{const}<1/m$: the limit of the ratio of the left- and right-hand sides equals $e^{im/8}$, $e^{im^3/3}$ or does not exist, respectively, rather than equal $1$. (The non-existence of the latter limit confirms that letting $\varepsilon$ tend to $0$ was implicit in the Feynman problem.) Corollary 5 solves the Feynman problem (and moreover, corrects the statement by revealing the required sharp assumptions). The main difficulty here is that it concerns the triple rather than iterated limit. We are not aware of any approach which could solve the problem without proving the whole of Theorem 2. For example, Darboux’s asymptotic formula for the Jacobi polynomials (see Remark 3) is suitable for the iterated limit as $t\to+\infty$ first and then $\varepsilon\to 0$, giving a (weaker) result, which is already independent on $x$. Neither Darboux’s, nor the Mehler–Heine, nor more recent asymptotic formulae [32] are applicable when $1/m\varepsilon$ or $x/\varepsilon$ is unbounded. The same concerns Proposition 2.2 in [49] because the error estimate there is not uniform in $m$ and $\varepsilon$. Conversely, the next theorem is suitable for the iterated limit as $\varepsilon\to 0$ first and then $x/t\to 0$, then $t\to+\infty$, but not for the triple limit because the error term blows up as $t\to+\infty$. 3.2.3. The continuum limitThe limit as $\varepsilon\to 0$ is described by the Bessel functions of the first kind:
$$
\begin{equation*}
J_0(z):=\sum_{k=0}^\infty(-1)^k\,\frac{(z/2)^{2k}}{(k!)^2}\quad\text{and}\quad J_1(z):=\sum_{k=0}^\infty(-1)^k\,\frac{(z/2)^{2k+1}}{k!\,(k+1)!}\,.
\end{equation*}
\notag
$$
Theorem 5 (asymptotic formula in the continuum limit). For each $m,\varepsilon,\delta>0$ and $(x,t)\in\varepsilon\mathbb{Z}^2$ such that $(x+t)/\varepsilon$ is even, $t-|x|\geqslant\delta$, and $\varepsilon< \delta \,e^{-3ms}/16$, where $s:=\sqrt{t^2-x^2}$, we have Recall that $f(x,t,m,\varepsilon)=O\bigl(g(x,t,m,\varepsilon)\bigr)$ means that there is a constant $C$ (independent of $x$, $t$, $m$, and $\varepsilon$) such that for any $x$, $t$, $m$, and $\varepsilon$ satisfying the assumptions of the theorem we have $|f(x,t,m,\varepsilon)|\leqslant Cg(x,t,m,\varepsilon)$. The main term in Theorem 5 was computed in [36], § 1. Numerical experiment shows that the error term decreases faster than asserted (see Table 5 computed in [45], § 14). Table 5.The approximation of the spin-$1/2$ retarded propagator (26) by Feynman checkers ($m=10$, $\delta=0.2$, $t=1$)
In the next corollary, we approximate a fixed point $(x,t)$ in the plane by the lattice point $\bigl(2\varepsilon\lceil x/(2\varepsilon)\rceil, 2\varepsilon\lceil t/(2\varepsilon)\rceil\bigr)$ (see Fig. 4). The factors of $2$ make the latter accessible for the checker. Corollary 6 (uniform continuum limit; see Fig. 4). For each fixed $m\geqslant 0$ we have The proof of pointwise convergence is simpler and is presented in Appendix B (see § 14). This result, although expected, has not been found in the literature. An elementary proof is given in § 12.6. We observe a sharp contrast between the continuum and the large-time limit here: by Corollary 1, there is no concentration of measure as $t\to+\infty$ for fixed $\varepsilon$. 3.3. A physical interpretationLet us discuss the meaning of the Feynman triple limit and the continuum limit. In this subsection we omit some technical definitions not used in the sequel. The right-hand side of (22) is, up to the factor $e^{-imt_n}$, the free-particle kernel
$$
\begin{equation}
K(x,t)=\sqrt{\frac{m}{2\pi t}}\,\exp\biggl(\frac{i m x^2}{2t}- \frac{i\pi}{4}\biggr),
\end{equation}
\tag{24}
$$
describing the motion of a non-relativistic particle emitted from the origin along a line. Limit (23) reproduces the spin-$1/2$ retarded propagator describing the motion of an electron along a line. More precisely, the spin-$1/2$ retarded propagator, or the retarded Green function for Dirac’s equation (1) is a matrix-valued tempered distribution $G^{\rm R}(x,t)=(G^{\rm R}_{kl}(x,t))$ on $\mathbb{R}^2$ vanishing for $t<|x|$ and satisfying
$$
\begin{equation}
\begin{pmatrix} m & \dfrac{\partial}{\partial x}-\dfrac{\partial}{\partial t} \\ \dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial t} & m \end{pmatrix} \begin{pmatrix} G^{\rm R}_{11}(x,t) & G^{\rm R}_{12}(x,t) \\ G^{\rm R}_{21}(x,t) & G^{\rm R}_{22}(x,t) \end{pmatrix}=\delta(x)\delta(t)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},
\end{equation}
\tag{25}
$$
where $\delta(x)$ is the Dirac delta function. For $|x|<t$ the propagator is given by
$$
\begin{equation}
\begin{gathered} \, G^{\rm R}(x,t)=\frac{m}{2}\,\begin{pmatrix} J_0(m\sqrt{t^2-x^2}\,) & -\sqrt{\dfrac{t+x}{t-x}}\,J_1(m\sqrt{t^2-x^2}\,) \\ \sqrt{\dfrac{t-x}{t+x}}\,J_1(m\sqrt{t^2-x^2}\,) & J_0(m\sqrt{t^2-x^2}\,) \end{pmatrix} \end{gathered}
\end{equation}
\tag{26}
$$
(cf. [23], (13), and [41], (3.117)). In addition, $G^{\rm R}(x,t)$ involves a generalized function supported on the straight lines $t=\pm x$, not observed in the limit (23) and not indicated there. A more common expression is
$$
\begin{equation}
G^{\rm R}(x,t)=\frac{1}{4\pi^2}\int_{-\infty}^{+\infty}\, \int_{-\infty}^{+\infty}\lim_{\delta\to+0}\begin{pmatrix} m & -ip-i\omega \\ -ip+i\omega & m \end{pmatrix} \frac{ e^{i p x-i\omega t}\,dp\,d\omega}{m^2+p^2-(\omega+i\delta)^2}
\end{equation}
\tag{27}
$$
(cf. Proposition 13), where the limit is taken in the weak topology of matrix- valued tempered distributions and the integral is understood as the Fourier transform of tempered distributions (cf. [14], (6.47)). The ‘square’ $G^{\rm R}_{11}(x,t)^2+G^{\rm R}_{12}(x,t)^2$ of the propagator is ill defined (because the square of the Dirac delta-function supported on the lines $t=\pm x$ is involved). Thus the propagator lacks a probabilistic interpretation, and global charge conservation (Proposition 6) has no continuum analogue. For instance,
$$
\begin{equation*}
\int_{(-t,t)} (G^{\rm R}_{11}(x,t)^2+G^{\rm R}_{12}(x,t)^2)\,dx\approx \frac{m^2t}{2}\ne\operatorname{const}
\end{equation*}
\notag
$$
paradoxically. A physical explanation: the line $t=x$ carries an infinite charge flowing inside the angle $|x|<t$. One can interpret the ‘square’ of the propagator for $|x|\ne t$ as a relative probability density or charge density (see Fig. 1, right). In the spin-chain interpretation, the propagator is the limit of the partition function for the one-dimensional Ising model at the inverse temperature $\beta=i\pi/4-\log(m\varepsilon)/2$. Those are essentially the values of $\beta$ for which a phase transition is possible [35]. The normalization factor $1/(2\varepsilon)$ before $a$ in (23) can be explained as division by the length associated to one black lattice point on a horizontal line. At a deeper level it comes from the normalization of $G^{\rm R}(x,t)$ arising from (25). Theorem 5 is a toy result in algorithmic quantum field theory: it determines the lattice step needed to compute the propagator with prescribed accuracy. So far, this is not a big deal, because the propagator has an explicit analytic expression and is not really experimentally-measurable; neither the efficiency of the algorithm is taken into account. But this is the first step. Algorithm 1 (approximation algorithm for the spin-$1/2$ retarded propagator (26)). Input: mass $m>0$, coordinates $|x|<t$, accuracy level $\Delta$. Output: an approximate value $G_{kl}$ of $G_{kl}^{\rm R}(x,t)$ within distance $\Delta$ from the true value (26). Algorithm: compute
$$
\begin{equation*}
G_{kl}=\frac{(-1)^{(k-1)l}}{2\varepsilon}\,{a}_{(k+l)\,\mathrm{mod}\,2\,+\,1} \biggl(2\varepsilon\biggl\lceil\frac{(-1)^{(k-1)l}x} {2\varepsilon}\biggr\rceil,2\varepsilon\!\biggl\lceil \frac{t}{2\varepsilon}\biggr\rceil,{m},{\varepsilon}\biggr)
\end{equation*}
\notag
$$
by (2) for Here we have used an explicit estimate for the constant $C$ understood in the big-O notation in Theorem 5; it is easily extracted from the proof. The theorem and the estimate remain true, if $a(x,t,m,\varepsilon)$ with $(x,t)\in\varepsilon\mathbb{Z}^2$ is replaced by $a\bigl(2\varepsilon\lceil x/(2\varepsilon)\rceil, 2\varepsilon\lceil t/(2\varepsilon)\rceil,{m},{\varepsilon}\bigr)$ with arbitrary $(x,t)\in\mathbb{R}^2$. Remark 4. A general (homogeneous) one-dimensional quantum walk essentially reduces to the upgrade studied in this section. Namely, consider the equation 4. SpinQuestion: what is the probability to find a right electron at $(x,t)$, if a right electron was emitted from $(0,0)$? Assumptions: the electron chirality is now taken into account. Results: the probability of a chirality flip. A feature of the model is that the electron spin emerges naturally rather than is added artificially. It goes almost without saying to view an electron as being in one of the two states depending on the last-move direction: right-moving or left-moving (or just ‘right’ or ‘left’ for brevity). The probability to find a right electron in the square $(x,t)$ if a right electron was emitted from the square $(0,0)$ is the square of the length of the vector $\sum_s a(s)$, where the sum is only over those paths from $(0,0)$ to $(x,t)$ that both start and finish with an upwards-right move. The probability to find a left electron is defined analogously, only the sum is taken over those paths which start with an upwards-right move but finish with an upwards-left move. Clearly, these probabilities equal $a_2(x,t)^2$ and $a_1(x,t)^2$, respectively, because the last move is directed upwards-right if and only if the number of turns is even. These right and left electrons are exactly the $(1+1)$-dimensional analogue of chirality states for a spin-$1/2$ particle (see [41], § 19.1). Indeed, it is known that the components $a_2(x,t)$ and $a_1(x,t)$ in Dirac equation in the Weyl basis (1) are interpreted as wave functions of right- and left-handed particles, respectively. The relation to the motion direction becomes transparent for $m=0$: then a general solution of (1) is $(a_2(x,t),a_1(x,t))=(a_2(x-t,0),a_1(x+t,0))$; thus the maxima of $a_2(x,t)$ and $a_1(x,t)$ (if any) move right and left, respectively, as $t$ increases. Beware that in three or more dimensions, spin is not the motion direction and cannot be explained in non-quantum terms. This gives a more conceptual interpretation of the model: an experiment outcome is a pair whereas the final $t$-coordinate is fixed. The probabilities to find a right/left electron are the fundamental ones. In further upgrades, $a_1(x,t)$ and $a_2(x,t)$ become complex numbers and $P(x,t)$ should be defined to be $|{a}_1(x,t)|^2+|{a}_2(x,t)|^2$ rather than by the above formula $P(x,t)=|{a}(x,t)|^2=|{a}_1(x,t)+i{a}_2(x,t)|^2$, which is a coincidence.See Fig. 13 for an illustration and a comparison with the upgrade from § 5. The physical interpretation of the theorem is limited: in continuum theory the probability of a chirality flip (for an electron emitted by a point source) is ill-defined similarly to the ‘square’ of the propagator (see § 3.3). A related more reasonable quantity is studied in [23], p. 381 (cf. Problem 5). Recently Bogdanov has generalized this theorem to an arbitrary mass and an arbitrary lattice step (see Definition 2): if $0\leqslant m \varepsilon\leqslant 1$, then
$$
\begin{equation*}
\lim_{t\to+\infty,\,t\in\varepsilon\mathbb{Z}}\,\sum_{x\in\varepsilon\mathbb{Z}} a_1(x,t,m,\varepsilon)^2=\frac{m\varepsilon}{2\sqrt{1+m^2\varepsilon^2}}
\end{equation*}
\notag
$$
(see [6], Theorem 2). This has confirmed a conjecture by Gaidai-Turlov, Kovalev, and Lvov.
5. External field (inhomogeneous quantum walk)Question: what is the probability to find an electron at $(x,t)$ if it moves in a given electromagnetic field $u$? Assumptions: the electromagnetic field vanishes outside the $xt$-plane; it is not affected by the electron. Results: ‘spin precession’ in a magnetic field (a qualitative explanation), charge conservation. Another feature of the model is that an external electromagnetic field emerges naturally rather than is added artificially. We start with an informal definition, then give a precise one, and finally demonstrate exact charge conservation. In the basic model the stopwatch hand did not rotate as the checker moved straight. It suggests itself to modify the model by rotating the hand uniformly during the motion. This does not change the model significantly: since all the paths from the initial to final position have the same length, their vectors rotate through the same angle, without affecting probabilities. A more interesting modification is when the current rotation angle depends on the checker position. This is exactly what electromagnetic field does. In what follows the rotation angle assumes only the two values $0^\circ$ and $180^\circ$ for simplicity, meaning multiplication by $\pm 1$. Thus an electromagnetic field is viewed as a fixed assignment $u$ of numbers $+1$ and $-1$ to all the vertices of the squares. For instance, in Fig. 14 the field equals $-1$ at the top-right vertex of each square $(x,t)$ with both $x$ and $t$ even. We modify the definition of the vector $a(s)$ by reversing the direction each time when the checker passes through a vertex with the field $-1$. Denote the resulting vector by $a(s,u)$. Define $a(x,t,u)$ and $P(x,t,u)$ analogously to $a(x,t)$ and $P(x,t)$ by replacing $a(s)$ by $a(s,u)$ in the definition. For instance, if $u=+1$ identically, then $P(x,t,u)=P(x,t)$. Let us slightly rephrase this construction by making the relation to lattice gauge theory more transparent. We introduce an auxiliary grid with vertices at the centres of black squares (see Fig. 11 to the right). It is the graph where the checker actually moves. Definition 3. An auxiliary edge is a line segment joining nearest-neighbour integer points with even sums of the coordinates. Let $u$ be a map from the set of all auxiliary edges to $\{+1,-1\}$. Denote by the sum over all checker paths $s=(s_0,s_1,\dots,s_t)$ with $s_0=(0,0)$, $s_1=(1,1)$, and $s_t=(x,t)$. Set $P(x,t,u):=|{a}(x,t,u)|^2$. Define $a_1(x,t,u)$ and $a_2(x,t,u)$ analogously but add the condition $s_{t-1}=(x+1,t-1)$ or $s_{t-1}=(x-1,t-1)$, respectively. For half-integers $x$ and $t$ denote by $u(x,t)$ the value of $u$ on the auxiliary edge with midpoint $(x,t)$. Remark 5. Here $u$ is a fixed external classical field not affected by the electron. This definition is analogous to one of the first constructions of gauge theory due to Weyl, Fock, and London; it gives a coupling of Feynman checkers with the Wegner–Wilson $\mathbb{Z}/2\mathbb{Z}$ lattice gauge theory. In particular, it reproduces the correct spin $1$ for the electromagnetic field: a function defined on the set of edges is a discrete analogue of a vector field, that is, a spin-$1$ field. Although this way of coupling with gauge field is classical, it has never been explicitly applied to Feynman checkers (cf. [16], p. 36), and it is very different from both the approach of [38] and Feynman-diagram intuition [11]. For instance, the field $u$ in Fig. 14 has the form
$$
\begin{equation*}
u(s_1s_2)=\exp\biggl(-i\int_{s_1}^{s_2}(A_0\,dt+A_1\,dx)\biggr)
\end{equation*}
\notag
$$
for each auxiliary edge $s_1s_2$, where
$$
\begin{equation*}
(A_0,A_1):=\frac{\pi}{2}\biggl(x+\frac{1}{2},x+\frac{1}{2}\biggr)
\end{equation*}
\notag
$$
is the vector potential of a constant homogeneous electromagnetic field. For an arbitrary gauge group, $a_1(x,t,u)$ and $a_2(x,t,u)$ are defined analogously, only $u$ becomes a map from the set of auxiliary edges to a matrix group, for instance, $\operatorname{U}(1)$ or $\operatorname{SU}(n)$. Then where the $(a_j)_{kl}$ are the entries of the matrix $a_j$. Example 5 (spin ‘precession’ in a magnetic field). Let The following propositions are proved analogously to Propositions 5 and 6, only a factor of $u(x \pm 1/2,t+1/2)$ is added due to the last step of the path passing through the vertex $(x \pm 1/2,t-1/2)$. Proposition 14 (Dirac’s equation in an electromagnetic field). For any integers $x$ and ${t\geqslant 1}$, Recently Ozhegov has found analogues of the ‘explicit’ formula (Proposition 11) and the continuum limit (Corollary 6) for the ‘homogeneous field’ in Example 5 (see [40]). 6. SourceQuestion: what is the probability to find an electron at $(x,t)$ if it was emitted by a source of wavelength $\lambda$? Assumptions: the source is now realistic. Results: wave propagation, dispersion relation. A realistic source produces a wave rather than electrons localized at $x=0$ (as in the basic model). This means solving Dirac’s equation (3), (4) with (quasi-)periodic initial conditions. To state the result, it is convenient to rewrite Dirac’s equation (3), (4) using the notation
$$
\begin{equation*}
\widetilde a_1(x,t)=a_1(x,t+\varepsilon,m,\varepsilon) \quad\text{and}\quad \widetilde a_2(x,t)=a_2(x+\varepsilon,t+\varepsilon,m,\varepsilon),
\end{equation*}
\notag
$$
so that it gets the form
$$
\begin{equation}
\widetilde a_1(x,t) =\frac{1}{\sqrt{1+m^2\varepsilon^2}} \bigl(\widetilde a_1(x+\varepsilon,t-\varepsilon)+ m\varepsilon\widetilde a_2(x,t-\varepsilon)\bigr),
\end{equation}
\tag{28}
$$
$$
\begin{equation}
\widetilde a_2(x,t) =\frac{1}{\sqrt{1+m^2\varepsilon^2}} \bigl(\widetilde a_2(x-\varepsilon,t-\varepsilon)- m\varepsilon\widetilde a_1(x,t-\varepsilon)\bigr).
\end{equation}
\tag{29}
$$
The following proposition is proved by direct checking (available in [45], § 12). Proposition 16 (wave propagation, dispersion relation). Equations (28) and (29) with the initial condition Remark 6. The solution of the continuum Dirac equation (1) is given by the same expressions (30) and (31), except that $2\pi/T$ and $\alpha$ are redefined by respectively. In both the continuum and discrete setups, these are the hypotenuse and an angle in a right triangle with one leg $2\pi/\lambda$ and the other leg either $m$ or $(\arctan m\varepsilon)/\varepsilon$, respectively, which lies in the plane or on a sphere of radius $1/\varepsilon$, respectively. This spherical-geometry interpretation is new and totally unexpected. From the physical point of view the functions (30) and (31) describe a wave with period $T$ and wavelength $\lambda$. A formula relating $T$ and $\lambda$ is called a dispersion relation. Planck’s and de Broglie’s formulae state that $E := 2\pi\hbar/T$ and $p := 2\pi\hbar/\lambda$ are the energy and momentum of the wave (recall that $\hbar= c = 1$ in our units). As $\varepsilon\to0$ and $\lambda\to\infty$ the above dispersion relation becomes Einstein’s formula $E=mc^2$. A comment for specialists: replacing $a$ and $b$ by creation and annihilation operators, that is, the second quantization of the lattice Dirac equation, leads to the model from § 9. For the upgrades below we just announce results to be discussed in subsequent publications, for instance, in [46]. 7. MediumQuestion: what fraction of light of given colour is reflected from a glass plate of given width? Assumptions: the right angle of incidence, no polarization nor dispersion of light; the mass depends now on $x$ but not on the colour. Results: thin-film reflection (a quantitative explanation). Feynman checkers can be applied to describe the propagation of light in transparent media such as glass. Inside the media light propagates as if it had acquired some non-zero mass plus potential energy (depending on the refractive index), while both remain zero outside. In general the model is inappropriate to describe light; partial reflection is a remarkable exception. Notice that similar classical phenomena are described by quantum models (see [51], § 2.7). As we show in a subsequent publication, in Feynman checkers, we can come up with a rigorous derivation of the following well-known formula for the fraction $P$ of light of wavelength $\lambda$ reflected from a transparent plate of width $L$ and refractive index $n$:
$$
\begin{equation*}
P=\frac{(n^2-1)^2}{(n^2+1)^2+4n^2\cot^2(2\pi Ln/\lambda)}\,.
\end{equation*}
\notag
$$
This makes Feynman’s popular-science discussion of partial reflection [11] completely rigorous and shows that his model has experimentally-confirmed predictions in the real world, not just a two-dimensional one. 8. Identical particlesQuestion: what is the probability to find electrons in squares $F$ and $F'$ if they were emitted from $A$ and $A'$? Assumptions: there are several moving electrons. Results: the exclusion principle, locality, charge conservation. We announce a simple-to-define upgrade describing the motion of several electrons, respecting the exclusion principle, locality, and probability conservation (cf. [52], § 4.2). Definition 4. Fix integer points The square of length $P(AB,A'B'\to EF,E'F'):=\bigl|{a}(AB,A'B'\to EF,E'F')\bigr|^2$ is called the probability to find right electrons at $F$ and $F'$ if they are emitted from $A$ and $A'$. We also define $P(AB,A'B'\to EF,E'F')$ analogously for $E=(x\pm 1,t-1)$, $E'=(x'\pm 1,t-1)$. Here we require that $x'\geqslant x$ if both signs are the same, and allow arbitrary $x'$ and $x$ otherwise. 9. AntiparticlesQuestion: what is the expected charge in the square $(x,t)$ if an electron was emitted from the square $(0,0)$? Assumptions: electron-positron pairs are now created and annihilated, the $t$-axis is time. Results: the spin-$1/2$ Feynman propagator in the continuum limit, an analytic expression for the large-time limit. 9.1. Identities and asymptotic formulaeFinally, we introduce a completely new upgrade (Feynman anti-checkers), allowing the creation and annihilation of electron-positron pairs during the motion. This upgrade is defined just by allowing odd $(x+t)/\varepsilon$ in the Fourier integral (Proposition 12), that is, by computing the same integral in white checkerboard squares in addition to black ones. This is equivalent to the second quantization of the lattice Dirac equation (28), (29), but we do not need to work out this procedure (cf. [2], § 9F, and [3], § IV, for the massless case). Anyway, the true motivation of the upgrade is consistency with the original model and the emergence of the spin-$1/2$ Feynman propagator (34) in the continuum limit (see Fig. 15). We also give a combinatorial definition (see Definition 6).  Figure 15.The plots of $b_1(x,6,4,0.03)/0.12$ ((a), the dots), $b_2(x,6,4,0.03)/0.12$ ((b), the dots), their analytic approximation from Theorem 7 (the light curves), the imaginary parts of matrix elements of the Feynman propagator $\operatorname{Im}G^{\rm F}_{11}(x,6)$ ((a), the dark curve) and $\operatorname{Im}G^{\rm F}_{12}(x,6)$ ((b), the dark curve) given by (34) for $m=4$ and $t=6$. Definition 5 (cf. Proposition 12; see Fig. 15). Fix $m,\varepsilon>0$. For each $(x,t)\in \varepsilon\mathbb{Z}^2$ denote One can show that $A_k(x,t,m,\varepsilon)$ is purely imaginary for $(x+t)/\varepsilon$ odd. Thus the real and the imaginary parts ‘live’ on the black and white squares, respectively, analogously to how discrete analytic functions are defined (see [8]). The sign convention for $t\leqslant 0$ is dictated by an analogy to continuum theory (see (33) and (35)). Example 6. The value The other values are even more complicated irrationalities (see Table 6). Table 6.The values $b_1(x,t,1,1)$ and $b_2(x,t,1,1)$ for small $x$ and $t$ (see Definition 5 and Example 6)
We announce that the analogues of Propositions 5–8 and 10 remain true literally if $a_1$ and $a_2$ are replaced by $b_1$ and $b_2$, respectively (the assumption $t>0$ can then be dropped). As a consequence, $2^{(t-1)/2}b_1(x,t,1,1)$ and $2^{(t-1)/2}b_2(x,t,1,1)$ for all $(x,t)\in \mathbb{Z}^2$ are rational linear combinations of the Gauss constant $G$ and the inverse lemniscate constant $L'$. In general, we announce that $b_1$ and $b_2$ can ‘explicitly’ be expressed through the Gauss hypergeometric function: for $m,\varepsilon>0$ and $(x,t)\in\varepsilon\mathbb{Z}^2$ with $(x+t)/\varepsilon$ odd we get
$$
\begin{equation*}
\begin{aligned} \, b_1(x,t,m,\varepsilon)&=(1+m^2\varepsilon^2)^{1/2-t/(2\varepsilon)} (-m^2\varepsilon^2)^{(t-|x|)/(2\varepsilon)-1/2} \begin{pmatrix} \dfrac{t+|x|}{2\varepsilon}-1 \\ \dfrac{|x|}{\varepsilon} \end{pmatrix} \\ &\qquad\times{}_2F_1\biggl(1+\frac{|x|-t}{2\varepsilon}\,, 1+\frac{|x|-t}{2\varepsilon}\,;1+\frac{|x|}{\varepsilon}\,; -\frac{1}{m^2\varepsilon^2}\biggr), \\ b_2(x+\varepsilon,t+\varepsilon,m,\varepsilon)&= (1+m^2\varepsilon^2)^{-t/(2\varepsilon)} (m\varepsilon)^{(t-|x|)/\varepsilon} \\ &\qquad\times(-1)^{(t-|x|)/(2\varepsilon)+1/2} \begin{pmatrix} \dfrac{t+|x|}{2\varepsilon}-1+\theta(x) \\ \dfrac{|x|}{\varepsilon} \end{pmatrix} \\ &\qquad\times{}_2F_1\biggl(\frac{|x|-t}{2\varepsilon}\,, 1+\frac{|x|-t}{2\varepsilon}\,;1+\frac{|x|}{\varepsilon}\,; -\frac{1}{m^2\varepsilon^2}\biggr), \end{aligned}
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\theta(x):=\begin{cases} 1 & \text{if}\ x\geqslant0, \\ 0 & \text{if}\ x<0, \end{cases}
\end{equation*}
\notag
$$
and ${}_2F_{1}(p,q;r;z)$ is the principal branch of the hypergeometric function. The idea of the proof is induction on $t/\varepsilon\geqslant 1$: the basis is given by formulae 9.112, 9.131.1, 9.134.3, and 9.137.15 in [19] and the step is given by the analogues of (3) and (4) for $b_1$ and $b_2$ plus formulae 9.137.11, 12, and 18 in [19]. Remark 7 (cf. Remark 3). These expressions can be rewritten as the Jacobi functions of the second kind of half-integer order (see the definition in [50], (4.61.1)). For instance, for each $(x,t)\in\varepsilon\mathbb{Z}^2$ such that $|x|>t$ and $(x+t)/\varepsilon$ is odd we have Remark 8. The number $b_1(x,\varepsilon,m,\varepsilon)$ equals $(1+\sqrt{1+m^2\varepsilon^2})/m\varepsilon$ times the probability that a planar simple random walk over white squares dies at $(x,\varepsilon)$, if it starts at $(0,\varepsilon)$ and dies with the probability $1-1/\sqrt{1+m^2\varepsilon^2}$ before each step. Nothing like that is known for $b_{1}(x,t,m,\varepsilon)$ and $b_{2}(x,t,m,\varepsilon)$ with $t\ne \varepsilon$. The following two results are proved almost literally as Proposition 13 and Theorem 2. The only difference is the change of the sign of the terms involving $f_-(p)$ in (41), (46) (50), (52); the analogues of Lemmas 5 and 11 are then obtained by direct checking. Proposition 17 (full space-time Fourier transform). Denote $\delta_{x\varepsilon}:=1$ if $x=\varepsilon$, and $\delta_{x\varepsilon}:=0$ if $x\ne\varepsilon$. For all $m>0$ and $(x,t)\in \varepsilon\mathbb{Z}^2$ we get Theorem 7 (large-time asymptotic formula between the peaks; see Fig. 15). For each $\delta>0$ there is $C_\delta>0$ such that for all $m,\varepsilon>0$ and each $(x,t)\in\varepsilon\mathbb{Z}^2$ satisfying (8) we have 9.2. A physical interpretationOne interprets
$$
\begin{equation*}
\dfrac{1}{2}|A_1(x,t,m,\varepsilon)|^2+\frac{1}{2}|A_2(x,t,m,\varepsilon)|^2
\end{equation*}
\notag
$$
as the expected charge in a square $(x,t)$ with $t>0$, in the units of electron charge. Numbers cannot be anymore interpreted as probabilities to find the electron in the square. The reason is that now the outcomes of the experiment are not mutually exclusive: one can detect an electron in two distinct squares simultaneously. There is nothing mysterious about that: any measurement influences the electron necessarily. This influence can be enough to create an electron-positron pair from the vacuum. Thus one can detect a newborn electron in addition to the initial one; and there is no way to distinguish one from another. (A more formal explanation for specialists: two states in the Fock space representing an electron localized in distant regions are not mutually orthogonal; their scalar product is essentially provided by the Feynman propagator.) We announce that the model reproduces the Feynman propagator rather than the retarded one in the continuum limit (see Fig. 15). The spin-$1/2$ Feynman propagator equals
$$
\begin{equation}
{\small G^{\rm F}(x,t)=\begin{cases} \dfrac{m}{4}\begin{pmatrix} J_0(ms)-iY_0(ms) & -\dfrac{t+x}{s}\bigl(J_1(ms)-iY_1(ms)\bigr) \\ \dfrac{t-x}{s}\bigl(J_1(ms)-iY_1(ms)\bigr) & J_0(ms)-iY_0(ms) \end{pmatrix} \text{ if } |x|<|t|, \\ \dfrac{im}{2\pi}\begin{pmatrix} K_0(ms) & \dfrac{t+x}{s}\,K_1(ms) \\ \dfrac{x-t}{s}\,K_1(ms) & K_0(ms) \end{pmatrix} \text{ if } |x|>|t|, \end{cases}}
\end{equation}
\tag{34}
$$
where the $Y_n(z)$ and $K_n(z)$ are Bessel functions of the 2nd kind and modified Bessel functions of the 2nd kind, and $s:=\sqrt{|t^2-x^2|}$. In addition, there is a generalized function supported on the lines $t=\pm x$, which we do not specify. The Feynman propagator satisfies (25). We see that it has an additional imaginary part (and an overall factor of $1/2$) compared to the retarded propagator (26). In particular, it does not vanish for $|x|>|t|$: the annihilation of electron at one point and creation of one at another point can result in an apparent motion faster than light. A more common expression for the Feynman propagator is the Fourier transform
$$
\begin{equation}
G^{\rm F}(x,t)= \frac{1}{4\pi^2}\int_{-\infty}^{+\infty}\, \int_{-\infty}^{+\infty}\lim_{\delta\to+0} \begin{pmatrix} m & -ip-i\omega \\ -ip+i\omega & m \end{pmatrix} \frac{e^{i p x-i\omega t}\,dp\,d\omega}{m^2+p^2-\omega^2-i\delta}
\end{equation}
\tag{35}
$$
(cf. (27) and [14], (6.51)). Overall, a small correction introduced by the upgrade reflects some fundamental limitations on measurement rather than adds something meaningful to the description of the motion. The upgrade should only be viewed as an ingredient for more realistic models with interaction. 9.3. Combinatorial definitionInformally, the combinatorial definition of Feynman anti-checkers (Definition 6) is obtained from the definition of Feynman checkers (Definition 2) by the following four-step modification.
Here Step 2 is completely new whereas the other steps are standard. It reflects the general principle that the real and imaginary parts of a quantity should always be put on dual lattices. Definition 6 (see Fig. 16). Fix $T\in\mathbb{Z}$ and $\varepsilon,m,\delta>0$ called the lattice size, the lattice step, the particle mass, and the small imaginary mass respectively. Assume that $T>0$ and $\delta<1$. The lattice is the quotient set (This is a finite subset of the torus obtained from the square $[0,T\varepsilon]^2$ by an identification of the opposite sides.) A lattice point $(x,t)$ is even (respectively, odd) if $2x/\varepsilon$ is even (respectively, odd). An edge is a vector starting from a lattice point $(x,t)$ and ending at the lattice point $(x+\varepsilon/2,t+\varepsilon/2)$ or $(x-\varepsilon/2,t+\varepsilon/2)$. A generalized checker path is a finite sequence of distinct edges such that the endpoint of each edge is the starting point of the next one. A cycle is defined analogously, except that the sequence contains a single repetition: the first and last edges coincide and there is at least one edge in between. (In particular, a generalized checker path such that the endpoint of the last edge is the starting point of the first one is not a cycle; the coincidence of the first and the last edges is required. The first and the last edges of a generalized checker path coincide only if the path has a single edge. Thus, in our setup, a path is never a cycle.) Changing the starting edge of a cycle means the removal of the first edge from the sequence, then a cyclic permutation, and then the addition of the last edge of the resulting sequence at the beginning. A loop is a cycle up to fixing the starting edge. A node of a path or loop $s$ is an ordered pair of consecutive edges in $s$ (the edges in the pair are ordered as in $s$). A turn is a node such that the two edges are orthogonal. A node or turn is even (odd) if the endpoint of the first edge in the pair is even (odd, respectively). Denote by $\operatorname{eventurns}(s)$, $\operatorname{oddturns}(s)$, $\operatorname{evennodes}(s)$, $\operatorname{oddnodes}(s)$ the numbers of even and odd turns and nodes in $s$. The arrow (or weight) of $s$ is
$$
\begin{equation*}
{A}(s,m,\varepsilon,\delta) :=\pm\frac{(-im\varepsilon)^{\operatorname{oddturns}(s)} (-\delta)^{\operatorname{eventurns}(s)}} {(1+m^2\varepsilon^2)^{\operatorname{oddnodes}(s)/2} (1-\delta^2)^{\operatorname{evennodes}(s)/2}}\,,
\end{equation*}
\notag
$$
where the overall minus sign is taken when $s$ is a loop. A set of checker paths or loops is edge-disjoint, if no two of them have a common edge. An edge-disjoint set of loops is a loop configuration. A loop configuration with source $a$ and sink $f$ is an edge-disjoint set of any number of loops and exactly one generalized checker path starting with the edge $a$ and ending with the edge $f$. The arrow $A(S,m,\varepsilon,\delta)$ of a loop configuration $S$ (possibly with a source and a sink) is the product of arrows of all loops and paths in the configuration. An empty product is set to be $1$. The arrow from an edge $ a$ to an edge $f$ (or finite-lattice propagator) is
$$
\begin{equation*}
{A}(a\to f,m,\varepsilon,\delta,T):= \frac{\sum_{\substack{\text{loop configurations } S\\ \text{with the source } e \text{ and the sink } f}}A(S,m,\varepsilon,\delta)} {\sum_{\substack{\text{loop configurations } S }} A(S,m,\varepsilon,\delta)}\,.
\end{equation*}
\notag
$$
Now take a point $(x,t)\in (\varepsilon\mathbb{Z})^2$ and set Let $a_0$, $f_1$, and $f_2$ be the edges starting at $(0,0)$, $(x',t')$, and $(x',t')$ and ending at $(\varepsilon/2,\varepsilon/2)$, $(x'-\varepsilon/2,t'+\varepsilon/2)$, and $(x'+\varepsilon/2,t'+\varepsilon/2)$ respectively. The arrow of the point $(x,t)$ (or infinite-lattice propagator) is the pair of complex numbers Example 7 (see Fig. 16, middle). The lattice of size $1$ lies on the square $[0,\varepsilon]^2$ with the opposite sides identified. The lattice has $2$ points: the midpoint and the identified vertices of the square. It has four edges $a$, $b$, $c$, and $d$. The generalized checker paths $abdc$, $acdb$, and $bacd$ are distinct although they contain the same edges. Their arrows are Informally, the loops form the Dirac sea of electrons filling the whole space, and the edges not participating in loops form paths of holes, that is, antiparticles. We announce that Definitions 5 and 6 are equivalent in the following sense: $\widetilde{A}_1(x,t,m,\varepsilon)={A}_1(x,t+\varepsilon,m,\varepsilon)$ and $\widetilde{A}_2(x,t,m,\varepsilon)={A}_2(x+\varepsilon,t+\varepsilon,m,\varepsilon)$; cf. § 6. The idea is that both definitions produce actually the matrix elements of the ‘inverse’ of the same lattice Dirac operator: the first does this via the Fourier transform, and the second via the ratio of determinants. (See details in [46].) 10. Towards $(1+1)$-dimensional quantum electrodynamicsQuestion: what is the probability to find electrons (or an electron and a positron) with momenta $q$ and $q'$ in the far future if they were emitted with momenta $p$ and $p'$ in the far past? Assumptions: interaction is now switched on; all simplifying assumptions are dropped except the default ones: no nuclear forces, no gravitation, electron moves along the $x$-axis only, and the $t$-axis is time. Results: repulsion of like charges and attraction of opposite charges (a qualitative explanation expected). The construction of the required model is a widely open problem because, in particular, it requires the missing mathematically rigorous construction of the Minkowskian lattice gauge theory. 11. Open problemsWe start with problems relying on Definition 1. Plots suggest that for fixed $t$, the most probable position of the electron is close to $x=t/\sqrt{2}$ (see Fig. 9 (top) and Theorems 1, (B), and 2). Although this was noticed 20 years ago, the following question is still open. Problem 1 (Daniyarkhodzhaev and Kuyanov; see Fig. 9, top). Denote by $x_{\max}(t)$ the point where $P(x,t)$ has a maximum for fixed $t$. Is $x_{\max}(t)-t/\sqrt{2}$ bounded as $t\to+\infty$? What makes the problem hard is that the behaviour of $P(x,t)$ is only known close to $x=t/\sqrt{2}$ (Theorem 3) and far from $x=t/\sqrt{2}$ (Theorems 2 and 4) but not at intermediate distances. Problem 2 (cf. [49]). Find an asymptotic formula for $P(x,t)$ as $t\to+\infty$ which is uniform in $x\in [-t;t]$. Problem 3 (Nechaev; see Fig. 9, top). Find the positions of ‘wide gaps’ (intervals, where oscillations are smaller) in the plot of $P(x,t)$ for fixed large $t$. (Cf. formulae (9) and (10).) The aim of the next two problems is to study phase transition by means of various order parameters (see p. 460). Specifically, we conjecture that the limiting ‘probability’ of equal signs at the endpoints of the spin chain, as well as the limiting ‘probability’ of equal signs at the endpoints and the midpoint, are not analytic at $v=\pm 1/\sqrt{2}$. Problem 4 (see Fig. 10). Prove that for each $0<v<1/\sqrt{2}$ we have Problem 5 (cf. [23], p. 381). Find the weak limit The next problem is on absorption probabilities; it relies on the definition before Example 1. Problem 6 (Minaev and Russkikh; cf. [1], § 5, [10], and [37], § 4). Find for all $n\in\mathbb{Z}$. Find the weak limit and an asymptotic formula for $P(x,t, \text{ bypass } \{x=0\})$ as $t\to+\infty$. (Cf. Theorems 1 and 2.) The following problem generalizes and specifies Problem 1 above; it relies on Definition 2. Problem 7 (Daniyarkhodzhaev and Kuyanov). Denote by $x_{\max}=x_{\max}(t,m,\varepsilon)$ the point where $P(x):=P(x,t,m,\varepsilon)$ has a maximum. Is $x_{\max}/\varepsilon-t/(\varepsilon\sqrt{1+m^2\varepsilon^2}\,)$ uniformly bounded? Does $P(x)$ decrease for $x>x_{\max}$? Find an asymptotic formula for $a(x,t,m,\varepsilon)$ as $t\to+\infty$ which is valid for all $x\in [-t,t]$ and uniform in $x$, $m$, and $\varepsilon$. Problem 8 (Blank and Shlosman). Is the number of times the function $a_1(x):=a_1(x,t,m,\varepsilon)$ changes sign on $[-t,t]$ bounded as $\varepsilon\to 0$ for fixed $t$ and $m$? Corollary 6 gives a uniform limit on compact subsets of the angle $|x|<t$, hence it misses the main contribution to the probability. Now we ask for the weak limit detecting the peak. Problem 9. Find the weak limits The following problem is to construct a continuum analogue of Feynman checkers. Problem 10 (Lifshits). Consider $(-im\varepsilon)^{\operatorname{turns}(s)}$ as a charge on the set of all checker paths $s$ from $(0,0)$ to $(x,t)$ starting and ending with an upwards-right move. Does the charge converge (weakly or in another sense) to a charge on the space of all continuous functions $[0,t]\to\mathbb{R}$ with boundary values $0$ and $x$, respectively, as ${\varepsilon\to 0}$? The following problem, relying on Definition 3, would demonstrate ‘spin precession’. Problem 11 (see Fig. 13, right; cf. [40]). Is a periodic function asymptotically as $t\to+\infty$ for Find the weak limit of $P(x,t,u)$ and asymptotic formulae for $a_k(x,t,u)$ as $t\to+\infty$. (Cf. Theorems 1 and 2.) We define $a(x,t,m,\varepsilon,u)$ analogously to $a(x,t,m,\varepsilon)$ and $a(x,t,u)$, by unifying Definitions 2, 3 and Remark 5. The next problem asks if this reproduces Dirac’s equation in an electromagnetic field. Problem (cf. [16]). Fix $A_0(x,t),A_1(x,t)\in C^2(\mathbb{R}^2)$. For each auxiliary edge $s_1s_2$ set The next two problems rely on Definition 5. Problem 13 (cf. Corollary 1). Prove that Problem (cf. Theorems 4 and 7). Find an asymptotic formula for $A_k(x,t,{m},{\varepsilon})$ for $|x|>|t|$ as $t\to+\infty$. The last problem is informal; it stays open for half a century. Problem (Feynman; cf. [15]). Generalize the model to four dimensions so that 12. ProofsLet us present a chart showing the dependence of the above results and further subsections (the statements proved in the corresponding subsections are indicated in parentheses, and the numbers of the propositions used are shown above arrows): Propositions 7, 10, and 13 are not used in the proofs of the main results.In the process of proofs we give zero-knowledge introductions to the methods used. Some proofs are simpler than the original ones. 12.1. Identities: elementary combinatorics (Propositions 1–13)Let us prove the identities from § 3; the ones from § 2 are the particular case $m=\varepsilon=1$. Proof of Propositions 1 and 5. We derive a recurrence relation for $a_2(x,t,m,\varepsilon)$. Take a path $s$ on $\varepsilon\mathbb{Z}^2$ from $(0,0)$ to $(x,t)$ with the first step to $(\varepsilon,\varepsilon)$. Set
The last move in the path $s$ is made either from $(x-\varepsilon,t-\varepsilon)$ or from $(x+\varepsilon,t-\varepsilon)$. If it is from $(x+\varepsilon,t-\varepsilon)$, then $\operatorname{turns}(s)$ must be odd, hence $s$ does not contribute to $a_2(x,t,m,\varepsilon)$. Assume further that the last move in $s$ is made from $(x-\varepsilon,t-\varepsilon)$. Denote by $s'$ the path $s$ without the last move. If the directions of the last moves in $s$ and $s'$ coincide, then
$$
\begin{equation*}
a(s,m\varepsilon)=\frac{1}{\sqrt{1+m^2\varepsilon^2}}a(s',m\varepsilon),
\end{equation*}
\notag
$$
otherwise
$$
\begin{equation*}
a(s,m\varepsilon)=\frac{-im\varepsilon}{\sqrt{1+m^2\varepsilon^2}} a(s',m\varepsilon)=\frac{m\varepsilon}{\sqrt{1+m^2\varepsilon^2}} \bigl[\operatorname{Im}a(s',m \varepsilon)- i\operatorname{Re}a(s',m \varepsilon)\bigr].
\end{equation*}
\notag
$$
Summation over all paths $s'$ gives the required equation
$$
\begin{equation*}
\begin{aligned} \, a_2(x,t,m,\varepsilon)&=\operatorname{Im} \sum_{s\ni(x-\varepsilon,t-\varepsilon)}a(s,m\varepsilon) \\ &=\sum_{s'\ni(x-2\varepsilon,t-2\varepsilon)} \frac{\operatorname{Im}a(s',m\varepsilon)}{\sqrt{1+m^2\varepsilon^2}}- \sum_{s'\ni(x,t-2\varepsilon)} \frac{m\varepsilon\operatorname{Re}a(s',m\varepsilon)} {\sqrt{1+m^2\varepsilon^2}} \\ &=\frac{1}{\sqrt{1+m^2\varepsilon^2}} \bigl[a_2(x-\varepsilon,t-\varepsilon,m,\varepsilon)- m\varepsilon a_1(x-\varepsilon,t-\varepsilon,m,\varepsilon)\bigr]. \end{aligned}
\end{equation*}
\notag
$$
The recurrence for $a_1(x,t,m,\varepsilon)$ is proved analogously. $\Box$ Proof of Propositions 2 and 6. The proof is by induction on $t/\varepsilon$. The base $t/\varepsilon=1$ is obvious. The step of induction follows immediately from the following computation using Proposition 5:
$$
\begin{equation*}
\begin{aligned} \, &\sum_{x\varepsilon\in \mathbb{Z}}P(x,t+\varepsilon,m,\varepsilon)= \sum_{x \in \varepsilon\mathbb{Z}}\bigl[a_1(x,t+\varepsilon,m,\varepsilon)^2+ a_2(x,t+\varepsilon,m,\varepsilon)^2\bigr] \\ &\qquad=\frac{1}{1+m^2\varepsilon^2} \biggl(\,\sum_{x \in \varepsilon\mathbb{Z}} \bigl[a_1(x+\varepsilon,t,m,\varepsilon)+ m\varepsilon a_2(x+\varepsilon,t,m,\varepsilon)\bigr]^2 \\ &\qquad\qquad+\sum_{x \in \varepsilon\mathbb{Z}} [a_2(x-\varepsilon,t,m,\varepsilon)- m\varepsilon a_1(x-\varepsilon,t,m,\varepsilon)]^2\biggr) \\ &\qquad=\frac{1}{1+m^2\varepsilon^2} \biggl(\,\sum_{x \in \varepsilon\mathbb{Z}}\bigl[a_1(x,t,m,\varepsilon)+ m\varepsilon a_2(x,t,m,\varepsilon)\bigr]^2 \\ &\qquad\qquad+\sum_{x\in\varepsilon\mathbb{Z}} \bigl[a_2(x,t,m,\varepsilon)- m\varepsilon a_1(x,t,m,\varepsilon)\bigr]^2\biggr) \\ &\qquad=\sum_{x \in \varepsilon\mathbb{Z}} \bigl[a_1(x,t,m,\varepsilon)^2+a_2(x,t,m,\varepsilon)^2\bigr] =\sum_{x \in \varepsilon\mathbb{Z}}P(x,t,m,\varepsilon). \end{aligned}
\end{equation*}
\notag
$$
Lemma 1 (adjoint Dirac equation). For each $(x,t)\in \varepsilon\mathbb{Z}^2$, where $t>\varepsilon$, we have Proof. The second equation is obtained from Proposition 5 by substituting $(x,t)$ by $(x-\varepsilon,t-\varepsilon)$ and $(x+\varepsilon,t-\varepsilon)$ in (3) and (4), respectively, and adding them with the coefficients $m\varepsilon/\sqrt{1+m^2\varepsilon^2}$ and $1/\sqrt{1+m^2\varepsilon^2}$. The first equation is obtained analogously. $\Box$ Proof of Proposition 7. The real part of the desired equation is the sum of the first equations of Lemma 1 and Proposition 5. The imaginary part is the sum of the second ones. $\Box$ Proof of Proposition 8. Let us prove the first identity. For a path $s$ denote by $s'$ the reflection of $s$ with respect to the $t$ axis, and by $s''$ the path consisting of the same moves as $s'$, but in the opposite order.
Take a path $s$ from $(0,0)$ to $(x,t)$ with the first move upwards-right such that $\operatorname{turns}(s)$ is odd (the paths with even $\operatorname{turns}(s)$ do not contribute to $a_1(x,t,m,\varepsilon)$). Then the last move in $s$ is upwards-left. Therefore, the last move in $s'$ is upwards- right, hence the first move in $s''$ is upwards-right. The endpoint of both $s'$ and $s''$ is $(-x,t)$, because reordering of moves does not affect the endpoint. Thus $s\mapsto s''$ is a bijection between the paths to $(x,t)$ and $(-x,t)$ with $\operatorname{turns}(s)$ odd. Thus $a_1(x,t,m,\varepsilon)=a_1(-x,t,m,\varepsilon)$. We prove the second identity by induction on $t/\varepsilon$ (this proof was found and written down by Kolpakov). The base of induction ($t/\varepsilon=1$ and $t/\varepsilon=2$) is obvious. Step of induction: take $t\geqslant 3\varepsilon$. Applying the inductive hypothesis for the three points $(x-\varepsilon,t-\varepsilon)$, $(x+\varepsilon,t-\varepsilon)$, and $(x,t-2\varepsilon)$ and the identity just proved, we get
$$
\begin{equation*}
\begin{aligned} \, (t-x)a_2(x-\varepsilon,t-\varepsilon,m,\varepsilon)&= (x+t-4\varepsilon)a_2(3\varepsilon-x,t-\varepsilon,m,\varepsilon), \\ (t-x-2\varepsilon)a_2(x+\varepsilon,t-\varepsilon,m,\varepsilon)&= (x+t-2\varepsilon)a_2(\varepsilon-x,t-\varepsilon,m,\varepsilon), \\ (t-x-2\varepsilon)a_2(x,t-2\varepsilon,m,\varepsilon)&= (x+t-4\varepsilon)a_2(2\varepsilon-x,t-2\varepsilon,m,\varepsilon), \\ a_1(x-\varepsilon,t-\varepsilon,m,\varepsilon)&= a_1(\varepsilon-x,t-\varepsilon,m,\varepsilon). \end{aligned}
\end{equation*}
\notag
$$
Summing up the four equations with the coefficients $1$, $1$, $-\sqrt{1+m^2\varepsilon^2}$, and $-2m\varepsilon^2$, respectively, we get
$$
\begin{equation*}
\begin{aligned} \, &(t-x)\bigl[a_2(x-\varepsilon,t-\varepsilon,m,\varepsilon)+ a_2(x+\varepsilon,t-\varepsilon, m, \varepsilon) \\ &\qquad\qquad- \sqrt{1+m^2\varepsilon^2}\,a_2(x,t-2\varepsilon,m,\varepsilon)\bigr] -2m\varepsilon^2\,a_1(x-\varepsilon,t-\varepsilon,m,\varepsilon) \\ &\qquad\qquad -2\varepsilon\, a_2(x+\varepsilon,t-\varepsilon,m,\varepsilon) +2\varepsilon\sqrt{1+m^2\varepsilon^2}\,a_2(x,t-2\varepsilon,m,\varepsilon) \\ &\qquad=-2m\varepsilon^2\,a_1(\varepsilon-x,t-\varepsilon,m,\varepsilon) -2\varepsilon\, a_2(3\varepsilon-x,t-\varepsilon,m,\varepsilon) \\ &\qquad\qquad+2\varepsilon\sqrt{1+m^2\varepsilon^2}\, a_2(2\varepsilon-x,t-2\varepsilon, m,\varepsilon) \\ &\qquad\qquad+(t+x-2\varepsilon) \bigl[a_2(3\varepsilon-x,t-\varepsilon,m,\varepsilon) \\ &\qquad\qquad+ a_2(\varepsilon-x,t-\varepsilon,m,\varepsilon)-\sqrt{1+m^2\varepsilon^2}\, a_2(2\varepsilon-x,t-2\varepsilon, m, \varepsilon)\bigr]. \end{aligned}
\end{equation*}
\notag
$$
Here the three last terms on the left-hand side cancel by Lemma 1, as well as the three first terms on the right-hand side. Applying the Klein–Gordon equation (Proposition 7) to the remaining expressions on the left- and right-hand sides and cancelling the common factor $\sqrt{1+m^2\varepsilon^2}$ we get the desired identity
$$
\begin{equation*}
(t-x)a_2(x,t,m,\varepsilon)= (t+x-2\varepsilon)a_2(2\varepsilon-x,t,m,\varepsilon).
\end{equation*}
\notag
$$
The third identity follows from the first and Proposition 5:
$$
\begin{equation*}
\begin{aligned} \, a_1(x,t,m,\varepsilon)+m\varepsilon\, a_2(x,t,m,\varepsilon) &=\sqrt{1+m^2\varepsilon^2}\, a_1(x-\varepsilon,t+\varepsilon,m,\varepsilon) \\ &=\sqrt{1+m^2\varepsilon^2}\, a_1(\varepsilon-x,t+\varepsilon,m,\varepsilon) \\ &=a_1(2\varepsilon-x,t,m,\varepsilon)+ m\varepsilon\, a_2(2\varepsilon-x,t,m,\varepsilon). \end{aligned}
\end{equation*}
\notag
$$
The first and third identities can also be proved simultaneously by induction on $t/\varepsilon$, using Proposition 5. $\Box$ Proof of Proposition 9. Take a checker path $s$ from $(0,0)$ to $(x,t)$. Denote by $(x',t')$ the point where $s$ intersects the line $t = t'$. Denote by $s_1$ the part of $s$ that joins $(0,0)$ with $(x',t')$. Denote by $s_2$ the part starting at the intersection point of $s$ with the line $t = t'-\varepsilon$ and ending at $(x,t)$ (see Fig. 17). Translate the path $s_2$ so that it starts at $(0,0)$.
Set
$$
\begin{equation*}
a(s,m \varepsilon):=i(-im \varepsilon)^{\operatorname{turns}(s)} (1+m^2\varepsilon^2)^{(1-t/\varepsilon)/2}.
\end{equation*}
\notag
$$
Since $\operatorname{turns}(s)=\operatorname{turns}(s_1)+\operatorname{turns}(s_2)$, it follows that
$$
\begin{equation*}
\operatorname{Re}a(s,m\varepsilon)= \begin{cases} \operatorname{Re}a(s_1,m\varepsilon) \operatorname{Im}a(s_2,m\varepsilon) & \text{if the move to } (x',t') \text{ is upwards-left}, \\ \operatorname{Im}a(s_1,m\varepsilon) \operatorname{Re}a(s_2,m\varepsilon) & \text{if the move to } (x',t') \text{ is upwards-right}. \end{cases}
\end{equation*}
\notag
$$
In the former case replace the path $s_2$ by the path $s_2'$ obtained by the reflection with respect to the line $x=0$ (and starting at the origin). Then we have $\operatorname{Im}a(s_2',m\varepsilon)= \operatorname{Im}a(s_2,m\varepsilon)$. Therefore,
$$
\begin{equation*}
\begin{aligned} \, a_1(x,t,m,\varepsilon)&=\sum_{s}\operatorname{Re}a(s,m\varepsilon) = \sum_{x'} \sum_{s \ni (x',t')} \operatorname{Re}a(s,m\varepsilon) \\ &=\sum_{x'} \biggl(\sum_{s \ni (x',t'),(x'-\varepsilon,t'-\varepsilon)} \operatorname{Im}a(s_1,m\varepsilon)\operatorname{Re}a(s_2,m\varepsilon) \\ &\qquad+\sum_{s \ni (x',t'),(x'+\varepsilon,t'-\varepsilon)} \operatorname{Re}a(s_1,m\varepsilon)\operatorname{Im} a(s_2',m\varepsilon)\biggr) \\ &=\sum_{x'} \bigl[a_2(x',t',m,\varepsilon) a_1(x-x'+\varepsilon,t-t'+\varepsilon,m,\varepsilon) \\ &\qquad+a_1(x',t',m,\varepsilon)a_2(x'-x+\varepsilon, t-t'+\varepsilon,m,\varepsilon)\bigr]. \end{aligned}
\end{equation*}
\notag
$$
The formula for $a_2(x,t,m,\varepsilon)$ is proved analogously. $\Box$ Proof of Proposition 10. Denote by $f(x,t)$ the difference between the left- and the right-hand side of (5). Introduce the operator
$$
\begin{equation*}
[\square_m f](x,t):=\sqrt{1+m^2\varepsilon^2}\,f(x,t+\varepsilon)+ \sqrt{1+m^2\varepsilon^2}\,f(x,t-\varepsilon)-f(x+\varepsilon,t)- f(x-\varepsilon,t).
\end{equation*}
\notag
$$
It suffices to prove that
$$
\begin{equation}
[\square_m^4 f](x,t)=0 \quad \text{for } t\geqslant 5\varepsilon.
\end{equation}
\tag{36}
$$
Then (5) will follow by induction on $t/\varepsilon$: (36) expresses $f(x,t+4\varepsilon)$ as a linear combination of $f(x',t')$ with smaller $t'$; it remains to check that $f(x,t)=0$ for $t\leqslant 8\varepsilon$, which was done in [45], § 11.
To prove (36), write
$$
\begin{equation*}
f(x,t)=p_1(x,t)a(x-2\varepsilon,t,m,\varepsilon)+ p_2(x,t)a(x+2\varepsilon,t,m,\varepsilon)+p_3(x,t)a(x,t,m,\varepsilon)
\end{equation*}
\notag
$$
for certain cubic polynomials $p_k(x,t)$ (see (5)), and use the Leibnitz rule
$$
\begin{equation*}
\begin{aligned} \, \square_m(fg)&=f\cdot\square_m g+\sqrt{1+m^2\varepsilon^2}\, (\nabla_{t+}f\cdot T_{t+}g-\nabla_{t-}f\cdot T_{t-}g) \\ &\qquad-\nabla_{x+}f\cdot T_{x+}g+\nabla_{x-}f\cdot T_{x-}g, \end{aligned}
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
[\nabla_{t\pm}f](x,t):=\pm(f(x,t\pm\varepsilon)-f(x,t))
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
[\nabla_{x\pm}f](x,t):=\pm(f(x\pm\varepsilon,t)-f(x,t))
\end{equation*}
\notag
$$
are the finite difference operators,
$$
\begin{equation*}
[T_{t\pm}g](x,t):=g(x,t\pm\varepsilon)\quad\text{and}\quad [T_{x\pm}g](x,t):=g(x\pm\varepsilon,t)
\end{equation*}
\notag
$$
are the shift operators. Since $\square_m\, a(x,t,m,\varepsilon)=0$ by Proposition 7, each operator $\nabla_{t\pm},\nabla_{x\pm}$ decreases $\deg p_k(x,t)$, and all the above operators commute, by the Leibnitz rule we get (36). This proves the first identity in the proposition; the second one is proved analogously (the induction base is checked in [45; §11]). $\Box$ Alternatively, Proposition 10 can be derived by applying Gauss’s contiguous relations (see [19], 9.137) to the hypergeometric expression in Remark 3 seven times. Proof of Propositions 3 and 11. Let us find $a_1(x,t,m,\varepsilon)$. Consider a path with an odd number of turns; the other paths do not contribute to $a_1(x,t,m,\varepsilon)$. Denote by $2r+1$ the number of turns in the path. Denote by $R$ and $L$ the number of upwards-right and upwards-left moves, respectively. Let $x_1, x_2, \dots, x_{r+1}$ be the number of upwards-right moves before the first, the third, …, the last turn, respectively. Let $y_1, y_2, \dots, y_{r+1}$ be the number of upwards-left moves after the first, the third, …, the last turn respectively. Then $x_k,y_k\geqslant 1$ for $1\leqslant k\leqslant r+1$ and
$$
\begin{equation*}
R=x_1+\cdots+x_{r+1}\quad\text{and}\quad L=y_1+\cdots+y_{r+1}.
\end{equation*}
\notag
$$
The problem now reduces to a combinatorial one: the number of paths with $2r+1$ turns equals the number of positive integer solutions of the resulting equations. For the first equation, this number is the number of ways to put $r$ sticks between $R$ coins in a row, that is, $\begin{pmatrix} R-1 \\ r\end{pmatrix}$. Thus,
$$
\begin{equation*}
a_1(x,t,m,\varepsilon)=(1+m^2\varepsilon^2)^{(1-t/\varepsilon)/2} \sum_{r=0}^{\min\{R,L\}}(-1)^r \begin{pmatrix} R-1 \\ r\end{pmatrix} \begin{pmatrix} L-1 \\ r\end{pmatrix}(m\varepsilon)^{2r+1}.
\end{equation*}
\notag
$$
Thus (6) follows from $L+R=t/\varepsilon$ and $R-L=x/\varepsilon$. Formula (7) is derived analogously. $\Box$ Proof of Proposition 12. The proof is by induction on $t/\varepsilon$.
The base $t/\varepsilon=1$ is obtained by the change of variable $p\mapsto p+\pi/\varepsilon$ so that the integrals over $[0;\pi/\varepsilon]$ and $[-\pi/\varepsilon;0]$ cancel for odd $x/\varepsilon$ and the remaining relation is
$$
\begin{equation*}
\frac{\varepsilon}{2\pi}\int_{-\pi/\varepsilon}^{\pi/\varepsilon} e^{ip(x-\varepsilon)}\,dp=\delta_{x\varepsilon}.
\end{equation*}
\notag
$$
The inductive step is the following computation and an analogous computation for $a_2(x, t+\varepsilon,m,\varepsilon)$:
$$
\begin{equation*}
\begin{aligned} \, a_1(x,t+\varepsilon,m,\varepsilon)&=\frac{1}{\sqrt{1+m^2\varepsilon^2}} \bigl[a_1(x+\varepsilon,t,m,\varepsilon)+m\varepsilon\, a_2(x+\varepsilon,t,m,\varepsilon)\bigr] \\ &=\frac{m\varepsilon^2}{2\pi\sqrt{1+m^2\varepsilon^2}} \int_{-\pi/\varepsilon}^{\pi/\varepsilon} \biggl(\frac{ie^{ip\varepsilon}} {\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}} +1 \\ &\qquad+\frac{\sin (p\varepsilon)} {\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}}\biggr) e^{i p x-i\omega_p(t-\varepsilon)}\,dp \\ &=\frac{m\varepsilon^2}{2\pi} \int_{-\pi/\varepsilon}^{\pi/\varepsilon} \frac{(i\cos(\omega_p\varepsilon) +\sin(\omega_p\varepsilon)) e^{i p x-i\omega_p(t-\varepsilon)}\,dp} {\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}} \\ &=\frac{im\varepsilon^2}{2\pi}\int_{-\pi/\varepsilon}^{\pi/\varepsilon} \frac{e^{ipx-i\omega_pt}\,dp}{\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}}\,. \end{aligned}
\end{equation*}
\notag
$$
Here the first equality is Proposition 5. The second one is the inductive hypothesis. The third follows from
$$
\begin{equation*}
\cos(\omega_p\varepsilon)= \frac{\cos(p\varepsilon)}{\sqrt{1+m^2\varepsilon^2}}\ \text{and}\ \sin(\omega_p\varepsilon)= \sqrt{1-\frac{\cos^2(p\varepsilon)}{1+m^2\varepsilon^2}}= \sqrt{\frac{m^2\varepsilon^2+\sin^2(p\varepsilon)}{1+m^2\varepsilon^2}}\,. \ \square
\end{equation*}
\notag
$$
Alternatively, Proposition 12 can be derived by integration of (30) and (31) over $p=2\pi/\lambda$ for $\widetilde a_1(0,0)=0$ and $\widetilde a_2(0,0)=1$. Proof of Proposition 13. To prove the formula for $a_1(x,t,m,\varepsilon)$ we do the $\omega$-integral:
$$
\begin{equation*}
\begin{aligned} \, &\frac{\varepsilon}{2\pi}\int_{-\pi/\varepsilon}^{\pi/\varepsilon} \frac{e^{-i\omega(t-\varepsilon)}\,d\omega} {\sqrt{1+m^2\varepsilon^2}\,\cos(\omega\varepsilon) -\cos(p\varepsilon)-i\delta} \\ &\quad\overset{(*)}{=}\frac{1}{2\pi i}\oint_{|z|=1} \frac{2\,z^{t/\varepsilon-1}\,dz} {\sqrt{1+m^2\varepsilon^2}\,z^2 -2(\cos(p\varepsilon)+i\delta)z+\sqrt{1+m^2\varepsilon^2}} \\ &\quad\overset{(**)}{=}\frac{\bigl([\cos(p\varepsilon)+i\delta -i\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)+\delta^2 -2i\delta\cos(p\varepsilon)}\,]/ \sqrt{1+m^2\varepsilon^2}\,\bigr)^{t/\varepsilon-1}} {-i\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)+\delta^2 -2i\delta\cos(p\varepsilon)}} \\ &\quad\overset{(***)}{\rightrightarrows} \frac{i\,e^{-i\omega_p(t-\varepsilon)}} {\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}} \end{aligned}
\end{equation*}
\notag
$$
as $\delta\to 0$ uniformly in $p$. Here we assume that $m,t,\delta>0$ and $\delta$ is sufficiently small. Equality $(*)$ is obtained by the change of variables $z=e^{-i\omega\varepsilon}$ and then the change of the contour direction to the counterclockwise one. To prove $(**)$ we find the roots
$$
\begin{equation*}
z_\pm=\frac{\cos(p\varepsilon)+ i\delta\pm i\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)+\delta^2 -2i\delta\cos(p\varepsilon)}}{\sqrt{1+m^2\varepsilon^2}}
\end{equation*}
\notag
$$
of the denominator, where $\sqrt{z}$ denotes the value of the square root with positive real part. Then $(**)$ follows from the residue formula: the expansion shows that $z_-$ lies inside the unit circle, whereas $z_+$ lies outside, for $\delta>0$ sufficiently small in terms of $m$, $\varepsilon$. In $(*{*}*)$ we denote and pass to the limit $\delta\to 0$, which is uniform in $p$ by the assumption ${m>0}$.
The resulting uniform convergence allows us to interchange the limit and the $p$-integral, and we arrive at Fourier integral for $a_1(x,t,m,\varepsilon)$ in Proposition 12. The formula for $a_2(x,t,m,\varepsilon)$ is proved analogously, with the case $t=\varepsilon$ considered separately. $\Box$ 12.2. The phase transition: the method of moments (Theorem 1)In this subsection we give a simple exposition of the proof of Theorem 1 in [20] using the method of moments. The theorem also follows from Corollary 1 obtained in § 12.4 by another method. We rely on the following well-known result. Lemma 2 (see [5], Theorems 30.1 and 30.2). Let $f_t\colon\mathbb{R}\to [0,+\infty)$, where $t=0,1,2,\dots$, be piecewise continuous functions such that $\alpha_{r,t}:=\displaystyle\int_{-\infty}^{+\infty}v^rf_t(v)\,dv$ is finite and $\alpha_{0,t}=1$ for each $r,t=0,1,2,\dots$ . If the series $\sum_{r=0}^{\infty}\alpha_{r,0}z^r/r!$ has a positive radius of convergence and $\lim_{t\to+\infty}\alpha_{r,t}=\alpha_{r,0}$ for each $r=0,1,2,\dots$, then $f_t$ converges to $f_0$ in distribution. Proof of Theorem 1. Let us prove (C); then (A) and (B) will follow from Lemma 2 for $f_0(v):=F'(v)$ and $f_t(v):=tP(\lceil vt \rceil,t)$ because $F'(v)=0$ for $|v|>1$, hence
Rewrite Proposition 12 in a form valid for all $x,t\in\mathbb{Z}$ independently on the parity:
$$
\begin{equation}
\begin{aligned} \, \nonumber \begin{pmatrix} a_1(x,t) \\ a_2(x,t) \end{pmatrix}&=\int_{-\pi}^{\pi}\begin{pmatrix} \widehat a_1(p,t) \\ \widehat a_2(p,t) \end{pmatrix}e^{ip(x-1)}\,\frac{dp}{2\pi} \\ &\!:=\int_{-\pi}^{\pi} \begin{pmatrix} \widehat a_{1+}(p,t)+\widehat a_{1-}(p,t) \\ \widehat a_{2+}(p,t)+\widehat a_{2-}(p,t) \end{pmatrix} e^{ip(x-1)}\,\frac{dp}{2\pi}, \end{aligned}
\end{equation}
\tag{37}
$$
where
$$
\begin{equation}
\begin{aligned} \, \widehat a_{1\pm}(p,t)&=\mp\frac{ie^{ip}}{2\sqrt{1+\sin^2 p}} e^{\pm i\omega_p(t-1)}, \\ \widehat a_{2\pm}(p,t)&=\frac{1}{2} \biggl(1\mp\frac{\sin p}{\sqrt{1+\sin^2 p}}\biggr) e^{\pm i\omega_p(t-1)}, \end{aligned}
\end{equation}
\tag{38}
$$
and $\omega_p:=\arccos\dfrac{\cos p}{\sqrt{2}}$ . Now (37) holds for all $x,t\in\mathbb{Z}$: indeed, the identity
$$
\begin{equation*}
\begin{aligned} \, &\exp\bigl(-i\omega_{p+\pi}(t-1)+i(p+\pi)(x-1)\bigr) \\ &\qquad=\exp\bigl(-i(\pi-\omega_{p})(t-1)+ip(x-1)+i\pi(x-1)\bigr) \\ &\qquad=(-1)^{(x+t)}\exp\bigl(i\omega_p(t-1)+ip(x-1)\bigr) \end{aligned}
\end{equation*}
\notag
$$
shows that the contributions of the two summands $\widehat a_{k\pm}(p,t)$ to integral (37) are equal for even $t+x$ and cancel for odd $t+x$. The summand $\widehat a_{k-}(p,t)$ contributes $a_k(x,t)/2$ by Proposition 12.
By the derivative property of Fourier series and Parseval’s theorem, we have
$$
\begin{equation}
\begin{aligned} \, \nonumber \sum_{x\in\mathbb{Z}}\frac{x^r}{t^r}P(x,t)&=\sum_{x\in\mathbb{Z}} \begin{pmatrix} a_1(x,t) \\ a_2(x,t) \end{pmatrix}^*\frac{x^r}{t^r}\begin{pmatrix} a_1(x,t) \\ a_2(x,t) \end{pmatrix} \\ &=\int_{-\pi}^{\pi}\begin{pmatrix} \widehat a_1(p,t) \\ \widehat a_2(p,t) \end{pmatrix}^*\frac{i^r}{t^r}\,\frac{\partial^r}{\partial p^r}\begin{pmatrix} \widehat a_1(p,t) \\ \widehat a_2(p,t) \end{pmatrix}\,\frac{dp}{2\pi}\,. \end{aligned}
\end{equation}
\tag{39}
$$
The derivative is estimated as follows:
$$
\begin{equation}
\begin{aligned} \, \nonumber \frac{\partial^r}{\partial p^r}\widehat a_{k\pm}(p,t)&= \biggl(\pm i(t-1)\frac{\partial \omega_p}{\partial p}\biggr)^r \widehat a_{k\pm}(p,t)+O_r(t^{r-1}) \\ &=\biggl(\pm \frac{i(t-1) \sin p}{\sqrt{1+\sin^2 p}}\biggr)^r \widehat a_{k\pm}(p,t)+O_r(t^{r-1}). \end{aligned}
\end{equation}
\tag{40}
$$
Indeed, let us differentiate (38) $r$ times using the Leibnitz rule. If we differentiate the exponential factor $e^{\pm i\omega_p(t-1)}$ each time, then we get the main term. If we differentiate a factor rather than the exponential $e^{\pm i\omega_p(t-1)}$ at least once, then we get less than $r$ factors of $(t-1)$, hence the resulting term is $O_r(t^{r-1})$ by compactness because it is continuous and $2\pi$-periodic in $p$.
Substituting (40) into (39) we arrive at
$$
\begin{equation*}
\begin{aligned} \, \sum_{x\in\mathbb{Z}}\frac{x^r}{t^r} P(x,t)&=\int_{-\pi}^{\pi} \begin{pmatrix} \widehat a_1(p,t) \\ \widehat a_2(p,t) \end{pmatrix}^*\biggl(\frac{\sin p}{\sqrt{1+\sin^2 p}}\biggr)^r \begin{pmatrix} (-1)^r \widehat a_{1+}(p,t)+\widehat a_{1-}(p,t) \\ (-1)^r \widehat a_{2+}(p,t)+\widehat a_{2-}(p,t) \end{pmatrix}\,\frac{dp}{2\pi} \\ &\qquad +O_r\biggl(\frac{1}{t}\biggr) \\ &=\int_{-\pi}^{\pi}\biggl(\frac{\sin p}{\sqrt{1+\sin^2 p}}\biggr)^r\, \frac{1}{2}\,\biggl((-1)^r\biggl(1-\frac{\sin p}{\sqrt{1+\sin^2 p}}\biggr)+1 \\ &\qquad+\frac{\sin p}{\sqrt{1+\sin^2 p}}\biggr)\,\frac{dp}{2\pi}+ O_r\biggl(\frac{1}{t}\biggr) \\ &=\int_{-\pi/2}^{\pi/2}\biggl(\frac{\sin p}{\sqrt{1+\sin^2 p}}\biggr)^r \biggl(1+\frac{\sin p}{\sqrt{1+\sin^2 p}}\biggr)\,\frac{dp}{\pi}+ O_r\biggl(\frac{1}{t}\biggr) \\ &=\int_{-1/\sqrt{2}}^{1/\sqrt{2}}\frac{v^r\,dv}{\pi(1-v)\sqrt{1-2v^2}}\, +\,O_r\biggl(\frac{1}{t}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Here the second equality follows from and $\widehat a_{1\pm }(p,t)^*\widehat a_{1\mp}(p,t)+\widehat a_{2\pm}(p,t)^*\widehat a_{2\mp}(p,t)=0$. The third is obtained by the changes of variables $p\mapsto -p$ and $p\mapsto \pi-p$ applied to the integral over $[-\pi/2,\pi/2]$. The fourth is obtained by the change of variables so that 12.3. The main result: the stationary phase method (Theorem 2)In this subsection we prove Theorem 2. First we outline the plan of the argument, then prove the theorem modulo some technical lemmas, and finally prove the lemmas themselves. The plan is to apply the Fourier transform and the stationary phase method to the resulting oscillatory integral. The proof has four steps, with the first two known previously:
Proof of Theorem 2 (modulo some lemmas). We derive an asymptotic formula for $a_1(x,t+\varepsilon,m,\varepsilon)$; the derivation for $a_2(x+\varepsilon,t+\varepsilon,{m},{\varepsilon})$ is analogous and is discussed at the end of the proof. By Proposition 12 and the identity
$$
\begin{equation*}
\exp\biggl(i\omega_{p+{\pi}/{\varepsilon}}t-i\biggl(p+ \frac{\pi}{\varepsilon}\biggr)x\biggr)= \exp\biggl(i\biggl(\frac{\pi}{\varepsilon}-\omega_{p}\biggr)t-ipx- \frac{i\pi x}{\varepsilon}\biggr)= -\exp(-i\omega_pt-ipx)
\end{equation*}
\notag
$$
for $(t+x)/\varepsilon$ odd, we get
$$
\begin{equation}
\begin{aligned} \, a_1(x,t+\varepsilon,m,\varepsilon)&=\frac{m\varepsilon^2}{2\pi i} \int_{-\pi/\varepsilon}^{\pi/\varepsilon}\frac{\exp(i\omega_pt-i p x)} {\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}}\,dp \nonumber \\ &=\int_{-\pi/(2\varepsilon)}^{\pi/(2\varepsilon)}g(p) [e(f_+(p))-e(f_-(p))]\,dp, \end{aligned}
\end{equation}
\tag{41}
$$
where $e(z):=e^{2\pi i z}$ and and
Step 1. We estimate the oscillatory integral (41) using the following version of the stationary phase method, which is the most precise one to our knowledge. Lemma 3 (weighted stationary phase integral; see [21], Lemma 5.5.6). Let $f(p)$ be a real function which is four times continuously differentiable for $\alpha\leqslant p\leqslant \beta$, and let $g(p)$ be a real function which is three times continuously differentiable for $\alpha\leqslant p\leqslant \beta$. Suppose that there are positive parameters $M$, $N$, $T$, and $U$, with Here the first term involving the values at the stationary point $\gamma$ is the main term, and the boundary terms involving the values at the endpoints $\alpha$ and $\beta$ are going to cancel out in Step 3. Lemma 4 (cf. [23], (25), and [1], § 4). Assume (8); then on $[-\pi/(2\varepsilon),\pi/(2\varepsilon)]$, the function $f_{\pm}(p)$ given by (42) has the unique critical point To estimate the integral (41) we are going to apply Lemma 3 twice, to the functions $f(p)=\pm f_\pm(p)$ in appropriate neighbourhoods of their critical points $\gamma_\pm$. In the case when $f(p)=-f_-(p)$ we perform complex conjugation of both sides of (44). Then the total contribution of the two resulting main terms is
$$
\begin{equation}
\operatorname{MainTerm}:= \frac{g(\gamma_+)e(f_{+}(\gamma_+)+1/8)}{\sqrt{f_{+}''(\gamma_+)}}- \frac{g(\gamma_-)e(f_{-}(\gamma_-)-1/8)}{\sqrt{-f_{-}''(\gamma_-)}}\,.
\end{equation}
\tag{46}
$$
The direct but long computation (see [45], § 2) gives the desired main term in the theorem. Step 2. To estimate the approximation error, we need to specify the particular values of parameters for which Lemma 3 is used: Lemma 6. If $\varepsilon\leqslant 1/m$ then functions (42) and (43) and parameters (47) satisfy the inequalities We also need to specify the interval
$$
\begin{equation}
[\alpha_\pm,\beta_\pm]:=\biggl[\gamma_\pm-\frac{m\delta}{2}\,, \gamma_\pm+\frac{m\delta}{2}\biggr].
\end{equation}
\tag{48}
$$
To estimate the derivative $|f_{\pm}''(p)|$ from below, we make sure that we are away from its roots $\pm\pi/(2\varepsilon)$. Lemma 7. Assume (8); then the interval (48), where $\gamma_\pm$ is as defined in (45), is contained in $[-\pi/(2\varepsilon)+m\delta/2,\pi/(2\varepsilon)-m\delta/2]$. The wise choice of the interval provides the following more technical estimate. Lemma 8. Assume (8), (42), (45), and (48). Then for each $p\in[\alpha_\pm,\beta_\pm]$ we have (the signs $\pm$ are compatible). This gives $|f_\pm''(p)|\geqslant T/C_2M^2$ for $C_2:=24\pi\delta^{-3/2}$ using the notation (47). Now all the assumptions of Lemma 3 have been verified ($M\geqslant \beta_\pm-\alpha_\pm$ and $N\geqslant M/\sqrt{T}$ are automatic because $\delta\leqslant 1$ and $t>C_\delta/m$ by (8)). We apply the lemma to $g(p)$ and $\pm f_\pm(p)$ on $[\alpha_\pm,\beta_\pm]$ (the minus sign before $f_-(p)$ guarantees the inequality $f''(p)>0$ and the factor of $i$ inside $g(p)$ is irrelevant for application of the lemma). Then we arrive at the following estimate for the approximation error on those intervals. Note that the error term in Lemma 9 is already of the same order as in Theorem 2. Step 3. To estimate the approximation error outside $[\alpha_\pm,\beta_\pm]$ we use another known technical result. Lemma 10 (weighted first derivative test; see [21], Lemma 5.5.5). Let $f(p)$ be a real function which is three times continuously differentiable for $\alpha\leqslant p\leqslant \beta$, and let $g(p)$ be a real function which is twice continuously differentiable for $\alpha\leqslant p\leqslant \beta$. Suppose that there are positive parameters $M$, $N$, $T$, and $U$, with $M\geqslant \beta-\alpha$, and positive constants $C_r$ such that, for $\alpha\leqslant p\leqslant \beta$, This lemma requires, in particular, the interval to be sufficiently small. For this reason we decompose the initial interval $[-\pi/(2\varepsilon),\pi/(2\varepsilon)]$ into a large number of intervals by the points
$$
\begin{equation*}
\begin{aligned} \, -\frac{\pi}{2\varepsilon}&=\alpha_{-K}<\beta_{-K}=\alpha_{-K+1}<\beta_{-K+1} =\alpha_{-K+2}<\cdots \\ &=\alpha_{i}<\widehat\beta_{i}=\alpha_\pm<\beta_\pm=\widehat\alpha_j<\beta_j=\cdots< \beta_{K-1}=\frac{\pi}{2\varepsilon}\,, \end{aligned}
\end{equation*}
\notag
$$
Here $\alpha_\pm$ and $\beta_\pm$ are given by (48) above. The other points are given by
$$
\begin{equation}
\begin{gathered} \, \alpha_k=\frac{k\pi}{2\varepsilon K}\quad\text{and}\quad \beta_k=\frac{(k+1)\pi}{2\varepsilon K}\,, \\ \text{where}\quad K=2\biggl\lceil\frac{\pi}{m\varepsilon}\biggr\rceil\quad\text{and}\quad k=-K,\dots,i,j+1,\dots,K-1. \end{gathered}
\end{equation}
\tag{49}
$$
The indices $i$ and $j$ are the smallest indices such that
$$
\begin{equation*}
\frac{(i+1)\pi}{2\varepsilon K}>\alpha_\pm \quad\text{and}\quad \frac{(j+1)\pi}{2\varepsilon K}>\beta_\pm.
\end{equation*}
\notag
$$
Thus all the resulting intervals except $[\alpha_\pm,\beta_\pm]$ and its neighbours have the same length $\pi/(2\varepsilon K)$. (Although it would be more conceptual to decompose the interval using a geometric sequence rather than an arithmetic one, this does not affect the final estimate here.) We have already applied Lemma 3 to $[\alpha_\pm,\beta_\pm]$ and we are going to apply Lemma 10 to each of the remaining intervals in the decomposition for $f(p)= f_\pm(p)$ (this time it is not necessary to change the sign of $f_-(p)$). After summation of the resulting estimates, all the terms involving $f_\pm(\alpha_k)$ and $f_\pm(\beta_k)$ at the endpoints, except the leftmost and rightmost ones, cancel. The remaining boundary terms give
$$
\begin{equation}
\begin{aligned} \, \nonumber \operatorname{BoundaryTerm}&:= \frac{g(\pi/(2\varepsilon))e(f_+(\pi/(2\varepsilon)))} {2\pi i f'_+(\pi/(2\varepsilon))}-\frac{g(-\pi/(2\varepsilon)) e(f_+(-\pi/(2\varepsilon)))}{2\pi i f'_+(-\pi/(2\varepsilon))} \\ &\qquad-\frac{g(\pi/(2\varepsilon))e(f_-(\pi/(2\varepsilon)))} {2\pi i f'_-(\pi/(2\varepsilon))}+ \frac{g(-\pi/(2\varepsilon))e(f_-(-\pi/(2\varepsilon)))} {2\pi i f'_-(-\pi/(2\varepsilon))}. \end{aligned}
\end{equation}
\tag{50}
$$
Lemma 11 (see [45], § 5). For $(x,t)\in\varepsilon\mathbb{Z}^2$ such that $(x+t)/\varepsilon$ is odd, the expression (50) vanishes. It remains to estimate the error terms. We start estimates with the central intervals $[\alpha_0,\beta_0]$ and $[\alpha_{-1},\beta_{-1}]$ (possibly without the parts cut out by $[\alpha_\pm,\beta_\pm]$): they require a special treatment. Apply Lemma 10 to these intervals for the same functions (42) and (43) and the same values (47) of the parameters $M$, $N$, $T$, and $U$ as in Step 2. All the assumptions of the lemma have been already verified in Lemma 6; we have
$$
\begin{equation*}
\beta_0-\alpha_0=\beta_{-1}-\alpha_{-1}= \frac{\pi}{2\varepsilon K}= \frac{\pi}{4\varepsilon\lceil\pi/(m\varepsilon)\rceil}<m=M \quad\text{and}\quad f''_\pm(p)\ne 0.
\end{equation*}
\notag
$$
We are thus left to estimate $|f_\pm'(p)|$ from below. Lemma 12. Assume (8), (42), (45), and (48); then for each $[-\pi/(2\varepsilon),\pi/(2\varepsilon)]\setminus [\alpha_\pm,\beta_\pm]$ we get Then the approximation error on the central intervals is estimated as follows. Lemma 13 (see [45], § 6). Assume (8), (45), and (48). Then the parameters (47) and functions (42) satisfy This value is $O_\delta(\varepsilon/m^{1/2}t^{3/2})$ because $t>C_\delta/m$ by (8). Hence the approximation error on the central intervals is within the error term in the theorem. Step 4. To estimate the approximation error on the other intervals $[\alpha_k,\beta_k]$, where we assume that $k>0$ without loss of generality, we apply Lemma 10 with slightly different parameters:
$$
\begin{equation}
T=\frac{mt}{k}\,,\quad M=mk, \quad U=\frac{\varepsilon}{k}\,, \quad N=mk.
\end{equation}
\tag{51}
$$
Lemma 14. For $0<k<K$ and $\varepsilon\leqslant 1/m$, the parameters (51), (49) and the functions (42), (43) on $[\alpha_k,\beta_k]$ satisfy all the assumptions of Lemma 10 except possibly the one on the sign of $f'(p)$. Since the neighbourhood $[\alpha_\pm,\beta_\pm]$ of the root of $f_\pm'(p)$ is cut out, it follows that $f_\pm'(p)$ has constant sign on the remaining intervals, and by Lemma 10 their contribution to the error term is estimated as follows. Lemma 15 (see [45], § 7). Assume (8), (45), (48), and let $0<k<K$. Then the functions (42) and parameters (51) satisfy Summation over all $k$ gives the approximation error
$$
\begin{equation*}
\sum_{k=1}^{K}O\biggl(\frac{\varepsilon}{k^2mt^2\delta^{15/2}}\biggr)= O\biggl(\frac{\varepsilon}{mt^2\delta^{15/2}} \sum_{k=1}^{\infty}\frac{1}{k^2}\biggr)= O_\delta\biggl(\frac{\varepsilon}{m^{1/2}t^{3/2}}\biggr),
\end{equation*}
\notag
$$
because the series $\sum_{k=1}^{\infty}\frac{1}{k^2}$ is convergent and $t>C_\delta/m$. Thus the total approximation error on all the intervals is within the error term in the theorem, which completes the proof of (9). The derivation of the asymptotic formula for $a_2(x+\varepsilon,t+\varepsilon,{m},{\varepsilon})$ is analogous. By Proposition 12 for even $(x+t)/\varepsilon$ we get
$$
\begin{equation}
\begin{aligned} \, \nonumber a_2(x+\varepsilon,t+\varepsilon,{m},{\varepsilon})&= \frac{\varepsilon}{2\pi}\int_{-\pi/\varepsilon}^{\pi/\varepsilon} \biggl(1+\frac{\sin (p\varepsilon)}{\sqrt{m^2\varepsilon^2+ \sin^2(p\varepsilon)}}\biggr)e^{i\omega_pt-i p x}\,dp \\ &=\int_{-\pi/(2\varepsilon)}^{\pi/(2\varepsilon)} \bigl[g_+(p)e(f_+(p))+g_-(p)e(f_-(p))\bigr]\,dp, \end{aligned}
\end{equation}
\tag{52}
$$
where $f_\pm(p)$ are the same as above (see (42)) and
$$
\begin{equation}
g_\pm(p)=\frac{\varepsilon}{2\pi}\biggl(1\pm\frac{\sin(p\varepsilon)} {\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}}\biggr).
\end{equation}
\tag{53}
$$
One repeats the argument of Steps 1–4 with $g(p)$ replaced by $g_\pm(p)$. The particular form of $g(p)$ was only used in Lemmas 5, 6, 11, and 14. The analogues of Lemmas 5 and 11 for $g_\pm(p)$ are obtained by direct checking (see [45], § 13). Lemma 6 holds for $g_\pm(p)$: one need not repeat the proof because
$$
\begin{equation*}
g_\pm(p)=\dfrac{\varepsilon}{t}\biggl(f_\pm'(p)+ \dfrac{x+t}{2\pi}\biggr)
\end{equation*}
\notag
$$
(see [45], § 1). But the parameters (51) in Lemmas 14–15 should be replaced by the following ones (then the analogues of Lemmas 14–15 hold); see Lemma 16:
$$
\begin{equation}
\begin{aligned} \, T=\frac{mt}{k}\,, \quad M=mk,\quad U=\varepsilon,\quad N=mk^{3/2}. \end{aligned}
\end{equation}
\tag{54}
$$
Lemma 16. For $0<k<K$ and $\varepsilon\leqslant 1/m$, parameters (54), (49) and functions (42), (53) on $[\alpha_k,\beta_k]$ satisfy all the assumptions of Lemma 10 except possibly the one on the sign of $f'(p)$. Again, we need not repeat the proof: Lemma 16 follows from Lemma 14 and the formula expressing $g_\pm(p)$ in terms of $f_\pm'(p)$. This completes the proof of Theorem 2 modulo the lemmas. $\Box$ Now we prove the lemmas. Lemmas 5, 9, 11, 13, and 15 are proved by direct checking [45]. The following expressions (see [45], §§ 1 and 3) are used frequently in the proofs of the other lemmas:
$$
\begin{equation}
f_\pm'(p) =\frac{1}{2\pi}\biggl(-x\pm\frac{t\sin(p\varepsilon)} {\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}}\biggr),
\end{equation}
\tag{55}
$$
$$
\begin{equation}
f_\pm''(p) =\pm\frac{m^2\varepsilon^3 t\cos(p\varepsilon)} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{3/2}}\,.
\end{equation}
\tag{56}
$$
Proof of Lemma 4. Using (55) and solving the quadratic equation $f_\pm'(p)=0$ in $\sin(p\varepsilon)$ we get (45). The assumption $|x|/t<1/\sqrt{1+m^2\varepsilon^2}$ from (8) guarantees that the arcsine exists. $\Box$ Proof of Lemma 6. By the computation of the derivatives in [45], § 3, and the assumption $m\varepsilon\leqslant 1$ we get
$$
\begin{equation*}
\begin{aligned} \, |g(p)|&=\frac{m\varepsilon^2} {2\pi\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}}\leqslant \frac{m\varepsilon^2}{2\pi\sqrt{m^2\varepsilon^2+0}}\leqslant \varepsilon=U, \\ |g'(p)|&=\frac{m\varepsilon^3|\sin(p\varepsilon)\cos(p\varepsilon)|} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{3/2}}\leqslant \frac{m\varepsilon^3|\sin(p\varepsilon)\cos(p\varepsilon)|} {2\pi(m^2\varepsilon^2+0)(0+\sin^2(p\varepsilon))^{1/2}}\leqslant \frac{\varepsilon}{m}=\frac{U}{N}\,, \\ |g''(p)|&=\frac{m\varepsilon^4|m^2\varepsilon^2+\sin^4(p \varepsilon)- 2(1+m^2\varepsilon^2)\sin^2(p \varepsilon)|} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{5/2}} \\ &\leqslant \frac{m\varepsilon^4(3m^2\varepsilon^2+3\sin^2(p \varepsilon))} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{5/2}}\leqslant \frac{\varepsilon}{m^2}=\frac{U}{N^2}\,, \\ |g'''(p)|&=\frac{1}{{2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{7/2}}} \,m\varepsilon^5|\sin(p\varepsilon)\cos(p\varepsilon)|\\ &\qquad\times|4m^4\varepsilon^4+9m^2\varepsilon^2+\sin^4(p \varepsilon)- (6+10m^2\varepsilon^2)\sin^2(p \varepsilon)| \leqslant \frac{3\varepsilon}{m^3}=\frac{3U}{N^3}\,, \\ |f''_\pm(p)|&=\frac{ m^2 \varepsilon^3 t\cos(p\varepsilon)} {2 \pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{3/2}}\leqslant \frac{t}{m}=\frac{T}{M^2}\,, \\ |f'''_\pm(p)|&=\frac{m^2\varepsilon^4 t|\sin(p\varepsilon)| (m^2\varepsilon^2+\cos (2 p \varepsilon)+2)} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{5/2}} \leqslant \frac{4m^2\varepsilon^4 t|\sin(p\varepsilon)|} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{5/2}} \\ &\leqslant\frac{t}{m^2}=\frac{T}{M^3}\,, \\ |f^{(4)}_\pm(p)|&=\frac{m^2\varepsilon^5t\cos(p\varepsilon) |m^4\varepsilon^4+3m^2\varepsilon^2+4\sin^4(p \varepsilon)- 2(6+5m^2\varepsilon^2)\sin^2(p\varepsilon)|} {2 \pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{7/2}} \\ &\leqslant \frac{3t}{m^3}=\frac{3T}{M^4}\,. \qquad\square \end{aligned}
\end{equation*}
\notag
$$
Proof of Lemma 7. The lemma follows from the sequence of estimates
$$
\begin{equation*}
\begin{aligned} \, \frac{\pi}{2\varepsilon}-|\gamma_\pm|&\overset{(*)}= \frac{\sin(\pi/2)- \sin|\gamma_\pm\varepsilon|}{\varepsilon\cos(\theta\varepsilon)} \overset{(**)}\geqslant\frac{\sin(\pi/2)-\sin|\gamma_\pm\varepsilon|} {\varepsilon\cos(\gamma_\pm\varepsilon)} \\ &\!\!\overset{(***)}=\frac{1-m\varepsilon |x|/\sqrt{t^2-x^2}} {\varepsilon\sqrt{1-m^2\varepsilon^2 x^2/(t^2-x^2)}} =\frac{\sqrt{1-x^2/t^2}-m\varepsilon |x|/t} {\varepsilon\sqrt{1-(1+m^2\varepsilon^2) x^2/t^2}} \\ &\geqslant\frac{1}{\varepsilon}\biggl(\sqrt{1-\frac{x^2}{t^2}}- \frac{m\varepsilon |x|}{t}\biggr) \geqslant\frac{1}{\varepsilon}\biggl(\sqrt{1-\frac{1}{1+m^2\varepsilon^2}}- \frac{m\varepsilon |x|}{t}\biggr) \\ &=m\biggl(\frac{1}{\sqrt{1+m^2\varepsilon^2}}- \frac{|x|}{t}\biggr)\geqslant {m\delta}. \end{aligned}
\end{equation*}
\notag
$$
Here the first equality, $(*)$, holds for some $\theta\in[|\gamma_\pm|,\pi/(2\varepsilon)]$ by Lagrange’s theorem. The next inequality, $(**)$, holds because the cosine is decreasing on the interval in question. The next one, $(*{*}*)$, is obtained by substituting (45) in. The rest is straightforward because $|x|/t<1/\sqrt{1+m^2\varepsilon^2}-\delta$ by (8). $\Box$ Proof of Lemma 8. Let us prove the lemma for $f_+(p)$ and $\gamma_+\geqslant 0$; for $\gamma_+<0$ and for $f_-(p)$ the proof is similar. We omit the index $+$ in the notation for $f_+$, $\alpha_+$, $\beta_+$, and $\gamma_+$. The lemma follows from the relations
$$
\begin{equation*}
\begin{aligned} \, |f''(p)|&\overset{(*)}{\geqslant} |f''(\beta)|= \frac{ m^2 \varepsilon^3 t \cos(\beta \varepsilon)} {2\pi(m^2\varepsilon^2+\sin^2(\beta\varepsilon))^{3/2}} \\ &\overset{(**)}{\geqslant}\frac{m^2\varepsilon^3 t\cos(\gamma\varepsilon)} {4\pi(m^2\varepsilon^2+\sin^2(\gamma\varepsilon)+ 2m^2\varepsilon^2t^2/(t^2-x^2))^{3/2}} \\ &\overset{(***)}{=}\frac{m^2\varepsilon^3 t\sqrt{t^2- (1+m^2\varepsilon^2)x^2}\,(t^2-x^2)} {4\pi(3m^2\varepsilon^2t^2)^{3/2}} \\ &\geqslant \frac{t\sqrt{1-(1+m^2\varepsilon^2)x^2/t^2}\,(1-x^2/t^2)} {24\pi m}\geqslant \frac{t\delta^{3/2}}{24 \pi m}\,. \end{aligned}
\end{equation*}
\notag
$$
Here inequality $(*)$ is proved as follows. By (56), $f''(p)$ is increasing on $[-\pi/(2\varepsilon),0]$ and decreasing on $[0,\pi/(2\varepsilon)]$, because it is even, the numerator is decreasing on $[0,\pi/(2\varepsilon)]$ and the denominator is increasing on $[0,\pi/(2\varepsilon)]$. Thus $|f''(p)|\geqslant \min\{|f''(\beta)|,|f''(\alpha)|\}$ for $p\in[\alpha,\beta]$ by Lemma 7. But since $f''(p)$ is even and $\gamma\geqslant 0$, by (48) we get
$$
\begin{equation*}
|f''(\alpha)|=\biggl|f''\biggl(\gamma-\frac{m\delta}{2}\biggr)\biggr|= \biggl|f''\biggl(\biggl|\gamma-\frac{m\delta}{2}\biggr|\biggr)\biggr|\geqslant \biggl|f''\biggl(\gamma+\frac{m\delta}{2}\biggr)\biggr|=|f''(\beta)|.
\end{equation*}
\notag
$$
Inequality $(**)$ follows from the following two estimates. First, by Lemma 7 and the convexity of the cosine on the interval $[\gamma\varepsilon,\pi/2]$ we obtain
$$
\begin{equation*}
\cos(\beta\varepsilon)\geqslant \cos\biggl(\frac{\gamma\varepsilon}{2}+ \frac{\pi}{4}\biggr)\geqslant \frac{1}{2}\biggl(\cos(\gamma\varepsilon)+ \cos\frac{\pi}{2}\biggr)=\frac{\cos (\gamma\varepsilon)}{2}\,.
\end{equation*}
\notag
$$
Second, using the inequality $\sin z-\sin w\leqslant z-w$ for $0\leqslant w\leqslant z\leqslant \pi/2$, and then $\delta\leqslant 1$ and (45)–(48), we get
$$
\begin{equation*}
\begin{aligned} \, \sin^2(\beta\varepsilon)-\sin^2(\gamma\varepsilon)&\leqslant \varepsilon(\beta-\gamma)\bigl[\sin(\beta\varepsilon)+ \sin(\gamma\varepsilon)\bigr]\leqslant \varepsilon(\beta-\gamma) \bigl[\varepsilon(\beta-\gamma)+2\sin(\gamma\varepsilon)\bigr] \\ &=\frac{m\varepsilon\delta}{2}\biggl(\frac{m\varepsilon\delta}{2}+ \frac{2m\varepsilon x}{\sqrt{t^2-x^2}}\biggr) \\ &\leqslant\frac{m\varepsilon t}{2\sqrt{t^2-x^2}} \biggl(\frac{2m\varepsilon t}{\sqrt{t^2-x^2}}+ \frac{2m\varepsilon t}{\sqrt{t^2-x^2}}\biggr)= \frac{2m^2\varepsilon^2t^2}{t^2-x^2}\,. \end{aligned}
\end{equation*}
\notag
$$
Equality $(*{*}*)$ is obtained from (45). The remaining estimates are straightforward. $\Box$ Proof of Lemma 12. By Lemmas 4 and 8, for $p\in [\beta_+,\pi/(2\varepsilon)]$ we have
$$
\begin{equation*}
f_+'(p)=f_+'(\gamma_+)+\int_{\gamma_+}^{p}f_+''(p)\,dp\geqslant 0+(p-\gamma_+)\,\frac{t\delta^{3/2}}{24 \pi m}\geqslant (\beta_+-\gamma_+)\,\frac{t\delta^{3/2}}{24 \pi m }= \frac{t\delta^{5/2}}{48 \pi}\,,
\end{equation*}
\notag
$$
because $f_+''(p)\geqslant 0$ by (56). For $p\in [-\pi/(2\varepsilon),\alpha_+]$ and for $f'_-(p)$ the proof is analogous. $\Box$ Proof of Lemma 14. Take $p\in [\alpha_k,\beta_k]$. By (49), the inequalities $\sin z\geqslant z/2$ for $z\in [0,\pi/2]$, and $m\varepsilon\leqslant 1$ we get
$$
\begin{equation*}
\begin{aligned} \, |g(p)|&=\frac{m\varepsilon^2} {2\pi\sqrt{m^2\varepsilon^2+\sin^2(p\varepsilon)}}\leqslant \frac{m\varepsilon^2}{2\pi\sin(p\varepsilon)}\leqslant \frac{m\varepsilon^2}{\pi p\varepsilon}\leqslant \frac{2m\varepsilon^2 K}{\pi^2 k}= \frac{4m\varepsilon^2}{\pi^2 k} \biggl\lceil\frac{\pi}{m\varepsilon}\biggr\rceil \\ &=O\biggl(\frac{\varepsilon}{k}\biggr)=O(U), \\ |g'(p)|&=\frac{m\varepsilon^3 |\sin(p\varepsilon)\cos(p\varepsilon)|} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{3/2}}\leqslant \frac{m\varepsilon^3}{2\pi\sin^2(p\varepsilon)}= {O}\biggl(\frac{\varepsilon}{mk^2}\biggr) ={O}\biggl(\frac{U}{N}\biggr), \\ |g''(p)|&=\frac{m\varepsilon^4|m^2\varepsilon^2+\sin^4(p \varepsilon)- 2(1+m^2\varepsilon^2)\sin^2(p \varepsilon)|} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{5/2}} \\ &\leqslant \frac{m\varepsilon^4(3m^2\varepsilon^2+3\sin^2(p \varepsilon))} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{5/2}}= {O}\biggl(\frac{U}{N^2}\biggr), \\ |g'''(p)|&=\frac{1}{2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{7/2}} \,m\varepsilon^5 |\sin(p\varepsilon)\cos(p\varepsilon)| \\ &\qquad \times |4m^4\varepsilon^4+9m^2\varepsilon^2+\sin^4(p \varepsilon)- (6+10m^2\varepsilon^2)\sin^2(p \varepsilon)| ={O}\biggl(\frac{U}{N^3}\biggr), \\ |f''_\pm(p)|&=\frac{m^2\varepsilon^3 t\cos(p\varepsilon)} {2\pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{3/2}}= {O}\biggl(\frac{t}{mk^3}\biggr)={O}\biggl(\frac{T}{M^2}\biggr), \\ |f'''_\pm(p)|&=\frac{m^2 \varepsilon^4 t|\sin(p \varepsilon)| (m^2\varepsilon^2+\cos (2 p \varepsilon)+2)} {2 \pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{5/2}}= {O}\biggl(\frac{t}{m^2k^4}\biggr)={O}\biggl(\frac{T}{M^3}\biggr), \\ |f^{(4)}_\pm(p)|&=\frac{m^2\varepsilon^5t\cos(p\varepsilon) |m^4\varepsilon^4+3m^2\varepsilon^2+4\sin^4(p \varepsilon)- 2(6+5m^2\varepsilon^2)\sin^2(p \varepsilon)|} {2 \pi(m^2\varepsilon^2+\sin^2(p\varepsilon))^{7/2}} \\ &={O}\biggl(\frac{t}{m^3k^5}\biggr)={O}\biggl(\frac{T}{M^4}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Further, $f_\pm''(p)$ does not change sign on the interval $[\alpha_k,\beta_k]$ because it vanishes only at $\pm\pi/(2\varepsilon)$. We also have Remark 9. Analogously to Steps 3 and 4 above (with a lot of simplifications because there are no stationary points), one can prove that for any $m,\varepsilon,\delta>0$ and each $(x,t)\in\varepsilon\mathbb{Z}^2$ satisfying $|x|/t>1/\sqrt{1+m^2\varepsilon^2}+\delta$ and $\varepsilon\leqslant 1/m$, we have ${a}(x,t,m,\varepsilon)=O(\varepsilon/(mt^{2}\delta^3))$ (see [44], Theorem 3B). 12.4. The large-time limit: the stationary phase method again (Corollaries 1–3)In this section we prove Corollaries 1–3. First we outline the plan of the argument, then prove Corollary 1 modulo a technical lemma, then the lemma itself, and finally Corollaries 2 and 3. The plan of the proof of Corollary 1 (and results such as Problems 4 and 5) consists of four steps (the reader can find an implementation of Steps 1–3 different from the one below in the arXiv version of [49]).
Proof of Corollary 1 (modulo some lemmas).
Step 1. Fix $m,\varepsilon,\delta>0$, denote $n:=1+m^2\varepsilon^2$, $F(v):=F(v,m,\varepsilon)$, and $V:=1/\sqrt{n}-\delta$, and fix $v$ such that $-V\leqslant v\leqslant V$. Let us prove that if $t$ is sufficiently large in terms of $\delta$, $m$, and $\varepsilon$, then
$$
\begin{equation}
\sum_{\substack{-Vt < x\leqslant vt\\x\in\varepsilon\mathbb{Z}}} P(x,t,m,\varepsilon)=F(v)-F(-V)+{O}_{\delta,m,\varepsilon} \biggl(\sqrt{\frac{\varepsilon}{t}}\,\biggr).
\end{equation}
\tag{57}
$$
This follows from the sequence of asymptotic formulae
$$
\begin{equation}
\begin{aligned} \, \nonumber &\sum_{\substack{-V(t+\varepsilon) < x\leqslant v(t+\varepsilon)\\ x\in\varepsilon\mathbb{Z}}}a_1^2(x,t+\varepsilon,m,\varepsilon) \\ \nonumber &\qquad\overset{(*)}=\sum_{\substack{-V(t+\varepsilon)<x\leqslant v(t+\varepsilon) \\(x+t)/\varepsilon\text{ odd}}}\biggl(\frac{2m\varepsilon^2}{\pi t} \biggl(1-\frac{nx^2}{t^2}\biggr)^{-1/2}\sin^2\theta(x,t,m,\varepsilon)+ {O}_{\delta}\biggl(\frac{\varepsilon^2}{t^{2}}\biggr)\biggr) \\ \nonumber &\qquad\overset{(**)}=\sum_{\substack{-Vt<x<vt\\(x+t)/\varepsilon \text{ odd}}} \frac{m\varepsilon^2}{\pi t}\biggl(1-\frac{nx^2}{t^2}\biggr)^{-1/2} \\ \nonumber &\qquad\qquad-\sum_{\substack{-Vt< x< vt\\(x+t)/\varepsilon\text{ odd}}} \frac{m\varepsilon^2}{\pi t}\biggl(1-\frac{nx^2}{t^2}\biggr)^{-1/2} \cos\bigl(2\theta(x,t,m,\varepsilon)\bigr)+ {O}_{\delta}\biggl(\frac{\varepsilon}{t}\biggr) \\ \nonumber &\qquad\overset{(***)}=\sum_{\substack{-Vt<x<vt\\(x+t)/\varepsilon \text{ odd}}} \frac{m\varepsilon\cdot 2\varepsilon/t}{2\pi\sqrt{1-nx^2/t^2}}+ {O}_{\delta,m,\varepsilon}\biggl(\sqrt{\frac{\varepsilon}{t}}\,\biggr) \\ \nonumber &\qquad\overset{(****)}= \int_{-V}^{v}\frac{m\varepsilon\,dv}{2\pi\sqrt{1-nv^2}}+ {O}_{\delta,m,\varepsilon}\biggl(\sqrt{\frac{\varepsilon}{t}}\,\biggr) \\ &\qquad=m\varepsilon\,\frac{\arcsin(\sqrt{n}\,v)-\arcsin(-\sqrt{n}\,V)} {2\pi\sqrt{n}}+ {O}_{\delta,m,\varepsilon}\biggl(\sqrt{\frac{\varepsilon}{t}}\,\biggr) \end{aligned}
\end{equation}
\tag{58}
$$
and the analogous asymptotic formula
$$
\begin{equation*}
\begin{aligned} \, &\sum_{\substack{-V(t+\varepsilon) < x\leqslant v(t+\varepsilon)\\ x\in\varepsilon\mathbb{Z}}}a_2^2(x,t+\varepsilon,m,\varepsilon)= \int_{-V}^{v}\frac{m\varepsilon(1+v)\, dv} {2\pi(1-v)\sqrt{1-nv^2}}+ {O}_{\delta,m,\varepsilon}\biggl(\sqrt{\frac{\varepsilon}{t}}\,\biggr) \\ &\qquad=F(v)-F(-V)-m\varepsilon\,\frac{\arcsin(\sqrt{n}\,v)- \arcsin(-\sqrt{n}\,V)}{2\pi\sqrt{n}}+{O}_{\delta,m,\varepsilon} \biggl(\sqrt{\frac{\varepsilon}{t}}\,\biggr). \end{aligned}
\end{equation*}
\notag
$$
Here $(*)$ follows from Theorem 2 because $\dfrac{|x|}{t}\leqslant \dfrac{V(t+\varepsilon)}{t}< V+\dfrac{\delta}{2}=\dfrac{1}{\sqrt{n}}-\dfrac{\delta}{2}$ for $t$ large enough; the product of the main term and the error term in (9) is estimated by $\varepsilon^2/t^2$. Asymptotic formula $(**)$ holds because the number of terms is less than $t/\varepsilon$ and the (possibly) dropped first and last summands are less than $m\varepsilon^2/(t\sqrt{\delta}\,)$.
Step 2. Let us prove formula $(*{*}*)$. We use the following simplified version of the stationary phase method. Lemma 17 (see [26], the corollary to Theorem 4). Under the assumptions of Lemma 3 (except the ones on $f'(p)$ and $g^{(3)}(p)$, and the inequality $N\geqslant M/\sqrt{T}$ ), if $M=N$ and $M/C\leqslant T\leqslant CM^2$ for some $C>0$, then For notational convenience, assume that $t/\varepsilon$ is odd; otherwise the proof is analogous. Then the summation index $x=2p\varepsilon$ for some integer $p$. We apply Lemma 17 to the functions
$$
\begin{equation}
f_{\pm}(p)= \pm\frac{1}{\pi}\theta(2p\varepsilon,t,m,\varepsilon)\quad\text{and}\quad g(p)=\frac{m\varepsilon^2}{\pi t} \biggl(1-\frac{4n\varepsilon^2p^2}{t^2}\biggr)^{-1/2}
\end{equation}
\tag{59}
$$
and the parameter values
$$
\begin{equation}
M=N=T=\frac{t}{\varepsilon}\,, \quad U=\frac{\varepsilon}{t}\,, \quad \alpha=-\frac{Vt}{2\varepsilon}\,, \quad \beta=\frac{vt}{2\varepsilon}\,.
\end{equation}
\tag{60}
$$
Lemma 18. For $\varepsilon\leqslant 1/m$ there exist $C,C_0,\dots,C_4$ depending on $\delta$, $m$, and $\varepsilon$ but not on $v$ and $p$ such that the parameters (60) and functions (59) satisfy all the assumptions of Lemma 17. Since the parameters (60) satisfy
$$
\begin{equation*}
\frac{(\beta-\alpha)U\sqrt{T}}{M}+\frac{UM}{\sqrt{T}}= {O}\biggl(\sqrt{\frac{\varepsilon}{t}}\,\biggr),
\end{equation*}
\notag
$$
formula $(*{*}*)$ follows. Step 3. Let us prove formula $(*{*}{*}*)$. We use yet another known result. Lemma 19 (Euler’s summation formula; see [26], the remark to Theorem 1). If $g(p)$ is continuously differentiable on $[\alpha,\beta]$ and $\rho(p):=1/2-\{p\}$, then Again assume without loss of generality that $t/\varepsilon$ is odd. We apply Lemma 19 to the same $\alpha$, $\beta$, and $g(p)$ (given by (59) and (60)) as in Step 2. By Lemma 18 we have $g(p)=O_{\delta,m,\varepsilon}(\varepsilon/t)$ and $g'(p)=O_{\delta,m,\varepsilon}(\varepsilon^2/t^2)$. Hence by Lemma 19 the difference between the sum and the integral in $(*{*}{*}*)$ is $O_{\delta,m,\varepsilon}(\varepsilon/t)$, and (57) follows. Step 4. Let us prove the corollary for arbitrary $v\in (-1/\sqrt{n}\,;1/\sqrt{n})$. By (57), for each $\delta$, $m$, and $\varepsilon$ there are $C_1(\delta,m,\varepsilon)$ and $C_2(\delta,m,\varepsilon)$ such that for each $v\in [-1/\sqrt{n}+\delta,1/\sqrt{n}-\delta]$ and each $t\geqslant C_1(\delta,m,\varepsilon)$ we have
$$
\begin{equation*}
\biggl|\,\sum_{(-1/\sqrt{n}+\delta)t< x\leqslant vt} P(x,t,m,\varepsilon)-F(v)\biggr|\leqslant F\biggl(-\frac{1}{\sqrt{n}}+\delta\biggr)+ C_2(\delta,m,\varepsilon)\sqrt{\frac{\varepsilon}{t}}\,.
\end{equation*}
\notag
$$
Clearly, we can assume that $C_1(\delta,m,\varepsilon)$ and $C_2(\delta,m,\varepsilon)$ are decreasing in $\delta$: the larger is the interval $[-1/\sqrt{n}+\delta,1/\sqrt{n}-\delta]$, the weaker is our error estimate in $(*)$–$(*{**}*)$. Let $\delta(t)$ tend to $0$ slowly enough so that $C_1(\delta(t),m,\varepsilon)\leqslant t$ for $t$ sufficiently large in terms of $m$ and $\varepsilon$, and $C_2(\delta(t),m,\varepsilon)\sqrt{\varepsilon/t}\to 0$ as $t\to+\infty$. Denote $V(t):=1/\sqrt{n}-\delta(t)$. Then since $F(-1/\sqrt{n}+\delta)\to F(-1/\sqrt{n}\,)=0$ as $\delta \to 0$ by the definition of $F(v)$, it follows that
$$
\begin{equation}
\sum_{-V(t)t< x\leqslant vt}P(x,t,m,\varepsilon)\rightrightarrows F(v)\quad\text{as } t\to+\infty
\end{equation}
\tag{61}
$$
uniformly in $v\in(-1/\sqrt{n}\,,1/\sqrt{n}\,)$. Similarly, since $F(1/\sqrt{n}-\delta)\to F(1/\sqrt{n}\,)=1$ as $\delta \to 0$, we get
$$
\begin{equation*}
\sum_{-V(t)t< x\leqslant V(t)t}P(x,t,m,\varepsilon)\to 1\quad\text{as } t\to+\infty.
\end{equation*}
\notag
$$
Then by Proposition 6 we get
$$
\begin{equation*}
\begin{aligned} \, \sum_{x\leqslant -V(t)t}P(x,t,m,\varepsilon)&= 1-\sum_{x>-V(t)t}P(x,t,m,\varepsilon) \\ &\leqslant 1-\sum_{-V(t)t< x\leqslant V(t)t}P(x,t,m,\varepsilon)\to 0. \end{aligned}
\end{equation*}
\notag
$$
With (61), this implies the corollary for $v\in(-1/\sqrt{n}\,;1/\sqrt{n}\,)$. For $v\leqslant -1/\sqrt{n}$ and similarly for $v\geqslant 1/\sqrt{n}$, the corollary follows from
$$
\begin{equation*}
\sum_{x\leqslant vt}P(x,t,m,\varepsilon)\leqslant \sum_{x\leqslant -V(t)t}P(x,t,m,\varepsilon)\to 0. \qquad\square
\end{equation*}
\notag
$$
Now we prove Lemma 18 and Corollaries 2 and 3. Proof of Lemma 18. The inequalities $M/C\leqslant T\leqslant CM^2$ and $M\geqslant \beta-\alpha$ are automatic for $C=1$ because $t/\varepsilon$ is a positive integer and $|V|,|v|\leqslant 1$. We estimate the derivatives (computed in [45], § 9) as follows, using the assumptions that $\varepsilon\leqslant 1/m$ and $\alpha\leqslant p\leqslant \beta$ and setting $C_2:=\max\{1/(m\varepsilon),2/\delta^{3/2}\}$:
$$
\begin{equation*}
\begin{aligned} \, |g(p)|&=\frac{m\varepsilon^2}{\pi t} \biggl(1-\frac{4n\varepsilon^2p^2}{t^2}\biggr)^{-1/2}\leqslant \frac{m\varepsilon^2}{\pi t\sqrt{1-nV^2}}\leqslant \frac{m\varepsilon^2}{t\sqrt{\delta}} \\ &\leqslant \frac{\varepsilon}{t\sqrt{\delta}}={O}_{\delta}(U), \\ |g'(p)|&=\frac{4m\varepsilon^4 n|p|}{\pi t^3} \biggl(1-\frac{4n\varepsilon^2p^2}{t^2}\biggr)^{-3/2}\leqslant \frac{2m\varepsilon^3 nVt}{\pi t^3(1-nV^2)^{3/2}}\leqslant \frac{m\varepsilon^3 n}{t^2\delta^{3/2}} \\ &\leqslant \frac{2\varepsilon^2}{t^2\delta^{3/2}}= {O}_{\delta}\biggl(\frac{U}{N}\biggr), \\ |g''(p)|&=\frac{4m\varepsilon^4 n(8\varepsilon^2 np^2+t^2)}{\pi t^5} \biggl(1-\frac{4n\varepsilon^2p^2}{t^2}\biggr)^{-5/2}\leqslant \frac{4m\varepsilon^4 n(2nV^2+1)t^2}{\pi t^5(1-nV^2)^{5/2}} \\ &={O}_{\delta}\biggl(\frac{\varepsilon^3}{t^3}\biggr)= {O}_{\delta}\biggl(\frac{U}{N^2}\biggr), \\ |f''(p)|&=\frac{4m\varepsilon^2}{\pi t} \biggl(1-\frac{4\varepsilon^2p^2}{t^2}\biggr)^{-1} \biggl(1-\frac{4n\varepsilon^2p^2}{t^2}\biggr)^{-1/2}\geqslant \frac{m\varepsilon^2}{t}\geqslant \frac{T}{C_2M^2}\,, \\ |f''(p)|&=\frac{4m\varepsilon^2}{\pi t} \biggl(1-\frac{4\varepsilon^2p^2}{t^2}\biggr)^{-1} \biggl(1-\frac{4n\varepsilon^2p^2}{t^2}\biggr)^{-1/2}\leqslant \frac{4m\varepsilon^2}{\pi t (1-V^2)\sqrt{1-nV^2}} \\ &\leqslant \frac{2m\varepsilon^2}{t\delta^{3/2}}\leqslant \frac{C_2T}{M^2}\,, \\ |f'''(p)|&=\frac{16m\varepsilon^4|(n+2)pt^2-12n\varepsilon^2 p^3|}{\pi t^5} \biggl(1-\frac{4\varepsilon^2p^2}{t^2}\biggr)^{-2} \biggl(1-\frac{4n\varepsilon^2p^2}{t^2}\biggr)^{-3/2} \\ &={O}_{\delta}\biggl(\frac{T}{M^3}\biggr), \\ |f^{(4)}(p)|&=\frac{1}{\pi t^9(1-{4\varepsilon^2p^2}/{t^2})^{3} (1-{4n\varepsilon^2p^2}/{t^2})^{5/2}} \\ &\qquad\times 16m\varepsilon^4|768n^2\varepsilon^6p^6-48n(2n+5) \varepsilon^4p^4 t^2 \\ &\qquad\qquad\qquad+8(n^2-n+3)\varepsilon^2 p^2 t^4+(n+2)t^6| =O_{\delta}\biggl(\frac{T}{M^4}\biggr). \qquad\square \end{aligned}
\end{equation*}
\notag
$$
Proof of Corollary 2. We have
$$
\begin{equation*}
n_+(h\times w)-n_-(h\times w)=-2^{(w+h-1)/2}a_1(w-h,w+h)
\end{equation*}
\notag
$$
by the obvious bijection between the Young diagrams with exactly $h$ rows and $w$ columns, and checker paths from $(0,0)$ to $(w-h,w+h)$ passing through $(1,1)$ and $(w-h+1,w+h-1)$ (see Fig. 11, left). Set $h:=\lceil rw\rceil$ and apply Theorem 2 and Remark 9 (or Theorem 4) to
$$
\begin{equation*}
\delta=\frac{1}{2}\biggl|\frac{1}{\sqrt{2}}-\frac{r-1}{r+1}\biggr|,\qquad m=\varepsilon=1, \quad x=w-h, \quad t=w+h-1.
\end{equation*}
\notag
$$
This completes the proof in the case when $r>3+2\sqrt{2}$. It remains to show that for $r<3+2\sqrt{2}$ the quantity (11) occurs in arbitrarily small neighbourhoods of points $\pi/2+\pi\mathbb{Z}$ as $w\to\infty$.
Denote $v:=(h-w)/(w+h-1)$ and $v_0:=(r-1)/(r+1)$. Write
$$
\begin{equation*}
\theta(vt,t,1,1)=t\biggl(\arcsin\frac{1}{\sqrt{2-2v^2}}- v\arcsin\frac{v}{\sqrt{1-v^2}}\biggr)+\frac{\pi}{4}=: t\theta(v)+\frac{\pi}{4}\,.
\end{equation*}
\notag
$$
Since $\theta(v)\in C^2[0;1/\sqrt{2}-\delta]$, by the Taylor expansion it follows that
$$
\begin{equation*}
\theta(vt,t,1,1)=\frac{\pi}{4}+t\theta(v_0)+t(v-v_0)\theta'(v_0)+ O_\delta\bigl(t(v-v_0)^2\bigr).
\end{equation*}
\notag
$$
Substituting
$$
\begin{equation*}
v-v_0=\frac{h-w}{w+h-1}-\frac{r-1}{r+1}= \frac{2h-2rw+r-1}{(r+1)(w+h-1)}=\frac{2\{-rw\}+r-1}{(r+1)t}\,,
\end{equation*}
\notag
$$
where $h=\lceil rw\rceil=rw+\{-rw\}$, we get
$$
\begin{equation*}
\begin{aligned} \, \theta(vt,t,1,1)&=\frac{\pi}{4}+(w+rw+\{-rw\}-1)\theta(v_0) \\ &\qquad+\frac{2\{-rw\}+r-1}{(r+1)}\,\theta'(v_0) +O_\delta\biggl(\frac{1}{w}\biggr) \\ &=\frac{\pi}{4}-\theta(v_0)+v_0\theta'(v_0)+w(r+1)\theta(v_0) \\ &\qquad+\{-rw\}\biggl(\theta(v_0)+\frac{2}{r+1}\theta'(v_0)\biggr)+ O_\delta\biggl(\frac{1}{w}\biggr) \\ &=:\pi\bigl(\alpha(r)w+\beta(r)\{-rw\}+\gamma(r)\bigr)+ O_\delta\biggl(\frac{1}{w}\biggr). \end{aligned}
\end{equation*}
\notag
$$
For almost every $r$, the numbers $1$, $r$, and $\alpha(r)$ are linearly independent over the rational numbers because the graph of the function $\alpha(r)=(r+1)\theta\bigl((r-1)/(r+1)\bigr)$ has just a countable number of points of intersection with the straight lines given by equations with rational coefficients. Hence by Kronecker’s theorem, for each $\Delta>0$ there are infinitely many $w$ such that
$$
\begin{equation*}
\{-rw\}<\Delta\quad\text{and}\quad \biggl|\{\alpha(r)w\}+\gamma(r)-\frac{1}{2}\biggr|<\Delta.
\end{equation*}
\notag
$$
By (9), Corollary 2 follows because these $w$ satisfy and We are going to deduce Corollary 3 from the results in [49]. Proof of Corollary 3. We apply Corollary 1.5 and Theorem 1.1 in [49] to
$$
\begin{equation*}
\begin{gathered} \, n=\frac{t}{\varepsilon}-1, \quad y_n=2\varepsilon\biggl\lceil \frac{vt}{2\varepsilon}\biggr\rceil-1,\quad \xi=v, \quad \phi=(0,1)^\top, \\ a=\frac{1}{\sqrt{1+m^2\varepsilon^2}}\,, \quad b=\frac{m\varepsilon}{\sqrt{1+m^2\varepsilon^2}}\,. \end{gathered}
\end{equation*}
\notag
$$
Case $|v|>1/\sqrt{1+m^2\varepsilon^2}$ follows from [49], Corollary 1.5. Case $|v|<1/\sqrt{1+m^2\varepsilon^2}$ follows from
$$
\begin{equation}
\begin{aligned} \, \nonumber \frac{2m\varepsilon^2}{\pi t(1-v)}\sqrt{\frac{a-|v|}{a+|v|}}+ {O}_{m,\varepsilon,v}\biggl(\frac{1}{t^2}\biggr)&\leqslant P\biggl(2\varepsilon\biggl\lceil\frac{vt}{2\varepsilon}\biggr\rceil, t,m,\varepsilon\biggr) \\ &\leqslant\frac{2m\varepsilon^2}{\pi t(1-v)} \sqrt{\frac{a+|v|}{a-|v|}}+{O}_{m,\varepsilon,v}\biggl(\frac{1}{t^2}\biggr), \end{aligned}
\end{equation}
\tag{62}
$$
where $t\in 2\varepsilon\mathbb{Z}$ and $a:=1/\sqrt{1+m^2\varepsilon^2}$. Estimate (62) follows from [49], Theorem 1.1, because the quantity $\mathrm{OSC}_n(\xi)$ in formula (1.11) in [49] satisfies
$$
\begin{equation*}
|\mathrm{OSC}_n(\xi)|\leqslant \sqrt{A(\xi)^2+B(\xi)^2}= \frac{|\xi|(1+\xi)}{|a|}
\end{equation*}
\notag
$$
(checked in [45], § 18).
Alternatively, (62) can be deduced from Theorem 2 using the method of § 12.5. $\Box$ 12.5. The solution of the Feynman problem: Taylor expansions (Corollaries 4 and 5)Here we deduce the solution of the Feynman problem from Theorem 2. For this purpose we approximate the functions in Theorem 2 by a few terms of their Taylor expansions. Proof of Corollary 4. First we derive an asymptotic formula for the function $\theta(x,t,m,\varepsilon)$ given by (11). Denote $n:=1+m^2\varepsilon^2$. Since
$$
\begin{equation*}
\frac{1}{\sqrt{1+z^2}}=1+O(z^2),\quad \arcsin z=z+O(z^3)\qquad\text{for } z\in[-1;1]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\frac{t}{\sqrt{t^2-x^2}}<\frac{1}{\sqrt{1-\sqrt{n}\,x/t}}< \frac{1}{\sqrt{\delta}}\,,
\end{equation*}
\notag
$$
we get
$$
\begin{equation*}
\begin{aligned} \, \arcsin\frac{m\varepsilon t}{\sqrt{n(t^2-x^2)}} &=\frac{m\varepsilon t}{\sqrt{1+m^2\varepsilon^2}\,\sqrt{t^2-x^2}}+ {O}\biggl(\frac{m^3\varepsilon^3}{n^{3/2}} \biggl(\frac{t}{\sqrt{t^2-x^2}}\biggr)^3\,\biggr) \\ &=\frac{m\varepsilon t}{\sqrt{t^2-x^2}}+ O_\delta(m^3\varepsilon^3). \end{aligned}
\end{equation*}
\notag
$$
Combining this with a similar asymptotic formula for $\arcsin(m\varepsilon x/\sqrt{t^2-x^2})$, we get
$$
\begin{equation*}
\begin{aligned} \, \theta(x,t,m,\varepsilon)&=\frac{m t^2} {\sqrt{t^2-x^2}}- \frac{m x^2} {\sqrt{t^2-x^2}}+\frac{\pi}{4}+ \frac{t+|x|}{\varepsilon}\,O_\delta(m^3\varepsilon^3) \\ &=m\sqrt{t^2-x^2}+\frac{\pi}{4}+O_\delta(m^3\varepsilon^2t). \end{aligned}
\end{equation*}
\notag
$$
Since
$$
\begin{equation*}
\biggl|\frac{\partial \sqrt{t^2-x^2}}{\partial t}\biggr|= \frac{t}{\sqrt{t^2-x^2}}<\frac{1}{\sqrt{\delta}}\quad\text{and}\quad \biggl|\frac{\partial \sqrt{t^2-x^2}}{\partial x}\biggr|= \frac{|x|}{\sqrt{t^2-x^2}}<\frac{1}{\sqrt{\delta}}\,,
\end{equation*}
\notag
$$
by Lagrange’s theorem it follows that
$$
\begin{equation*}
\theta(x,t-\varepsilon,m,\varepsilon)=m\sqrt{t^2-x^2}+\frac{\pi}{4}+ O_\delta(m\varepsilon+m^3\varepsilon^2t)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\theta(x-\varepsilon,t-\varepsilon,m,\varepsilon)= m\sqrt{t^2-x^2}+\frac{\pi}{4}+O_\delta(m\varepsilon+m^3\varepsilon^2t).
\end{equation*}
\notag
$$
Consider the remaining factors in (9) and (10). By Lagrange’s theorem, for some $\eta\in[0,nx^2/t^2]$ we get
$$
\begin{equation*}
\begin{aligned} \, \biggl(1-\frac{nx^2}{t^2}\biggr)^{-1/4}-1&=\frac{nx^2}{t^2}\, \frac{(1-\eta)^{-5/4}}{4} \leqslant \frac{nx^2}{t^2} \biggl(1-\frac{nx^2}{t^2}\biggr)^{-5/4} \\ &\leqslant \frac{x^2}{t^2}\biggl(\frac{1}{\sqrt{n}}-\frac{x}{t}\biggr)^{-5/4} \biggl(\frac{1}{\sqrt{n}}+\frac{x}{t}\biggr)^{-5/4} \\ &\leqslant\frac{x^2}{t^2}\,\delta^{-5/2}=O_\delta\biggl(\frac{|x|}{t}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Hence for $t\geqslant 2\varepsilon$ we get
$$
\begin{equation}
\begin{aligned} \, \biggl(1-\frac{nx^2}{(t-\varepsilon)^2}\biggr)^{-1/4}&= 1+O_\delta\biggl(\frac{|x|}{t}\biggr), \\ \biggl(1-\frac{n(x-\varepsilon)^2}{(t-\varepsilon)^2}\biggr)^{-1/4}&= 1+O_\delta\biggl(\frac{|x|+\varepsilon}{t}\biggr). \end{aligned}
\end{equation}
\tag{63}
$$
We also have
$$
\begin{equation}
\sqrt{\frac{t-\varepsilon+x-\varepsilon}{t-x}}= \sqrt{1+2\frac{x-\varepsilon}{t-x}}= 1+O\biggl(\frac{x-\varepsilon}{t-x}\biggr)= 1+O_\delta\biggl(\frac{|x|+\varepsilon}{t}\biggr).
\end{equation}
\tag{64}
$$
Substituting all the resulting asymptotic formulae into (9) and (10), we get
$$
\begin{equation*}
\begin{aligned} \, \operatorname{Re}{a}(x,t,m,\varepsilon)&=\varepsilon\sqrt{\frac{2m}{\pi t}}\, \biggl(\sin\biggl(m\sqrt{t^2-x^2}+\frac{\pi}{4}\biggr) \\ &\qquad+O_\delta\biggl(\frac{1}{mt}+\frac{|x|+\varepsilon}{t}+m\varepsilon+ m^3\varepsilon^2t\biggr)\biggr) \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \operatorname{Im}{a}(x,t,m,\varepsilon)&=\varepsilon\sqrt{\frac{2m}{\pi t}}\, \biggl(\cos\biggl(m\sqrt{t^2-x^2}+\frac{\pi}{4}\biggr) \\ &\qquad+O_\delta\biggl(\frac{1}{mt}+ \frac{|x|+\varepsilon}{t}+m\varepsilon+m^3\varepsilon^2t\biggr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
Since $m\varepsilon\leqslant 1/(mt)+m^3\varepsilon^2t$ and $\varepsilon/t\leqslant 1/(mt)$ by the assumption $\varepsilon\leqslant 1/m$, it follows that the error terms can be rewritten in the required form. $\Box$ Proof of Corollary 5. Equality (22) follows directly from Corollary 4 by plugging in the asymptotic formula Proof of the assertion in Example 4. The case when
$$
\begin{equation*}
(x_n,t_n,\varepsilon_n)=\biggl(n^3,n^4,\frac{1}{n^4}\biggr)
\end{equation*}
\notag
$$
follows from Corollary 4 by plugging in the Taylor expansion
In the remaining cases we need to estimate $\theta(\varepsilon,t,m,\varepsilon)$ given by (11) for $t\geqslant 2\varepsilon$ and $\varepsilon\leqslant 1/m$. Since
$$
\begin{equation*}
\frac{m\varepsilon t}{\sqrt{(1+m^2\varepsilon^2)(t^2-\varepsilon^2)}} \leqslant \sqrt{\frac{2}{3}}
\end{equation*}
\notag
$$
for such $t$, $m$, and $\varepsilon$, and for $0<w<z<\sqrt{2/3}$, and for $|z|\leqslant 1/2$, we get
$$
\begin{equation*}
\begin{aligned} \, &\theta(\varepsilon,t,m,\varepsilon)-\theta(0,t,m,\varepsilon) \\ &\qquad=\frac{t}{\varepsilon}\arcsin\frac{m\varepsilon t} {\sqrt{(1+m^2\varepsilon^2)(t^2-\varepsilon^2)}}- \frac{t}{\varepsilon}\arcsin\frac{m\varepsilon}{\sqrt{1+m^2\varepsilon^2}}- \arcsin\frac{m\varepsilon^2}{\sqrt{t^2-\varepsilon^2}} \\ &\qquad={O}\biggl(\frac{m\varepsilon^2}{t}\biggr)=O\biggl(\frac{1}{mt}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Then by Theorem 2 and (63), (64) for $x=0$ and $\delta=1/2$, we get
$$
\begin{equation}
a(0,t,m,\varepsilon)=\varepsilon\sqrt{\frac{2m}{\pi t}}\, \exp\biggl(-i\biggl(\frac{t}{\varepsilon}-1\biggr) \arctan(m\varepsilon)+\frac{i\pi}{4}\biggr) \biggl(1+{O}\biggl(\frac{1}{mt}\biggr)\biggr).
\end{equation}
\tag{65}
$$
In the case when $\varepsilon=\varepsilon_n=1/(2n)$ and $t=t_n=(2n)^2$ the right-hand side of (65) is equivalent to the right-hand side of (22) times $e^{im^3/3}$ because $\arctan(m\varepsilon)=m\varepsilon-m^3\varepsilon^3/3 +O(m^5\varepsilon^5)$. In the case when $\varepsilon=\varepsilon_n=\mathrm{const}$ and $t=t_n=2n\varepsilon$ the ratio of the right-hand sides of (65) and (22) has no limit because $\arctan(m\varepsilon)-m\varepsilon$ is not an integer multiple of $\pi$ for $0<\varepsilon<1/m$. $\Box$ 12.6. The continuum limit: the tail-exchange method (Theorem 5 and Corollaries 6 and 7) Proof of Theorem 5. The proof is based on the tail-exchange method and consists of five steps.
Let us derive the asymptotic formula for $a_1(x,t,m,\varepsilon)$; the argument for $a_2(x,t,m,\varepsilon)$ is analogous. Step 1. Consider the first factor in (6). We have $0\geqslant (1-t/\varepsilon)/2\geqslant -t/\varepsilon$ because $t\geqslant \delta$. Exponentiating, we get
$$
\begin{equation*}
1\geqslant (1+m^2\varepsilon^2)^{(1-t/\varepsilon)/2}\geqslant (1+{m^2}{\varepsilon^2})^{-t/\varepsilon}\geqslant e^{-{m^2}{\varepsilon^2}t/\varepsilon}\geqslant 1-m^2 t\varepsilon,
\end{equation*}
\notag
$$
where in the last two inequalities we used that $e^a\geqslant 1+a$ for each $a\in\mathbb{R}$. Thus
$$
\begin{equation*}
(1+{m^2}{\varepsilon^2})^{(1-t/\varepsilon)/2}=1+O(m^2t\varepsilon).
\end{equation*}
\notag
$$
Step 2. Consider the $T$th partial sum in (6) with $T=\lceil\log(\delta/\varepsilon)\rceil$ terms. The total number of terms is indeed at least $T$ because
$$
\begin{equation*}
\frac{t-|x|}{2\varepsilon}\geqslant \frac{\delta}{2\varepsilon}\geqslant \log\frac{\delta}{\varepsilon}
\end{equation*}
\notag
$$
by the inequalities $t-|x|\geqslant \delta$ and $e^a\geqslant 1+a+a^2/2\geqslant 2a$ for each $a\geqslant 0$.
For $r\geqslant T$ the ratio of consecutive terms in (6) equals
$$
\begin{equation*}
\begin{aligned} \, &(m\varepsilon)^2\frac{\bigl((t+x)/(2\varepsilon)-1-r\bigr) \bigl((t-x)/(2\varepsilon)-1-r\bigr)}{(r+1)^2}& \\ &\qquad<(m\varepsilon)^2\cdot \frac{t+x}{2\varepsilon T}\, \frac{t-x}{2\varepsilon T} =\frac{m^2s^2}{4T^2}<\frac{1}{2}\,, \end{aligned}
\end{equation*}
\notag
$$
where the latter inequality follows from
$$
\begin{equation*}
T=\biggl\lceil\log\frac\delta\varepsilon\biggr\rceil> \lceil\log e^{3ms}\rceil\geqslant 3ms.
\end{equation*}
\notag
$$
Therefore, the error term (that is, the sum over $r\geqslant T$) is less then the sum of geometric series with ratio $1/2$. Thus by Proposition 11 we get
$$
\begin{equation*}
\begin{aligned} \, a_1(x,t,m,\varepsilon)&=m\varepsilon\bigl(1+O(m^2 t\varepsilon)\bigr) \\ &\qquad\times\biggl[\,\sum_{r=0}^{T-1}(-1)^r \begin{pmatrix} (t+x)/(2\varepsilon)-1 \\ r \end{pmatrix} \begin{pmatrix}(t-x)/(2\varepsilon)-1 \\ r \end{pmatrix} (m\varepsilon)^{2r} \\ &\qquad+O\biggl(\begin{pmatrix} (t+x)/(2\varepsilon)-1 \\ T\end{pmatrix} \begin{pmatrix} (t-x)/(2\varepsilon)-1 \\ T \end{pmatrix} (m\varepsilon)^{2T}\biggr)\biggr]. \end{aligned}
\end{equation*}
\notag
$$
Step 3. To approximate the sum, take the integers $L:=(t\pm x)/(2\varepsilon)$ and $r<T$ and transform binomial coefficients as follows:
$$
\begin{equation*}
\begin{pmatrix} L-1 \\ r \end{pmatrix}=\frac{(L-1)\cdots(L-r)}{r!}= \frac{L^r}{r!}\biggl(1-\frac{1}{L}\biggr)\cdots\biggl(1-\frac{r}{L}\biggr).
\end{equation*}
\notag
$$
Here
$$
\begin{equation*}
\frac{r}{L}=\frac{2r\varepsilon}{t\pm x}<\frac{2T\varepsilon}{\delta}= \frac{2\varepsilon}{\delta} \biggl\lceil\log\frac{\delta}{\varepsilon}\biggr\rceil\leqslant \frac{2\varepsilon}{\delta}\biggl(\log\frac{\delta}{\varepsilon}+1\biggr) < \frac{1}{2}\,,
\end{equation*}
\notag
$$
because $\delta/\varepsilon\geqslant 16$, and $2(\log a+1)/a$ is decreasing for $a\geqslant 16$ and is less than $1/2$ for ${a=16}$. Using the inequality $1-a\geqslant e^{-2a}$ for $0\leqslant a\leqslant 1/2$ and then the inequalities $1-a\leqslant e^{-a}$ and $L\geqslant\delta/(2\varepsilon)$, we get
$$
\begin{equation*}
\begin{aligned} \, \biggl(1-\frac{1}{L}\biggr)\cdots\biggl(1-\frac{r}{L}\biggr)&\geqslant e^{{-2/L}}e^{{-4/L}}\cdots e^{{-2r/L}}=e^{{-r(r+1)/L}} \\ &\geqslant e^{-T^2/L}\geqslant 1-\frac{T^2}{L}\geqslant 1-\frac{2T^2\varepsilon}{\delta}\,. \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\frac{(t\pm x)^r}{r!\,(2\varepsilon)^r} \geqslant \begin{pmatrix}(t\pm x)/(2\varepsilon)-1 \\ r\end{pmatrix}\geqslant \frac{(t\pm x)^r}{r!\,(2\varepsilon)^r} \biggl(1-\frac{2T^2 \varepsilon}{\delta}\biggr).
\end{equation*}
\notag
$$
Substituting the result into the expression for $a_1(x,t,m,\varepsilon)$ from Step 2, we get
$$
\begin{equation*}
\begin{aligned} \, a_1(x,t,m,\varepsilon)&=m\varepsilon\bigl(1+O(m^2t\varepsilon)\bigr) \\ &\qquad\times\biggl[\,\sum_{r=0}^{T-1}(-1)^r\biggl(\frac{m}{2}\biggr)^{2r} \frac{(t^2-x^2)^r}{(r!)^2} \biggl(1+O\biggl(\frac{T^2 \varepsilon}{\delta}\biggr)\biggr) \\ &\qquad+O\biggl(\biggl(\frac{m}{2}\biggr)^{2T}\frac{(t^2-x^2)^T}{(T!)^2} \biggl(1+\frac{T^2 \varepsilon}{\delta}\biggr)\biggr)\biggr]. \end{aligned}
\end{equation*}
\notag
$$
The latter error term in the formula is estimated as follows. Since
$$
\begin{equation*}
T!\geqslant \biggl(\frac{T}{3}\biggr)^T\quad\text{and}\quad T\geqslant \log \frac{\delta}{\varepsilon}\geqslant 3m\sqrt{t^2-x^2}\geqslant \frac{3m}{2}\sqrt{t^2-x^2}\,\sqrt{e}\,,
\end{equation*}
\notag
$$
it follows that
$$
\begin{equation*}
\frac{(t^2-x^2)^T}{(T!)^2}\,\biggl(\frac{m}{2}\biggr)^{2T}\leqslant \frac{(t^2-x^2)^T}{(T)^{2T}}\,\biggl(\frac{3m}{2}\biggr)^{2T}\leqslant e^{-T}\leqslant \frac{\varepsilon}{\delta}\,.
\end{equation*}
\notag
$$
We have $(\varepsilon/\delta)(1+T^2 \varepsilon/\delta)= O(T^2 \varepsilon/\delta)$ because $T\geqslant 1$ and $\varepsilon<\delta$. Thus the error term in question can be absorbed into $O(T^2\varepsilon/\delta)$. We get
$$
\begin{equation*}
a_1(x,t,m,\varepsilon)=m\varepsilon\bigl(1+O(m^2t\varepsilon)\bigr)\, \sum_{r=0}^{T-1}(-1)^r\biggl(\frac{m}{2}\biggr)^{2r}\frac{(t^2-x^2)^r}{(r!)^2} \biggl(1+O\biggl(\frac{T^2 \varepsilon}{\delta}\biggr)\biggr).
\end{equation*}
\notag
$$
Notice that by our notational convention the constant implicit in $O(T^2 \varepsilon/\delta)$ does not depend on $r$.
Step 4. Now we can replace the sum with $T$ terms by an infinite sum because the ‘tail’ of alternating series with decreasing absolute value of the summands can be estimated by the first summand (which has just been estimated):
$$
\begin{equation*}
\biggl|\,\sum_{r=T}^{\infty}(-1)^r\biggl(\frac{m}{2}\biggr)^{2r} \frac{(t^2-x^2)^r}{(r!)^2}\biggr|\leqslant\frac{(t^2-x^2)^T}{(T!)^2}\, \biggl(\frac{m}{2}\biggr)^{2T}\leqslant \frac\varepsilon\delta= O\biggl(\frac{T^2 \varepsilon}{\delta}\biggr).
\end{equation*}
\notag
$$
Since the constant implicit in each term $O(T^2 \varepsilon/\delta)$ is the same (see Step 3), we get
$$
\begin{equation*}
\begin{aligned} \, a_1(x,t,m,\varepsilon)&=m\varepsilon\bigl(1+O(m^2t\varepsilon)\bigr)\, \sum_{r=0}^{\infty}(-1)^r\biggl(\frac{m}{2}\biggr)^{2r} \frac{(t^2-x^2)^r}{(r!)^2} \biggl[1+O\biggl(\frac{T^2\varepsilon}{\delta}\biggr)\biggr] \\ &=m\varepsilon\bigl(1+O({m^2t\varepsilon})\bigr)\, \biggl(J_0(ms)+O\biggl(\frac{T^2 \varepsilon}{\delta}I_0(ms)\biggr)\biggr), \end{aligned}
\end{equation*}
\notag
$$
where we use the modified Bessel functions of the first kind:
$$
\begin{equation*}
I_0(z):=\sum_{k=0}^\infty\frac{(z/2)^{2k}}{(k!)^2}\quad\text{and}\quad I_1(z):=\sum_{k=0}^\infty\frac{(z/2)^{2k+1}}{k!\,(k+1)!}\,.
\end{equation*}
\notag
$$
Step 5. We have $m^2t\delta\leqslant m^2(t+|x|)(t-|x|)= m^2s^2\leqslant 9m^2s^2\leqslant T^2$. Thus
$$
\begin{equation*}
{m^2t\varepsilon}\, J_0(ms)\leqslant \frac{T^2\varepsilon I_0(ms)}{\delta}\quad\text{and}\quad {m^2t\varepsilon}\leqslant \frac{T^2\varepsilon}{\delta}< \frac{(\log(\delta/\varepsilon)+1)^2{\varepsilon}}{\delta}< 2,
\end{equation*}
\notag
$$
because $(a+1)^2/2<e^a$ for $a\geqslant 0$. We arrive at the formula
$$
\begin{equation*}
a_1(x,t,{m},{\varepsilon}) ={m}{\varepsilon}\biggl(J_0(ms)+O\biggl(\frac{\varepsilon}{\delta} \log^2\biggl(\frac{\delta}{\varepsilon}\biggr) I_0(ms)\biggr)\biggr).
\end{equation*}
\notag
$$
Analogously,
$$
\begin{equation*}
\begin{aligned} \, a_2(x,t,m,\varepsilon)&=m\varepsilon\bigl(1+O({m^2t\varepsilon})\bigr) \frac{t+x}{\sqrt{t^2-x^2}}\sum_{r=1}^{T-1}(-1)^r \biggl(\frac{m}{2}\biggr)^{2r-1} \\ &\qquad\times\frac{(t^2-x^2)^{(2r-1)/2}}{(r-1)!\,r!} \biggl[1+O\biggl(\frac{T^2\varepsilon}{\delta}\biggr)\biggr] \\ &=-m{\varepsilon}\,\frac{t+x}{s}\biggl(J_1(ms)+ O\biggl(\frac{\varepsilon}{\delta} \log^2\biggl(\frac{\delta}{\varepsilon}\biggr)I_1(ms)\biggr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
This gives the required asymptotic formula for $a(x,t,m,\varepsilon)$ because
$$
\begin{equation*}
\begin{aligned} \, I_0(ms)&\leqslant \sum_{k=0}^\infty \frac{(ms/2)^{2k}}{k!}= e^{m^2s^2/4}\leqslant e^{m^2t^2}, \\ \frac{t+x}{s}I_1(ms)&\leqslant \frac{t+x}{s}\,\frac{ms}{2} \sum_{k=0}^\infty \frac{(ms/2)^{2k}}{k!}=m\,\frac{t+x}{2}e^{m^2s^2/4} \\ &\leqslant mt\,e^{m^2t^2/4}\leqslant e^{m^2t^2/2}e^{m^2t^2/4}\leqslant e^{m^2t^2}. \qquad\square\end{aligned}
\end{equation*}
\notag
$$
Proof of Corollary 6. This follows from Theorem 5 because the right-hand side of (23) is uniformly continuous on each compact subset of the angle $|x|<t$. Proof of Corollary 7. Since the right-hand side of (23) is continuous on $[-t+\delta; t-\delta]$, it is bounded there. Since a sequence of bounded functions uniformly converging to a bounded function is uniformly bounded, by Corollary 6 the absolute value of the left-hand side of (23) is less than some constant $C_{t,m,\delta}$ depending on $t$, $m$, and $\delta$ but not on $x$ and $\varepsilon$. Then by Proposition 6 for $t/(2\varepsilon)\in \mathbb{Z}$ we get
$$
\begin{equation*}
\begin{aligned} \, 1-\sum_{x\in\varepsilon\mathbb{Z}:\,|x|\geqslant t-\delta} P(x,t,m,\varepsilon)&=\sum_{x\in\varepsilon\mathbb{Z}:\, |x|<t-\delta} P(x,t,m,\varepsilon) \\ &=\sum_{x\in\varepsilon\mathbb{Z}:\,|x|<t-\delta} 4\varepsilon^2 \biggl|\frac{1}{2\varepsilon}a(x,t,m,\varepsilon)\biggr|^2 \\ &< 4\varepsilon^2C_{t,m,\delta}^2\,\frac{t-\delta}{\varepsilon}\to 0 \quad\text{as }\varepsilon\to 0. \qquad\square\end{aligned}
\end{equation*}
\notag
$$
12.7. The probability of chirality flip: combinatorial identities (Theorem 6)Although Theorem 6 can be deduced from (58), we give a direct proof relying on § 12.1 only. Proof of Theorem 6. Denote
$$
\begin{equation*}
S_1(t)=\sum_{x\in\mathbb{Z}}a_1^2(x,t),\qquad S_2(t)=\sum_{x\in\mathbb{Z}}a_2^2(x,t),\quad\text{and}\quad S_{12}(t)=\sum_{x\in\mathbb{Z}}a_1(x,t)a_2(x,t).
\end{equation*}
\notag
$$
By Propositions 1, 8, and 9 we have
$$
\begin{equation*}
\begin{aligned} \, a_1(0,2t)&=\frac{1}{\sqrt{2}}\sum_{x\in\mathbb{Z}}a_1(x,t)(a_2(x,t)-a_1(x,t))+ a_2(x,t)(a_2(x,t)+a_1(x,t)) \\ &=\frac{1}{\sqrt{2}}(S_2(t)+2S_{12}(t)-S_1(t)). \end{aligned}
\end{equation*}
\notag
$$
By definition and Proposition 1 we have Hence Since $S_1(t)+S_2(t)=1$ by Proposition 2, we have the recurrence relation $S_1(t+1)=S_1(t)+(1/\sqrt{2}\,)a_1(0,2t)$; cf. [23], (33). Then Proposition 4 implies by induction that
By Newton’s binomial theorem we get
$$
\begin{equation*}
\sum_{k=0}^{\infty}\begin{pmatrix} 2k \\ k\end{pmatrix}x^k= \dfrac{1}{\sqrt{1-4x}}
\end{equation*}
\notag
$$
for each $x\in [-{1}/{4},1/4)$. Setting $x=-1/4$ we obtain
$$
\begin{equation*}
\lim_{t\to+\infty}\frac{1}{2}\sum_{k=0}^{\lfloor t/2\rfloor-1} \begin{pmatrix} 2k \\ k\end{pmatrix}\biggl(-\frac{1}{4}\biggr)^k= \frac{1}{2\sqrt{2}}\,.
\end{equation*}
\notag
$$
Using Stirling’s formula we estimate the convergence rate:
$$
\begin{equation*}
\biggl|\,\sum_{x\in\mathbb{Z}}a_{1}(x,t)^2-\frac{1}{2\sqrt{2}}\biggr|< \frac{1}{2\cdot 4^{\lfloor t/2\rfloor}}\begin{pmatrix} 2\lfloor t/2 \rfloor \\ \lfloor t/2 \rfloor\end{pmatrix} < \frac{e}{2\pi\sqrt{2\lfloor t/2 \rfloor}}<\frac{1}{2\sqrt{t}}\,. \qquad\square
\end{equation*}
\notag
$$
Underwater rocksFinally, let us warn a mathematically-oriented reader. The outstanding papers [1], [29], and [30] are well-written, insomuch that the physical theorems and proofs there could be taken for mathematical ones, although some of them are wrong as written. The main source of these issues is actually the application of a wrong result, Theorem 3.3, from the mathematical paper [9]. A simple counterexample to Theorem 3.3 in [9] is $a=b=\alpha=\beta=x=0$ and $n$ odd. These values satisfy automatically the assumptions of the theorem, that is, condition (ii) of Lemma 3.1 in [9]. Then by Remark 3 and Proposition 4 the left-hand side of (2.16) in [9] vanishes. Thus it cannot be equivalent to the nonvanishing sequence on the right-hand side. Here we interpret the ‘$\approx$’ sign in (2.16) in [9] as the equivalence of sequences, following [29]. An attempt to interpret the sign so that the difference between the left- and the right-hand sides of (2.16) in [9] tends to zero would void the result because each side tends to zero separately. Although [1], [29], and [30] report minor errors in [9], the issue is more serious. The known asymptotic formulae for Jacobi polynomials are never stated as an equivalence but rather contain an additive error term. Estimating the error term is hard even in particular cases studied in [32], and the case from Remark 3 is reported as more difficult (see bottom of p. 198 in [32]). Thus Theorem 3.3 in [9] should be viewed as an interesting physical rather than a mathematical result. 13. Appendix A: A. Kudryavtsev. Alternative ‘explicit’ formulaeSet $\begin{pmatrix} n \\ k \end{pmatrix}:=0$ for integers $k<0<n$ or $k>n>0$. Denote
$$
\begin{equation*}
\theta(x):=\begin{cases} 1 & \text{if}\ x\geqslant0, \\ 0 & \text{if}\ x<0. \end{cases}
\end{equation*}
\notag
$$
Proposition 18 (‘explicit’ formula). For any integers $|x|<t$ such that $x+t$ is even we have: Proof of Proposition 18, (A). Introduce the generating functions
$$
\begin{equation*}
\widehat a_1(p,q):=2^{n/2}\sum_{n>k\geqslant 0}a_1(2k-n+1,n+1)p^kq^n
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\widehat a_2(p,q):=2^{n/2}\sum_{n>k\geqslant 0}a_2(2k-n+1,n+1)p^kq^n.
\end{equation*}
\notag
$$
By Proposition 1 we get
$$
\begin{equation*}
\begin{cases} \widehat a_1(p,q)-\widehat a_1(p,0)= q\cdot(\widehat a_2(p,q)+\widehat a_1(p,q)), \\ \widehat a_2(p,q)-\widehat a_2(p,0)= pq\cdot(\widehat a_2(p,q)-\widehat a_1(p,q)). \end{cases}
\end{equation*}
\notag
$$
Since $\widehat a_1(p,0)=0$ and $\widehat a_2(p,0)=1$, the solution of this system is and
$$
\begin{equation*}
\widehat a_1(p,q)=\frac{q}{1-q-pq+2pq^2}= q+q^2(1+p-2pq)+q^3(1+p-2pq)^2+\cdots\,.
\end{equation*}
\notag
$$
The coefficient at $p^kq^n$ in $\widehat a_1(p,q)$ equals
$$
\begin{equation*}
\sum_{j=\max(k,n-k)}^n (-2)^{n-j-1} \begin{pmatrix} & j & \\ n-j-1 & k-n+j+1 & j-k\end{pmatrix},
\end{equation*}
\notag
$$
because we must take exactly one combination of factors from every term of the form $q^{j+1}(1+p-2pq)^j$:
The formula for $a_2(x,t)$ follows from the one for $a_1(x,t)$, Proposition 1, and the Pascal rule:
$$
\begin{equation*}
\begin{aligned} \, a_2(x,t)&=\sqrt{2}\,a_1(x-1,t+1)-a_1(x,t) \\ &=2^{(1-t)/2}\sum_{r=0}^{(t-|x|)/2} (-2)^{r}\begin{pmatrix} (x+t-2)/2 \\ r \end{pmatrix} \\ &\qquad \times \biggl(\begin{pmatrix} t-r-1 \\ (x+t-2)/2\end{pmatrix} -\begin{pmatrix} t-r-2 \\ (x+t-2)/2\end{pmatrix}\biggr) \\ &=2^{(1-t)/2}\sum_{r=0}^{(t-|x|)/2}(-2)^{r} \begin{pmatrix} (x+t-2)/2 \\ r\end{pmatrix} \begin{pmatrix} t-r-2 \\ (x+t-4)/2\end{pmatrix}. \qquad\square\end{aligned}
\end{equation*}
\notag
$$
Proof of Proposition 18, (B) (by Voropaev). By Proposition 4, for each $|x|<t$ the numbers $a_1(x,t)$ and $a_2(x,t)$ are the coefficients at $z^{(t-x-2)/2}$ and $z^{(t-x)/2}$, respectively, in the expansion of the polynomial For $x<0$ this implies the required proposition immediately. For $x\geqslant 0$, we first change the summation variable to $r'=(t-x-2)/2-r$ or $r'=(t-x)/2-r$ for $a_1(x,t)$ and $a_2(x,t)$, respectively. $\Box$
14. Appendix B: A. Lvov. Pointwise continuum limit Theorem 8 (pointwise continuum limit). For each real $m\geqslant 0$ and $|x|<t$ we have Proof of Theorem 8. Denote $A:=\lfloor nx/2 \rfloor+\lfloor nt/2 \rfloor$ and $B:=\lfloor nt/2 \rfloor-\lfloor nx/2 \rfloor$. The first limit is computed as follows:
$$
\begin{equation}
\begin{aligned} \, \nonumber &n\,a_1\biggl(\frac{2}{n}\biggl\lfloor \frac{nx}{2}\biggr\rfloor, \frac{2}{n}\biggl\lfloor \frac{nt}{2}\biggr\rfloor,m,\frac{1}{n}\biggr) \\ \nonumber &\qquad\overset{(*)}=n\biggl(1+\frac{m^2}{n^2}\biggr)^{\lfloor nt/2 \rfloor-1/2} \sum_{r=0}^{\infty}(-1)^{r} \begin{pmatrix} A-1 \\ r\end{pmatrix} \begin{pmatrix} B-1 \\ r\end{pmatrix}\biggl(\frac{m}{n}\biggr)^{2r+1} \\ \nonumber &\qquad\!\overset{(**)}\sim\sum_{r=0}^{\infty}(-1)^{r} \begin{pmatrix} A-1 \\ r\end{pmatrix}\begin{pmatrix} B-1 \\ r\end{pmatrix} \frac{m^{2r+1}}{n^{2r}} \\ \nonumber &\qquad\!\!\overset{(***)}=\sum_{r=0;\, 2\mid r}^{\infty} \begin{pmatrix} A-1 \\ r\end{pmatrix}\begin{pmatrix} B-1 \\ r\end{pmatrix} \frac{m^{2r+1}}{n^{2r}}-\sum_{r=0;\, 2 \nmid r}^{\infty} \begin{pmatrix} A-1 \\ r\end{pmatrix} \begin{pmatrix} B-1 \\ r\end{pmatrix}\frac{m^{2r+1}}{n^{2r}} \\ \nonumber &\qquad\to \sum_{r=0;\, 2\mid r}^{\infty} \frac{(x+t)^r(t-x)^r m^{2r+1}}{2^{2r}(r!)^2}- \sum_{r=0;\, 2 \nmid r}^{\infty} \frac{(x+t)^r(t-x)^r m^{2r+1}}{2^{2r}(r!)^2} \\ &\qquad=m\, J_0(m\sqrt{t^2-x^2}\,) \quad\text{as } n \to \infty. \end{aligned}
\end{equation}
\tag{66}
$$
Here equality $(*)$ is Proposition 11. Equivalence $({*}{*})$ follows from
$$
\begin{equation*}
\begin{aligned} \, 1 &\leqslant \biggl(1+\frac{m^2}{n^2}\biggr)^{\lfloor nt/2 \rfloor-1/2} \leqslant \biggl(1+\frac{m^2}{n^2}\biggr)^{nt} \\ &= \sqrt[n]{\biggl(1+\frac{m^2}{n^2}\biggr)^{n^2t}} \sim \sqrt[n]{e^{m^2t}} \to 1 \quad\text{as } n \to \infty \end{aligned}
\end{equation*}
\notag
$$
by the squeeze theorem. Equality $({*}{*}{*})$ holds because
$$
\begin{equation*}
\begin{pmatrix} A-1 \\ r\end{pmatrix}\begin{pmatrix} B-1 \\ r\end{pmatrix} \frac{m^{2r+1}}{n^{2r}}=0\quad\text{for } r > \max\{A,B\},
\end{equation*}
\notag
$$
hence all the three sums involved are finite. The convergence in (66) is established in Lemmas 20–22 below. The second limit in the theorem is computed analogously. $\Box$ Proof. We have
$$
\begin{equation*}
\begin{aligned} \, \begin{pmatrix} A-1 \\ r\end{pmatrix}\begin{pmatrix} B-1 \\ r\end{pmatrix} \frac{m^{2r+1}}{n^{2r}}&= \frac{(A-1)\cdots(A-r)\cdot(B-1)\cdots(B-r)}{(r!)^2}\,\frac{m^{2r+1}}{n^{2r}} \\ & \to \biggl(\frac{x+t}{2}\biggr)^r\biggl(\frac{t-x}{2}\biggr)^r \frac{m^{2r+1}}{(r!)^2}\quad\text{as } n \to \infty, \end{aligned}
\end{equation*}
\notag
$$
because for any $1\leqslant i\leqslant r$ and analogously, Proof. This follows analogously because for each $1\leqslant i\leqslant r<\min\{A,B\}$ we have
$$
\begin{equation*}
\begin{aligned} \, (A-i)(B-i) &\leqslant \biggl(\biggl\lfloor \frac{nx}{2} \biggr\rfloor+ \biggl\lfloor \frac{nt}{2} \biggr\rfloor-1\biggr) \biggl(\biggl\lfloor \frac{nt}{2} \biggr\rfloor- \biggl\lfloor \frac{nx}{2} \biggr\rfloor-1\biggr) \\ &\leqslant\biggl(\frac{nx}{2}+\frac{nt}{2}\biggr) \biggl(\frac{nt}{2}-\frac{nx}{2}\biggr). \qquad\square\end{aligned}
\end{equation*}
\notag
$$
Lemma 22. Suppose $\{a_k(n)\}_{k=0}^\infty$ is a sequence of nonnegative sequences such that (i) $\lim_{n \to \infty} a_k(n) = b_k$ for each $k$; (ii) $a_k(n) \leqslant b_k$ for each $k$ and $n$; and (iii) $\sum_{k = 0}^{\infty}b_k$ is finite. Then Proof. Denote $b := \sum_{k=0}^{\infty}b_k$. Then for each $n$ we have $\sum_{k=0}^{\infty}a_k(n)\leqslant b$. Take any $\varepsilon > 0$. Take such $N$ that $\sum_{k=0}^{N}b_k > b - \varepsilon$. For each $k \leqslant N$ take $M_k$ such that for each $n \geqslant M_k$ we have $a_k(n)>b_k-\dfrac{\varepsilon}{2^{k + 1}}$. Then for each $n > \max_{0\leqslant k\leqslant N}M_k$ we have $\sum_{k=0}^{\infty}a_k(n)>b-2\varepsilon$. So $\lim_{n\to\infty}\sum_{k=0}^{\infty}a_k(n)=b$. $\Box$ AcknowledgementsThis work was been presented as courses at the Summer Conference of Tournament of Towns in Arandelovac, Summer School in Contemporary Mathematics in Dubna, Faculty of Mathematics in HSE University in Moscow, and Independent University of Moscow in 2019. The authors are grateful to all the participants of these events for their contribution, especially to M. Dmitriev, I. Novikov, F. Ozhegov, and A. Voropaev for numerous remarks and the translation into Russian, to E. Akhmedova and R. Valieva for typesetting some parts of the text, to I. Bogdanov, A. Daniyarkhodzhaev, M. Fedorov, I. Gaidai-Turlov, T. Kovalev, F. Kuyanov, G. Minaev, I. Russkikh, and V. Skopenkova for some of the figures, to A. Kudryavtsev and A. Lvov for writing appendices (those two authors were less than 16 years old at that time). The authors are grateful to V. Akulin, T. Batenev, A. Belavin, M. Bershtein, M. Blank, A. Borodin, V. Buchstaber, G. Chelnokov, V. Chernyshev, I. Dynnikov, I. Ibragimov, I. Ivanov, T. Jacobson, D. U. Kim, M. Khristoforov, E. Kolpakov, A.B.J. Kuijlaars, S. Lando, M. Lifshits, M. Maeda, V. Nazaikinskii, S. Nechaev, S. Novikov, G. Olshanski, Yu. Petrova, I. Polekhin, F. Popelensky, P. Pylyavskyy, A. Rybko, I. Sabitov, L. Schulman, A. Semenov, S. Shlosman, A. Slizkov, T. Tate, S. Tikhomirov, D. Treschev, L. Velázquez, A. Vershik, and P. Zakorko for useful discussions.
Citation in format AMSBIB
\Bibitem{SkoUst22} |
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