Feynman Checkers: the Probability to Find an Electron Vanishes Nowhere Inside the Light Cone

Feynman checkers: the probability to find an electron vanishes nowhere inside the light cone

Novikov Ivan

Abstract We study Feynman checkers, the most elementary model of electron motion introduced by R. Feynman. For the model, we prove that the probability to find an electron vanishes nowhere inside the light cone. We also prove several results on the average electron velocity. In addition, we present a lot of identities related to the model. Keywords: Feynman checkerboard, quantum mechanics, average velocity, Dirac equation.

1 Introduction

This paper is on the most elementary model of one-dimensional electron motion that is known as “Feynman checkers”. This model was introduced by R. Feynman around the 1950s and published in 1965; see [3, Problem 2.6]. Afterwards, a large amount of physical articles on the model appeared (see, for example, [1, 5, 6, 7, 9]). But the first mathematical work [10] on the subject appeared, apparently, only in 2020. We use the inessential modification of the model from [3] that was presented in [10]. Many properties of our model are parallel to usual quantum mechanics: there are analogues of Dirac equation (Proposition 1), probability/charge conservation (Proposition 4), Klein–Gordon equation [10, Proposition 7], Fourier integral [10, Propositions 12-13], concentration of measure on the light cone [10, Corollary 6] etc. Other striking properties have sharp contrast with both continuum quantum theory and the classical random walks: the (essentially) maximal electron velocity is strictly less than the speed of light [1, Theorem 1], [10, Theorem 1(B)]; adding absorbing boundary increases the probability of returning to the initial point [1, Theorems 8 and 10]. It should be noted that Feynman checkers almost completely identical to one-dimensional quantum walk and Hadamard walk. These notions are discussed in [1, 6]; see [11] for a comprehensive survey. Our main (new) result states that the probability to find an electron at a lattice point is nonzero if there is at least one checker path from the origin to that point (Theorem 1), answering a question by A. Ustinov. Also we present several results on the average velocity of the electron (S5). We prove that the expectation of the average electron velocity equals the time-average of the expectation of the instantaneous velocity (Proposition 8), answering a question by D. Treschev. Also we compute the arXiv:2010.05088v1 [math-ph] 10 Oct 2020 limit value of the average electron velocity when time tends to infinity (Theorem 2). This result has been proven in [4, Theorem 1] (see also an exposition in [10, S12.2]), but we present a short elementary proof. In addition, we state a lot of new identities related to the model, which were found in numeric experiments (See S6). A few of these identities are proven and the rest are interesting open problems.

2 Definition and examples

In this section we present the definition and physical interpretation of Feynman checkers from [10].

Definition 1 ([10, Definition 2]). Fix 휀 > 0 and 푚 > 0 called lattice step and particle mass respectively. Consider the lattice 휀Z2 = { (푥, 푡): 푥/휀, 푡/휀 ∈ Z }. The elements of 휀Z2 are called lattice points.A checker path is a finite sequence of points of 휀Z2 such that the vector from each point (except the last

1 one) to the next one equals either (휀, 휀) or (−휀, 휀).A turn is a point of the path (not the first and not the last one) such that the vectors from the point to the next and to the previous ones are orthogonal. For (푥, 푡) ∈ 휀Z2, where 푡 > 0, denote ∑︁ 푎(푥, 푡, 푚, 휀) := (1 + 푚2휀2)(1−푡/휀)/2 푖 (−푖푚휀)turns(푠), 푠 where the sum is over all checker paths 푠 from (0, 0) to (푥, 푡) with the first step to (휀, 휀), and turns(푠) is the number of turns in 푠. Denote

푃 (푥, 푡, 푚, 휀) := |푎(푥, 푡, 푚, 휀)|2.

Denote by 푎1(푥, 푡, 푚, 휀) and 푎2(푥, 푡, 푚, 휀) the real and the imaginary part of 푎(푥, 푡, 푚, 휀) respectively. Remark 1 (Physical interpretation of the model, [10]). We use the natural system of units, where both the speed of light and the Plank constant equal 1. The 푡− and 푥−coordinates are interpreted as time and position of the particle of spin 1/2 and mass 푚 respectively. Thus any checker path is interpreted as motion of a particle in 1D space with the speed of light (with change of direction). The line 푥 = 푡 represents motion in one direction with the speed of light. In what follows, we consider the motion of an electron (so that 푚 is the mass of an electron). The number 푃 (푥, 푡, 푚, 휀) is called the probability to find an electron at the lattice point (푥, 푡), if the electron was emitted from the point (0, 0). Such terminology is confirmed by the fact that all the numbers 푃 (푥, 푡, 푚, 휀) on one horizontal sum up to 1 (see Proposition 4). Figure 1 shows 푃 (푥, 1000, 1, 1) for |푥| 6 1000. Note that if 푥 is greater than 750, then the probability is very small but still non-zero (see Theorem 1).

Figure 1: The graph of 푃 (푥, 1000, 1, 1) (A. Daniyarkhodzhaev–F. Kuyanov). Note that for |푥| > 750 the probability is very small (but still non-zero).

Example 1. Let us compute 푎(0, 4휀, 푚, 휀). Figure 2 shows all three checker paths from (0, 0) to (0, 4휀) 푚2휀2 starting with an upward-right move. Thus by definition 푎(0, 4휀, 푚, 휀) = − (1+푚2휀2)3/2 푖.

2 Figure 2: Checker paths contributing to 푎(0, 4휀, 푚, 휀)

Example 2. It is easy to show that for each 푥 ∈ 휀Z+ we have: 1) 푎(푥, 푥, 푚, 휀) = (1 + 푚2휀2)(1−푥/휀)/2 푖; 2) 푎(−푥, 푥 + 2휀, 푚, 휀) = 푚휀 (1 + 푚2휀2)−(1+푥/휀)/2.

Let us present several tables that show 푎(푥, 푡, 푚, 휀) and 푃 (푥, 푡, 푚, 휀) for small 푥 and 푡. In Table 1, the number in a cell (푥, 푡) is 푎(푥, 푡, 푚, 휀), and an empty cell means that 푎(푥, 푡, 푚, 휀) = 0. Analogously, in Table 2, the number in a cell (푥, 푡) is 푃 (푥, 푡, 푚, 휀), and an empty cell means that 푃 (푥, 푡, 푚, 휀) = 0. Note that for fixed 푡 the sum of the probabilities equals 1.

푚휀 (푚휀−푚3휀3)−푚2휀2푖 푚휀−2푚2휀2푖 1 4휀 (1+푚2휀2)3/2 (1+푚2휀2)3/2 (1+푚2휀2)3/2 (1+푚2휀2)3/2 푖 푚휀 푚휀−푚2휀2푖 푚2휀2 3휀 1+푚2휀2 1+푚2휀2 (1+푚2휀2) 푖 2휀 √ 푚휀 √ 1 푖 1+푚2휀2 1+푚2휀2 휀 푖 푡 푥 −2휀 −휀 0 휀 2휀 3휀 4휀

Table 1: 푎(푥, 푡, 푚, 휀) for small 푥 and 푡.

푚2휀2 푚2휀2(1−푚2휀2+푚4휀4) 푚2휀2(1+4푚2휀2) 1 4휀 (1+푚2휀2)3 (1+푚2휀2)3 (1+푚2휀2)3 (1+푚2휀2)3 푚2휀2 푚2휀2 1 3휀 (1+푚2휀2)2 1+푚2휀2 (1+푚2휀2)2 푚2휀2 1 2휀 1+푚2휀2 1+푚2휀2 휀 1 푡 푥 −2휀 −휀 0 휀 2휀 3휀 4휀

Table 2: 푃 (푥, 푡, 푚, 휀) for small 푥 and 푡.

3 Known results

In this section, we state some properties of the model, Propositions 1-5 being folklore. They (and Propositions 6-7) are proved and discussed in [10] and [2].

3 Notation 1. In what follows, we use the following notation:

2 휀Z = { (푥, 푡): 푥/휀, 푡/휀 ∈ Z }; 휀Z = { 푡 : 푡/휀 ∈ Z }; 휀Z+ = { 푡 > 0 : 푡/휀 ∈ Z }; Z+ = { 푡 > 0 : 푡 ∈ Z }. 2 Proposition 1 (Dirac equation; [10, Proposition 5]). For each (푥, 푡) ∈ 휀Z , where 푡 > 2휀, we have: 1 1) 푎1(푥, 푡, 푚, 휀) = √ (푎1(푥 + 휀, 푡 − 휀, 푚, 휀) + 푚휀 푎2(푥 + 휀, 푡 − 휀, 푚, 휀)); 1 + 푚2휀2 1 2) 푎2(푥, 푡, 푚, 휀) = √ (푎2(푥 − 휀, 푡 − 휀, 푚, 휀) − 푚휀 푎1(푥 − 휀, 푡 − 휀, 푚, 휀)). 1 + 푚2휀2 2 Proposition 2 ([10, Lemma 1]). For each (푥, 푡) ∈ 휀Z , where 푡 > 2휀, we have: 1 1) 푎1(푥, 푡 − 휀, 푚, 휀) = √ (푎1(푥 − 휀, 푡, 푚, 휀) − 푚휀 푎2(푥 + 휀, 푡, 푚, 휀)); 1 + 푚2휀2 1 2) 푎2(푥, 푡 − 휀, 푚, 휀) = √ (푎2(푥 + 휀, 푡, 푚, 휀) + 푚휀 푎1(푥 − 휀, 푡, 푚, 휀)). 1 + 푚2휀2 2 Proposition 3 (Formulae for 푎1(푥, 푡, 푚, 휀) and 푎2(푥, 푡, 푚, 휀); [10, Proposition 11]). For all (푥, 푡) ∈ 휀Z such that 푡 > |푥| and (푥 + 푡)/휀 is even we have:

(푡−|푥|)/2휀 ∑︁ (︂(푡 + 푥 − 2휀)/2휀)︂(︂(푡 − 푥 − 2휀)/2휀)︂ 푎 (푥, 푡, 푚, 휀) = (1 + 푚2휀2)(1−푡/휀)/2 (−1)푟 (푚휀)2푟+1; 1 푟 푟 푟=0

(푡−|푥|)/2휀 ∑︁ (︂(푡 + 푥 − 2휀)/2휀)︂(︂(푡 − 푥 − 2휀)/2휀)︂ 푎 (푥, 푡, 푚, 휀) = (1 + 푚2휀2)(1−푡/휀)/2 (−1)푟 (푚휀)2푟. 2 푟 푟 − 1 푟=1

Proposition 4 (Probability conservation law; [10, Proposition 6]). For each 푡∈휀Z+ we get ∑︁ 푃 (푥, 푡, 푚, 휀) = 1. 푥∈휀Z Proposition 5 (Symmetry; [10, Proposition 8]). For all (푥, 푡) ∈ 휀Z2 with 푡 > 0 we have: 1) 푎1(푥, 푡, 푚, 휀) = 푎1(−푥, 푡, 푚, 휀); 2) (푡 − 푥)푎2(푥, 푡, 푚, 휀) = (푡 + 푥 − 2휀)푎2(2휀 − 푥, 푡, 푚, 휀).

Proposition 6 ([10, Theorem 5, its proof, and Remark 1]). If 푡 ∈ 휀Z+ and 푚휀 = 1, then ⌊푡/2휀⌋−1 (︂ )︂ (︂√︂ )︂ ∑︁ 2 1 ∑︁ 1 2푘 1 휀 푎1(푥, 푡, 푚, 휀) = = √ + O . 2 (−4)푘 푘 2 2 푡 푥∈휀Z 푘=0 Hereafter notation 푓(푡, 푚, 휀) = O(푔(푡, 푚, 휀)) means there is a constant 퐶 (not depending on 푡, 푚, 휀) such that for each 푡, 푚, 휀 satisfying the assumptions of the proposition we have |푓(푡, 푚, 휀)| 6 퐶푔(푡, 푚, 휀). The following genealization of Proposition 6 was conjectured by I. Gaidai-Turlov, T. Kovalev, A. Lvov in 2019, and proved by I. Bogdanov in 2020.

Proposition 7 ([2, Theorem 2]). If 0 6 푚휀 6 1, then ∑︁ 2 푚휀 lim 푎1(푥, 푡, 푚, 휀) = √ . 푡→+∞ 2 2 푥∈휀 2 1 + 푚 휀 푡∈휀Z+ Z

4 4 Main result: the probability to find an electron vanishes nowhere inside the light cone

The goal of this section is to prove the following theorem.

Theorem 1. For each 푚 > 0 and a point (푥, 푡) ∈ 휀Z2 such that (푥 + 푡)/휀 is even and 푡 > |푥| we have 푃 (푥, 푡, 푚, 휀) ̸= 0.

Remark 2. Thus 푃 (푥, 푡, 푚, 휀) ̸= 0 if and only if (푥 + 푡)/휀 is even and either 푡 > |푥| or 푡 = 푥 > 0. In other words, 푃 (푥, 푡, 푚, 휀) ̸= 0 if and only if there exists at least one checker path from (0, 0) to (푥, 푡).

Remark 3. Note that the assertion of the theorem is not obvious at all. It follows neither from Definition 1 nor from the explicit formulae from Proposition 3. Figure 1 shows that for some pairs (푥, 푡) the probability can be a very small number. But Proposition 5 and a simple trick helps us to prove the theorem.

Proof of Theorem 1. Denote 푀 = {(푥, 푡) ∈ 휀Z2 :(푥 + 푡)/휀 is even, 푡 > |푥|, 푃 (푥, 푡, 푚, 휀) = 0}. If 푀 = ∅, then there is nothing to prove. Assume that 푀 ̸= ∅. Among the points of 푀, select the one with the minimal 푡−coordinate (if there are several such points, select any of them). Denote by (푥0, 푡0) 2 2 2 2 (1−푡/휀) the selected point. By Example 2, for all 푡 ∈ 휀Z+ we have 푃 (−푡+2휀, 푡) = 푚 휀 (1+푚 휀 ) ̸= 0, and thus 푥0 ̸= −푡0+2휀. Subsequently, by Proposition 5 it follows that 푎1(−푥0, 푡0, 푚, 휀) = 푎1(푥0, 푡0, 푚, 휀) = 0 푎2(푥0,푡0,푚,휀) and 푎2(2휀 − 푥0, 푡0, 푚, 휀) = (푡0 − 푥0) = 0. See Figure 3. By Proposition 2 we get 푡0+푥0−2휀

푎1(−푥0, 푡0, 푚, 휀) − 푚휀 푎2(−푥0 + 2휀, 푡0, 푚, 휀) 푎1(−푥0 + 휀, 푡0 − 휀, 푚, 휀) = √ = 0 and 1 + 푚2휀2 푎2(−푥0 + 2휀, 푡0, 푚, 휀) + 푚휀 푎1(−푥0, 푡0, 푚, 휀) 푎2(−푥0 + 휀, 푡0 − 휀, 푚, 휀) = √ = 0. 1 + 푚2휀2

Thus 푃 (−푥0 + 휀, 푡0 − 휀, 푚, 휀) = 0. This contradicts to the minimality of 푡0, because (푥0 − 휀 + 푡0 − 휀)/휀 is even and 푡0 − 휀 > |푥0 − 휀| by the condition 푥0 ̸= −푡0 + 2휀 above.

Figure 3: The pair in a cell (푥, 푡) is (푎1(푥, 푡), 푎2(푥, 푡))

5 On the electron velocity

In this section we prove Proposition 8 and Theorem 2 stated below. The former answers a question by D. Treschev. For the statements, we need the following definitions.

5 5.1 The expectation of the average electron velocity equals the time-average of the expectation of the instantaneous electron velocity It goes almost without saying to consider the 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). Actually, these two states are exactly (1+1)-dimensional analogue of chirality states for a spin 1/2 particle. See, for example, a discussion on this topic in [10]. Let us give a few new definitions.

2 Definition 2 ([10]). The probability 푃푡,푚,휀(푥, +) to find a right electron at the lattice point (푥, 푡) ∈ 휀Z , if the right electron was emitted from the point (0, 0), is the length square of the vector 푎(푥, 푡, 푚, 휀, +) := 2 2 (1−푡/휀)/2 ∑︀ turns(푠) (1 + 푚 휀 ) 푖 푠(−푖푚휀) , where the sum is over only those checker paths from (0, 0) to (푥, 푡) that both start and finish with an upwards-right move, and turns(푠) is the number of turns in 푠. The probability 푃푡,푚,휀(푥, −) to find a left electron is defined analogously, only the sum is taken over checker paths that start with an upwards-right move but finish with an upwards-left move.

2 2 Remark 4 ([10]). Clearly, the above probabilities equal 푎2(푥, 푡, 푚, 휀) and 푎1(푥, 푡, 푚, 휀) respectively, because the last move is directed upwards-right if and only if the number of turns is even. By Propo- sition 4, 푃푡,푚,휀(푥, ±) is indeed a probability measure on the set {(푥, ±): 푥 ∈ 휀Z}.

Definition 3. The average electron velocity is the random variable on the set {(푥, ±): 푥 ∈ 휀Z} given by 푣푡(푥, ±) = 푣푡(푥) = 푥/푡. The instantaneous electron velocity is the random variable on the same set given by {︃ +1, if 휎 = +, 푢푡(푥, 휎) = −1, if 휎 = −.

Remark 5. The random variables in the above definition depend on 푡, 푚, 휀. To emphasize that, we use the notation E푡,푚,휀 for their expectation. Proposition 8. The expectation of the average electron velocity equals the time-average of the expectation of the instantaneous electron velocity, i.e., for each 푇 ∈ 휀Z+ we have [︃ ]︃ 휀 ∑︁ (푣 ) = (푢 ) . E푇,푚,휀 푇 푇 E푡,푚,휀 푡 푡=휀,...,푇

Remark 6. This is not at all automatic because the model does not give any probability distribution on ∑︀ the set of checker paths. Moreover, notice that in the expression E푡,푚,휀 (푢푡) we take expectations 푡=휀,...,푇 in 푇/휀 different probability spaces.

Proof of Proposition 8. If 푇 = 휀, then there is nothing to prove. Thus we assume that 푇 > 2휀.

6 By Proposition 1 for each 푡 > 휀 we have:

∑︁ 2 ∑︁ 2 푎2(푥, 푡, 푚, 휀) − 푎1(푥, 푡, 푚, 휀) = 푥∈휀Z 푥∈휀Z 1 [︂ ∑︁ ∑︁ = 푎 (푥, 푡 − 휀, 푚, 휀)2 + 푚2휀2 푎 (푥, 푡 − 휀, 푚, 휀)2− 1 + 푚2휀2 2 1 푥∈휀Z 푥∈휀Z ∑︁ ]︂ − 2푚휀 푎1(푥, 푡 − 휀, 푚, 휀)푎2(푥, 푡 − 휀, 푚, 휀) − 푥∈휀Z 1 [︂ ∑︁ ∑︁ − 푎 (푥, 푡 − 휀, 푚, 휀)2 + 푚2휀2 푎 (푥, 푡 − 휀, 푚, 휀)2+ 1 + 푚2휀2 1 2 푥∈휀Z 푥∈휀Z ∑︁ ]︂ + 2푚휀 푎1(푥, 푡 − 휀, 푚, 휀)푎2(푥, 푡 − 휀, 푚, 휀) = 푥∈휀Z (︃ )︃ 1 − 푚2휀2 ∑︁ ∑︁ = 푎 (푥, 푡 − 휀, 푚, 휀)2 − 푎 (푥, 푡 − 휀, 푚, 휀)2 − 1 + 푚2휀2 2 1 푥∈휀Z 푥∈휀Z 4푚휀 ∑︁ − 푎 (푥, 푡 − 휀, 푚, 휀)푎 (푥, 푡 − 휀, 푚, 휀). 1 + 푚2휀2 1 2 푥∈휀Z This implies that (︃ )︃ ∑︁ ∑︁ ∑︁ 2 ∑︁ 2 휀 E푡,푚,휀 (푢푡) = 휀 푎2(푥, 푡, 푚, 휀) − 푎1(푥, 푡, 푚, 휀) = 푡=휀,...,푇 푡=휀,...,푇 푥∈휀Z 푥∈휀Z (︃ )︃ 1 − 푚2휀2 ∑︁ ∑︁ ∑︁ = 휀 + 휀 푎 (푥, 푡 − 휀, 푚, 휀)2 − 푎 (푥, 푡 − 휀, 푚, 휀)2 − 1 + 푚2휀2 2 1 푡=2휀,...,푇 푥∈휀Z 푥∈휀Z 4푚휀 ∑︁ ∑︁ − 휀 푎 (푥, 푡 − 휀, 푚, 휀)푎 (푥, 푡 − 휀, 푚, 휀), (1) 1 + 푚2휀2 1 2 푡=2휀,...,푇 푥∈휀Z (︂ )︂ ∑︀ 2 ∑︀ 2 where we use the obvious equality 휀 푎2(푥, 휀, 푚, 휀) − 푎1(푥, 휀, 푚, 휀) = 휀. On the other hand, 푥∈휀Z 푥∈휀Z

7 by Proposition 1 we have:

∑︁ 2 ∑︁ 2 푇 · E푇,푚,휀 (푣푇 ) = 푥푎1(푥, 푇, 푚, 휀) + 푥푎2(푥, 푇, 푚, 휀) = 푥∈휀Z 푥∈휀Z ∑︁ 2 ∑︁ 2 = (푥 − 휀)푎1(푥 − 휀, 푇, 푚, 휀) + (푥 + 휀)푎2(푥 + 휀, 푇, 푚, 휀) = 푥∈휀Z 푥∈휀Z 1 ∑︁ [︂ = (푥 − 휀)푎 (푥, 푇 − 휀, 푚, 휀)2 + (푥 − 휀)푚2휀2 푎 (푥, 푇 − 휀, 푚, 휀)2+ 1 + 푚2휀2 1 2 푥∈휀Z ]︂ + 2(푥 − 휀)푚휀 푎1(푥, 푇 − 휀, 푚, 휀)푎2(푥, 푇 − 휀, 푚, 휀) +

1 ∑︁ [︂ + (푥 + 휀)푎 (푥, 푇 − 휀, 푚, 휀)2 + (푥 + 휀)푚2휀2 푎 (푥, 푇 − 휀, 푚, 휀)2− 1 + 푚2휀2 2 1 푥∈휀Z ]︂ − 2(푥 + 휀)푚휀 푎1(푥, 푇 − 휀, 푚, 휀)푎2(푥, 푇 − 휀, 푚, 휀) =

∑︁ 2 ∑︁ 2 = 푥푎1(푥, 푇 − 휀, 푚, 휀) + 푥푎2(푥, 푇 − 휀, 푚, 휀) + 푥∈휀Z 푥∈휀Z (︃ )︃ 1 − 푚2휀2 ∑︁ ∑︁ + 휀 푎 (푥, 푇 − 휀, 푚, 휀)2 − 푎 (푥, 푇 − 휀, 푚, 휀)2 − 1 + 푚2휀2 2 1 푥∈휀Z 푥∈휀Z 4푚휀2 ∑︁ − 푎 (푥, 푇 − 휀, 푚, 휀)푎 (푥, 푇 − 휀, 푚, 휀) = 1 + 푚2휀2 1 2 푥∈휀Z (︃ )︃ 1 − 푚2휀2 ∑︁ ∑︁ ∑︁ = 휀 + 휀 푎 (푥, 푡 − 휀, 푚, 휀)2 − 푎 (푥, 푡 − 휀, 푚, 휀)2 − 1 + 푚2휀2 2 1 푡=2휀,...,푇 푥∈휀Z 푥∈휀Z 4푚휀2 ∑︁ ∑︁ − 푎 (푥, 푡 − 휀, 푚, 휀)푎 (푥, 푡 − 휀, 푚, 휀), (2) 1 + 푚2휀2 1 2 푡=2휀,...,푇 푥∈휀Z where the latter equality is obtained by repeating the same transformation 푇 − 2휀 times. Comparing (1) and (2), we get the required result.

5.2 Analogue of Proposition 8 for classical random walks To show that the statement of Proposition 8 is natural, let us give its analogue for classical random walks. Let us consider a flea that makes a random walk on the integer number line starting from 0. If the flea is situated in the number 푁 at the time 푡0, then at the time 푡0 + 1 it is situated in the number 푁 + 1 with the probability 푝 or in the number 푁 − 1 with the probability 푞 = 1 − 푝. If we denote the probability to find the flea at 푥 ∈ N at the time 푡 by 푃̂︀(푥, 푡), then 푃̂︀(푥, 푡) = 푝푃̂︀(푥 − 1, 푡 − 1) + 푞푃̂︀(푥 + 1, 푡 − 1).

2 Definition 4. For a point (푥, 푡) ∈ Z (푡 > 0), we define the probability 푃̂︀(푥, 푡) to find the flea at the lattice point (푥, 푡), if the flea makes a random walk by induction on 푡: {︃ 1, if 푥 = 0, 푃̂︀(푥, 0) = 0, if 푥 ̸= 0. 푃̂︀(푥, 푡) = 푝푃̂︀(푥 − 1, 푡 − 1) + 푞푃̂︀(푥 + 1, 푡 − 1) for 푡 > 1. Remark 7. As in Definition 1, 푥 is interpreted as position and 푡 is interpreted as time.

8 Remark 8. It is easy to prove that for fixed 푡 we have ∑︀ 푃 (푥, 푡) = 1. 푥∈Z ̂︀

The average flea velocity 푣̂︀푡 and the instantaneous flea velocity 푢̂︀푡 are defined literally as in Defini- tion 3. The random variables 푣̂︀푡 and 푢̂︀푡 depend on 푡 ∈ Z+; we write E푡 for their expectations. The following easy well-known proposition is an analogue of Proposition 8. Proposition 9 (An analogue of Proposition 8 for classical random walks). The expectation of the average flea velocity equals the time-average of the expectation of the instantaneous flea velocity, i.e., for each 푇 ∈ Z+ we have [︃ 푇 ]︃ 1 ∑︁ (푣 ) = (푢 ) . E푇 ̂︀푇 푇 E푡 ̂︀푡 푡=1

Proof. Clearly, for each 푇 ∈ Z+ we have E푇 (푢̂︀푇 ) = 푝 − 푞. Let us prove by induction on 푇 ∈ Z+ that for each 푇 we have E푇 (푣̂︀푇 ) = 푝 − 푞. The base is obvious. To perform the induction step, suppose that for 푇 ∈ Z+ we have E푇 (푣̂︀푇 ) = 푝 − 푞. Then 1 ∑︁ 1 ∑︁ 1 ∑︁ 푇 +1(푣푇 +1) = 푥푃̂︀(푥, 푇 + 1) = 푥푝푃̂︀(푥 − 1, 푇 ) + 푥푞푃̂︀(푥 + 1, 푇 ) E ̂︀ 푇 + 1 푇 + 1 푇 + 1 푥∈Z 푥∈Z 푥∈Z 1 ∑︁ 1 ∑︁ = (푥 + 1)푝푃̂︀(푥, 푇 ) + (푥 − 1)푞푃̂︀(푥, 푇 ) = 푇 + 1 푇 + 1 푥∈Z 푥∈Z 1 ∑︁ (︁ )︁ 1 ∑︁ 푥푝푃̂︀(푥, 푇 ) + 푥푞푃̂︀(푥, 푇 ) + (푝 − 푞)푃̂︀(푥, 푇 ) = 푇 + 1 푇 + 1 푥∈Z 푥∈Z (푎) 1 푝 − 푞 (푏) 푇 (푝 − 푞) 푝 − 푞 = (푣 ) + = + = 푝 − 푞, 푇 + 1E푇 ̂︀푇 푇 + 1 푇 + 1 푇 + 1 where (a) follows from the identity ∑︀ 푃 (푥, 푇 ) = 1 for fixed 푇 , and (b) follows from the inductive 푥∈Z ̂︀ hypothesis. Thus E푇 (푣̂︀푇 ) = 푝 − 푞 for all 푇 ∈ Z+.

5.3 The limit value of the average electron velocity The following theorem gives us the limit value of the average electron velocity when time tends to infinity. This theorem and Proposition 10 are corollaries of a more general result [4, (18)],butwe present short elementary proofs.

Theorem 2. If 0 6 푚휀 6 1, then we have 푚휀 lim E푇,푚,휀 (푣푇 ) = 1 − √ . 푇 →+∞ 1 + 푚2휀2 푇 ∈휀Z+ Proof. By Propositions 8 and 4 we have (︃ )︃ (︃ )︃ 휀 ∑︁ 휀 ∑︁ ∑︁ ∑︁ (푣 ) = (푢 ) = 푎 (푥, 푡, 푚, 휀)2 − 푎 (푥, 푡, 푚, 휀)2 = E푇,푚,휀 푇 푇 E푡,푚,휀 푡 푇 2 1 푡=휀,...,푇 푡=휀,...,푇 푥∈휀Z 푥∈휀Z (︃ )︃ 휀 ∑︁ ∑︁ 휀 푇 2휀 ∑︁ ∑︁ = 1 − 2 푎 (푥, 푡, 푚, 휀)2 = − 푎 (푥, 푡, 푚, 휀)2 = 푇 1 푇 휀 푇 1 푡=휀,...,푇 푥∈휀Z 푡=휀,...,푇 푥∈휀Z (︃ )︃ 푚휀 2휀 ∑︁ ∑︁ 2 푚휀 푚휀 = 1 − √ − 푎1(푥, 푡, 푚, 휀) − √ =: 1 − √ − Δ(푇, 푚, 휀). 1 + 푚2휀2 푇 2 1 + 푚2휀2 1 + 푚2휀2 푡=휀,...,푇 푥∈휀Z

9 Thus it remains to prove that Δ(푇, 푚, 휀) → 0 as 푇 → +∞. Take 훿 > 0. By Proposition 7 there exists 푡0 ∈ 휀Z+ such that for any 푡 > 푡0 we have ⃒ ⃒ ⃒∑︁ 2 푚휀 ⃒ 훿 ⃒ 푎1(푥, 푡, 푚, 휀) − √ ⃒ < . ⃒ 2 1 + 푚2휀2 ⃒ 3 ⃒푥∈휀Z ⃒ Then by Proposition 4 we have

2휀 ∑︁ ∑︁ 2 2휀 ∑︁ 푚휀 |Δ(푇, 푚, 휀)| 6 푎1(푥, 푡, 푚, 휀) + √ + 푇 푇 2 1 + 푚2휀2 푡=휀,...,푡0 푥∈휀Z 푡=휀,...,푡0 ⃒ ⃒ 2휀 ∑︁ ⃒∑︁ 2 푚휀 ⃒ + ⃒ 푎1(푥, 푡, 푚, 휀) − √ ⃒ 6 푇 ⃒ 2 1 + 푚2휀2 ⃒ 푡=푡0+휀,...,푇 ⃒푥∈휀Z ⃒ 2휀 푡 2휀 푡 2휀 푇 훿 4푡 2훿 12푡 0 + 0 + 0 + < 훿 for 푇 > 0 . 6 푇 휀 푇 휀 푇 휀 3 6 푇 3 훿

Since 훿 is arbitrary, it follows that lim Δ(푇, 푚, 휀) = 0 and the theorem follows. 푇 →+∞ Theorem 2 provides the limit value, but tells nothing about the convergence rate. But the following proposition provides both for the particular case 푚휀 = 1.

Proposition 10. For each 휀 > 0 an each 푇 ∈ 휀Z+ we have 1 (︁ 휀 )︁ E푇,1/휀,휀 (푣푇 ) = 1 − √ + O . 2 푇

Proof. Analogously to the proof of Theorem 2, for each 푇 ∈ 휀Z+ we have (︃ )︃ 1 2휀 ∑︁ ∑︁ 2 1 E푇,1/휀,휀 (푣푇 ) = 1 − √ − 푎1(푥, 푡, 1/휀, 휀) − √ . 2 푇 2 2 푡=휀,...,푇 푥∈휀Z It remains to show that (︃ )︃ ∑︁ ∑︁ 2 1 ∑︁ 푎1(푥, 푡, 1/휀, 휀) − √ =: 푐푡 = O(1). 2 2 푡=휀,...,푇 푥∈휀Z 푡=휀,...,푇 This follows from the following facts:

(i) If 푡 = 2푛휀, where 푛 ∈ Z, then 푐푡 = 푐푡+휀;

(ii) 푐푡 → 0 as 푡 → ∞;

(iii) For each 푡 ∈ 휀Z, 푡 > 휀, the signs of 푐푡 and 푐푡+2휀 are opposite;

(iv) For each 푡 ∈ 휀Z, 푡 > 휀, we have |푐푡| > |푐푡+2휀|. Here (i) and (ii) follow directly from Proposition 6. Assertion (iii) follows from Proposition 6 and the 1 (︀2푘)︀ 1 (︀2(푘+1))︀ fact that 4푘 푘 > 4푘+1 푘+1 for 푘 ∈ 푍+. Also note that from Proposition 6 and (iii) we know that 푐푡 is positive if 푡 = (4푛 + 2)휀 for some 푛 ∈ Z+ ∪ {0}, and 푐푡 is negative if 푡 = 4푛휀 for some 푛 ∈ Z+.

10 Let us prove (iv). Fix any 푡 = (4푛 + 2)휀, where 푛 ∈ Z+. By Proposition 6 we have

2푛+푚 1 1 ∑︁ 1 (︂2푘)︂ 푐 = − √ + (here 푚 ∈ ). 푡+2푚휀 2 (−4)푘 푘 Z+ 2 2 푘=0 Note that 1 (︂ 1 (︂4푛 + 2)︂ 2 (︂4푛 + 4)︂ 1 (︂4푛 + 6)︂)︂ 푐 + 푐 = 푐 + 푐 + − + − . 푡+6휀 푡+4휀 푡+2휀 푡 2 42푛+1 2푛 + 1 42푛+2 2푛 + 2 42푛+3 2푛 + 3

It easy to show that the latter expression in parentheses is negative for any 푛 ∈ Z+. Thus

푐푡+2휀 + 푐푡 > 푐푡+6휀 + 푐푡+4휀 > 푐푡+10휀 + 푐푡+8휀 > . . . (3)

To prove (iv), we want to show that |푐푡| > |푐푡+2휀|. Equivalently, we want to prove that 푐푡 + 푐푡+2휀 > 0.

Suppose that this is not true. Then by (3) the sequence {푐푡+(2푚+2)휀 + 푐푡+2푚휀}푚∈Z+ does not have 0 as its limit. On the other hand, by (ii) this sequence should have 0 as its limit. This is a contradiction. One can obtain a similar contradiction⃒ ⃒ for the case when 푡 = 4푛휀. ⃒ ⃒ ⃒ ∑︀ ⃒ ∑︀ Hence by (i)-(iv) we have ⃒ 푐푡⃒ 6 2푐2휀 for each 푇 , and thus 푐푡 = O(1). ⃒푡=휀,...,푇 ⃒ 푡=휀,...,푇

6 Identities

In this section, we present several identities in Feynman checkers. Not all of them pretend to be new, but we could not find them in the literature. All the identities were first discovered inWolfram Mathematica, sometimes with the help of the On-Line Encyclopedia of Integer Sequences [8]. The identities in this subsection should be considered as just some combinatorial equalities, which have absolutely no physical meaning. In all the identities below we assume that 푚 > 0.

6.1 Linear identities ∑︀ 2 2 From Proposition 4 we know that for each 푡 ∈ 휀Z+ we have (푎1(푥, 푡, 푚, 휀) + 푎2(푥, 푡, 푚, 휀) ) = 1. 푥∈휀Z But what if we consider not the sum of squares, but the sum of 푎1-s and 푎2-s themselves? The following proposition gives the answer.

Proposition 11. For each 푡 ∈ 휀Z+ we have ∑︁ (︂푡 − 휀 )︂ ∑︁ (︂푡 − 휀 )︂ 푎 (푥, 푡, 푚, 휀) = sin arctan 푚휀 and 푎 (푥, 푡, 푚, 휀) = cos arctan 푚휀 . 1 휀 2 휀 푥∈휀Z 푥∈휀Z Proof. We prove both formulae simultaneously by induction on 푡. The base 푡 = 휀 is obvious. To perform the induction step, suppose that for 푡 ∈ 휀Z+ the formulae from the statement hold. Then by

11 Proposition 1 and by the summation formulae for the sine and the cosine we have

∑︁ 1 ∑︁ 푚휀 ∑︁ 푎1(푥, 푡 + 휀, 푚, 휀) = √ 푎1(푥, 푡, 푚, 휀) + √ 푎2(푥, 푡, 푚, 휀) = 1 + 푚2휀2 1 + 푚2휀2 푥∈휀Z 푥∈휀Z 푥∈휀Z sin ((푡/휀 − 1) arctan 푚휀) 푚휀 cos ((푡/휀 − 1) arctan 푚휀) = √ + √ = 1 + 푚2휀2 1 + 푚2휀2 [︂sin ((푡/휀) arctan 푚휀) 푚휀 cos ((푡/휀) arctan 푚휀)]︂ = − + 1 + 푚2휀2 1 + 푚2휀2 [︂푚휀 cos ((푡/휀) arctan 푚휀) 푚2휀2 sin ((푡/휀) arctan 푚휀)]︂ + + = 1 + 푚2휀2 1 + 푚2휀2 (︂ 푡 )︂ = sin arctan 푚휀 . 휀 ∑︀ The formula for 푎2(푥, 푡, 푚, 휀) is proved analogously. 푥∈휀Z

Corollary 1. For each 푡 ∈ 휀Z+ we have ∑︁ (︂ 푡 − 휀 )︂ 푎(푥, 푡, 푚, 휀) = 푖 exp −푖 arctan 푚휀 . 휀 푥∈휀Z Proof. Follows directly from Proposition 11 and Euler’s formula.

푡+푥 푡−푥 Now, perform a change of the coordinates: (푥, 푡) ↦→ (휆, 휇) = ( 2휀 , 2휀 ). We just rotate the coordinate axes through 90∘ about zero, and afterwards we scale everything 1/휀 times (see Figure 4). For positive integers 휆, 휇 denote 푏(휆, 휇, 푚, 휀) := 푎(휀(휆 − 휇), 휀(휆 + 휇), 푚, 휀). We use the notation 푏1(휆, 휇, 푚, 휀) and 푏2(휆, 휇, 푚, 휀) for the real and the imaginary part of 푏(휆, 휇, 푚, 휀).

Change of the coordinates −−−−−−−−−−−−−−→

Figure 4

The following table shows 푏(휆, 휇, 푚, 휀) for small 휆 and 휇. The number in a cell (휆, 휇) is 푏(휆, 휇, 푚, 휀). (Compare this table with Table 1.)

12 푚휀 (푚휀−2푚3휀3)−푚2휀2푖 (푚휀−4푚3휀3+푚5휀5)+2(푚4휀4−푚2휀2)푖 (푚휀−6푚3휀3+3푚5휀5)−(3푚2휀2−6푚4휀4+푚6휀6)푖 3 (1+푚2휀2)3/2 (1+푚2휀2)2 (1+푚2휀2)5/2 (1+푚2휀2)3 푚휀 (푚휀−푚3휀3)−푚2휀2푖 (푚휀−푚3휀3)+(푚4휀4−2푚2휀2)푖 (푚휀−푚3휀3)+3(푚4휀4−푚2휀2)푖 2 1+푚2휀2 (1+푚2휀2)3/2 (1+푚2휀2)2 (1+푚2휀2)5/2 2 2 2 2 2 2 1 √ 푚휀 푚휀−푚 휀 푖 푚휀−2푚 휀 푖 푚휀−3푚 휀 푖 1+푚2휀2 1+푚2휀2 (1+푚2휀2)3/2 (1+푚2휀2)2 0 푖 √ 1 푖 1 푖 1 푖 1+푚2휀2 1+푚2휀2 (1+푚2휀2)3/2 휇 휆 1 2 3 4

Table 3: 푏(휆, 휇, 푚, 휀) for small 휆 and 휇.

Proposition 12. 1) For each fixed 휇 ∈ Z+ we have

∞ √ ∞ √ ∑︁ 1 + 푚2휀2 + 1 ∑︁ 2 + 푚2휀2 + 2 푚2휀2 + 1 푏 (휆, 휇, 푚, 휀) = (−1)휇+1 and 푏 (휆, 휇, 푚, 휀) = (−1)휇 . 1 푚휀 2 푚2휀2 휆=1 휆=1

2) For each fixed 휆 ∈ Z+ we have

∞ √ ∞ ∑︁ 1 + 푚2휀2 + 1 ∑︁ 푏 (휆, 휇, 푚, 휀) = (−1)휆+1 and 푏 (휆, 휇, 푚, 휀) = (−1)휆+1. 1 푚휀 2 휇=0 휇=0

Proposition 12 easily follows from Proposition 1. We omit the proof.

6.2 Quadratic identities Proposition 13. 1) For each fixed 휇 ∈ Z+ we have:

∞ ∑︁ 2 푏1(휆, 휇, 푚, 휀) = 1. 휆=1

2) For each fixed 휆 ∈ Z+ we have:

∞ ∑︁ 2 푎) 푏1(휆, 휇, 푚, 휀) = 1; 휇=0 ∞ ∑︁ 2 푏) 푏2(휆, 휇, 푚, 휀) = 1. 휇=0

In the proof of quadratic identities, Proposition 1 does not help much. To prove Proposition 13, we need a generalization of Proposition 4. For this purpose, we need an auxiliary definition.

Definition 5 ([10]). For a set 푇 ⊂ 휀Z2 and a point (푥, 푡) ∈ 휀Z2 with 푡 > 0, we define 푎(푥, 푡 bypass 푇 ; 푚, 휀) analogously to 푎(푥, 푡, 푚, 휀), only the sum is over those checker paths that do not pass through the points of the set 푇 . Denote 푃 (푥, 푡 bypass 푇 ; 푚, 휀) = |푎(푥, 푡 bypass 푇 ; 푚, 휀)|2. We denote by 푎1(푥, 푡 bypass 푇 ; 푚, 휀) and 푎2(푥, 푡 bypass 푇 ; 푚, 휀) the real and the imaginary part of 푎(푥, 푡 bypass 푇 ; 푚, 휀) respectively.

13 Example 3. Let us compute 푃 (0, 4 bypass (2휀, 2휀); 푚, 휀). Now, we must not consider the leftmost checker path from Figure 2. Thus

−푚2휀2 − 푚3휀3푖 푎(0, 4휀 bypass (2휀, 2휀); 푚, 휀) = ; (1 + 푚2휀2)3/2 푚4휀4 + 푚6휀6 푚4휀4 푃 (0, 4휀 bypass (2휀, 2휀); 푚, 휀) = = . (1 + 푚2휀2)3 (1 + 푚2휀2)2 The following proposition generalizes Proposition 1. We do not give the proof because it is essentially the same as the proof of Proposition 1 (See proof of Proposition 4 from [10]).

2 2 Proposition 14. For each set 푇 ⊂ 휀Z and each point (푥, 푡) ∈ 휀Z such that (푥, 푡) ∈/ 푇 and 푡 > 2휀 we have: 1 1) 푎1(푥, 푡 bypass 푇 ; 푚, 휀) = √ (푎1(푥 + 휀, 푡 − 휀 bypass 푇 ; 푚, 휀) + 푚휀 푎2(푥 + 휀, 푡 − 휀 bypass 푇 ; 푚, 휀); 1 + 푚2휀2 1 2) 푎2(푥, 푡 bypass 푇 ; 푚, 휀) = √ (푎2(푥 − 휀, 푡 − 휀 bypass 푇 ; 푚, 휀) − 푚휀 푎1(푥 − 휀, 푡 − 휀 bypass 푇 ; 푚, 휀)). 1 + 푚2휀2

Remark 9. If we apply the above proposition for 푇 = ∅, then we obtain Proposition 1. The following proposition was first stated and proven by G. Minaev and I. Russkikh, buttheir proof was quite complicated and technical. We present a simple alternative proof. Proposition 15 (Generalized probability conservation law, Minaev-Russkikh, private communica- tion). For each finite set 푇 ∈ 휀Z2 such that there is no infinite checker path from (0, 0) bypassing the points of 푇 we have ∑︁ 푃 (푥, 푡 bypass 푇 ∖ (푥, 푡); 푚, 휀) = 1. (푥,푡)∈푇

2 Remark 10. If we apply the above proposition for the set 푇푡 = {(푥, 푡) ∈ 휀Z : −푡 6 푥 6 푡)} (where 푡 is fixed), then we obtain Proposition 4.

Proof of Proposition 15. (See Figure 5a) Join each point (푥, 푡) ∈ 휀Z2, where 푡 > 0 and (푡 + 푥)/휀 is 2 even, with the points (푥−휀, 푡−휀) and (푥+휀, 푡−휀). Assign the numbers 푎2(푥, 푡 bypass 푇 ∖(푥, 푡); 푚, 휀) 2 and 푎1(푥, 푡 bypass 푇 ∖ (푥, 푡); 푚, 휀) respectively to the resulting edges. An edge joining (푥 ± 휀, 푡 − 휀) and (푥, 푡) is painted red, if (푥, 푡) ∈ 푇 and (푥 ± 휀, 푡 − 휀) ∈/ 푇 (see Figure 5b). It is clear that ∑︁ ∑︁ 푃 (푥, 푡 bypass 푇 ∖ (푥, 푡); 푚, 휀) = 푗(푒), (푥,푡)∈푇 red edges 푒 where 푗(푒) is the number assigned to the edge 푒. Now, the required assertion follows from the following two observations: 1) 푗((0, 0)-(휀, 휀)) = 1 (We denote by 푎-푏 the edge joining points 푎 and 푏); 2) (see Figure 5c) For each point (푥, 푡) ∈ 휀Z2 (with positive 푡 and even (푥 + 푡)/휀) we have 푗((푥, 푡)-(푥 + 휀, 푡 − 휀)) + 푗((푥, 푡)-(푥 − 휀, 푡 − 휀)) = 푗((푥, 푡)-(푥 + 휀, 푡 + 휀)) + 푗((푥, 푡)-(푥 − 휀, 푡 + 휀)). (4)

The latter observation easily follows from Proposition 14. Remark 11. Suppose that in Figure 5c there is electrical current, which flows into the point (푥, 푡) from the points (푥 ± 휀, 푡 − 휀), and out of the point (푥, 푡) to the points (푥 ± 휀, 푡 + 휀). Then equation (4) is just Kirchhoff’s current law.

14 (b) (c) (a)

Figure 5: In (b) the red (dash) edges for a particular set 푇 consisting of the red (square) points

Using the latter proposition, we finally prove Proposition 13.

Proof of Proposition 13. Let us prove assertion 1). Fix 휇̂︀ ∈ Z+. Consider two sequences of sets of 2 2 points in Z , 푆푛 and 푇푛, defined as follows: 푆푛 = {(휆, 휇) ∈ Z | 1 6 휆 6 푛, 휇 = 휇̂︀}, 푇푛 = {(휆, 휇) ∈ 2 Z | 0 6 휇 < 휇,̂︀ 휆 + 휇 = 푛 + 휇̂︀}. Table 4 shows sets 푇푛 and 푆푛 for small 푛, 휇̂︀. By Proposition 15 for each 푛 ∈ Z+ we have

푛 휇̂︀−1 ∑︁ ∑︁ 2 ∑︁ (︀ 2 2)︀ 1 = 푃 (휆, 휇 bypass (푇푛∪푆푛)∖(휆, 휇); 푚, 휀) = 푏1(휆, 휇,̂︀ 푚, 휀) + 푏1(푛, 휇, 푚, 휀) + 푏2(푛, 휇, 푚, 휀) , (휆,휇)∈푇푛∪푆푛 휆=1 휇=0 where the last equality holds because any checker path to a point of 푇푛 cannot pass through other points of 푇푛 ∪ 푆푛, and also because a checker path to a point of 푆푛 does not pass through other points of 푇푛 ∪ 푆푛 if and only if the checker path finishes with an upwards-left move. Thus it remains to prove that for each fixed 휇 we have 푏1(푛, 휇, 푚, 휀) → 0 and 푏2(푛, 휇, 푚, 휀) → 0 as 푛 → +∞. By Proposition 3 for 휆 > 휇 we have

휇−1 ∑︁ (︂휇 − 1)︂(︂푛 − 1)︂ 푏 (푛, 휇, 푚, 휀) = (1 + 푚2휀2)(1−푛−휇)/2(−1)푟(푚휀)2푟+1 . 1 푟 푟 푟=0 Each summand in the sum tends to 0 as 푛 → +∞. Since for each fixed 휇 the number of summands is finite, it follows that 푏1(푛, 휇, 푚, 휀) → 0 as 푛 → +∞. Analogously 푏2(푛, 휇, 푚, 휀) → 0. Assertion 1) is proved. Assertion 2a) follows from 1) by the first identity of Proposition 5, and assertion 2b) is proved analogously to 1).

15 Table 4: Sets 푇푛 and 푆푛 for small 푛, 휇̂︀

1 휇 ̂︀ 푛 1 2 3

In conclusion, we state a conjecture.

Conjecture 1. For each fixed 휇 ∈ Z+ we have

∞ ∑︁ 푚2휀2 + 2 1) 푏 (휆, 휇, 푚, 휀)2 = ; 2 푚2휀2 휆=1 ∞ ∑︁ 2 2) 휆푏1(휆, 휇, 1/휀, 휀) = 3휇 − 1; 휆=1 ∞ ∑︁ 2 2 2 3) 휆 푏1(휆, 휇, 1/휀, 휀) = 13휇 − 10휇 + 3; 휆=1 ∞ 휇−1 ∑︁ 1 ∑︁ 1 4) 푏 (휆, 휇, 1/휀, 휀)2 = 2휇−1 log 2 − ; 휆 1 푗 · 2푗 휆=1 푗=1 ∞ 휇−1(︀2휇−2)︀ ∑︁ 1 2 휇−1 5) 푏 (휆, 휇, 1/휀, 휀)2 = . 2휆 1 32휇+1 휆=1 Remark 12. Despite the fact that the first equality in the conjecture is similar to the equalities in Proposition 13, it cannot be proved analogously.

7 Conclusions

In this work, we presented a lot of combinatorial identities in Feynman checkers. We used some of these identities to prove the following results: if there is at least one checker path to the point, then the probability to find an electron at this point is non-zero (Theorem 1); the expectation of theaverage electron velocity equals the time-average of the expectation of the instantaneous electron velocity

16 (Proposition 8). We also found the limit value of the average electron velocity (Theorem 2). There are several identities that are yet to be proved and possibly there are many interesting identities that are yet to be discovered.

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