Feynman checkers: lattice quantum field theory with real time

Abstract

We present a new completely elementary model that describes the creation, annihilation, and motion of non-interacting electrons and positrons along a line. It is a modification of the model known under the names Feynman checkers or one-dimensional quantum walk. It can be viewed as a six-vertex model with certain complex weights of the vertices. The discrete model is consistent with the continuum quantum field theory, namely, reproduces the known expected charge density as the lattice step tends to zero. It is exactly solvable in terms of hypergeometric functions. We introduce interaction resembling Fermi’s theory and establish perturbation expansion.

Appendices

Alternative definitions and proofs

Here we give alternative combinatorial proofs of Propositions 12–16 and alternative definitions of Feynman anticheckers in the spirit of the six-vertex model and exclusion process. The proofs are elementary and rely only on the assertion that the finite-lattice propagator is well-defined (see Theorem 4). The proofs of Propositions 14 and 16 are especially simple and are presented first. Cf. [21, Proof of Lemma 3.4] and [8]. The proofs of Propositions 12 and 13 require some auxiliary definitions and assertions. Proposition 15 follows from Propositions 12–14.

Second proof of Proposition 14

Case 1: $a\ne f$. Define $A(a\rightarrow f\text { pass }e)$ (respectively, $A(a\rightarrow f\text { bypass }e)$) analogously to ${A}(a\rightarrow f)$, only the sum in the numerator of (20) is now over loop configurations with the source a and the sink f containing the edge e (respectively, not containing e). In this case the proposition follows from the two identities:

$$\begin{aligned} A(a\rightarrow e\text { bypass }f)A(ef)+A(a\rightarrow e'\text { bypass }f)A(e'f)&=A(a\rightarrow f), \end{aligned}$$

(31)

$$\begin{aligned} A(a\rightarrow e\text { pass }f)A(ef)+A(a\rightarrow e'\text { pass }f)A(e'f)&= 0. \end{aligned}$$

(32)

Let us prove (31). For each loop configuration S with the source a and the sink f, remove the last edge of the path from a to f. Since $a\ne f$, we get a loop configuration $S'$ not containing f, with the source a and the sink either e or $e'$. We have decreased the number of nodes in S by 1, hence either $A(S)=A(S')A(ef)$ or $A(S)=A(S')A(e'f)$ depending on if the sink is e or $e'$. Summing over all S we get (31) because the map $S\mapsto S'$ is clearly invertible.

Let us prove (32). To each loop configuration S with the source a and the sink e containing f, assign a loop configuration $S'$ with the source a and the sink $e'$ containing f as follows. If f in contained in a loop $f\dots e'f$ from S then combine the loop with the path $a\dots e$ from S into the new path $a\dots ef\dots e'$. If f in contained in the path $a\dots e'f\dots e$ from S, then decompose the path into a new path $a\dots e'$ and a new loop $f\dots ef$. All the other loops in S remain unmodified. The resulting loop configuration is $S'$. The map $S\mapsto S'$ changes the parity of the number of loops and preserves the nodes except for that the node $(e',f)$ is replaced by (e, f). Hence $A(S')A(e'f)=-A(S)A(ef)$. Summing over all S we get (32) because the map $S\mapsto S'$ is invertible.

Case 2: $a=f$. To each loop configuration S with the source f and the sink $f,e,e'$ respectively, assign a loop configuration $S'$ (without sources and sinks) as follows. If S has the sink f, then remove the path f from S. If S has the sink either e or $e'$, then close up the path in S into a new loop. The resulting loop configuration is $S'$. In the former case, $S'$ has the same loops and nodes as S, and in the latter case, we have added one loop and one node. Thus, if the sink is f, or e, or $e'$, then $A(S')=A(S)$, or $-A(S)A(ef)$, or $-A(S)A(e'f)$ respectively. Summing over all S and dividing by the denominator of (20), we get

$$\begin{aligned} A(f\rightarrow f) -A(f\rightarrow e)A(ef)-A(f\rightarrow e')A(e'f)=1, \end{aligned}$$

because the map $S\mapsto S'$ is invertible. $\square$

Second proof of Proposition 16

The proof is by induction on n. The base $n=1$ is the trivial assertion $A(f)=1$. To perform the induction step, take a path $e\dots f$ of length $n-1$. Let d and $d'$ be the two edges with the endpoint at the starting point of e. By Proposition 14 we get

$$\begin{aligned} A(a\rightarrow e)A(e\dots f)= & {} \left( A(a\rightarrow d)A(de)+A(a\rightarrow d')A(d'e)+\delta _{ae}\right) A(e\dots f)\\= & {} \!A(a\!\rightarrow \! d)A(de\dots f)\!+\!A(a\!\rightarrow \! d')A(d'e\dots f)\!+\!\delta _{ae}A(e\dots f). \end{aligned}$$

Here d and $d'$ are distinct from all the edges in $e\dots f$ by the assumption $n\le 2T$, and hence can be added to the path. Summing over all such paths $e\dots f$ we get

$$\begin{aligned} \sum _{e\dots f\text { of length }n-1}A(a\rightarrow e)A(e\dots f)= & {} \sum _{d\dots f\text { of length }n}A(a\rightarrow d)A(d\dots f)\\{} & {} +\sum _{a\dots f\text { of length }n-1}A(a\dots f). \end{aligned}$$

Adding $-A(a\rightarrow f)+\sum _{a\dots f\text { of length }<n-1}A(a\dots f)$ to both sides and applying the inductive hypothesis, we get the required identity. $\square$

The next definitions and a lemma are needed for combinatorial proofs of Propositions 12–13.

Definition 8

(See Fig. 5 to the right) Let (e, f) be a pair of edges such that the endpoint of e is the starting point of f. The complementary pair $(e',f')$ is formed by the other edge $e'\ne e$ with the same endpoint as e and the other edge $f'\ne f$ with the same starting point as f.

Let S be a loop configuration containing both nodes (e, f) and $(e',f')$. The flip of S (at the endpoint of e) is the loop configuration obtained as follows. If the nodes (e, f) and $(e',f')$ belong to distinct loops $ef\dots e$ and $e'f'\dots e'$ of S, then combine them into one new loop $ef'\dots e'f\dots e$. If the nodes (e, f) and $(e',f')$ belong to the same loop $ef\dots e'f'\dots e$, then decompose the latter into two new loops $e'f\dots e'$ and $ef'\dots e$. All the other loops in S remain unmodified. The flip of a loop configuration with sources and sinks is defined analogously.

Lemma 15

If a loop configuration S (without sources and sinks) contains all the edges, then the number of loops in S has the same parity as one half of the total number of turns in S.

Proof

The proof is by induction over the total number of turns.

Base: If s has no turns, then each loop entirely consists of the edges of the same direction. The reflection with respect to the vertical line $x=T\varepsilon /2$ shows that there is equal number of loops consisting of upwards-left and upwards-right edges. Hence the number of loops is even.

Step: Assume that s has a loop with a turn (e, f). Since s contains all the edges, it has also a loop with the complementary turn $(e',f')$. Then a flip of s changes the parity of the number of loops and reduces the total number of turns by 2. By induction, the lemma follows. $\square$

Now we give a definition of Feynman anticheckers as a six-vertex model with complex weights (see Fig. 1). Instead of assigning a direction to each edge, as in the common six-vertex model, we speak of choosing a set of edges, and the edges are always oriented in the time direction. To each configuration, i.e., set of edges, we also assign an overall sign defined globally in terms of a loop decomposition of the set. Our loop decomposition is very different from the one in [10].

Definition 9

A set s of edges is a current, if for each lattice point the number of edges in s starting at the point equals the number of edges in s ending at the point (recall that each edge is oriented in the direction where time t increases). A set s of edges is a current with sources $a_1,\dots ,a_n$ and sinks $f_1,\dots ,f_n$, if edges $a_1,\dots ,a_n$ are distinct, egdes $f_1,\dots ,f_n$ are distinct, s contains $a_1,\dots ,a_n,f_1,\dots ,f_n$, and for each lattice point the number of edges in s starting at the point and distinct from $a_1,\dots ,a_n$ equals the number of edges in s ending at the point and distinct from $f_1,\dots ,f_n$.

A lattice point is a singularity of s, if it is the starting point of two edges of s, distinct from the sources. Clearly, for each s there exists a unique loop configuration (called the loop decomposition of s) having the same sources and sinks, consisting of the same edges, and having no turns at the singularities of s. If the loop decomposition has exactly l loops and n paths joining $a_1,\dots ,a_n$ with $f_{\sigma (1)},\dots ,f_{\sigma (n)}$ respectively for some permutation $\sigma$, then denote $\textrm{sgn}(s):=(-1)^l\textrm{sgn}(\sigma )$. Here we set $\textrm{sgn}(\sigma )=+1$ for $n=0$.

A node of s is an ordered pair (e, f) of edges of s such that the endpoint of e is the starting point of f and is not a singularity of s, the edge e is not a sink, and f is not a source. The numbers $\textrm{eventurns}(s)$, $\textrm{oddturns}(s)$, $\textrm{evennodes}(s)$, $\textrm{oddnodes}(s)$, and A(s) are defined literally as for a path or loop (see Definition 3), with the overall sign in (19) set to be $\textrm{sgn}(s)$. Denote

$$\begin{aligned} {A}^{\textrm{6V}}(a_1,\dots ,a_n\rightarrow f_1,\dots ,f_n ):=\frac{\sum \limits _{\begin{array}{c} \text {currents }s\\ \text {with the sources }a_1,\dots ,a_n\text { and the sinks }f_1,\dots ,f_n \end{array}}A(s )}{\sum \limits _{\begin{array}{c} \text {currents } s \end{array}}A(s )}. \end{aligned}$$

If all $a_1,\dots ,a_n,f_1,\dots ,f_n$ are distinct, then the complement to s is the current ${{\bar{s}}}$ with sources $f_1,\dots ,f_n$ and sinks $a_1,\dots ,a_n$ formed by $f_1,\dots ,f_n, a_1,\dots ,a_n$ and exactly those other edges that do not belong to s.

Example 5

(Empty and complete currents) We have $A(\emptyset )=A({\overline{\emptyset }} )=1$, where ${\overline{\emptyset }}$ is the current consisting of all the edges. Indeed, the currents $\emptyset$ and ${\overline{\emptyset }}$ have no nodes, and $\textrm{sgn} ({\overline{\emptyset }})=\textrm{sgn}(\emptyset )=+1$ by Lemma 15 because the loop decomposition of ${\overline{\emptyset }}$ has no turns.

Proposition 26

(Equivalence of definitions) For any edges $a_1,\dots ,a_n,f_1,\dots ,f_n$ we have

$$\begin{aligned} {A}^{\textrm{6V}}(a_1,\dots ,a_n\rightarrow f_1,\dots ,f_n ) = {A}(a_1,\dots ,a_n\rightarrow f_1,\dots ,f_n ). \end{aligned}$$

Proof of Proposition 26

To each loop configuration S (possibly with sources and sinks), assign the set of all edges contained in the loops and paths of S. Clearly, we get a current with the same sources and sinks. A current s with K even and J odd singularities has $2^{K+J}$ preimages, obtained from the loop decomposition S of s by flips at any subset of the set of singularities.

It suffices to prove that $A(s)=\sum _{S'}A(S')$, where the sum is over all $2^{K+J}$ preimages $S'$ of s. Take a loop configuration $S'$ obtained from S by flips at k even and j odd singularities. Since each such flip increases the number of turns by 2 and changes either the parity of the number of loops or the sign of the permutation $\sigma$ from Definition 5, it follows that $A(S')=(-1)^{k+j}(-im\varepsilon )^{2j}(-\delta \varepsilon )^{2k}A(S)= (m^2\varepsilon ^2)^{j}(-\delta ^2\varepsilon ^2)^{k}A(S)$. Summing over all the subsets of the set of singularities, we get the required equality

$$\begin{aligned} \sum _{S'}A(S')= & {} \sum _{k=0}^{K}\sum _{j=0}^{J} \left( {\begin{array}{c}K\\ k\end{array}}\right) \left( {\begin{array}{c}J\\ j\end{array}}\right) (m^2\varepsilon ^2)^{j}(-\delta ^2\varepsilon ^2)^{k}A(S) \\= & {} (1+m^2\varepsilon ^2)^{J}(1-\delta ^2\varepsilon ^2)^{K}A(S) = A(s), \end{aligned}$$

where the factor before A(S) in the latter equality compensates the contribution of the $2K+2J$ nodes of the loop decomposition S which are not nodes of the current s. $\square$

The following proposition demonstrates a symmetry between particles and antiparticles.

Proposition 27

(Complement formula) For each current s, possibly with sources $a_1,\dots ,a_n$ and sinks $f_1,\dots ,f_n$, where all $a_1,\dots ,a_n,f_1,\dots ,f_n$ are distinct, we have ${A}(s ) = (-1)^{|\{k:a_k\parallel f_k\}|} {A}({{\bar{s}}} ).$

Example 6

In Example 3, if the set $s=\{a,c\}$ is viewed as a current without sources and sinks, then it has the complement ${\bar{s}}=\{b,d\}$, so that ${A}(s ) =-1/\sqrt{1+m^2\varepsilon ^2}\sqrt{1-\delta ^2\varepsilon ^2} ={A}({{\bar{s}}} )$. If the same set $s=\{a,c\}$ is viewed as a current with the source a and the sink c, then the complement ${\bar{s}}=\{a,b,c,d\}$ has the source c and the sink a, so that ${A}(s ) =1/\sqrt{1+m^2\varepsilon ^2}= -{A}({{\bar{s}}} )$.

Proof of Proposition 27

First let us show that $A(s)=A({{\bar{s}}})$ up to sign, namely, $A(s)\textrm{sgn}(s)=A({{\bar{s}}})\textrm{sgn}({{\bar{s}}})$. To each node (e, f) of s, assign the complementary pair $(e',f')$. The latter is a node of ${{\bar{s}}}$. Indeed, since the starting point of f (equal to the endpoint of e) is not a singularity, it follows that either $f'\notin s$ or $f'$ is a source of s. Thus $f'\in {{\bar{s}}}$ and it is not a source of ${{\bar{s}}}$. Analogously, $e'\in {{\bar{s}}}$ and it is not a sink of ${{\bar{s}}}$. Since f is not a source, it follows that either $f\notin {{\bar{s}}}$ or f is a source of ${{\bar{s}}}$. This means that the starting point of $f'$ is not a singularity of ${{\bar{s}}}$. Thus $(e,f)\mapsto (e',f')$ is a bijection between the sets of nodes of s and ${{\bar{s}}}$. This bijection preserves the parity of nodes and takes turns to turns. Thus $A(s)\textrm{sgn}(s)=A({{\bar{s}}})\textrm{sgn}({{\bar{s}}})$.

Second let us show that $\textrm{sgn}(s)\textrm{sgn}({{\bar{s}}})=(-1)^{n-\textrm{turns}(s)}$, where $\textrm{turns}(s)$ is the total number of turns in the current s. Let S and ${\overline{S}}$ be the loop decompositions of s and ${{\bar{s}}}$. Let S have exactly l loops and n paths $a_1\dots f_{\sigma (1)}$, ..., $a_n\dots f_{\sigma (n)}$ for some permutation $\sigma$. Let ${{\bar{S}}}$ have exactly ${{\bar{l}}}$ loops and n paths $f_1\dots a_{{{\bar{\sigma }}}(1)}$, ..., $f_n\dots a_{{{\bar{\sigma }}}(n)}$ for some permutation ${{\bar{\sigma }}}$. Form the loop

$$\begin{aligned} a_1\dots f_{\sigma (1)}\dots a_{{{\bar{\sigma }}}\circ \sigma (1)}\dots f_{\sigma \circ {{\bar{\sigma }}}\circ \sigma (1)}\dots a_1, \end{aligned}$$

starting from $a_1$ and alternating the paths of S and ${{\bar{S}}}$ until the first return to $a_1$. Form analogous loops starting from the other not yet visited edges $a_k$. The resulting loops are in bijection with the loops in the loop decomposition of the permutation ${{\bar{\sigma }}}\circ \sigma$. Hence their total number is even if and only if $(-1)^n\textrm{sgn}(\sigma )\textrm{sgn}({{\bar{\sigma }}})=+1$. Consider the set consisting of the resulting loops (obtained by gluing the paths of S and ${\overline{S}}$) and the loops of S and ${\overline{S}}$. The number of loops in the set is even if and only if $(-1)^n\textrm{sgn}(s)\textrm{sgn}({{\bar{s}}})=+1$ because $\textrm{sgn}(s)=(-1)^l\textrm{sgn}(\sigma )$. On the other hand, by Lemma 15 this number has the same parity as $\textrm{turns}(s)$, because the total number of turns in S and the total number of turns in ${\overline{S}}$ both equal $\textrm{turns}(s)$. We get $\textrm{sgn}(s)\textrm{sgn}({{\bar{s}}})=(-1)^{n-\textrm{turns}(s)}$.

It remains to notice that $n-\textrm{turns}(s)=|\{k:a_k\parallel f_k\}|\mod 2$. Indeed, each loop in S has an even number of turns, a path $a_k\dots f_{\sigma (k)}$ has an even number of turns if and only if $a_k\parallel f_{\sigma (k)}$, and the parity of $|\{k:a_k\parallel f_k\}|$ is invariant under a permutation of $f_1,\dots ,f_n$. $\square$

Second proof of Proposition 12

Use Definition 9 and Proposition 26. The result follows from

$$\begin{aligned} \sum _{s\text { with the source and sink }a}A(s )= & {} \sum _{s\not \ni a}A(s ) =\sum _{s\ni a}A(s ) =\frac{1}{2}\left( \sum _{s\not \ni a}A(s )+\sum _{s\ni a}A(s )\right) \\= & {} \frac{1}{2}\sum _{s}A(s ). \end{aligned}$$

Here the sums are over currents s (in the first sum — with the source and the sink a). The first equality holds because $s\mapsto s-\{a\}$ is a bijection between currents with the source and sink a and currents (without sources and sinks) not containing a. This bijection preserves A(s) because a does not belong to any node of s. The second equality holds because $s\mapsto {\overline{s}}$ is a bijection between the currents containing and not containing a. This bijection preserves A(s) by Proposition 27. The third equality follows from the second one. $\square$

Second proof of Proposition 13

The map $s\mapsto {\overline{s}}$ is a bijection between the currents with the source a and sink f and the currents with the source f and sink a. By Proposition 27, this bijection preserves A(s) for $a\perp f$ and changes the sign of A(s) for $a\parallel f$. Summing over all s and diving by the sum over all the currents, we get the required assertion by Proposition 26. $\square$

We conclude this section by restating Definition 9 informally in a self-contained way resembling exclusion process. (Cf. a different quantum exclusion process [4] defined by a continuous-time stochastic differential equation.)

Definition Sketch (See Fig. 7) Fix $T\in {\mathbb {Z}},\mu ,\nu >0$ called half-period, particle mass, small imaginary mass respectively. Take a checkered stripe $1\times 2T$ closed in a ring. Enumerate the 2T squares by the numbers $0,\dots , 2T-1$ consecutively.

Define a realization of the exclusion process inductively. At time $t=0$ some squares are occupied by identical particles, at most one per square.

At time $t=k$ decompose the stripe into T rectangles $1\times 2$ so that squares k and $k+1$ form a rectangle. In each rectangle with exactly 1 particle, the particle is allowed to jump into the empty square of the same rectangle. In rectangles with 2 or 0 particles, nothing is changed.

Fig. 7

A realization

Finally at time $t=2T$ it is requested that the particles occupy the same set of squares as at $t=0$. The resulting sequence of 2T configurations of particles at times $t=0,1,\dots ,2T-1$ is a realization of the exclusion process.

A realization with a source at (0, 0) and a sink at is defined analogously, only:

• At time 0 before any jumps the square 0 is empty, and a particle is added to the square;

• At time t after all jumps the square x is occupied, and the particle is removed from it.

To each realization s (possibly with a source and a sink), assign a complex number A(s) as follows. Start with $A(s)=(-1)^n$, where n is the number of particles at time 0. For each moment $t=0,1,\dots 2T-1$ and each $1\times 2$ rectangle containing two particles at time t, multiply the current value of A(s) by $-1$. For each moment $t=0,1,\dots 2T-1$ and each $1\times 2$ rectangle containing exactly one particle at time t, multiply the current value of A(s) by

$$\begin{aligned} {\left\{ \begin{array}{ll} 1/\sqrt{1+\mu ^2}, &{}\quad \text {if the particle jumps and }t\text { is even};\\ -i\mu /\sqrt{1+\mu ^2}, &{}\quad \text {if the particle does not jump and }t\text { is even};\\ 1/\sqrt{1-\nu ^2}, &{}\quad \text {if the particle jumps and }t\text { is odd};\\ -\nu /\sqrt{1-\nu ^2}, &{}\quad \text {if the particle does not jump and }t\text { odd}. \end{array}\right. } \end{aligned}$$

Finally, multiply A(s) by the sign of the permutation of the particles obtained at time $t=2T$ (if the particles added at the source and removed at the sink are distinct, then they are identified, and A(s) is multiplied by an additional $-1$).

The two-point function is then .

Using Proposition 26, one can see that the two-point function actually equals the finite-lattice propagator; for instance, we conjecture that for x, t even it equals $A(a_0\!\rightarrow \! f_{2}; \mu /2,2,\nu /2,T)$. Notice that if we restrict to just realizations without particles at time 0, drop space- and time-periodicity requirements, and take $\nu =0$, then the definition becomes equivalent to Definition 1.

Wightman axioms

To put the new model in the general framework of quantum theory, we define the Hilbert space describing the states of the model along with the Hamiltonian and the field operators acting on this space. The definition is similar to (and simpler than) the continuum free spin 1/2 field [14, §5.2], only we have unusual dispersion relation (3) and smaller number of spin components (coming from smaller spacetime dimension). It goes along the lines of [6, §IV], where the massless case is considered. For the massive case, we have not found the detailed construction in the literature (cf. [3, §2.3]). Although the definition is self-contained, familiarity with the continuum analogue is desirable. We use notation $f^*$, $f^\dagger$, ${{\bar{f}}}$, $\langle f| g\rangle$ from [14, §1.1] (introduced below) unusual in mathematics but common in physics. For simplicity, we first perform the construction for the model with a fixed spatial size, then for the infinite lattice, and finally discuss which Wightman axioms of quantum field theory are satisfied.

1.1 Informal motivation

In quantum theory, a system is described by a Hilbert space encoding all possible states of the system. Examples of states of a free field (in a box of fixed spatial size) are: the vacuum state without any particles at all; the state with one particle of given momentum p; the state with two particles of momenta $p_1$ and $p_2$; the state with one particle of momentum p and one anti-particle of momentum q; and so on. States of this kind actually form a basis of the Hilbert space. In general, a state is an arbitrary unit vector of the Hilbert space up to scalar multiples.

What quantum theory can compute is the expectation of observables such as the total energy of the system (the Hamiltonian) or charge density at a particular point. In general, an observable is a self-adjoint operator on the Hilbert space. The expectation of the observable in a given state equals the inner product of the state with its image under the operator.

Field operators are not observables but are building blocks for those. They are used to construct states such as the state with one right electron at position x and time t, and useful functions such as the propagator.

1.2 Definition for fixed spatial size

Definition 10

Fix $X\in {\mathbb {Z}}$ called lattice spatial size and $\varepsilon ,m>0$. Assume $X>0$. Define ${\widetilde{A}}_k(x,t,m,\varepsilon ,X)$ analogously to ${\widetilde{A}}_k(x,t,m,\varepsilon )$ (see Definition 2), only take the quotient

The momentum space is

$$\begin{aligned} {\mathcal {P}}_X:=\left\{ \frac{2\pi k}{X\varepsilon }:-\frac{X}{2}<k\le \frac{X}{2}, k\in {\mathbb {Z}}\right\} . \end{aligned}$$

Denote by $L_2({\mathcal {P}}_X)$ the Hilbert space with the finite orthonormal basis formed by the functions $\chi _p:{\mathcal {P}}_X\rightarrow {\mathbb {C}}$ equal to 1 at a particular element $p\in {\mathcal {P}}_X$ and vanishing at all the other elements. Equip it with the natural inner product antilinear in the first argument. Let $\otimes$ and $\wedge$ be respectively the tensor and exterior product over ${\mathbb {C}}$. An empty exterior product of vectors (respectively, spaces) is set to be 1 (respectively, ${\mathbb {C}}$).

The Hilbert space of X-periodic Feynman anticheckers is the $2^{2X}$-dimensional Hilbert space

$$\begin{aligned} {\mathcal {H}}_X:=\left( \bigoplus _{n=0}^{X}\bigwedge ^n L_2\left( {\mathcal {P}}_X \right) \right) ^{\otimes 2}. \end{aligned}$$

It has an orthonormal basis formed by the vectors

$$\begin{aligned} \sqrt{n!l!}(\chi _{p_1}\wedge \dots \wedge \chi _{p_n})\otimes (\chi _{q_1}\wedge \dots \wedge \chi _{q_l}) \end{aligned}$$

(33)

for all integers n, l from 0 to X and all $p_1,\dots ,p_n,q_1,\dots ,q_l \in {\mathcal {P}}_X$ such that $p_1<\dots <p_n$ and $q_1<\dots <q_l$. (Physically, the vectors mean the states with n electrons of momenta $p_1,\dots ,p_n$ and l positrons of momenta $q_1,\dots ,q_l$.) The basis vector $1\in {\mathbb {C}}=\bigwedge ^0 L_2\left( {\mathcal {P}}_X \right)$ obtained for $n=l=0$ is the vacuum vector. It is denoted by $|0\rangle$. The dual vector is denoted by $\langle 0|\in {\mathcal {H}}_X^*$.

The Hamiltonian of X-periodic Feynman anticheckers is the linear operator on ${\mathcal {H}}_X$ such that all basis vectors (33) are eigenvectors with the eigenvalues (see notation (3))

$$\begin{aligned} \omega _{p_1}+\dots +\omega _{p_n}+\omega _{q_1}+\dots +\omega _{q_l}. \end{aligned}$$

(Physically, $\omega _{p}$ is viewed as the energy of a particle with momentum p; hence the Hamiltonian eigenvalues mean total energy of the eigenstates.)

For $p\in {\mathcal {P}}_X$, the creation operators $a_p^\dagger$ and $b_q^\dagger$ of particles and antiparticles with momenta p and q respectively are the linear operators on ${\mathcal {H}}_X$ defined on basis vectors (33) by

$$\begin{aligned}&a_p^\dagger \left( \sqrt{n!l!}(\chi _{p_1}\wedge \dots \wedge \chi _{p_n})\otimes (\chi _{q_1}\wedge \dots \wedge \chi _{q_l})\right) \\&\quad := \sqrt{(n+1)!l!}(\chi _p\wedge \chi _{p_1}\wedge \dots \wedge \chi _{p_n})\otimes (\chi _{q_1}\wedge \dots \wedge \chi _{q_l}),\\&b_q^\dagger \left( \sqrt{n!l!}(\chi _{p_1}\wedge \dots \wedge \chi _{p_n})\otimes (\chi _{q_1}\wedge \dots \wedge \chi _{q_l})\right) \\&\quad := (-1)^n\sqrt{n!(l+1)!}(\chi _{p_1}\wedge \dots \wedge \chi _{p_n})\otimes (\chi _q\wedge \chi _{q_1}\wedge \dots \wedge \chi _{q_l}). \end{aligned}$$

Their adjoint operators are denoted by $a_p$ and $b_q$ respectively. For each $t\in \varepsilon {\mathbb {Z}}$ and define the field operator $\psi _X(x,t):{\mathcal {H}}_X\rightarrow {\mathcal {H}}_X\oplus {\mathcal {H}}_X$ by

$$\begin{aligned} \psi _X(x,t)\!:=\!\frac{1}{\sqrt{X}}\! \sum _{p\in {\mathcal {P}}_X }\! \left( \! \begin{pmatrix} -i\cos (\alpha _p/2) \\ \sin (\alpha _p/2) \end{pmatrix} e^{ipx-i\omega _pt}a_p \!+\! \begin{pmatrix} i\cos (\alpha _p/2) \\ \sin (\alpha _p/2) \end{pmatrix} e^{i\omega _pt-ipx}b_p^\dagger \right) , \end{aligned}$$

where $\alpha _p\in [0,\pi ]$ is determined by the condition $\cot \alpha _p=\frac{\sin (p\varepsilon )}{m\varepsilon }.$ (Informally, the first component of the field operator creates a right positron or annihilates a right electron at position x and time t. The second component does the same for left particles.)

The propagator is defined through field operators as follows. Denote by $\psi ^\dagger _X(x,t):{\mathcal {H}}_X\oplus {\mathcal {H}}_X\rightarrow {\mathcal {H}}_X$ the adjoint of the operator $\psi _X(x,t)$. Define the Dirac adjoint by ${{\bar{\psi }}}_X(x,t):=\psi _X^\dagger (x,t) \left( {\begin{matrix} 0 &{} -i \\ i &{} 0 \end{matrix}}\right)$. Let $\psi _X(0,0)^T$ be the transpose of $\psi _X(0,0)$. Define the time-ordered product

$$\begin{aligned} \textrm{T}\psi _X(x,t){{\bar{\psi }}}_X(0,0):= {\left\{ \begin{array}{ll} \psi _X(x,t){{\bar{\psi }}}_X(0,0), &{}\quad \text {if }t\ge 0;\\ -{{\bar{\psi }}}_X(0,0)^T\psi _X(x,t)^T, &{}\quad \text {if }t<0. \end{array}\right. } \end{aligned}$$

The Feynman propagator for X-periodic Feynman anticheckers is $\langle 0|\textrm{T}\psi _X(x,t){{\bar{\psi }}}_X(0,0)|0\rangle$.

A direct checking using an analogue of Proposition 1 shows that the two constructions of the propagator are consistent: for each $x,t\in \varepsilon {\mathbb {Z}}$ and positive $X\in {\mathbb {Z}}$ the propagator for X-periodic Feynman anticheckers equals (cf. Proposition 28 below)

$$\begin{aligned} -\frac{i}{2} \begin{pmatrix} {\widetilde{A}}_1(-x,t,m,\varepsilon ,X)&{} {\widetilde{A}}_2(x,t,m,\varepsilon ,X) \\ -{\widetilde{A}}_2(-x,t,m,\varepsilon ,X) &{} {\widetilde{A}}_1(x,t,m,\varepsilon ,X) \end{pmatrix}. \end{aligned}$$

Remark 7

In 1 space dimension, unlike 3 dimensions, there are no states such as “one right electron of momentum p” or “one left electron of momentum p”. What we do have is the state $a_p^\dagger |0\rangle$, “one electron of momentum p”, which is right with the probability $\cos ^2(\alpha _p/2)$ and left with the probability $\sin ^2(\alpha _p/2)$. But there are states “one right (or one left) electron at position x and time t”; they are the two components of $\psi _X(x,t)|0\rangle$. The reason is that the Dirac equation in 1 space- and 1 time-dimension (both lattice and continuum) has just one (up to proportionality) solution for given momentum p and positive energy [32, Proposition 16].

1.3 Definition for the infinite lattice

Definition 11

Denote by $L_2[a;b]$ the Hilbert space of square-integrable functions $[a;b]\rightarrow {\mathbb {C}}$ with respect to the Lebesque measure up to changing on a set of measure zero. Equip it with the inner product $\langle f| g\rangle :=\int _{[a;b]}f^*(p)g(p)\,dp$ antilinear in the first argument, where $^*$ denotes complex conjugation. Denote by $\bigoplus$, $\bigotimes$, and $\bigwedge$ the orthogonal direct sum, the tensor and exterior product of Hilbert spaces, that is, completions of the orthogonal direct sum, the tensor and exterior product of Hermitian spaces over ${\mathbb {C}}$.

Fix $\varepsilon ,m>0$. The Hilbert space of Feynman anticheckers is the Hilbert space

$$\begin{aligned} {\mathcal {H}}:=\left( \bigoplus _{n=0}^{\infty }\bigwedge ^n L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right] \right) ^{\otimes 2}. \end{aligned}$$

The vector $1\in {\mathbb {C}}=\bigwedge ^0 L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon }\right]$ is denoted by $|0\rangle$. The dual vector is denoted by $\langle 0|\in {\mathcal {H}}^*$. Denote by ${\mathcal {H}}^0\subset {\mathcal {H}}$ the (incomplete) linear span of $\bigwedge ^n L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon }\right] \otimes \bigwedge ^l L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon }\right]$ for all $n,l$.

The Hamiltonian of Feynman anticheckers is the linear operator on ${\mathcal {H}}^0$ given by (see (3))

$$\begin{aligned} H(u_1\wedge \dots \wedge u_n \otimes v_1\wedge \dots \wedge v_l)&:= (\omega _p u_1)\wedge \dots \wedge u_n \otimes v_1\wedge \dots \wedge v_l+\dots \\&\quad + u_1\wedge \dots \wedge u_n \otimes v_1\wedge \dots (\omega _pv_l) \end{aligned}$$

for all integers $n,l\ge 0$ and all $u_1,\dots ,u_n,v_1,\dots ,v_l\in L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right]$, where $\omega _p$ is understood as a function in p. The evolution operator is the bounded linear operator on ${\mathcal {H}}$ given by

$$\begin{aligned} e^{-iHt}(u_1\wedge \dots \wedge u_n \otimes v_1\wedge \dots \wedge v_l)&:= (e^{-i\omega _pt} u_1)\wedge \dots \wedge (e^{-i\omega _pt}u_n)\\&\quad \otimes (e^{-i\omega _pt}v_1)\wedge \dots \wedge (e^{-i\omega _pt}v_l). \end{aligned}$$

The operators P and $e^{-iPx}$ are defined analogously, only $\omega _p$ and t are replaced by p and x.

For $f\in L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right]$, the creation operators $a(f)^\dagger$ and $b(f)^\dagger$ of particles and antiparticles respectively with momentum distribution f are the linear operators on ${\mathcal {H}}$ defined by

$$\begin{aligned} a(f)^\dagger (u\otimes v)&:= \sqrt{n+1}\,(f\wedge u)\otimes v,\\ b(f)^\dagger (u\otimes v)&:= (-1)^n\sqrt{l+1}\,u \otimes (f^*\wedge v) \end{aligned}$$

for all $u\in \bigwedge ^n L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right]$ and $v\in \bigwedge ^l L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right]$. Their adjoint operators are denoted by a(f) and b(f) respectively. For each $x,t\in \varepsilon {\mathbb {Z}}$ define the field operator $\psi (x,t):{\mathcal {H}}\rightarrow {\mathcal {H}}\oplus {\mathcal {H}}$ by

$$\begin{aligned}{} & {} \psi (x,t):= \left( \begin{matrix} \psi _1(x,t)\\ \psi _2(x,t) \end{matrix} \right) := \left( \begin{matrix} a(f_{1,x,t})+b(f^*_{1,x,t})^\dagger \\ a(f_{2,x,t})+b(f^*_{2,x,t})^\dagger \end{matrix} \right) , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} f_{1,x,t}&=\sqrt{{\varepsilon }/{2\pi }}\,i\cos (\alpha _p/2)e^{i\omega _pt-ipx},\\ f_{2,x,t}&=\sqrt{{\varepsilon }/{2\pi }}\,\sin (\alpha _p/2)e^{i\omega _pt-ipx}. \end{aligned} \end{aligned}$$

(The creation, annihilation, and field operators are bounded; see [14, (4.57)].) The Feynman propagator for Feynman anticheckers is defined through them analogously to Definition 10.

This construction of the propagator is consistent with the ones from Definitions 2, 3, and 9:

Proposition 28

Let $\textrm{T}\langle f_{k,x,t}|f_{l,0,0}\rangle :={\left\{ \begin{array}{ll} \langle f_{k,x,t}|f_{l,0,0}\rangle , &{} \text{ if } t\ge 0; \\ -\langle f_{l,0,0}|f_{k,x,t}\rangle , &{} \text{ if } t<0. \end{array}\right. }$

Then for all $x,t\in \varepsilon {\mathbb {Z}}$ we get

$$\begin{aligned} \langle 0|\textrm{T}\psi (x,t){{\bar{\psi }}}(0,0)|0\rangle= & {} i\,\textrm{T}\begin{pmatrix} \langle f_{1,x,t}|f_{2,0,0}\rangle &{} -\langle f_{1,x,t}|f_{1,0,0}\rangle \\ \langle f_{2,x,t}|f_{2,0,0}\rangle &{} -\langle f_{2,x,t}|f_{1,0,0}\rangle \end{pmatrix}\\= & {} -\frac{i}{2}\begin{pmatrix} {\widetilde{A}}_1(-x,t,m,\varepsilon )&{} {\widetilde{A}}_2(x,t,m,\varepsilon ) \\ -{\widetilde{A}}_2(-x,t,m,\varepsilon ) &{} {\widetilde{A}}_1(x,t,m,\varepsilon ) \end{pmatrix}. \end{aligned}$$

Proof

Case 1: $t\ge 0$. By definition, for each $k,l\in \{1,2\}$ we have

$$\begin{aligned} \langle 0|\textrm{T}\psi _k(x,t){{\bar{\psi }}}_l(0,0)|0\rangle \!= & {} \! \langle 0|\psi _k(x,t){{\bar{\psi }}}_l(0,0)|0\rangle \!=\! -i(-1)^{l}\langle 0|\psi _k(x,t)\psi _{3-l}^\dagger (0,0)|0\rangle \\= & {} -i(-1)^{l}\psi _{k}^\dagger (x,t)|0\rangle \cdot \psi _{3-l}^\dagger (0,0)|0\rangle \\= & {} -i(-1)^{l}\langle f_{k,x,t}|f_{3-l,0,0}\rangle , \end{aligned}$$

where $\cdot$ denotes the inner product in ${\mathcal {H}}$. The latter equality follows from

$$\begin{aligned} \psi _{k}^\dagger (x,t)|0\rangle&= \left( a(f_{k,x,t})^\dagger +b(f^*_{k,x,t}) \right) |0\rangle \\&=f_{k,x,t}\otimes 1 \in \bigwedge ^1 L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right] \otimes \bigwedge ^0 L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right] , \end{aligned}$$

where $b(f)|0\rangle =0$ because for each $u\in \bigwedge ^n L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right]$ and $v\in \bigwedge ^l L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right]$ we have

$$\begin{aligned} b(f)|0\rangle \cdot (u\otimes v)= |0\rangle \cdot b(f)^\dagger (u\otimes v)= |0\rangle \cdot (-1)^n\sqrt{l+1}\,u \otimes (f^*\wedge v)=0 \end{aligned}$$

by the condition $\bigwedge ^0 L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right] \otimes \bigwedge ^0 L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right] \perp \bigwedge ^n L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right] \otimes \bigwedge ^{l+1} L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon } \right]$. Thus by the definitions of $f_{k,x,t}$, $\alpha _p$ and Proposition 1 we get

$$\begin{aligned}{} & {} \langle 0|\textrm{T}\psi (x,t){{\bar{\psi }}}(0,0)|0\rangle \\{} & {} \quad = \frac{-i\varepsilon }{2\pi } \int _{-\pi /\varepsilon }^{\pi /\varepsilon } \begin{pmatrix} i\,\cos (\alpha _p/2)\sin (\alpha _p/2) &{} \cos ^2(\alpha _p/2) \\ -\sin ^2(\alpha _p/2) &{} i\,\cos (\alpha _p/2)\sin (\alpha _p/2) \end{pmatrix}e^{ipx-i\omega _pt}\,dp\\{} & {} \quad = \frac{-i\varepsilon }{4\pi } \int \limits _{-\pi /\varepsilon }^{\pi /\varepsilon } \begin{pmatrix} \frac{im\varepsilon }{\sqrt{m^2\varepsilon ^2+\sin ^2 p\varepsilon }} &{} 1+\frac{\sin p\varepsilon }{\sqrt{m^2\varepsilon ^2+\sin ^2 p\varepsilon }} \\ \frac{\sin p\varepsilon }{\sqrt{m^2\varepsilon ^2+\sin ^2 p\varepsilon }}-1 &{} \frac{im\varepsilon }{\sqrt{m^2\varepsilon ^2+\sin ^2 p\varepsilon }} \end{pmatrix}e^{ipx-i\omega _pt}\,dp\\{} & {} \quad = -\frac{i}{2} \begin{pmatrix} {\widetilde{A}}_1(-x,t,m,\varepsilon )&{} {\widetilde{A}}_2(x,t,m,\varepsilon ) \\ -{\widetilde{A}}_2(-x,t,m,\varepsilon ) &{} {\widetilde{A}}_1(x,t,m,\varepsilon ) \end{pmatrix}. \end{aligned}$$

Case 2: $t<0$. An analogous computation shows that

$$\begin{aligned} \langle 0|\textrm{T}\psi _k(x,t){{\bar{\psi }}}_l(0,0)|0\rangle= & {} i(-1)^{l}\psi _{3-l}(0,0)|0\rangle \cdot \psi _{k}(x,t)|0\rangle = i(-1)^{l}\langle f_{3-l,0,0}|f_{k,x,t}\rangle \\= & {} \left( -i(-1)^{l}\langle f_{k,x,t}|f_{3-l,0,0}\rangle \right) ^*. \end{aligned}$$

Thus we get the same integral formula as in Case 1, only the whole expression is conjugated. The change of the variables $p\mapsto \pi /\varepsilon -p$ and Proposition 1 complete the proof. $\square$

Formula (2) for the expected charge is consistent with the expression through field operators, which we briefly recall now (this paragraph is for specialists). Under notation from [14, §6.4], $:\!\psi ^\dagger (x,t)\psi (x,t)\!:$ is the charge density operator for the field [29, (3.113)]. Thus minus the expected charge density in the state $\psi _1^\dagger (0,0)|0\rangle$ (meaning one right electron at the origin) is

$$\begin{aligned}{} & {} \frac{\langle 0|\psi _1(0,0):\!\psi ^\dagger (x,t)\psi (x,t)\!: \psi _1^\dagger (0,0)|0\rangle }{\langle 0|\psi _1(0,0)\psi _1^\dagger (0,0)|0\rangle } \\{} & {} \quad =\frac{\left| \langle 0|\psi (x,t)\psi _1^\dagger (0,0)|0\rangle \right| ^2}{\langle 0|\psi _1(0,0)\psi _1^\dagger (0,0)|0\rangle } =\frac{1}{2}|{\widetilde{A}}_1\left( x,t,m,\varepsilon \right) |^2+ \frac{1}{2}|{\widetilde{A}}_2\left( x,t,m,\varepsilon \right) |^2, \end{aligned}$$

which coincides with (2). Here the first equality can be deduced, for instance, from a version of Wick’s theorem [14, (6.42)], and the second one — from Proposition 28.

1.4 Wightman axioms

The continuum limit of the new model is the well-known free spin-1/2 quantum field theory which of course satisfies Wightman axioms [14, §5.5]. One cannot expect the discrete model to satisfy all the axioms before passing to the limit because they are strongly tied to continuum spacetime and Lorentz transformations. Remarkably, some of them still hold on the lattice.

Proposition 29

(Checking of Wightman axioms) The objects introduced in Definition 11 satisfy the following conditions:

Axiom 1:

$\psi _k(x,t)$ is a bounded linear operator on ${\mathcal {H}}$ for each $(x,t)\in \varepsilon {\mathbb {Z}}^2$ and $k\in \{1,2\}$;

Axiom 2:

$|0\rangle$ is the unique up to proportionality vector in ${\mathcal {H}}$ such that $e^{iH\varepsilon }|0\rangle =e^{iP\varepsilon }|0\rangle =|0\rangle$;

Axiom 3:

the vectors $\psi _{k_1}(x_1,t_1)\dots \psi _{k_l}(x_l,t_l) \psi ^\dagger _{k_{l+1}}(x_{l+1},t_{l+1})\dots \psi ^\dagger _{k_{l+n}}(x_{l+n},t_{l+n})|0\rangle$ for all $0\le n,l\in {\mathbb {Z}}$, $k_j\in \{1,2\}$, $(x_j,t_j)\in \varepsilon {\mathbb {Z}}^2$ span a dense linear subspace in ${\mathcal {H}}$;

Axiom 4 (weakened):

$e^{biH-aiP}\psi _k(x,t)e^{aiP-biH}=\psi _k(x+a,t+b)$ for each $(a,b)\in \varepsilon {\mathbb {Z}}^2$;

Axiom 5 (weakened):

$H\ge 0$ and $H^2-(1-\omega _0\varepsilon /\pi )^2P^2\ge 0$ (here $\omega _0$ is $\omega _p$ for $p=0$);

Axiom 6:

$[\psi _{k}(x,t),\psi _{k'}(x',t')]_+ =[\psi _{k}(x,t),\psi ^\dagger _{k'}(x',t')]_+ =[\psi ^\dagger _{k}(x,t),\psi ^\dagger _{k'}(x',t')]_+=0$ for all $k,k'\in \{1,2\}$ and $(x,t),(x',t')\in \varepsilon {\mathbb {Z}}^2$ such that $|x-x'|>|t-t'|$, where $[a,b]_+:=ab+ba$.

Here Axiom 1 is weaker than the continuum one in the sense that the field operators are defined only on the lattice, but stronger in the sense that they are genuine bounded operators rather than distributions. The vectors in Axiom 3 mean states with electrons at the points $(x_{l+1},t_{l+1}),\dots ,(x_{l+n},t_{l+n})$ and positrons at the points $(x_1,t_1),\dots ,(x_l,t_l)$. Axiom 4 is much weaker than the continuum one, which involves general Lorentz transformations. On the lattice, only translations remain, because (almost all) the other Lorentz transformations do not preserve the lattice. Axiom 5 in continuum theory asserts that $H\ge 0$ and $H^2-P^2\ge 0$, i.e. the energy is positive in any frame of reference. The inequality $H^2-P^2\ge 0$ is violated on the lattice (but this does not mean negative energy because Lorentz transformations do not preserve the lattice). The weakened Axiom 5 shows that it still holds “in the continuum limit”. Axiom 6 is equivalent to vanishing of the real part of the Feynman propagator outside the light cone; this is obvious in the original Feynman model but nontrivial in the new one.

Proof

Axioms 1 and 2 hold by definition; recall that the operators are bounded by [14, (4.57)]. Axiom 3 holds because linear span of the functions $(\psi _1(x,0)+\psi _2(x,0))|0\rangle =i\sqrt{\varepsilon /\pi }\,e^{-i\alpha _p/2-ipx}$, where x runs through $\varepsilon {\mathbb {Z}}$, is dense in $L_2\left[ -\frac{\pi }{\varepsilon };\frac{\pi }{\varepsilon }\right]$. Axiom 4 is checked directly. Axiom 5 follows from the inequality $\omega _p\ge |p|(1-\omega _0\varepsilon /\pi )$.

To check Axiom 6, recall that all the anticommutators of the operators $a(f),a(f)^\dagger ,b(f),b(f)^\dagger$ vanish except $[a(f),a(g)^\dagger ]_+=[b(f^*),b(g^*)^\dagger ]_+=\langle f|g\rangle I$ [14, (4.56) and p.96]. This immediately implies that $[\psi _{k}(x,t),\psi _{k'}(x',t')]_+ =[\psi ^\dagger _{k}(x,t),\psi ^\dagger _{k'}(x',t')]_+=0$. Finally, by Proposition 28, Theorem 1, and Definition 1 for $|x-x'|>|t-t'|$ we have

$$\begin{aligned}{}[\psi _{k}(x,t),\psi ^\dagger _{k'}(x',t')]_+= & {} [a(f_{k,x,t}),a(f_{k',x',t'})^\dagger ]_+ + [b(f^*_{k,x,t})^\dagger ,b(f^*_{k',x',t'})]_+\\= & {} \langle f_{k,x,t}|f_{k',x',t'}\rangle I +\langle f_{k',x',t'}|f_{k,x,t}\rangle I\\= & {} 2I\,\textrm{Re}\langle f_{k,x-x',t-t'}|f_{k',0,0}\rangle =0. \end{aligned}$$

$\square$