Spin on a 4D Feynman Checkerboard

Abstract

We discretize the Weyl equation for a massless, spin-1/2 particle on a time-diagonal, hypercubic spacetime lattice with null faces. The amplitude for a step of right-handed chirality is proportional to the spin projection operator in the step direction, while for left-handed it is the orthogonal projector. Iteration yields a path integral for the retarded propagator, with matrix path amplitude proportional to the product of projection operators. This assigns the amplitude i ^±T 3^−B/2 2^−N to a path with N steps, B bends, and T right-handed minus left-handed bends, where the sign corresponds to the chirality. Fermion doubling does not occur in this discrete scheme. A Dirac mass m introduces the amplitude i 𝜖 m to flip chirality in any given time step 𝜖, and a Majorana mass similarly introduces a charge conjugation amplitude.

Appendix A: Norm of the Amplification Matrix

In this appendix we prove that the norm of the matrix A(𝜃) defined in (28) is less than unity unless at least three of the 𝜃 _i coincide, in which case the norm is unity. The proof is due to Eli Hawkins.

Let |ν〉 be any unit spinor. The squared norm of A|ν〉 is ∥A|ν〉∥²=〈ν|A ^‡ A|ν〉, whose maximum is the larger eigenvalue of A ^‡ A. This value defines the squared norm ∥A∥².

Using the definition of the spin projection operators (8) and the inner products of the unit vectors $\hat {n}_{i}$ (1 if i = j and $-\frac 13$ if i≠j) we find

$$ \text{tr} (A^{\dagger} A) = 1 + \tfrac16 \sum\limits_{(ij)} \cos(\theta_i-\theta_j) $$

(44)

where the sum is over the 6 choices of {i,j}⊂{1,2,3,4}. This trace is at most 2, and therefore the smaller eigenvalue of A ^‡ A is less than 1 unless A ^‡ A=1. Hence

$$ {\Phi} := \det \left( A^{\dag} A -1\right) $$

(45)

has the same sign as 1−∥A∥.

The matrix A ^‡ A is linear in terms of $e^{i(\theta _{i}-\theta _{j})}$, therefore Φ is quadratic. Because Φ is invariant under all permutations of the 𝜃’s, it can be written as a quadratic function of the cosines cos(𝜃 _i −𝜃 _j ). Because Φ vanishes when the 𝜃’s are all equal, it is convenient to write it in terms of the cosines minus 1. It thus takes the form,

$$\begin{array}{@{}rcl@{}} {\Phi} &=& a \sum\limits_{(ij)} \left( 1 - \cos[\theta_i-\theta_j]\right) + b \sum\limits_{(ij)} \left( 1 - \cos[\theta_i-\theta_j]\right)^2 \\ &&+ c\!\!\sum\limits_{(ij)(kl)} \left( 1 - \cos[\theta_i-\theta_j]\right) \left( 1 - \cos[\theta_k-\theta_l]\right) . \end{array} $$

(46)

The last sum is over the 3 partitions of {1,2,3,4} into pairs. When 𝜃 ₂ = 𝜃 ₃ = 𝜃 ₄, A takes the form (16), which obviously has an eigenvalue of unit modulus. Therefore Φ=0 in this case, which shows that a = b=0. To determine the value of c, consider the case that 𝜃 ₁ = 𝜃 ₂=0 and 𝜃 ₃ = 𝜃 ₄ = π. Then $A=\frac 12(\hat {n}_{1}+\hat {n}_{2})\cdot \vec {\sigma }$, so A ^‡ A=1/3, hence Φ=4/9. The last sum in (46) is 8, so c=1/18.

The determinant Φ is thus given by

$$ {\Phi} = \tfrac1{18}\sum\limits_{(ij)(kl)} \left( 1 - \cos[\theta_i-\theta_j]\right) \left( 1 - \cos[\theta_k-\theta_l]\right). $$

(47)

This satisfies Φ≥0, therefore ∥A∥≤1. Each term is non-negative, therefore Φ=0 only if every term vanishes. This occurs only if at least three of the 𝜃’s are equal.