# Fermion doubling

The fermion doubling problem is a problem that is encountered when naively trying to put fermionic fields on a lattice. It consists in the appearance of spurious states, such that one ends up having 2d fermionic particles (with d the number of discretized dimensions) for each original fermion. In order to solve this problem, several strategies are in use, such as Wilson fermions and staggered fermions.

## Mathematical overview

The action of a free Dirac fermion in d dimensions,[note 1] of mass m, and in the continuum (i.e. without discretization) is commonly given as

$S = \int d^dx \; \bar\psi (\partial\!\!\!/ + m) \psi \;.$

Here, the Feynman slash notation was used to write

$\gamma_\mu a^{\mu} = a\!\!\!/$

where γμ are the gamma matrices. When this action is discretized on a cubic lattice, the fermion field ψ(x) is replaced with a discretized version ψx, where x now denotes the lattice site. The derivative is replaced by the finite difference. The resulting action is now:[1]

$S = a^d \sum_{x,\mu} \frac1{2a} (\bar\psi_x\gamma_\mu\psi_{x+\hat\mu} - \bar\psi_{x+\hat\mu}\gamma_\mu\psi_x) + a^d \sum_x m\bar\psi_x\psi_x \;,$

where a is the lattice spacing and $\hat\mu$ is the vector of length a in the μ direction. When one computes the inverse fermion propagator in momentum space, one readily finds:[1]

$S^{-1}(p) = m + \frac ia \sum_\mu \gamma_\mu \sin(p_\mu a) \;.$

Due to the finite lattice spacing the momenta pμ have to be inside the Brillouin zone, which is typically taken to be the interval [−π/a,+π/a].

When simply taking the limit a → 0 in the above inverse propagator, one recovers the correct continuum result. However, when instead expanding this expression around a value of pμ where one or more of the components are at the corners of the Brillouin zone (i.e. equal to π/a), one finds the same continuum form again, although the sign in front of the gamma matrix can change.[2][note 2] This means that, when one of the components of the momentum is near π/a, the discretized fermion field will again behave like a continuum fermion. This can happen with all d components of the momentum, leading to —with the original fermion with momentum near the origin included— 2d different "tastes" (in analogy to flavor).[note 3]

## The Nielsen–Ninomiya theorem

Nielsen and Ninomiya proved a theorem[3] stating that a local, real, free fermion lattice action, having chiral and translational invariance, necessarily has fermion doubling. The only way to get rid of the doublers is by violating one of the presuppositions of the theorem —for example: