Doubling symmetry

The naive fermion action possesses a new taste-exchange symmetry not found in the continuum theory acting on the fermion fields as^[2]

${\displaystyle \psi _{n}\rightarrow e^{-in\cdot \pi _{A}}S_{A}\psi _{n},\ \ \ \ \ \ {\bar {\psi }}_{n}\rightarrow {\bar {\psi }}_{n}S_{A}^{\dagger }e^{in\cdot \pi _{A}},}$

where the vectors ${\displaystyle \pi _{A}}$ are the sixteen vectors with non-zero entries of ${\displaystyle \pi }$ specified by ${\displaystyle A}$. For example, ${\displaystyle \pi _{0}=(0,0,0,0)}$, ${\displaystyle \pi _{2}=(0,\pi ,0,0)}$, ${\displaystyle \pi _{14}=(\pi ,0,0,\pi )}$, and ${\displaystyle \pi _{1234}=(\pi ,\pi ,\pi ,\pi )}$. The Dirac structure in the symmetry is similarly defined by the indices of ${\displaystyle A}$ as ${\displaystyle S_{A}=S_{\nu _{1}}S_{\nu _{2}}S_{\nu _{3}}S_{\nu _{4}}}$ where ${\displaystyle S_{0}=I}$ and ${\displaystyle S_{\nu }=i\gamma _{5}\gamma _{\nu }}$; for example with ${\displaystyle S_{14}=(i\gamma _{5}\gamma _{1})(i\gamma _{5}\gamma _{4})}$.

The presence of these sixteen symmetry transformations implies the existence of sixteen identical fermion states rather than just one. Starting with a fermion field ${\displaystyle \psi _{n}}$, the symmetry maps it to another field ${\displaystyle \psi '_{n}=e^{-in\cdot \pi _{A}}S_{A}\psi _{n}}$. Fourier transforming this shows that its momentum has been shifted as ${\displaystyle p^{\mu }\rightarrow p^{\mu }+\pi _{A}^{\mu }}$. Therefore, a fermion with momentum near the center of the Brillouin zone is mapped to one of its corners while one of the corner fermions comes in to replace the center fermion, showing that the transformation acts to exchange the tastes of the fermions. Since this is a symmetry of the action, the different tastes must be physically indistinguishable from each other. Here the Brillouin momentum ${\displaystyle k^{\mu }=p^{\mu }+\pi _{A}^{\mu }}$ for small ${\displaystyle p^{\mu }}$ is not the physical momentum of the particle, rather that is ${\displaystyle p^{\mu }}$. Instead ${\displaystyle \pi _{A}^{\mu }}$ acts more as an additional quantum number specifying the taste of a fermion.

The ${\displaystyle S_{A}}$ term is responsible for changing the representation of the ${\displaystyle \gamma }$-matrices of the doublers to ${\displaystyle \gamma _{\mu }^{(A)}=S_{A}^{\dagger }\gamma _{\mu }S_{A}}$, which has the effect of changing the signs of the matrices as ${\displaystyle (\gamma _{1},\gamma _{2},\gamma _{3},\gamma _{4})\rightarrow (\pm \gamma _{1},\pm \gamma _{2},\pm \gamma _{3},\pm \gamma _{4})}$. Since any such sign change results in a set of matrices still satisfying the Dirac algebra, the resulting matrices form a valid representation. It is also the term that enters the wave function of the doublers given by ${\displaystyle S_{A}^{\dagger }u({\boldsymbol {p}})}$ and ${\displaystyle S_{A}^{\dagger }v({\boldsymbol {p}})}$, where ${\displaystyle u({\boldsymbol {p}})}$ and ${\displaystyle v({\boldsymbol {p}})}$ are the usual Dirac equation solutions with momentum ${\displaystyle {\boldsymbol {p}}}$.^[3]

Propagator and dispersion relation

In the continuum theory, the Dirac propagator has a single pole as the theory describes only a single particle. However, calculating the propagator from the naive action yields

${\displaystyle S(p)={\frac {m-ia^{-1}\sum _{\mu }\gamma _{\mu }\sin(p^{\mu }a)}{m^{2}+a^{-2}\sum _{\mu }\sin(p^{\mu }a)^{2}}},}$

for a fermion with momentum ${\displaystyle p^{\mu }}$.^[4] For low momenta ${\displaystyle \sin(p^{\mu }a)\approx p^{\mu }a}$ this still has the expected pole at ${\displaystyle ap^{\mu }=(am,0,0,0)}$, but there are fifteen additional poles when ${\displaystyle ap^{\mu }=(am,0,0,0)+\pi _{A}^{\mu }}$. Each of these is a new fermion species with doubling arising because the ${\displaystyle \sin(p^{\mu }a)}$ function has two poles over the range ${\displaystyle p^{\mu }\in [-\pi /a,\pi /a]}$. This is in contrast to what happens when particles of different spins are discretized. For example, scalars acquire propagators taking a similar form except with ${\displaystyle \sin(p^{\mu }a/2)}$, which only has a single pole over the momentum range and so the theory does not suffer from a doubling problem.^[5]

Comparison between the continuum and lattice dispersion relations of a free fermion, with the latter having a doubler at the Brillouin zone boundary.

The necessity of fermion doubling can be deduced from the fact that the massless fermion propagator is odd around the origin.^[6] That is, in the continuum limit it is proportional to ${\displaystyle \gamma _{\mu }p^{\mu }}$, which must still be the case on the lattice in the small momentum limit. But since any local lattice theory that can be constructed must have a propagator that is continuous and periodic, it must cross the zero axis at least once more, which is exactly what occurs on the Brillouin zone corners where ${\displaystyle ap_{\mu }=\pi _{A}^{\mu }}$ for the naive fermion propagator. This is in contrast to the bosonic propagator which is quadratic around the origin and so does not have such problem. Doubling can be avoided if a discontinuous propagator is used, but this results in a non-local theory.

The presence of doublers is also reflected in the fermion dispersion relation. Since this is a relation between the energy ${\displaystyle \omega }$ of the fermion and its momentum, it requires performing an inverse Wick transformation ${\displaystyle p_{4}=\pm i\omega }$, with the dispersion relation arising from the pole of the propagator^[7]

${\displaystyle \sinh \omega ({\boldsymbol {p}})={\sqrt {m^{2}+\sum _{j=1}^{3}\sin ^{2}p_{j}}}.}$

The zeros of this dispersion relation are local energy minima around which excitations correspond to different particle species. The above has eight different species arising due to doubling in the three spatial directions. The remaining eight doublers occur due to another doubling in the Euclidean temporal direction, which seems to have been lost. But this is due to a naive application of the inverse Wick transformation. The theory has an obstruction that does not allow for the simple replacement of ${\displaystyle p_{4}=\pm i\omega }$ and instead requires performing the full contour integration. Doing this for the position space propagator results in two separate terms, each of which has the same dispersion relation of eight fermion species, giving a total of sixteen.^[8] The obstruction between the Minkowski and Euclidean naive fermion lattice theories occurs because doubling does not occur in the Minkowski temporal direction, so the two theories differ in their particle content.

Derivative discretization

Another possible although impractical solution to the doubling problem is to adopt a derivative discretization different from the symmetric difference

${\displaystyle \partial _{\mu }f(x)\rightarrow \lim _{a\rightarrow 0}{\frac {f(x+a{\hat {\mu }})-f(x-a{\hat {\mu }})}{2a}},}$

used in the naive fermion action. Instead it is possible to use the forward difference

${\displaystyle \partial _{\mu }f(x)\rightarrow \lim _{a\rightarrow 0}{\frac {f(x+a{\hat {\mu }})-f(x)}{a}},}$

or a backward difference discretizations. The effect of the derivative discretizations on doubling is seen by considering the one-dimensional toy problem of finding the eigensolutions of ${\displaystyle -i\partial _{x}f(x)=\lambda f(x)}$.^[27] In the continuum this differential equation has a single solution. However, implementing the symmetric difference derivative leads to the presence of two distinct eigensolutions, while a forward or backward difference derivative has one eigensolution. This effect carries forward to the fermion action where fermion doubling is absent with forward or backward discretizations.

The reason for this particle content disparity is that the symmetric difference derivative maintains the hermiticity property of the continuum ${\displaystyle i\partial _{\mu }}$ operator, while the forward and backward discretizations do not. These latter discretizations lead to non-hermitian actions, breaking the assumptions of the Nielsen–Ninomiya theorem, and so avoid the fermion doubling problem. Developing an interacting theory with a non-hermitian derivative discretization leads to a theory with non-covariant contributions to the fermion self-energy and vertex function, rendering the theory non-renormalizable and difficult to work with.^[28] For this reason such a resolution to the fermion doubling problem is generally not implemented.