# Spectrum of the momentum operator for free Dirac fermions on a lattice

I am studying lattice field theory and would like to understand the momentum operator for free Dirac fermions on a square lattice.

In this case one needs to discretize the momentum operator (which otherwise would have a continuous spectrum):

$$P_\mu^\text{(free)} = - i\hbar \int \text{d}x^3 \psi(x)^\dagger\partial_\mu \psi(x) \to -i\hbar\sum_x \psi_x^\dagger \left( \dfrac{\psi_{x+\hat{\mu}} - \psi_{x-\hat{\mu}}}{2 a} \right)$$

Where $$\psi$$ is a Dirac spinor, $$x$$ denotes the lattice site, $$a$$ the lattice constant, $$\hat{\mu}$$ the unit vector in direction $$\mu$$ and I used central differences to discretize the derivative.

My question now is: What is the spectrum of this $$P_\mu$$ on the lattice (assuming periodic boundary conditions)?

My take on it: I would expect that since the phase of the Dirac Fermion is ~$$e^{ip \cdot x}$$ ($$\hbar = 1$$), then the periodic boundary conditions on the lattice require for each dimension: $$e^{ip_\mu L} = 1$$. Here $$L = N_\mu a$$ denotes the size of my lattice.

From this I would deduce $$p_\mu \in \left( - \frac{\pi}{a}, -\frac{\pi}{Na} (n-1), ... , \frac{\pi}{a} \right)$$ with integers $$n$$ as is typical in condensed matter physics. However, when I try to calculate the spectrum of the discretized $$P_\mu$$ numerically for small lattices, I get different results (sometimes the spectrum is not even linear). Where did I go wrong? Is my expected spectrum correct?

• As stated in @Gec's answer, the momentum operator is the generator of translations. On a lattice, there are no infinitesimal translations; only finite translations. So there is no "generator" in the conventional Lie-algebra sense. As a substitute, you can consider the log of the spectrum of the unitary discrete-translation operator, as Gec did. – Chiral Anomaly Jul 23 '19 at 1:45
• Thank you for your comment! So then what would my momentum $P^\text{free}$ above correspond to? Is it possible to make any statement about it's spectrum? – B. Croydon Jul 24 '19 at 9:04

The real question here, in my opinion, is: how do you define the momentum operator? If the momentum operator is the generator of translations of the lattice, then it's spectrum is $$2\pi n/L (\mbox{mod} 2\pi)$$, where $$n = 0,1,2,\dots,N-1$$, $$L=Na$$. Your observation shows that the discretized version of the $$P^{\mbox{(free)}}$$ operator does not coincide with the generator of translations. These operators commute but have different spectra.