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?

  • $\begingroup$ 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. $\endgroup$ – Chiral Anomaly Jul 23 '19 at 1:45
  • $\begingroup$ 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? $\endgroup$ – 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.

Update. My words about the generator of translations of the lattice are not mathematically accurate. Infinitesimal translations of a lattice are not possible. It is better to speak about the log of the unitary discrete-translation operator, as @Chiral Anomaly pointed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.