1
$\begingroup$

I am trying to work out a simple 1-D lattice model of non interacting fermions. In the simple case we have the Hamiltonian, $$ H = -J \sum_{x} a_x^\dagger a_{x+1} + a^\dagger_{x+1} a_{x}. $$

This can be diagonalized using a Fourier transform and the spectrum is found to be a cosine function.But for a single fermion I would expect the spectrum to be a quadratic function. But that's alright, because the Hamiltonian says nothing about the total number of particles in the system. I would like to enforce the condition that the total number of particles is some fixed number (say $\mu$). I thought I could do it using a Lagrange multiplier resulting in the Hamiltonian, $$ H = -J \sum_{x} a_x^\dagger a_{x+1} + a^\dagger_{x+1} a_{x} + \lambda\left( \sum _ x a^\dagger_x a_x - \mu\right). $$

But I don't know how to proceed with this. Should the Lagrange multiplier be treated as a dynamical variable? Is this the correct way to solve for a fixed number of fermions on a lattice ? Or am I missing something here ?

Improve this question
$\endgroup$
3
  • $\begingroup$ A few comments. 1) The dispersion relation of each fermion is a cosine, not quadratic. This is because of the lattice. Quadratic dispersion corresponds to the continuum limit of zero lattice spacing. 2) The Hamiltonian is already correct for an arbitrary (possibly fixed) number of fermions, no modifications are needed. Fixing the fermion number corresponds to choosing a particular basis. 3) It would help if you specify exactly what you want to compute or solve for. Do you want the ground state for fixed fermion number? Or the full spectrum of energies and eigenstates? Or something else? $\endgroup$
    – Mark Mitchison
    Commented Oct 26, 2016 at 23:47
  • $\begingroup$ I would like to compute the full spectrum of a fixed number of fermions. I can solve and get the spectrum for the first Hamiltonian but I don't see how to get the spectrum for a fixed number of fermions. Can you explain how to do it? $\endgroup$
    – biryani
    Commented Oct 27, 2016 at 1:51
  • 1
    $\begingroup$ It's simpler than you might think. You already have the cosine-like spectrum of single-particle energies. If the lattice has $N$ sites then there are $N$ such single-particle states. To construct a $k$-particle energy eigenstate you simply fill up $k$ different single-particle states with one fermion each. This is represented by a Slater determinant, or simply a string of creation operators acting on the vacuum in second quantised notation. There are $N$ choose $k$ such states in total, and their energies are given by the sum of the energies of each occupied orbital. That's the full solution. $\endgroup$
    – Mark Mitchison
    Commented Oct 27, 2016 at 2:11

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.