I have a (simple?) question about Fourier transforms.

Consider a 1D Hamiltonian of the form \begin{equation} H = -JS\sum_{j = 1}^{N-1}a_{j+1}^\dagger a_j + a_j^\dagger a_{j+1} - a^\dagger_{j+1}a_{j+1} - a^\dagger_j a_j \end{equation} where $J$ is a coupling between two nearest neighbours, and $S$ is the spin projection along some z-axis, i.e a standard ferromagnetic chain with $N$ lattice sites and lattice spacing $d$.

To diagonalize this one typically introduces the fourier transformed creation/annhilation operators \begin{equation} a_j = \frac{1}{\sqrt{N}} \sum_k e^{ik jd}a_{k}. \end{equation} This is fine as long as we assume periodic boundary conditions such that $a_{j+N}=a_{j}$.

Now consider the case when we let $N\rightarrow \infty$. In this case, it no longer makes sense to use periodic boundary conditions. How then do we define a Fourier transform in order to diagonalize such a problem?

Is it as simple as just writing \begin{equation} a_j = \int dk e^{ikjd}a_{k} \end{equation}?


1 Answer 1


"Now consider the case when we let N→∞. In this case, it no longer makes sense to use periodic boundary conditions."

Why doesn't it make sense? I mean, from a physical point of view, taking this you just assume that the number of lattice sites is pretty large, that is you take the continuum limit(mathematical infinity is not acceptable and let's say is impractical for physics). So why can't you impose the same boundary conditions to solve your problem for large N?

And the Fourier transformation becomes exactly what you have written.

  • $\begingroup$ Right, ok. Let me make matters more interesting. Consider that at $j = 1$ there is some boundary. For instance a normal-metal obeying a tight-binding Hamiltonian. We're not interested in the lattice points $j < 1$, I only introduced it such that we can not "cheat" and use periodic boundary conditions. Do you know if we can still diagonalize the Hamiltonian for $j \geq 1$ by some ansatz similar to a Fourier transform? I was thinking of using the ansatz $ A^\dagger_k = \sum_{a=1}^{N}\alpha_{ak}a \dagger_a$ in the finite case, but I'm unsure about how to deal with the infinite case. $\endgroup$
    – MOOSE
    Commented Nov 3, 2017 at 18:30
  • $\begingroup$ My previous comment was too long, but I was thinking of letting \begin{equation}\alpha_{ak} = r_k e^{i\pi ka /N} + l_k e^{-i\pi ka/N}.\end{equation} As you see one runs into trouble if we let $N$ approach infinity. $\endgroup$
    – MOOSE
    Commented Nov 3, 2017 at 18:36
  • $\begingroup$ You cannot use Fourier transforms at this case. You need to have translation invariance to use it, and that's why we impose periodic bountary conditions. By considering a boundary, you need to take into account surface effects and I think the whole analysis change drastically. Hope I helped. $\endgroup$
    – S.M.
    Commented Nov 3, 2017 at 23:14

Your Answer

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

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