# Fourier transform in a semi-infinite Ferromagnet

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

Consider a 1D Hamiltonian of the form $$H = -JS\sum_{j = 1}^{N-1}a_{j+1}^\dagger a_j + a_j^\dagger a_{j+1} - a^\dagger_{j+1}a_{j+1} - a^\dagger_j a_j$$ 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 $$a_j = \frac{1}{\sqrt{N}} \sum_k e^{ik jd}a_{k}.$$ 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 $$a_j = \int dk e^{ikjd}a_{k}$$?

• 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. – MOOSE Nov 3 '17 at 18:30
• My previous comment was too long, but I was thinking of letting $$\alpha_{ak} = r_k e^{i\pi ka /N} + l_k e^{-i\pi ka/N}.$$ As you see one runs into trouble if we let $N$ approach infinity. – MOOSE Nov 3 '17 at 18:36