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I got a small, rather technical question concerning the Heisenberg model. (It is technical indeed.)

Consider the Heisenberg Hamiltonian: $H = \sum_{(i,j)} S_{i} S_{j}$ = $- \frac{J}{V} \sum_{q} \gamma_{q} S_{q} S_{-q}$.

with $\gamma_{q} = 2 \sum_{\alpha=x,y,...}cos(q_{\alpha})$

and the Fourier transform $S_{i}=1/V \cdot \sum_{q} S_{q} e^{iqr_{i}}$

Where we assume a lattice constant of a=1 and impose periodic boundary conditions on a hypercube of edge length L and respective volume V.

Here comes my question: I don't see how one arrives at the equation for the fourier transformed H in detail. I mean...i see where the cosines are coming from if you only have -q and q left and suspect that one has to apply an identity like $\delta_{ij} = \sum_{q} e^{iq(r_{i}-r_{j}}$ but somehow it doesn't work out for me and even though it has nothing to do with the physics it leaves a bad feeling behind.

I'd be really thankful if someone could help me.

Best regards and thanks in advance.

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1 Answer 1

The thing to keep in mind is that that formula only holds because you are coupling nearest-neighbor spins. Let's pretend we're in 1D (the generalization for higher dimensions is trivial). When you replace the operators by their fourier transformed representations that you provided, you get $$ J\sum_i S_i \cdot S_{i+1} = \frac{J}{V^2}\sum_i \sum_{k,q}S_q e^{iqr_i} \cdot S_k e^{ikr_{i+1}} $$

Since I'm only coupling nearest neighbors, the spatial coordinates in the exponential are always just one site away from one another, so I can write $$ S_q e^{iqr_i} \cdot S_k e^{ikr_{i+1}} = e^{ik}(S_q \cdot S_k)e^{i(q+k)r_i} $$

Now, you almost noted the identity you need in your comment, but what you really need is $$ \sum_{i}e^{i(q-k)r_i}=V\delta_{qk} $$

I hope this helps

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