# Hamiltonian for the Periodic Kitaev Model

The Hamiltonian for a system of spinless fermions on a 1D chain (with chemical potential $\mu=0$) is given by $$H=-\sum_j\left( c^\dagger_{j+1} c_j+h.c.\right)+\Delta \sum_j \left( c^\dagger_{j+1}c^\dagger_j+h.c.\right)$$ where $\Delta$ is some number. If we introduce $$c_j=\frac{1}{\sqrt{N}}\sum_k e^{ikj}c_k$$ We obtain the result below:

$$H=\sum_k \xi(k) c_k^\dagger c_k+\Delta\sum_k \left(e^{-ik}c_k^\dagger c_{-k}^\dagger +e^{ik}c_k c_{-k}\right)$$

where $\xi(k)=-2\cos(k)$I am trying to represent this Hamiltonian in matrix form by using the Nambu operator

$$\phi_k=\begin{pmatrix} c_k \\ c_{-k}^\dagger \end{pmatrix}$$

Numerous texts give it as $$H=\sum_k \phi_k^\dagger \begin{pmatrix} \xi(k) & 2i\Delta \sin(k)\\ -2i\Delta \sin(k ) & -\xi(k)\end{pmatrix}\phi(k)$$ However, when I expand the above out, I do not get my original coupling term back--instead, I get $$\Delta \sum_k \left( e^{-ik}c_k^\dagger c_{-k}^\dagger -e^{ik}c_k^\dagger c_{-k}^\dagger+e^{ik}c_k c_{-k}-e^{-ik}c_k c_{-k}\right)$$

I see that, to obtain my old coupling term, I have to let $e^{ik}c_k^\dagger c_{-k}^\dagger=e^{-ik}c_k c_{-k}=0$, but I can't explain why. Can someone please help me with this step? Here is a similar question posed in a problem set from a German university for your reference: http://users.physik.fu-berlin.de/~romito/qft2011/set6.pdf

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First, watch out for the factors of 2 and $\sin(k)$s in your line 3 (after doing the fourier transform).
Instead, remember that $k$ is just a dummy index. I could consider each term as a separate sum, and for some of them, I'll set $k \rightarrow -k$. Then $$-e^{ik}c_{k}^{\dagger}c_{-k}^{\dagger} \rightarrow -e^{-ik}c_{-k}^{\dagger}c_{k}^{\dagger}= +e^{-ik}c_{k}^{\dagger}c_{-k}^{\dagger}$$
Another way to think about this is that we should, strictly speaking, only consider the sum in the Nambu hamiltonian as only counting modes with $k \geq 0$, and then we need both kinds of terms since one ends up then also counting the original terms with $k \leq 0$. People tend to be very sloppy with this notation however.