Hamiltonian for the Periodic Kitaev Model

Question

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

Answer 1

First, watch out for the factors of 2 and $\sin(k)$s in your line 3 (after doing the fourier transform).

Second, you do not want to set those terms to zero.

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 $$-e^{ik}c_{k}^{\dagger}c_{-k}^{\dagger} \rightarrow -e^{-ik}c_{-k}^{\dagger}c_{k}^{\dagger}= +e^{-ik}c_{k}^{\dagger}c_{-k}^{\dagger}$$

... and this guy just gets absorbed into the first term.

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.

Answer 2

This is due to the diagonalization with Nambu state. If you need $(c_k,c^\dagger_{-k})$, you also need to handle $c_{-k} c^\dagger_{-k}$ term in the Hamiltonian. $c^\dagger_{j}c^\dagger_{j+1}$ and its h.c is also to be considered, which can be determined by anticommutation relationship. However, $$H=\sum_k \xi(k) c_k^\dagger c_k+\Delta\sum_k \left(e^{-ik}c_k^\dagger c_{-k}^\dagger +e^{ik}c_k c_{-k}\right)$$ have no such terms in Nambu state. So we need to use anti-commutation properties of c and $c^\dagger$ to get the correct form of diagonalization in Nambu state:

change the Hamiltonian from: $$H=-\sum_j\left( c^\dagger_{j+1} c_j+h.c.\right)+\Delta \sum_j \left( c^\dagger_{j+1}c^\dagger_j+h.c.\right) $$

to be

$$H=-(1/2)\sum_j\left( c^\dagger_{j+1} c_j-c_jc^\dagger_{j+1} +h.c.\right)+(1/2)\Delta \sum_j \left( c^\dagger_{j+1}c^\dagger_j-c^\dagger_jc^\dagger_{j+1}+h.c.\right). $$ Then you can work it out in k space and Nambu state for forming matrix form....

Here, do Fourier transforem on the second term: $$ (1/2)\Delta \sum_k \left( e^{-ika}c_k^\dagger c_{-k}^\dagger -e^{ika}c_k^\dagger c_{-k}^\dagger+e^{ik}c_k c_{-k}-e^{-ik}c_k c_{-k}\right), $$

leading to:

$$ i\Delta \sum_k \left( -sin(ka)c_k^\dagger c_{-k}^\dagger+sin(ka)c_k c_{-k}\right). $$

However, if we use one more anticommutation on $ e^{ik}c_k^\dagger c_{-k}^\dagger $ and $ e^{-ik}c_k c_{-k} $ term, we will end up with: $$ \Delta \sum_k \left( e^{-ka} c_k^\dagger c_{-k}^\dagger+e^{ka}c_k c_{-k}\right). $$