0
$\begingroup$

I encounter a problem when I use Fourier transformation to transform the real space Kitaev Chain to momentum space. Suppose the real space Kitaev Chain can be written as follow: \begin{equation} H_{KM} = -\sum^{N-1}_{i} (t c^{\dagger}_{i} c_{i+1} + tc^{\dagger}_{i+1} c_{i} + \Delta c^{\dagger}_{i} c^{\dagger}_{i+1} + \Delta^{*}c_{i+1}c_{i}) - \mu \sum^{N}_{i}c^{\dagger}_{i} c_{i} \end{equation} And the expected result should be like this (Hamiltonian for the Periodic Kitaev Model) : \begin{equation} H_{k} = -\sum_{k}( 2t \cos(k) + \mu) c^{\dagger}_{k} c_{k} + \Delta e^{-ik}c^{\dagger}_{k} c^{\dagger}_{-k} + \Delta^{*} e^{ik} c_{k}c_{-k}) \end{equation} However, when I use the Fourier transform $ c_{j} = \frac{1}{\sqrt{N}} \sum_{k} e^{-ikx_{j}} c_{k}$ to manipulate the midterm two terms $c^{\dagger}_{i} c^{\dagger}_{i+1}$ and $c_{i}c_{i+1}$, I got a trouble there since I cannot get the correct phase $e^{\pm ik}$. My calculation steps are as follow: \begin{equation} \begin{split} \sum_{i} c^{\dagger}_{i} c^{\dagger}_{i+1} &= \frac{1}{N} \sum_{kqi} c^{\dagger}_{k} c^{\dagger}_{q} e^{ix_{i}k} e^{iqx_{i+1}} \\ &=\frac{1}{N} \sum_{kqi} c^{\dagger}_{k} c^{\dagger}_{q} e^{ix_{i}k} e^{iqx_{I}} e^{iq} ~~~~ \text{($x_{i+1} = x_{i} + 1$)} \\ &= \sum_{kq} c^{\dagger}_{k} c^{\dagger}_{q}e^{iq} \big(\frac{1}{N} \sum_{i} e^{ix_{i}(k+q)} \big) \\ &= \sum_{kq} c^{\dagger}_{k}c^{\dagger}_{q} e^{iq}\delta_{k,-q}\\ &= \sum_{k} c^{\dagger}_{k}c^{\dagger}_{-k} e^{-ik} \end{split} \end{equation}

Similarly for $c_{i+1}c_{i}$ term \begin{equation} \begin{split} \sum_{i} c_{i+1} c_{i} &= \frac{1}{N} \sum_{kqi} c_{k} c_{q} e^{-ix_{i+1}k} e^{-iqx_{i}} \\ &= \frac{1}{N} \sum_{kqi} c_{k} c_{q} e^{-ik} e^{-ikx_{i}} e^{-iqx_{i}} ~~~~ \text{($e^{-ikx_{i+1}} =e^{-ik(x_{i} +1)} $)} \\ &=\sum_{kq} c_{k} c_{q} e^{-ik} \big( \frac{1}{N} \sum_{i} e^{-i(k+q)x_{i}}\big) \\ &= \sum_{kq} c_{k}c_{-k} e^{-ik} \delta_{k,-q} \\ &= \sum_{k} c_{k}c_{-k} e^{-ik} \end{split} \end{equation}

Therefore, could anyone help me to point out the mistakes that I made in my calculation? Thank you.

$\endgroup$
6
  • $\begingroup$ For the Fourier Transform of $c^{\dagger}$, I directly take the conjugation of the Fourier transform of $c$. Therefore, I use $c^{\dagger}_{i} = \frac{1}{N} \sum_{k} e^{ikx_{i}} c^{\dagger}_{k}$ as the Fourier transform of $c^{\dagger}_{i}$ $\endgroup$
    – Ricky Pang
    Mar 28, 2021 at 13:37
  • $\begingroup$ Are you sure that $H_k$ is correct? It doesn't look hermitian. $\endgroup$ Mar 28, 2021 at 14:10
  • $\begingroup$ I am not sure whether the $H_{k}$ is correct. I saw some examples in BCS theory, the $H_{k}$ should contain $c_{-k}c_{k}$ instead of $c_{k}c_{-k}$. This is also a problem that confused me a lot. Besides, may I ask how to check Hermitian of such Hamiltonian? $\endgroup$
    – Ricky Pang
    Mar 28, 2021 at 14:24
  • $\begingroup$ For it to be hermitian - unless I am missing sth. - the last term should have $c_{-k}c_k$. And then your result is correct (substitute $k$ with $-k$). $\endgroup$ Mar 28, 2021 at 14:26
  • $\begingroup$ This is a comment but: you don't need to work out the second $\propto \Delta$ term like this. The two pairing terms are hermitian conjugates in real space and will be in momentum space - just do the FT of one of them and then write the h.c. of it for the second. $\endgroup$
    – jacob1729
    Mar 28, 2021 at 14:50

1 Answer 1

0
$\begingroup$

For it to be hermitian, the Hamiltonian should read \begin{equation} H_{k} = -\sum_{k}( 2t \cos(k) + \mu) c^{\dagger}_{k} c_{k} + \Delta e^{-ik}c^{\dagger}_{k} c^{\dagger}_{-k} + \Delta^{*} e^{ik} c_{-k}c_{k}) \ , \end{equation} or something the like.

This is what your derivations gives: Substitute $k\to-k$ and use $\sum_k=\sum_{-k}$, and you get $$ \sum k c_kc_{-k} = \sum_q c_{-k} c_{k} e^{ik}\ . $$

$\endgroup$
1
  • $\begingroup$ Thank you for your reply. I now understand how to how to check the hermitian. $\endgroup$
    – Ricky Pang
    Mar 28, 2021 at 14:31

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