# Fourier transform of creation and annihilation operators in Kitaev Chain [closed]

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: $$$$H_{KM} = -\sum^{N-1}_{i} (t c^{\dagger}_{i} c_{i+1} + tc^{\dagger}_{i+1} c_{i} + \Delta c^{\dagger}_{i} c^{\dagger}_{i+1} + \Delta^{*}c_{i+1}c_{i}) - \mu \sum^{N}_{i}c^{\dagger}_{i} c_{i}$$$$ And the expected result should be like this (Hamiltonian for the Periodic Kitaev Model) : $$$$H_{k} = -\sum_{k}( 2t \cos(k) + \mu) c^{\dagger}_{k} c_{k} + \Delta e^{-ik}c^{\dagger}_{k} c^{\dagger}_{-k} + \Delta^{*} e^{ik} c_{k}c_{-k})$$$$ However, when I use the Fourier transform $$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}$$. My calculation steps are as follow: $$$$\begin{split} \sum_{i} c^{\dagger}_{i} c^{\dagger}_{i+1} &= \frac{1}{N} \sum_{kqi} c^{\dagger}_{k} c^{\dagger}_{q} e^{ix_{i}k} e^{iqx_{i+1}} \\ &=\frac{1}{N} \sum_{kqi} c^{\dagger}_{k} c^{\dagger}_{q} e^{ix_{i}k} e^{iqx_{I}} e^{iq} ~~~~ \text{(x_{i+1} = x_{i} + 1)} \\ &= \sum_{kq} c^{\dagger}_{k} c^{\dagger}_{q}e^{iq} \big(\frac{1}{N} \sum_{i} e^{ix_{i}(k+q)} \big) \\ &= \sum_{kq} c^{\dagger}_{k}c^{\dagger}_{q} e^{iq}\delta_{k,-q}\\ &= \sum_{k} c^{\dagger}_{k}c^{\dagger}_{-k} e^{-ik} \end{split}$$$$

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

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

• 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 $c^{\dagger}_{i}$ Mar 28, 2021 at 13:37
• Are you sure that $H_k$ is correct? It doesn't look hermitian. Mar 28, 2021 at 14:10
• 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? Mar 28, 2021 at 14:24
• 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 $-k$). Mar 28, 2021 at 14:26
• 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. Mar 28, 2021 at 14:50

For it to be hermitian, the Hamiltonian should read $$$$H_{k} = -\sum_{k}( 2t \cos(k) + \mu) c^{\dagger}_{k} c_{k} + \Delta e^{-ik}c^{\dagger}_{k} c^{\dagger}_{-k} + \Delta^{*} e^{ik} c_{-k}c_{k}) \ ,$$$$ or something the like.
This is what your derivations gives: Substitute $$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}\ .$$