Basis transformation of eigenvectors of Hamiltonian written in different basis

There is a very famous topological model, Su-Schrieffer Heeger (SSH) model, according to which the hopping strength between even and odd sites is different. i.e. Now, there are two ways one can write Hamiltonain for this system:

First way:

$c_n^\dagger$ are creating operator at site $n$. Hamiltonian ($\tilde{H_1}$) can be written as: $$\tilde{H_1}=\sum_{n=1}^L[\frac{t+\delta t (-1)^n}{2}]c_n^\dagger c_{n+1}+H.C.$$

After converting this into k-space one can find following:

$$\psi^\dagger H_1\psi=\begin{bmatrix}c_k^\dagger&c_{k+\pi}^\dagger \end{bmatrix} \begin{bmatrix}A&iB\\-iB&-A \end{bmatrix} \begin{bmatrix}c_k\\c_{k+\pi} \end{bmatrix}$$ with $A=-t\cos(k)$ and $B=-\delta t \sin(k)$

Second way:

Second way to write Hamiltonian for the same system is to consider solid sites (in above picture) as $a_n^\dagger$ and open sites as $b_n^\dagger$. Now Hamiltonian ($\tilde{H_2}$) is: $$\tilde{H_2} = \sum_{n=1}^L [(t+\delta t)a_n^\dagger b_n+(t-\delta t)b_n^\dagger a_{n+1}]+H.C.$$ in k-space it looks like following: $$\phi^\dagger H_2\phi=\begin{bmatrix}a_k^\dagger&b_{k}^\dagger \end{bmatrix} \begin{bmatrix}0&A-iB\\A+iB&0 \end{bmatrix} \begin{bmatrix}a_k\\b_{k} \end{bmatrix} \, .$$ with $A=-t\cos(k)$ and $B=-\delta t \sin(k)$.

I have not written all the steps to convert Hamiltonian into k-space becuase it will be very messing to write all equations here but I am 100% sure that I have done everything correct till here.

My main goal:

My main goal is to first write wavefunction for this system in $\psi^\dagger=\begin{bmatrix}c_k^\dagger &c_{k+\pi}^\dagger \end{bmatrix}$ basis i.e. (One can find eigen-vector easily but let's say it looks like following:) $$|WF_1\rangle = (X_1c_k^\dagger + X_2c_{k+\pi}^\dagger)|0\rangle$$

with $X_i$'s as some known terms. After that I want to convert this $WF_1$ into $\phi_k^\dagger=\begin{bmatrix}a_k^\dagger &b_{k}^\dagger \end{bmatrix}$ basis. i.e. I want to write it as: $$|WF_2\rangle = (Y_1a_k^\dagger + Y_2b_{k}^\dagger)|0\rangle$$ What will be $Y_i$'s?

In previous version of this question I didn't give this much details becuase I thought finding a unitary matrix to convert one basis into other is enough but after some kind comments I find that I needed to write all details.

I know I can find $WF_2$ directly from $H_2$ but for some reason I don't want to do that.

Sorry for confusing notations.

The question you are asking is actually not quite related to the physical system (Hamiltonian). Instead, you are only asking how to change a state from basis $\{ c \}$ to $\{ a, b \}$.

First, the 1st way you wrote down is actually slightly different from the 2nd way: if you imagine you have L unit cells (2L sites), then your "$n$" in the first way should range from 1 to $2L$. Thus, the range of your "$k$" is also different from the k in the second way.

Thus, your true question is: "for $a_{n}=c_{2n-1}, b_{n}=c_{2n}$, how is the Fourier mode related to each others?"

I will just do one of them here,

\begin{align} c_{\tilde{n}}:=& \frac{1}{\sqrt{2L}}\sum_{p=1}^{2L} c_{\frac{\pi}{L}p} e^{i(\frac{\pi}{L}p)\tilde{n}} \\ a_{n}:=& \frac{1}{\sqrt{L}}\sum_{q=1}^{L} a_{\frac{2 \pi}{L}q} e^{i(\frac{2\pi}{L}q)n} \\ =&c_{2n-1}\\ =&\frac{1}{\sqrt{2L}}\sum_{p=1}^{2L} c_{\frac{\pi}{L}p} e^{i(\frac{\pi}{L}p)(2n-1)} \\ =&\frac{1}{\sqrt{2L}}\sum_{q=1}^{L} c_{\frac{\pi}{L}q} e^{i(\frac{\pi}{L}q)(2n-1)}+\frac{1}{\sqrt{2L}}\sum_{q=1}^{L} c_{\frac{\pi}{L}(q+L)} e^{i(\frac{\pi}{L}(q+L))(2n-1)} \\ =&\frac{1}{\sqrt{2L}}\sum_{q=1}^{L} \left[ c_{\frac{\pi}{L}q} e^{-i(\frac{\pi}{L}q)}- c_{\frac{\pi}{L}q + \pi} e^{-i(\frac{\pi}{L}q)} \right] e^{i(\frac{2\pi}{L}q)n} \\ \implies & a_{\frac{2 \pi}{L}q} =\frac{1}{\sqrt{2}}\left[ c_{\frac{\pi}{L}q} - c_{\frac{\pi}{L}q + \pi} \right]e^{-i(\frac{\pi}{L}q)} \\ \text{or } & a_{2 \tilde{k}} =\frac{1}{\sqrt{2}}\left[ c_{\tilde{k}} - c_{\tilde{k} + \pi} \right]e^{-i \tilde{k}} \end{align}

This means $a_{2 \tilde{k}}$ is the superposition between the $\tilde{k}$ and $\tilde{k}+\pi$ in the original basis, with some phase shift. I believe you can figure the same relation for $b$, do the inverse, and plug it into your state to find the relation between $X$ and $Y$.

• Thank you so much. Please have a look if I am doing everything correctly: $b_{n}=c_{2n} \implies b_{2\tilde{k}}=\frac{1}{\sqrt{2}}\left[ c_{\tilde{k}} + c_{\tilde{k} + \pi} \right]$ which gives $c_{\tilde{k}}=\frac{1}{\sqrt{2}}\left[ a_{\tilde{2k}} e^{ik} + b_{\tilde{2k}} \right]$ and $c_{\tilde{k}+\pi}=\frac{1}{\sqrt{2}}\left[ -a_{\tilde{2k}} e^{ik} + b_{\tilde{2k}} \right]$. So, $|WF_2\rangle=\frac{1}{\sqrt{2}}\left[ (X_1-X_2) e^{-ik}a_{\tilde{2k}}^\dagger +(X_1+X_2) b_{\tilde{2k}}^\dagger \right]|0\rangle$ Is there anyway to get $a_{k}^\dagger(b_{k}^\dagger)$ instead of $2k$. – Luqman Saleem Apr 11 '18 at 20:10
• What if I take $H_1=\sum_{n=1}^L [(t+\delta t)c_{2n-1}^\dagger c_{2n}+(t-\delta t)c_{2n}^\dagger c_{2n+1}]+H.C.$ – Luqman Saleem Apr 11 '18 at 20:10
• $\tilde{k}$ in $c_{\tilde{k}}$ is denser than k in $a_k$, so $a_{2\tilde{k}}$ include all the allowed k already. – Yen-Ta Huang Apr 11 '18 at 22:58
• For your more staggered Hamilonian, a unit cell will include more sites, let's call them $a_1, a_2, a_3, a_4$. It's easy to follow the same recipe, and each $a_i$ will be a linear superposition of $c(k),c(k+\pi/2),c(k+\pi), c(k+3\pi/2)$. – Yen-Ta Huang Apr 11 '18 at 23:03