2020-02-25 Kitaev model on a hex

作者: 低维量子系统 | 来源:发表于2020-02-25 11:52 被阅读0次

note:

参考资料见前面的笔记。
原文的附录标题是Solution of a BCS Hamiltonian。
文献中某些公式中的打印错误得以修正。

Solution of a BCS Hamiltonian

$H=\sum_k \epsilon_k d_k^{\dagger}d_k + \frac {\Delta_k} {2} (id_k^{\dagger}d^{\dagger}_{-k}-id_{-k}d_k)$

With

$d_k$ : fermion operators
$\epsilon_{-k}=\epsilon_k$
$\Delta_{-k}=-\Delta_{k}$

By a global gauge transformation

$d_k^{\dagger} \rightarrow e^{-i\frac{\pi}{4}} d_k^{\dagger}$
$d_k \rightarrow e^{+i\frac{\pi} {4}} d_k$

note:

$d_k^{\dagger} d_k \rightarrow e^{-i \frac{\pi} {4} } d_k^{\dagger} e^{+i\frac{\pi} {4}} d_k = d_k^{\dagger} d_k$

$d_k^{\dagger} d^{\dagger}_{-k} \rightarrow e^{-i\frac{\pi}{4}} d_k^{\dagger} e^{-i\frac{\pi}{4}} d_{-k}^{\dagger} = e^{-i\frac{\pi} {2} }d_k^{\dagger} d_{-k}^{\dagger} = - i d_k^{\dagger} d_{-k}^{\dagger}$

$d_{-k} d_k \rightarrow e^{+i\frac{\pi} {4}} d_{-k} e^{+i\frac{\pi} {4}} d_{k} = e^{+i\frac{\pi} {2}} d_{-k} d_{k} = i d_{-k} d_{k}$

We shall have

$H=\sum_k \epsilon_k d_k^{\dagger}d_k + \frac {\Delta_k} {2} (d_k^{\dagger} d^{\dagger}_{-k}+d_{-k}d_k)$

Using the property of $\epsilon_k$ , $H$ can be expressed in a more symmetric form as

$H=\sum_k \frac{\epsilon_k}{2} (d_k^{\dagger}d_k+d_{-k}^{\dagger}d_{-k}) + \frac{\Delta_k}{2}(d_k^{\dagger}d^{\dagger}_{-k}+d_{-k} d_k)$

Let us denote

$d_k \rightarrow a_k$
$d_{-k} \rightarrow b_k$

And then we obtain

$H=\sum_k [ \frac{\epsilon_k} {2} (a^{\dagger}_k a_k+b^{\dagger}_k b_k)+\frac{\Delta_k} {2} (a^{\dagger}_k b^{\dagger}_k +b_k a_k) ]$

It is clear that

$a_{-k}=b_k$ , $a_{-k}^{\dagger}=b_k^{\dagger}$
$b_{-k}=a_k$ , $b_{-k}^{\dagger}=a_k^{\dagger}$

Bogoliubov transformations

can help us to get a diagonal Hamiltonian.

$A_k = u_k a_k + v_k b_k^{\dagger}$ , $A_k^{\dagger} = u_k a_k^{\dagger} + v_k b_k$
$B_k = v_k a_k^{\dagger} - u_k b_k$ , $B_k^{\dagger} = v_k a_k - u_k b_k^{\dagger}$

$u_k$ and $v_k$ are real numbers. We demand that $A_k$ and $B_k$ are fermion operators with

$\{ A_k, A_k^{\dagger} \} = \{ B_k, B_k^{\dagger} \} =1$

Thereby, it is straightforward to find that $u_k$ and $v_k$ should satisfy the condition

$u_k^2+v_k^2=1$ .

We shall parameterize them as

$u_k=\cos \theta_k$ , $v_k = \sin \theta_k$
Some useful equations
- $B_{-k}^{\dagger}=-A_k^{\dagger}$
- $B_{-k}=-A_k$
Proof
$\Leftarrow$ $B_{-k}= v_{-k} a_{-k}^{\dagger} - u_{-k} b_{-k}$
$\Leftarrow$ $B_{-k}=- v_{k} a_{-k}^{\dagger} - u_{k} b_{-k}$
$\Leftarrow$ $B_{-k}=- v_{k} b_{k}^{\dagger} - u_{k} a_{k} = - A_k$

Bogoliubov inverse transformations

$a_k=u_kA_k + v_k B_k^{\dagger}$ , $a_k^{\dagger}=u_kA_k ^{\dagger}+ v_k B_k$
$b_k=v_kA_k^{\dagger} - u_k B_k$ , $b_k^{\dagger}=v_kA_k - u_k B_k^{\dagger}$

We substitute the above transformations into the Hamiltonian,

and will get

$H=\sum_k[\frac{\epsilon_k}{2}(u_k^2-v_k^2)+\Delta_k u_k v_k](A_k^{\dagger}A_k +B_k^{\dagger} B_k)$

$+[\epsilon_ku_kv_k-\frac{\Delta_k}{2}(u_k^2-v_k^2)](A_k^{\dagger}B_k^{\dagger} + B_k A_k)$

solutions of $u_k$ and $v_k$

We need to choose some $u_k$ and $v_k$ so that the non-diagonal term vanishes. In other words,

$\epsilon_k u_k v_k - \frac {\Delta_k} {2} (u_k^2-v_k^2)=0$

$\Rightarrow \epsilon_k \sin 2\theta_k - \Delta_k \cos 2\theta_k = 0$

$\Rightarrow \tan 2\theta_k = \frac{\Delta_k}{\epsilon_k}$
$u_k^2-v_k^2 = \cos 2\theta_k = \frac{1} {\sqrt{1+\tan^2 2\theta_k}}=\frac{\epsilon_k}{\sqrt{\epsilon_k^2+\Delta_k^2}}$
$u_kv_k=\frac{1}{2}\sin 2\theta_k = \frac{1}{2}\tan 2\theta_k \cos 2\theta_k = \frac{1}{2}\frac{\Delta_k}{\sqrt{\epsilon_k^2+\Delta_k^2}}$
$u_k^2 = \frac{1} {2}(1+\frac{\epsilon_k}{\sqrt{\epsilon_k^2+\Delta_k^2}})$
$v_k^2 = \frac{1}{2}(1-\frac{\epsilon_k}{\sqrt{\epsilon_k^2+\Delta_k^2}})$

solution of the spectrum

$H=\sum_k[ \frac{\epsilon_k} {2} (u_k^2-v_k^2)+\Delta_k u_k v_k](A_k^{\dagger}A_k +B_k^{\dagger} B_k)$

$+[\epsilon_ku_kv_k-\frac{\Delta_k}{2}(u_k^2-v_k^2)](A_k^{\dagger}B_k^{\dagger} + B_k A_k)$

$\Rightarrow$ $H=\sum_k[\frac{\epsilon_k}{2}(u_k^2-v_k^2)+\Delta_k u_k v_k](A_k^{\dagger}A_k +B_k^{\dagger}B_k)$

$\Rightarrow$ $H=\sum_k \frac {\sqrt{\epsilon_k^2 + \Delta^2_k}} {2} (A_k^{\dagger}A_k +B_k^{\dagger}B_k)$

Since $B_{-k}^{\dagger}=-A_k^{\dagger}$ and $B_{-k}=-A_k$ ,
the Hamiltonian can be further simplified as
$\Rightarrow$ $H=\sum_k \sqrt{\epsilon_k^2 + \Delta^2_k}A_k^{\dagger}A_k$

The ground state is the one without any quasiparticle, thus

$A_k \vert g\rangle=0$ for any $k$ .

2020-02-25 Kitaev model on a hex

Solution of a BCS Hamiltonian

Bogoliubov transformations

Bogoliubov inverse transformations

solutions of $u_k$ and $v_k$

solution of the spectrum

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2020-02-25 Kitaev model on a hex

Solution of a BCS Hamiltonian

Bogoliubov transformations

Bogoliubov inverse transformations

solutions of and

solution of the spectrum

相关文章

网友评论

延伸阅读

深度阅读

栏目导航

热点阅读

solutions of $u_k$ and $v_k$