2020-02-14 Kitaev model on a hex

作者: 低维量子系统 | 来源:发表于2020-02-14 23:00 被阅读0次

Kitaev model on the hexagonal lattice

这两天摈弃了所有杂念，踏踏实实学习Kitaev model。找到的一篇能够follow的文献是Chen在2008年发表的论文。

标题：Exact results of the Kitaev model on a hexagonal lattice: spin states, string and brane correlators, and anyonic excitations
作者：Han-Dong Chen and Zohar Nussinov
全文链接：论文

笔记的目的
- 日后关于Jordan-Wigner变换，Wick定理，傅立叶变换等，可以在这里温习。
- 如果日后恰好有人学到这篇文献，希望这篇笔记能够帮到他（她）。

Hamiltonian

$H=-J_x\sum_{x-links}\sigma_j^x\sigma_k^x-J_y\sum_{y-links}\sigma_j^y\sigma_k^y-J_z\sum_{z-links}\sigma_j^z\sigma_k^z$

pauli operators

$\sigma^x=[0,1;1,0]$

$\sigma^y=[0,-i;i,0]$

$\sigma^z=[1,0;0,-1]$
$\sigma_x^2=\sigma_y^2=\sigma_z^2=I$
$\sigma_x\sigma_y\sigma_z=I$

$\sigma_y\sigma_z\sigma_x=I$

$\sigma_z\sigma_x\sigma_y=I$
$\sigma_z\sigma_y\sigma_x=-i$

$\sigma_y\sigma_x\sigma_z=-i$

$\sigma_x\sigma_z\sigma_y=-i$

fermion operators

$\{a^{\dagger}_j,a_j\}=1$

$\{a^{\dagger}_j,a_k\}=0$

$\{a_j,a_k\}=0$

$\{a^{\dagger}_j,a^{\dagger}_k\}=0$

$a_j^2=0$

${a^{\dagger2}_j}=0$

Jordan-Wigner transformation from spin operators to fermion operators

$\sigma^x_{\ell}=[\Pi_{m=1...\ell-1}\sigma^z_m ] \cdot(c^{\dagger}_{\ell}+c_{\ell})$

$\sigma^y_{\ell}=-i\cdot[\Pi_{m=1...\ell-1}\sigma^z_m ] \cdot(c^{\dagger}_{\ell}-c_{\ell})$

$\sigma^z_{\ell}=2c^{\dagger}_{\ell}c_{\ell}-1$
communication relations

$[\sigma^z_{\ell},c^{\dagger}_j]$

$=[2c^{\dagger}_{\ell}c_{\ell}-1,c^{\dagger}_j]$

$=(2c^{\dagger}_{\ell}c_{\ell}-1)\cdot c^{\dagger}_j - c^{\dagger}_j \cdot (2c^{\dagger}_{\ell}c_{\ell}-1)$

$=(2c^{\dagger}_{\ell}c_{\ell}\cdot c^{\dagger}_j- c^{\dagger}_j) - (2 c^{\dagger}_j \cdot c^{\dagger}_{\ell}c_{\ell}-c^{\dagger}_j )$

$=2c^{\dagger}_{\ell}c_{\ell}\cdot c^{\dagger}_j - 2 c^{\dagger}_j \cdot c^{\dagger}_{\ell}c_{\ell}$

$=0$

$[\sigma^z_{\ell},c^{\dagger}_{\ell}]$

$=[2c^{\dagger}_{\ell}c_{\ell}-1,c^{\dagger}_{\ell}]$

$=(2c^{\dagger}_{\ell}c_{\ell}-1)\cdot c^{\dagger}_{\ell} - c^{\dagger}_{\ell} \cdot (2c^{\dagger}_{\ell}c_{\ell}-1)$

$=(2c^{\dagger}_{\ell}c_{\ell}\cdot c^{\dagger}_{\ell}- c^{\dagger}_{\ell}) - (2 c^{\dagger}_{\ell} \cdot c^{\dagger}_{\ell}c_{\ell}-c^{\dagger}_{\ell} )$

$=2c^{\dagger}_{\ell}c_{\ell}\cdot c^{\dagger}_{\ell}$

$=2c^{\dagger}_{\ell}\cdot (1-c^{\dagger}_{\ell} c_{\ell})$

$=2c^{\dagger}_{\ell}$
Jordan-Wigner transformation for $\sigma^x_{\ell}\sigma^x_{\ell+1}$

$\sigma^x_{\ell}\sigma^x_{\ell+1}$

$=[\Pi_{m=1...\ell-1}\sigma^z_m ] \cdot(c^{\dagger}_{\ell}+c_{\ell}) \cdot [\Pi_{m=1...\ell}\sigma^z_m ] \cdot(c^{\dagger}_{\ell+1}+c_{\ell+1})$

$=(c^{\dagger}_{\ell}+c_{\ell}) \cdot[\Pi_{m=1...\ell-1}\sigma^z_m ] \cdot [\Pi_{m=1...\ell}\sigma^z_m ]\cdot(c^{\dagger}_{\ell+1}+c_{\ell+1})$

$=(c^{\dagger}_{\ell}+c_{\ell}) \cdot \sigma^z_{\ell}\cdot(c^{\dagger}_{\ell+1}+c_{\ell+1})$ ------(1)

Since $\sigma^z_{\ell}=2c^{\dagger}_{\ell}c_{\ell}-1$ , one will have

$(c^{\dagger}_{\ell}+c_{\ell})\cdot \sigma^z_{\ell}$

$= (c^{\dagger}_{\ell}+c_{\ell})\cdot(2c^{\dagger}_{\ell}c_{\ell}-1)$

$= c^{\dagger}_{\ell} \cdot(2c^{\dagger}_{\ell}c_{\ell}-1) +c_{\ell}\cdot(2c^{\dagger}_{\ell}c_{\ell}-1)$

$=- c^{\dagger}_{\ell} +c_{\ell}\cdot(2c^{\dagger}_{\ell}c_{\ell}-1)$

$=- c^{\dagger}_{\ell} +2c_{\ell}\cdot c^{\dagger}_{\ell}c_{\ell}-c_{\ell}$

$=- c^{\dagger}_{\ell} +2c_{\ell}\cdot (1- c_{\ell}c^{\dagger}_{\ell})-c_{\ell}$

$= - c^{\dagger}_{\ell} + c_{\ell}$ ------(2)

According to (1) and (2), we finally have

$\sigma^x_{\ell}\sigma^x_{\ell+1}=- (c^{\dagger}_{\ell}- c_{\ell})\cdot(c^{\dagger}_{\ell+1}+c_{\ell+1})$
Jordan-Wigner transformation for $\sigma^y_{\ell}\sigma^y_{\ell+1}$

$\sigma^y_{\ell}\sigma^y_{\ell+1}$

$=-[\Pi_{m=1...\ell-1}\sigma^z_m ] \cdot(c^{\dagger}_{\ell}-c_{\ell})\cdot [\Pi_{m=1...\ell}\sigma^z_m ] \cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-(c^{\dagger}_{\ell}-c_{\ell})\cdot [\Pi_{m=1...\ell-1}\sigma^z_m ] \cdot [\Pi_{m=1...\ell}\sigma^z_m ] \cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-(c^{\dagger}_{\ell}-c_{\ell})\cdot \sigma^z_{\ell} \cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-(c^{\dagger}_{\ell}-c_{\ell})\cdot (2c^{\dagger}_{\ell}c_{\ell}-1)\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-[c^{\dagger}_{\ell}\cdot (2c^{\dagger}_{\ell}c_{\ell}-1)-c_{\ell}\cdot (2c^{\dagger}_{\ell}c_{\ell}-1)]\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-[ (c^{\dagger}_{\ell}\cdot2c^{\dagger}_{\ell}c_{\ell}-c^{\dagger}_{\ell})-(2c_{\ell}\cdot c^{\dagger}_{\ell}c_{\ell}-c_{\ell})]\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-[ (0-c^{\dagger}_{\ell})-(2c_{\ell}\cdot c^{\dagger}_{\ell}c_{\ell}-c_{\ell})]\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-[ -c^{\dagger}_{\ell}-2c_{\ell}\cdot c^{\dagger}_{\ell}c_{\ell}+c_{\ell}]\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-[ -c^{\dagger}_{\ell}-2c_{\ell}\cdot (1-c_{\ell}c^{\dagger}_{\ell})+c_{\ell}]\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-[ -c^{\dagger}_{\ell}-2c_{\ell}+c_{\ell}]\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=-[ -c^{\dagger}_{\ell}-c_{\ell}]\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$=(c^{\dagger}_{\ell}+c_{\ell})\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$
Jordan-Wigner transformation for $\sigma^z_{\ell}\sigma^z_{\ell+1}$

$\sigma^z_{\ell}\sigma^z_{\ell+1}=(2c^{\dagger}_{\ell}c_{\ell}-1)(2c^{\dagger}_{\ell+1}c_{\ell+1}-1)$
Finally, according to the sequence in Fig. 2 of Chen2008, we have

$H=-J_x\sum_{x-links}[- (c^{\dagger}_{\ell}- c_{\ell})\cdot(c^{\dagger}_{\ell+1}+c_{\ell+1})]$

$-J_y\sum_{y-links}(c^{\dagger}_{\ell}+c_{\ell})\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$

$-J_z\sum_{z-links}(2c^{\dagger}_{\ell}c_{\ell}-1)(2c^{\dagger}_{\ell+1}c_{\ell+1}-1)$

$H=J_x\sum_{x-links}(c^{\dagger}_{\ell}- c_{\ell})\cdot(c^{\dagger}_{\ell+1}+c_{\ell+1})$
$-J_y\sum_{y-links}(c^{\dagger}_{\ell}+c_{\ell})\cdot(c^{\dagger}_{\ell+1}-c_{\ell+1})$
$-J_z\sum_{z-links}(2c^{\dagger}_{\ell}c_{\ell}-1)(2c^{\dagger}_{\ell+1}c_{\ell+1}-1)$

$H=J_x\sum_{x-links}(c^{\dagger}- c)_w\cdot(c^{\dagger}+c)_b$
$-J_y\sum_{y-links}(c^{\dagger}+c)_b\cdot(c^{\dagger}-c)_w$
$-J_z\sum_{z-links}(2c^{\dagger}c-1)_b(2c^{\dagger} c -1)_w$ ------(4)