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--- 물리:요르단-위그너_변환_jordan-wigner_transformation [2024/09/08 17:19] – minwoo
+++ 물리:요르단-위그너_변환_jordan-wigner_transformation [2024/09/10 12:58] (current) – minwoo
@@ Line 249: / Line 249: @@
 처음에 언급한 1차원 양자 스핀 모형인 '가로장이 걸려있는 XY 스핀 사슬 (XY spin chain in a transverse field)'에 요르단-위그너 변환을 적용해보도록 하자.
-우선, 기존의 Hamiltonian은 다음과 같다.
+우선, 기존의 해밀토니안은 다음과 같다.
 $$
 \hat{H} =  \sum_{i=1}^N g_i \hat{\sigma}^z_i - \sum_{i=1}^N (J_x \hat{\sigma}^x_{i} \hat{\sigma}^x_{i+1} + J_y \hat{\sigma}^y_{i}\hat{\sigma}^y_{i+1}).
@@ Line 273: / Line 273: @@
 $$
 이를 적용하면,
-$$
 \begin{align}
 & \sum_{i=1}^N (J_x \hat{\sigma}^x_{i} \hat{\sigma}^x_{i+1} + J_y \hat{\sigma}^y_{i}\hat{\sigma}^y_{i+1})\\
@@ Line 279: / Line 279: @@
 &=\sum_{j=1}^N \left[J_x  (\hat{\sigma}_j^+ + \hat{\sigma}_j^-) (\hat{\sigma}_{j+1}^+ + \hat{\sigma}_{j+1}^-) - J_y (\hat{\sigma}_j^+ - \hat{\sigma}_j^-)(\hat{\sigma}_{j+1}^+ - \hat{\sigma}_{j+1}^-)\right]\\
-&=\sum_{j=1}^N \big[(J_x + J_y)\left(\hat{\sigma}_{j+1}^+\hat{\sigma}_j^-+\hat{\sigma}_{j+1}^-\hat{\sigma}_j^+\right) \\
+&=\sum_{j=1}^N \Big[(J_x + J_y)\left(\hat{\sigma}_j^-\hat{\sigma}_{j+1}^+ + \hat{\sigma}_j^+ \hat{\sigma}_{j+1}^-\right) \\
-&\qquad \quad +(J_x-J_y)(\hat{\sigma}_{j+1}^+\hat{\sigma}_j^+ +\hat{\sigma}_{j+1}^-\hat{\sigma}_j^-)\big].\\
+&\qquad \quad +(J_x-J_y)\left(\hat{\sigma}_j^+\hat{\sigma}_{j+1}^+ +\hat{\sigma}_j^-\hat{\sigma}_{j+1}^-\right)\Big].\\
 \end{align}
-$$
 위의 식에 요르단-위그너 변환을 아래와 같이 적용해보자.
-$$
 \begin{align}
-&\hat{\sigma}_{j+1}^+ \hat{\sigma}_j^- \\
+& \hat{\sigma}_j^- \hat{\sigma}_{j+1}^+ \\
-&= \hat{f}_{j+1}^\dagger e^{i\pi \sum_{l<j+1} n_l} \hat{f}_j  e^{-i\pi \sum_{l<j} n_l}.\\
+&=\hat{f}_j  e^{-i\pi \sum_{l<j} n_l} \hat{f}_{j+1}^\dagger e^{i\pi \sum_{l<j+1} n_l} .\\
-&= \hat{f}_{j+1}^\dagger e^{i\pi n_j}\hat{f}_j=\hat{f}_{j+1}^\dagger \hat{f}_j,
+&= \hat{f}_j \hat{f}_{j+1}^\dagger e^{i\pi n_j}= -\hat{f}_j\hat{f}_{j+1}^\dagger, \\
+&\\
+& \hat{\sigma}_j^+ \hat{\sigma}_{j+1}^- \\
+&=\hat{f}_j^\dagger  e^{i\pi \sum_{l<j} n_l} \hat{f}_{j+1}  e^{-i\pi \sum_{l<j+1} n_l} .\\
+&= \hat{f}_j^\dagger \hat{f}_{j+1}  e^{-i\pi n_j}= \hat{f}_j^\dagger \hat{f}_{j+1}, \\
+&\\
+& \hat{\sigma}_j^+ \hat{\sigma}_{j+1}^+\\
+&=  \hat{f}_j^\dagger e^{i\pi \sum_{l<j} n_l}\hat{f}_{j+1}^\dagger  e^{i\pi \sum_{l<j+1} n_l}.\\
+&= \hat{f}_{j}^\dagger \hat{f}_{j+1}^\dagger e^{i\pi n_j} =\hat{f}_{j}^\dagger \hat{f}_{j+1}^\dagger. \\
+&\\
+& \hat{\sigma}_j^- \hat{\sigma}_{j+1}^-\\
+&=  \hat{f}_j^- e^{-i\pi \sum_{l<j} n_l}\hat{f}_{j+1} e^{-i\pi \sum_{l<j+1} n_l}.\\
+&= \hat{f}_{j}\hat{f}_{j+1} e^{-i\pi n_j} =-\hat{f}_{j} \hat{f}_{j+1}.
+\end{align}
-\\
-\\
-&\hat{\sigma}_{j+1}^+ \hat{\sigma}_j^+ \\
-&= \hat{f}_{j+1}^\dagger e^{i\pi \sum_{l<j+1} n_l} \hat{f}_j^\dagger  e^{i\pi \sum_{l<j} n_l}.\\
-&= \hat{f}_{j+1}^\dagger e^{i\pi n_j}\hat{f}_j^\dagger=\hat{f}_{j+1}^\dagger \hat{f}_j^\dagger.
-\end{align}
-$$
 위의 두 번째 결과에서는 정수 $m$에 대해 $e^{2\pi i m}=1$을 사용했다.
 $\\$
-따라서, 변환된 Hamiltonian은 다음과 같다.
+따라서, 해밀토니안은 다음과 같이 쓰여진다.
-$$
 \begin{align}
 \hat{H}&=-\sum_{i=1}^N g_i (1-2\hat{f}_j^\dagger \hat{f}_j)
--\sum_{j=1}^{N-1} (J_x + J_y)\left(\hat{f}_{j+1}^\dagger \hat{f}_j+\hat{f}_{j+1} \hat{f}_j^\dagger \right) \\
+-\sum_{j=1}^{N-1} (J_x + J_y)\left(-\hat{f}_j\hat{f}_{j+1}^\dagger + \hat{f}_j^\dagger \hat{f}_{j+1} \right) \\
-& -\sum_{j=1}^{N-1} (J_x-J_y)\left(\hat{f}_{j+1}^\dagger \hat{f}_j^\dagger+\hat{f}_{j+1}\hat{f}_j\right).
+& -\sum_{j=1}^{N-1} (J_x-J_y)\left(\hat{f}_j^\dagger \hat{f}_{j+1}^\dagger-\hat{f}_j\hat{f}_{j+1}\right).\\ \\
+&=-\sum_{i=1}^N g_i (1-2\hat{f}_j^\dagger \hat{f}_j)
+-\sum_{j=1}^{N-1} (J_x + J_y)\left(\hat{f}_j^\dagger \hat{f}_{j+1} + \hat{f}_{j+1}^\dagger\hat{f}_j  \right) \\
+& -\sum_{j=1}^{N-1} (J_x-J_y)\left(\hat{f}_j^\dagger \hat{f}_{j+1}^\dagger+\hat{f}_{j+1}\hat{f}_j\right).
 \end{align}
-$$
+위에서 마지막 식에 도달할 때는 anticommutation relation을 사용하였다.
-(편의상, 주기적 경계 조건(periodic boundary condtiion)이 아닌 열린 경계 조건(open boudnary condition)을사용함으로써 두 번째와 세 번째 항의 합은 $j=N$이 아닌 $j=N-1$이다.)
+(편의상, 주기적 경계 조건(periodic boundary condtiion)이 아닌 열린 경계 조건(open boudnary condition)을 사용함으로써 두 번째와 세 번째 항의 합은 $j=N$이 아닌 $j=N-1$이다.)
 $\\$
@@ Line 326: / Line 337: @@
 $$
-이를 아래와 같이 풀이하여 확인해보자.
+이를 아래와 같이 풀이하여 직접 확인해보자.
 $$
 \hat{H}(t) = i \sum_{j=1}^N g_j(t)\{\hat{f}_j^\dagger + \hat{f}_j\}\{i(\hat{f}_j^\dagger - \hat{f}_j)\}\\
@@ Line 350: / Line 361: @@
 \\
 \to\
-\hat{H}(t) = - \sum_{j=1}^N g_j(t)\Big[ 1-2\hat{f}_{j}^\dagger \hat{f}_{j}  \Big]
+\hat{H}(t) = - \sum_{j=1}^N g_j(t)\left( 1-2\hat{f}_{j}^\dagger \hat{f}_{j} \right)
 \\
-- \sum_{j=1}^{N-1}\Big[
+-\sum_{j=1}^{N-1} (J_x + J_y)\left(\hat{f}_j^\dagger \hat{f}_{j+1} + \hat{f}_{j+1}^\dagger\hat{f}_j  \right) \\
-J_x\{\hat{f}_{j}^\dagger \hat{f}_{j+1}^\dagger + \hat{f}_{j}^\dagger \hat{f}_{j+1} - \hat{f}_{j}\hat{f}_{j+1}^\dagger - \hat{f}_{j}\hat{f}_{j+1} \}\\
+ -\sum_{j=1}^{N-1} (J_x-J_y)\left(\hat{f}_j^\dagger \hat{f}_{j+1}^\dagger+\hat{f}_{j+1}\hat{f}_j\right).
--J_y \{\hat{f}_{j}^\dagger \hat{f}_{j+1}^\dagger - \hat{f}_{j}^\dagger \hat{f}_{j+1} + \hat{f}_{j}\hat{f}_{j+1}^\dagger - \hat{f}_{j}\hat{f}_{j+1}\} \Big].
 $$
+즉, 앞서 본 Majorana fermion의 표현식이 실제로 원래의 식을 준다는 것을 확인하였다.
 ====== 참고 문헌 ======