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--- 물리:이징_모형_husimi_트리_kagome_격자 [2023/09/21 15:38] – minwoo
+++ 물리:이징_모형_husimi_트리_kagome_격자 [2023/10/20 23:40] – minwoo
@@ Line 37: / Line 37: @@
 $$
-g_n(\sigma_0) = \sum_{\{\sigma_1\}} \exp\Bigg[\beta\Bigg(J_3\sum_\Delta \sigma_0\sigma_1^{(1)} \sigma_2^{(2)} + J_2 \sum_{\text{n.n}}\sigma_0 \sigma_1 + h \sum _{j=1,2}\sigma_1^{(j)} \Bigg) \Bigg][g_{n-1}(\sigma_1^{(1)})]^{\gamma-1}[g_{n-1}(\sigma_1^{(2)})]^{\gamma-1}
+g_n(\sigma_0) = \sum_{\{\sigma_1\}} \exp\Bigg[\beta\Bigg(J_3\sum_\Delta \sigma_0\sigma_1^{(1)} \sigma_1^{(2)} + J_2 \sum_{\text{n.n}}\sigma_0 \sigma_1 + h \sum _{j=1,2}\sigma_1^{(j)} \Bigg) \Bigg][g_{n-1}(\sigma_1^{(1)})]^{\gamma-1}[g_{n-1}(\sigma_1^{(2)})]^{\gamma-1}
 $$
@@ Line 45: / Line 45: @@
 이러한 방식의 정의는 Husimi 트리가 트리 구조를 가지기에 가능한 것이다. 그에 따라, 분배함수 $Z=\sum_\sigma P(\sigma)$를 다음과 같이 적을 수 있다.
-$$ Z = \sum_{\sigma_0} \exp(\beta h\sigma_0) [g_n(\sigma_0)]^{(\gamma-1)} $$
+$$ Z = \sum_{\sigma_0} \exp(\beta h\sigma_0) [g_n(\sigma_0)]^{\gamma} $$
 따라서,
 $$
-\langle \sigma_0 \rangle = Z^{-1} \sum_{\sigma_0} \sigma_0 \exp (\beta h\sigma_0 )[g_n(\sigma_0)]^{(\gamma-1)}
+\langle \sigma_0 \rangle = Z^{-1} \sum_{\sigma_0} \sigma_0 \exp (\beta h\sigma_0 )[g_n(\sigma_0)]^{\gamma}
 $$
 로 쓸 수 있다.
@@ Line 66: / Line 66: @@
 $$
 \begin{align}
-\langle \sigma_0 \rangle &= \frac{\sum_{\sigma_0}\sigma_0 \exp(\beta h\sigma_0) [g_n(\sigma_0)]^{(\gamma-1)}}{Z} \\
+\langle \sigma_0 \rangle &= \frac{\sum_{\sigma_0}\sigma_0 \exp(\beta h\sigma_0) [g_n(\sigma_0)]^{\gamma}}{Z} \\
-&= \frac{e^{\beta h}g_n(+)^{(\gamma-1)}\  - \ e^{-\beta h}g_n(-)^{(\gamma-1)}}{e^{\beta h}g_n(+)^{(\gamma-1)}\  + \ e^{-\beta h}g_n(-)^{(\gamma-1)}} \\
+&= \frac{e^{\beta h}g_n(+)^{\gamma}\  - \ e^{-\beta h}g_n(-)^{\gamma}}{e^{\beta h}g_n(+)^{\gamma}\  + \ e^{-\beta h}g_n(-)^{\gamma}} \\
-&= \frac{e^{2\beta h}g_n(+)^{(\gamma-1)} \ -g_n(-)^{(\gamma-1)}}{e^{2\beta h}g_n(+)^{(\gamma-1)} \ +g_n(-)^{(\gamma-1)}} \\
+&= \frac{e^{2\beta h}g_n(+)^{\gamma} \ -g_n(-)^{\gamma}}{e^{2\beta h}g_n(+)^{\gamma} \ +g_n(-)^{\gamma}} \\
-&= \frac{az_n^{(\gamma-1)} \ -1}{az_n^{(\gamma-1)} \ +1}
+&= \frac{az_n^{\gamma} \ -1}{az_n^{\gamma} \ +1}
 \end{align}
 $$
@@ Line 83: / Line 83: @@
 $$
-g_n(\sigma_0) = \sum_{\{\sigma_1\}} \exp\Bigg[\beta\Bigg(J_3\sum_\Delta \sigma_0\sigma_1^{(1)} \sigma_2^{(2)} + J_2 \sum_{\text{n.n}}\sigma_0 \sigma_1 + h \sum _{j=1,2}\sigma_1^{(j)} \Bigg) \Bigg][g_{n-1}(\sigma_1^{(1)})]^{\gamma-1}[g_{n-1}(\sigma_1^{(2)})]^{\gamma-1}
+g_n(\sigma_0) = \sum_{\{\sigma_1\}} \exp\Bigg[\beta\Bigg(J_3\sum_\Delta \sigma_0\sigma_1^{(1)} \sigma_1^{(2)} + J_2 \sum_{\text{n.n}}\sigma_0 \sigma_1 + h \sum _{j=1,2}\sigma_1^{(j)} \Bigg) \Bigg][g_{n-1}(\sigma_1^{(1)})]^{\gamma-1}[g_{n-1}(\sigma_1^{(2)})]^{\gamma-1}
 $$
@@ Line 180: / Line 180: @@
 이를 통해, $$
 \begin{align}
-\langle \sigma_0 \rangle= \frac{az_n^{(\gamma-1)} \ -1}{az_n^{(\gamma-1)} \ +1}
+\langle \sigma_0 \rangle= \frac{az_n^{\gamma} \ -1}{az_n^{\gamma} \ +1}
 \end{align}
 $$