물리:복제_트릭

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
물리:복제_트릭 [2023/09/05 15:46] – external edit 127.0.0.1물리:복제_트릭 [2026/02/20 14:47] (current) – [보조장의 의미] admin
Line 178: Line 178:
 분모에서 $\text{Tr}_n \ \exp \left( -\beta\sum_{\gamma} H^{\gamma} \right) = Z^n$이고 $n\to 0$에서 이는 $1$로 수렴할 것이므로 우리가 고려해야 할 것은 다음과 같은 양이다: 분모에서 $\text{Tr}_n \ \exp \left( -\beta\sum_{\gamma} H^{\gamma} \right) = Z^n$이고 $n\to 0$에서 이는 $1$로 수렴할 것이므로 우리가 고려해야 할 것은 다음과 같은 양이다:
 \begin{align} \begin{align}
-&\left[ \text{Tr}_n \ S_{i}^{\gamma} S_{i}^{\delta} \exp \left( -\beta \sum_{\alpha} H^{\alpha} \right) \right]\\+&\left[ \text{Tr}_n \ S_{k}^{\gamma} S_{k}^{\delta} \exp \left( -\beta \sum_{\alpha} H^{\alpha} \right) \right]\\
     &= \int \left[ \prod_{i < j} dJ_{ij} \ P(J_{ij}) \right] \text{Tr}_n S_{k}^{\gamma} S_{k}^{\delta}     &= \int \left[ \prod_{i < j} dJ_{ij} \ P(J_{ij}) \right] \text{Tr}_n S_{k}^{\gamma} S_{k}^{\delta}
     \exp \left[ \beta \sum_{i < j} J_{ij} \sum_{\alpha = 1}^n S_{i}^{\alpha} S_{j}^{\alpha} + \beta h \sum_{i=1}^N \sum_{\alpha=1}^n S_{i}^{\alpha} \right] \\     \exp \left[ \beta \sum_{i < j} J_{ij} \sum_{\alpha = 1}^n S_{i}^{\alpha} S_{j}^{\alpha} + \beta h \sum_{i=1}^N \sum_{\alpha=1}^n S_{i}^{\alpha} \right] \\
Line 184: Line 184:
     \exp \left( -\frac{N \beta ^2 J^2}{2} \sum_{\alpha < \beta} q_{\alpha \beta}^2 - \frac{N \beta J_0}{2} \sum_{\alpha} m_{\alpha}^2 \right) \text{Tr}_n S_{k}^{\gamma} S_{k}^{\delta} \exp \left[ \beta^2 J^2 \sum_{\alpha < \beta} q_{\alpha \beta} \sum_i S_{i}^{\alpha} S_{i}^{\beta} + \beta \sum_{\alpha} \left( J_0 m_{\alpha} + h \right) \sum_{i} S_{i}^{\alpha} \right].     \exp \left( -\frac{N \beta ^2 J^2}{2} \sum_{\alpha < \beta} q_{\alpha \beta}^2 - \frac{N \beta J_0}{2} \sum_{\alpha} m_{\alpha}^2 \right) \text{Tr}_n S_{k}^{\gamma} S_{k}^{\delta} \exp \left[ \beta^2 J^2 \sum_{\alpha < \beta} q_{\alpha \beta} \sum_i S_{i}^{\alpha} S_{i}^{\beta} + \beta \sum_{\alpha} \left( J_0 m_{\alpha} + h \right) \sum_{i} S_{i}^{\alpha} \right].
 \end{align} \end{align}
-여기서 $\text{Tr}_n$ 부분을 계산할 때, $i\neq k$이면 앞과 똑같이 $\text{Tr}^{\prime} e^L$만큼을 내어놓고 $i=k$일 때에만 $\text{Tr}^{\prime} \left( S_{k}^{\gamma} S_{k}^{\delta} e^L \right)$을 주므로+여기서 $\text{Tr}_n$ 부분을 계산할 때, $i\neq k$이면 앞과 똑같이 $\text{Tr}^{\prime} e^L$만큼을 내어놓고 $i=k$일 때에만 $\text{Tr}^{\prime} \left( S_{k}^{\gamma} S_{k}^{\delta} e^L \right)$을 주므로, 즉 
 +\begin{align} 
 +\text{Tr}_n S_{k}^{\gamma} S_{k}^{\delta} \exp \left[ \beta^2 J^2 \sum_{\alpha < \beta} q_{\alpha \beta} \sum_i S_{i}^{\alpha} S_{i}^{\beta} + \beta \sum_{\alpha} \left( J_0 m_{\alpha} + h \right) \sum_{i} S_{i}^{\alpha} \right] 
 +&= \text{Tr}_n S_{k}^{\gamma} S_{k}^{\delta} \prod_{i=1}^N \exp \left[ \beta^2 J^2 \sum_{\alpha < \beta} q_{\alpha \beta} S_{i}^{\alpha} S_{i}^{\beta} + \beta \sum_{\alpha} \left( J_0 m_{\alpha} + h \right) S_{i}^{\alpha} \right]\\ 
 +&= \text{Tr}_n S_{k}^{\gamma} S_{k}^{\delta} \exp \left[ \beta^2 J^2 \sum_{\alpha < \beta} q_{\alpha \beta} S_{k}^{\alpha} S_{k}^{\beta} + \beta \sum_{\alpha} \left( J_0 m_{\alpha} + h \right) S_{k}^{\alpha} \right] \prod_{i\neq k} \exp \left[ \beta^2 J^2 \sum_{\alpha < \beta} q_{\alpha \beta} S_{i}^{\alpha} S_{i}^{\beta} + \beta \sum_{\alpha} \left( J_0 m_{\alpha} + h \right) S_{i}^{\alpha} \right]\\ 
 +\end{align} 
 +이므로
 \begin{align} \begin{align}
     \therefore \left[ \big\langle S_{i}^{\alpha} S_{i}^{\beta} \big\rangle \right]     \therefore \left[ \big\langle S_{i}^{\alpha} S_{i}^{\beta} \big\rangle \right]
  • 물리/복제_트릭.txt
  • Last modified: 2026/02/20 14:47
  • by admin