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--- 물리:구면_p-스핀_유리_모형 [2026/05/10 18:42] – [무작위 평균] admin
+++ 물리:구면_p-스핀_유리_모형 [2026/05/10 22:09] (current) – [무작위 평균] admin
@@ Line 504: / Line 504: @@
 ===리거(1992), 그리고 크리산티와 좀머스(1995)의 계산(작성 중)===
 여기에서는 $q$를 $m_i$들과 독립적인 변수로 취급하고 나중에 [[수학:디락_델타_함수|디락 델타 함수]]로써 $q = N^{-1}\sum_i m_i^2$의 제약을 둔다.
-$\zeta \equiv 1/(1-q) + \beta^2 p (p-1)(1-q) q^{p-2}/2$로 정의할 때 [[물리:tap_방정식|TAP 방정식]]을 다음처럼 적게 되고
+$\rho \equiv \beta^2 p(p-1) q^{p-2}$이고
+$\zeta \equiv 1/(1-q) + (1-q) \rho/2$로 정의할 때 [[물리:tap_방정식|TAP 방정식]]을 다음처럼 적게 되며,
 $$\mathcal{T}_i = \zeta m_i - \frac{\beta}{(p-1)!} \sum_{k_2, \ldots, k_p} J_{i k_2 \ldots k_p} m_{k_2} \cdots m_{k_p} = 0$$
 헤세 행렬의 원소는 이렇게 주어진다:
@@ Line 513: / Line 514: @@
 &=& N\int \frac{dq ~d\hat{q}}{2\pi} \int \left( \prod_i \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\hat{q} \left(Nq - \sum_i m_i^2 \right) \right] \exp \left[ I\zeta \sum_i \hat{m}_i m_i - \frac{I\beta}{(p-1)!} \sum_{i,k_2, \ldots, k_p} J_{i k_2 \ldots k_p} \hat{m}_i m_{k_2} \cdots m_{k_p}\right] \det \mathcal{H}.\\
 \end{eqnarray*}
-$\mathcal{N}$에 대해 곧바로 무작위 평균을 취하도록 하자. 지수 함수뿐만 아니라 $\det \mathcal{H}$에도 $J_{ijk_3\ldots k_p}$가 포함되어 있지만, 평균을 취하는 과정에서 생겨나는 교차항을 무시할 수 있다고 하면 아래처럼 따로 평균을 취한 후 곱할 수 있다:
+$\mathcal{N}$에 대해 곧바로 무작위 평균을 취하도록 하자. 지수 함수뿐만 아니라 $\det \mathcal{H}$에도 $J_{ijk_3\ldots k_p}$가 포함되어 있지만, 평균을 취하는 과정에서 생겨나는 교차항을 무시하면 아래처럼 따로 평균을 취한 후 곱할 수 있다:
 \begin{eqnarray*}
 \langle \mathcal{N} \rangle &\approx& N\int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_i \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_i \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_i m_i^2 \right) \right] \Biggl< \exp \left[- \frac{I\beta}{(p-1)!} \sum_{i,k_2, \ldots, k_p} J_{i k_2 \ldots k_p} \hat{m}_i m_{k_2} \cdots m_{k_p}\right] \Biggr>
@@ Line 542: / Line 543: @@
 이제 이 결과들을 앞의 식에 대입하면
 \begin{eqnarray*}
-\langle \mathcal{N} \rangle &\approx& N\int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_i \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_i \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_i m_i^2 \right) \right] \exp \left[ -\frac{\beta^2 p!}{4N^{p-1}} \frac{1}{(p-1)!^2} \left( \Sigma_1 + \Sigma_2 \right) \right]
+\langle \mathcal{N} \rangle &\approx& N\int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_{i=1}^N \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_{i=1}^N \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_{i=1}^N m_i^2 \right) \right] \exp \left[ -\frac{\beta^2 p!}{4N^{p-1}} \frac{1}{(p-1)!^2} \left( \Sigma_1 + \Sigma_2 \right) \right]
 \langle \det \mathcal{H} \rangle\\
-&=& N\int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_i \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_i \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_i m_i^2 \right) -\frac{\beta^2 p q^{p-1}}{4} \sum_{i=1}^N \hat{m}_i^2 -\frac{\beta^2 p(p-1) q^{p-2}}{4N} \left(\sum_{i=1}^N \hat{m}_i m_i \right)^2 \right]
+&=& N\int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_{i=1}^N \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_{i=1}^N \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_{i=1}^N m_i^2 \right) -\frac{\beta^2 p q^{p-1}}{4} \sum_{i=1}^N \hat{m}_i^2 -\frac{\rho}{4N} \left(\sum_{i=1}^N \hat{m}_i m_i \right)^2 \right]
 \langle \det \mathcal{H} \rangle\\
-&\propto& \int dy \int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_i \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_i \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_i m_i^2 \right) -\frac{\beta^2 p q^{p-1}}{4} \sum_{i=1}^N \hat{m}_i^2 + Iy\sum_i m_i \hat{m}_i -\frac{Ny^2}{\beta^2 p(p-1) q^{p-2}} \right]
+&\propto& \int dy \int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_{i=1}^N \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_{i=1}^N \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_{i=1}^N m_i^2 \right) -\frac{\beta^2 p q^{p-1}}{4} \sum_{i=1}^N \hat{m}_i^2 + Iy\sum_i m_i \hat{m}_i -\frac{Ny^2}{\rho} \right]
+\langle \det \mathcal{H} \rangle\\
+&\propto& \int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_{i=1}^N \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_{i=1}^N \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_{i=1}^N m_i^2 \right) -\frac{\beta^2 p q^{p-1}}{4} \sum_{i=1}^N \hat{m}_i^2 + Iy_\ast\sum_i m_i \hat{m}_i -\frac{Ny_\ast^2}{\rho} \right]
 \langle \det \mathcal{H} \rangle.
 \end{eqnarray*}
-마지막 줄로 넘어올 때에는 [[수학:허바드-스트라토노비치_변환|허바드-스트라토노비치 변환]]을 사용하여 변수 $y$를 도입했다.
+중간에 [[수학:허바드-스트라토노비치_변환|허바드-스트라토노비치 변환]]을 사용하여 변수 $y$를 도입했고, $y_\ast$는 [[수학:안장점_근사|안장점 근사]]의 의미에서 가장 많은 기여를 하는 지점이다.
 ++++
 ++++행렬식의 평균 계산|
@@ Line 580: / Line 584: @@
 \begin{eqnarray*}
 \langle \det \mathcal{H} \rangle &\approx& \lim_{n\to -2} \int \left(\prod_{i,\alpha} \frac{d\xi_{i\alpha}}{\sqrt{2\pi}}\right) \exp\left(-\frac12 \zeta \sum_{i\alpha} \xi_{i\alpha}^2 \right) \exp\left[ \frac{\beta^2 p!}{16N^{p-1} [(p-2)!]^2} 2(p-2)! q^{p-2} N^p \sum_{\alpha\gamma} \left( \frac{1}{N} \sum_i \xi_{i\alpha}\xi_{i\gamma} \right)^2 \right]\\
-&=& \lim_{n\to -2} \int \left(\prod_{i,\alpha} \frac{d\xi_{i\alpha}}{\sqrt{2\pi}}\right) \exp\left(-\frac12 \zeta \sum_{i\alpha} \xi_{i\alpha}^2 \right) \exp\left[ \frac{1}{8} N\beta^2 p(p-1) q^{p-2} \sum_{\alpha\gamma} \left( \frac{1}{N} \sum_i \xi_{i\alpha}\xi_{i\gamma} \right)^2 \right]\\
+&=& \lim_{n\to -2} \int \left(\prod_{i,\alpha} \frac{d\xi_{i\alpha}}{\sqrt{2\pi}}\right) \exp\left(-\frac12 \zeta \sum_{i\alpha} \xi_{i\alpha}^2 \right) \exp\left[ \frac{1}{8} N\rho \sum_{\alpha\gamma} \left( \frac{1}{N} \sum_i \xi_{i\alpha}\xi_{i\gamma} \right)^2 \right]\\
-&\propto& \lim_{n\to -2} \int \left(\prod_{\alpha\gamma} dz_{\alpha\gamma}\right) \int \left(\prod_{i,\alpha} \frac{d\xi_{i\alpha}}{\sqrt{2\pi}}\right) \exp\left(-\frac12 \zeta \sum_{i\alpha} \xi_{i\alpha}^2 - N\sum_{\alpha\gamma} \frac{z_{\alpha\gamma}^2}{2\beta^2 p(p-1) q^{p-2}} + \frac12 \sum_{\alpha\gamma} \sum_i z_{\alpha\gamma} \xi_{i\alpha}\xi_{i\gamma} \right).
+&\propto& \lim_{n\to -2} \int \left(\prod_{\alpha\gamma} dz_{\alpha\gamma}\right) \int \left(\prod_{i,\alpha} \frac{d\xi_{i\alpha}}{\sqrt{2\pi}}\right) \exp\left(-\frac12 \zeta \sum_{i\alpha} \xi_{i\alpha}^2 - N\sum_{\alpha\gamma} \frac{z_{\alpha\gamma}^2}{2\rho} + \frac12 \sum_{\alpha\gamma} \sum_i z_{\alpha\gamma} \xi_{i\alpha}\xi_{i\gamma} \right).
 \end{eqnarray*}
 마지막 줄에서 [[수학:허바드-스트라토노비치_변환|허바드-스트라토노비치 변환]]으로 변수 $z_{\alpha\gamma}$를 도입했다. 만일 $z_{\alpha\gamma} = z\delta_{\alpha\gamma}$라면,
 \begin{eqnarray*}
 \langle \det \mathcal{H} \rangle &\propto& \lim_{n\to -2} \int \left(\prod_{i,\alpha} \frac{d\xi_{i\alpha}}{\sqrt{2\pi}}\right)
-\prod_{\alpha=1}^n \int dz \exp \left[ -\frac12 \sum_{i} (\zeta - z) \xi_{i\alpha}^2 - N \frac{z^2}{2\beta^2 p(p-1) q^{p-2}} \right]\\
+\prod_{\alpha=1}^n \int dz \exp \left[ -\frac12 \sum_{i} (\zeta - z) \xi_{i\alpha}^2 - N \frac{z^2}{2\rho} \right]\\
 &\propto& \lim_{n\to -2} \int \left(\prod_{i,\alpha} \frac{d\xi_{i\alpha}}{\sqrt{2\pi}}\right)
-\prod_{\alpha=1}^n \exp \left[ -\frac12 \sum_{i} (\zeta - z_\ast) \xi_{i\alpha}^2 - N \frac{z_\ast^2}{2\beta^2 p(p-1) q^{p-2}} \right]\\
+\prod_{\alpha=1}^n \exp \left[ -\frac12 \sum_{i} (\zeta - z_\ast) \xi_{i\alpha}^2 - N \frac{z_\ast^2}{2\rho} \right]\\
-&\propto& \lim_{n\to -2} \exp\left[ -\frac{Nnz_\ast^2}{2\beta^2 p(p-1) q^{p-2}} \right] \prod_{i=1}^N (\zeta - z_\ast)^{-n/2}\\
+&\propto& \lim_{n\to -2} \exp\left[ -\frac{Nnz_\ast^2}{2\rho} \right] \prod_{i=1}^N (\zeta - z_\ast)^{-n/2}\\
-&=& \exp\left[ \frac{Nz_\ast^2}{\beta^2 p(p-1) q^{p-2}} \right] (\zeta - z_\ast)^N.
+&=& \exp\left[ \frac{Nz_\ast^2}{\rho} \right] (\zeta - z_\ast)^N.
 \end{eqnarray*}
 이때 $z_\ast$는, [[수학:안장점_근사|안장점 근사]]의 의미에서 적분에 가장 많은 기여를 하는 지점이다.
 ++++
 \begin{eqnarray*}
-\langle \mathcal{N} \rangle &\propto& \int dy \int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_i \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_i \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_i m_i^2 \right) -\frac{\beta^2 p q^{p-1}}{4} \sum_{i=1}^N \hat{m}_i^2 + Iy\sum_i m_i \hat{m}_i -\frac{Ny^2}{\beta^2 p(p-1) q^{p-2}} \right]
+\langle \mathcal{N} \rangle &\propto& \int \frac{dq ~d\hat{q}}{2\pi} \int \left(\prod_{i=1}^N \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_{i=1}^N \hat{m}_i m_i + I\hat{q} \left(Nq - \sum_{i=1}^N m_i^2 \right) -\frac{\beta^2 pq^{p-1}}{4} \sum_{i=1}^N \hat{m}_i^2 + Iy_\ast\sum_{i=1}^N m_i \hat{m}_i -\frac{Ny_\ast^2}{\rho} \right] \times \exp\left[ \frac{Nz_\ast^2}{\rho} \right] (\zeta - z_\ast)^N\\
-\times
+&\propto& \int \frac{dq ~d\hat{q}}{2\pi} \exp \left[N \left( I\hat{q}q - \frac{y_\ast^2}{\rho} + \frac{z_\ast^2}{\rho} \right) \right]\int \left(\prod_{i=1}^N \frac{dm_i d\hat{m}_i}{2\pi}\right) \exp\left[I\zeta \sum_{i=1}^N \hat{m}_i m_i - I\hat{q}\sum_{i=1}^N m_i^2 -\frac{\beta^2 pq^{p-1}}{4} \sum_{i=1}^N \hat{m}_i^2 + Iy_\ast\sum_{i=1}^N m_i \hat{m}_i \right] (\zeta - z_\ast)^N\\
-\exp\left[ \frac{Nz_\ast^2}{\beta^2 p(p-1) q^{p-2}} \right] (\zeta - z_\ast)^N
+&=& \int \frac{dq ~d\hat{q}}{2\pi} \exp \left[N \left( I\hat{q}q - \frac{y_\ast^2}{\rho} + \frac{z_\ast^2}{\rho} \right) \right] \left\{ \int \frac{dm ~d\hat{m}}{2\pi} \exp\left[I\zeta \hat{m} m - I\hat{q} m^2 -\frac{\beta^2 pq^{p-1}}{4} \hat{m}^2 + Iy_\ast m \hat{m} \right] (\zeta - z_\ast) \right\}^N\\
+\end{eqnarray*}
+다음의 두 변수를 정의하자:
+\begin{eqnarray*}
+B &\equiv& z_\ast - \rho(1-q)/2\\
+D &\equiv& y_\ast + \rho(1-q)/2.
+\end{eqnarray*}
+그러면 $z_\ast^2 - y_\ast^2 = B^2-D^2 + \rho(1-q) (B+D)$이고 $\zeta-z_\ast = 1/(1-q)-B$, 그리고 $\zeta+y_\ast = 1/(1-q) + D$이다. 이를 대입하면
+$$\langle \mathcal{N} \rangle \propto \int d\hat{q} dB ~dD \exp(N\Xi)$$
+로서,
+\begin{eqnarray*}
+\Xi &\equiv& I\hat{q}{q} + (B+D)(1-q) + \frac{1}{\rho}\left(B^2 - D^2 \right) + \ln\left( \frac{1}{1-q} - B \right) + \ln \mathcal{I}\\
+\mathcal{I} &\equiv& \int \frac{dm~d\hat{m}}{2\pi} \exp\left[ -\frac{\beta pq^{p-1}}{4} \hat{m}^2 + I\hat{m} \left( \frac{1}{1-q}+D \right)m - I\hat{q} m^2 \right]
+= \left[ \left(\frac{1}{1-q} + D\right)^2 + I\beta pq^{p-1} \hat{q} \right]^{-1/2}
 \end{eqnarray*}
+[[수학:안장점_근사|안장점 근사]]로 $B$에 관한 적분을 계산하기 위해 방정식 $\partial \Xi / \partial B=0$을 풀어보면 $B=0$이 해 중의 하나로 등장하며, 우리는 이것을 택한다.
 ===카바냐 등(1999)의 계산(작성 중)===