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--- 물리:p-스핀_유리_모형 [2026/03/03 11:27] – [복제 대칭 해] admin
+++ 물리:p-스핀_유리_모형 [2026/03/21 21:40] (current) – [함께 보기] admin
@@ Line 125: / Line 125: @@
 =====복제 대칭 해=====
-모든 복제본의 쌍 $a,b$에 대해 $Q_{ab}=Q$, $\lambda_{ab} = \lambda$로 가정하고 생략했던 지수 함수를 되살리면 자유에너지는
+모든 복제본의 쌍 $a,b$에 대해 $Q^{ab}=Q$, $\lambda^{ab} = \lambda$로 가정하고 생략했던 지수 함수를 되살리면 자유에너지는
 \begin{align*}
 G(Q,\lambda)&=-\frac14 \beta^2 J^2 n + \frac{n(n-1)}4 \frac{\lambda^2}{4\beta^2 J^2 Q^2} + \frac{n(n-1)}4\lambda Q-\ln\text{Tr}_{\sigma^a}\exp\left[\lambda\sum_{a< b}\sigma^a\sigma^b+\beta h\sum_a\sigma^a\right]
@@ Line 205: / Line 205: @@
 &=& n^2 \lambda_0 q_0 - n(1-x)\lambda_1 q_1 - nx \lambda_0 q_0.
 \end{eqnarray*}
+이때 $n^2$에 비례하는 항들은 $\lim_{n\to 0} G/n$에서 사라질 것이므로 무시한다.
+$G(Q, \lambda)$의 마지막 항을 생각해보면
+\begin{eqnarray*}
+\sum_{a<b} \lambda^{ab} \sigma^a \sigma^b &=& \sum_{a \neq b} \frac{1}{2} \lambda^{ab} \sigma^a \sigma^b\\
+&=& \sum_{ab} \frac{1}{2} \lambda^{ab} \sigma^a \sigma^b - \frac{1}{2} \sum_a \lambda^{aa} \left(\sigma^a \right)^2\\
+&=& \sum_{ab} \frac{1}{2} \lambda^{ab} \sigma^a \sigma^b - \frac{1}{2}n \lambda_1\\
+&=& \frac{1}{2} \lambda_0 \sum_{a=1}^n \sum_{b=1}^n \lambda^{ab} \sigma^a \sigma^b + \frac{1}{2} (\lambda_1-\lambda_0) \sum_{\text{same group}} \sigma^a \sigma^b - \frac{1}{2}n \lambda_1
+\end{eqnarray*}
+이므로,
+\begin{eqnarray*}
+\exp\left( \sum_{a<b} \lambda^{ab} \sigma^a \sigma^b + \beta h \sum_a \sigma^a \right)
+&\approx& \exp\left( - \frac{1}{2}n \lambda_1 + \beta h \sum_{a=1}^n \sigma^a \right) \exp\left(\frac{1}{2}\lambda_0 \sum_{a=1}^n \sum_{b=1}^n \sigma^a \sigma^b \right) \left\{ \exp \left[ \frac{1}{2} (\lambda_1-\lambda_0) \sum_{a=1}^x \sum_{b=1}^x \sigma^a \sigma^b \right] \right\}^y\\
+&=& \exp\left( - \frac{1}{2}n \lambda_1 + \beta h \sum_{a=1}^n \sigma^a \right) \exp\left[\frac{1}{2}\lambda_0 \left( \sum_{a=1}^n \sigma^a \right)^2 \right] \left\{ \exp \left[ \frac{1}{2} (\lambda_1-\lambda_0) \left( \sum_{a=1}^x \sigma^a \right)^2 \right] \right\}^y\\
+&=& \exp\left( - \frac{1}{2}n \lambda_1 + \beta h \sum_{a=1}^n \sigma^a \right) \left[ \int dz_0 \sqrt{\frac{\lambda_0}{2\pi}} \exp\left(-\frac{\lambda_0}{2}z_0^2 + \lambda_0 z_0 \sum_{a=1}^n \sigma^a \right) \right] \left\{ \exp \left[ \frac{1}{2} (\lambda_1-\lambda_0) \left( \sum_{a=1}^x \sigma^a \right)^2 \right] \right\}^y\\
+&=& \exp\left( - \frac{1}{2}n \lambda_1 + \beta h \sum_{a=1}^n \sigma^a \right) \left[ \int dz_0 \sqrt{\frac{\lambda_0}{2\pi}} \exp\left(-\frac{\lambda_0}{2}z_0^2 + \lambda_0 z_0 \sum_{a=1}^n \sigma^a \right) \right] \left\{ \int dz_1 \sqrt{\frac{\delta}{2\pi}} \exp\left(-\frac{\delta}{2}z_1^2 + \delta z_1 \sum_{a=1}^x \sigma^a \right) \right\}^y\\
+&=& \exp\left( - \frac{1}{2}n \lambda_1 + \beta h \sum_{a=1}^n \sigma^a \right) \left[ \int dz_0 \sqrt{\frac{1}{2\pi}} \exp\left(-\frac{1}{2}z_0^2 + \sqrt{\lambda_0} z_0 \sum_{a=1}^n \sigma^a \right) \right] \left\{ \int dz_1 \sqrt{\frac{1}{2\pi}} \exp\left(-\frac{1}{2}z_1^2 + \sqrt{\delta} z_1 \sum_{a=1}^x \sigma^a \right) \right\}^y
+\end{eqnarray*}
+로서 이때 $\delta \equiv \lambda_1 - \lambda_0$이다. 식을 정리해서 적어보면,
+\begin{eqnarray*}
+\exp\left( \sum_{a<b} \lambda^{ab} \sigma^a \sigma^b + \beta h \sum_a \sigma^a \right)
+&\approx& \exp\left( - \frac{1}{2}n \lambda_1 + \beta h \sum_{a=1}^n \sigma^a \right) \left[ \int Dz_0 Dz_1 \cdots Dz_y \exp\left( \sqrt{\lambda_0} z_0 \sum_{a=1}^n \sigma^a \right) \exp\left(\sqrt{\delta} z_1 \sum_{a_1=1}^x \sigma^{a_1} \right) \cdots \exp\left(\sqrt{\delta} z_y \sum_{a_y=1}^x \sigma^{a_y} \right) \right]\\
+&\approx& \exp\left( - \frac{1}{2}n \lambda_1 \right) \int Dz_0 Dz_1 \cdots Dz_y \exp \left[ \left( \sqrt{\lambda_0} z_0 + \sqrt{\delta} z_1 + \beta h \right) \sum_{a_1=1}^x \sigma^a \right] \cdots \exp \left[ \left( \sqrt{\lambda_0} z_0 + \sqrt{\delta} z_y + \beta h \right) \sum_{a_y=1}^x \sigma^a \right].
+\end{eqnarray*}
+$\left\{ \sigma^1, \ldots, \sigma^n \right\}$의 모든 조합에 대해 더하는 $\text{Tr}$ 연산을 취해보자.
+\begin{eqnarray*}
+\text{Tr} \exp\left( \sum_{a<b} \lambda^{ab} \sigma^a \sigma^b + \beta h \sum_a \sigma^a \right)
+&\approx& \exp\left( - \frac{1}{2}n \lambda_1 \right) \int Dz_0 Dz_1 \cdots Dz_y \left[ 2\cosh \left( \sqrt{\lambda_0} z_0 + \sqrt{\delta} z_1 + \beta h \right) \right]^x \cdots \left[ 2\cosh \left( \sqrt{\lambda_0} z_0 + \sqrt{\delta} z_y + \beta h \right) \right]^x\\
+&=& \exp\left( - \frac{1}{2}n \lambda_1 \right) \int Dz_0 \left\{ \int Dz_1 \left[ 2\cosh \left( \sqrt{\lambda_0} z_0 + \sqrt{\delta} z_1 + \beta h \right) \right]^x \right\}^y.
+\end{eqnarray*}
+모두 합쳐서 적어보면,
+\begin{eqnarray*}
+G\left( Q^{\text{1RSB}}, \lambda^{\text{1RSB}} \right) &=& -\frac{1}{4} \beta^2 J^2 n + \frac{1}{4} \beta^2 J^2 n \left[ (1-x)q_1^p + xq_0^p \right] - \frac{n}{2} \left[ (1-x)q_1 \lambda_1 + x q_0 \lambda_0 \right] + \frac{1}{2}\lambda_1 n - \ln \left\{ \int Dz_0 \left\{ \int Dz_1 \left[ 2\cosh \left( \sqrt{\lambda_0} z_0 + \sqrt{\delta} z_1 + \beta h \right) \right]^x \right\}^y \right\}.
+\end{eqnarray*}
+$q_0=\lambda_0=0$이라고 하자. $\int Dz = 1$이므로,
+\begin{eqnarray*}
+G\left( Q^{\text{1RSB}}, \lambda^{\text{1RSB}} \right) &=& -\frac{1}{4} \beta^2 J^2 n + \frac{1}{4} \beta^2 J^2 n (1-x)q_1^p - \frac{n}{2} (1-x)q_1 \lambda_1 + \frac{1}{2}\lambda_1 n - \ln \left\{ \int Dz_1 \left[ 2\cosh \left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x \right\}^y\\
+&=& -\frac{1}{4} \beta^2 J^2 n + \frac{1}{4} \beta^2 J^2 n (1-x)q_1^p - \frac{n}{2} (1-x)q_1 \lambda_1 + \frac{1}{2}\lambda_1 n - y \ln \left\{ \int Dz_1 \left[ 2\cosh \left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x \right\}.
+\end{eqnarray*}
+$\partial G/\partial q_1 = 0$로부터
+$$\lambda_1 = \frac{1}{2} \beta^2 J^2 p q_1^{p-1}$$
+을 얻고,
+\begin{eqnarray*}
+= \frac{\partial G}{\partial \lambda_1} &=& -\frac{n}{2} (1-x)q_1 + \frac12 n - \frac{y \int Dz_1 ~x \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^{x-1} \left[ 2\sinh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right] \frac{z_1}{2\sqrt{\lambda_1}}}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+&=& -\frac{n}{2} (1-x)q_1 + \frac12 n - \frac{n}{2\sqrt{\lambda_1}} \frac{ \int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x \tanh\left( \sqrt{\lambda_1} z_1 + \beta h \right) z_1}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+&=& -\frac{n}{2} (1-x)q_1 + \frac12 n + \frac{n}{2\sqrt{\lambda_1}} \frac{ \int \frac{dz_1}{\sqrt{2\pi}} \left( \frac{d}{dz_1} e^{-z_1^2/2} \right) \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x \tanh\left( \sqrt{\lambda_1} z_1 + \beta h \right)}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+&=& -\frac{n}{2} (1-x)q_1 + \frac12 n - \frac{n}{2\sqrt{\lambda_1}} \frac{ \int \frac{dz_1}{\sqrt{2\pi}} e^{-z_1^2/2} \left\{ \frac{d}{dz_1} \left[ 2^x\cosh^x\left( \sqrt{\lambda_1} z_1 + \beta h \right) \tanh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right] \right\}}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+&=& -\frac{n}{2} (1-x)q_1 + \frac12 n - \left(\frac{n}{2}\right) \frac{ \int Dz_1 \left[ x 2^x\cosh^x\left( \sqrt{\lambda_1} z_1 + \beta h \right) \tanh^2\left( \sqrt{\lambda_1} z_1 + \beta h \right) + 2^x\cosh^x\left( \sqrt{\lambda_1} z_1 + \beta h \right) \text{sech}^2\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+&=& -\frac{n}{2} (1-x)q_1 + \frac12 n - \left(\frac{n}{2}\right) \frac{ \int Dz_1 2^x\cosh^x\left( \sqrt{\lambda_1} z_1 + \beta h \right) \left[ x \tanh^2\left( \sqrt{\lambda_1} z_1 + \beta h \right) + \text{sech}^2\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+\end{eqnarray*}
+임의의 $z$에 대해 $\tanh^2 z + \text{sech}^2 z = 1$임을 이용하면 $x\tanh^2 z + \text{sech}^2 z = 1- (1-x)\tanh^2 z$이므로
+\begin{eqnarray*}
+(1-x)q_1 &=& 1 - \frac{ \int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x \left[ 1-(1-x) \tanh^2\left( \sqrt{\lambda_1} z_1 + \beta h \right)\right]}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+&=& (1-x) \frac{ \int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x \tanh^2\left( \sqrt{\lambda_1} z_1 + \beta h \right)}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+\therefore q_1 &=& \frac{ \int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x \tanh^2\left( \sqrt{\lambda_1} z_1 + \beta h \right)}{\int Dz_1 \left[ 2\cosh\left( \sqrt{\lambda_1} z_1 + \beta h \right) \right]^x}\\
+\end{eqnarray*}
+그러므로 주어진 역온도 $\beta$에 놓여있는 $p$-스핀 유리라면 먼저 적당한 $x \in (0,1]$를 택해 위의 두 방정식을 자기일관된 방식으로 풀어주는 $(q_1, \lambda_1)$을 구하고, 이를 아래 식에 대입하여
+$$\lim_{n\to 0}\frac{1}{n}G\left( x \right) = -\frac{1}{4} \beta^2 J^2 + \frac{1}{4} \beta^2 J^2 (1-x)q_1^p - \frac{1}{2} (1-x)q_1 \lambda_1 + \frac{1}{2}\lambda_1 - \frac{1}{x} \ln \left\{ \int Dz_1 \left[ 2\cosh \left( \sqrt{\lambda_1} z_1 \right) \right]^x \right\}$$
+값을 구한다. 편의상 $h=0$으로 놓았다. 이제 $x$를 바꾸어가며 위 식의 최솟값을 찾고 그 때를 $x_s$라고 하자. 계산 결과는 아래 그림처럼 주어진다.
+{{:물리:pspin_rsb.png?400|}}
+$x_s=0$은 복제 대칭해에 해당한다. $p=2$인 [[물리:셰링턴-커크패트릭_모형|셰링턴-커크패트릭 모형]]에서는 알려진 대로 $\beta=1$에서 복제 대칭성이 깨지는 것을 볼 수 있다.
+$p\to \infty$에서 [[물리:무작위_에너지_모형|무작위 에너지 모형]]을 따라 다음과 같은 결과가 예상된다:
+$$x_s = \left\{
+\begin{array}{ll}
+& \text{if }\beta<2\sqrt{\ln 2}\\
+\sqrt{\ln2}/\beta & \text{otherwise.}
+\end{array}
+\right.$$
 ======함께 보기======
   * [[물리:무작위_에너지_모형|무작위 에너지 모형]]
   * [[물리:셰링턴-커크패트릭_모형|셰링턴-커크패트릭 모형]]
+  * [[물리:구면_p-스핀_유리_모형|구면 $p$-스핀 유리 모형]]
 ======참고문헌======
@@ Line 214: / Line 284: @@
   * Haiping Huang, //Statistical Mechanics of Neural Networks// (Springer, Singapore, 2021).
   * E. Gardner, //Spin glasses with $p$-spin interactions//, Nuclear Physics B257, 747 (1985).
+  * Florent Krzakala and Lenka Zdeborová, //Performance of simulated annealing in p-spin glasses//, [[https://doi.org/10.1088/1742-6596/473/1/012022|J. Phys. Conf. Ser. 473, 012022 (2013)]].
+  * Stefan Boettcher and Ginger E. Lau, //Ground States of the Mean-Field Spin Glass with 3-Spin Couplings//, [[https://doi.org/10.1103/j159-lpfx|Phys. Rev. Lett. 135, 037402 (2025)]].
+  * Viviane M de Oliveira and J F Fontanari, //Replica analysis of the p-spin interaction Ising spin-glass model//, [[https://doi.org/10.1088/0305-4470/32/12/004|J. Phys. A: Math. Gen. 32 2285]].