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--- 물리:2차원_이징_모형 [2022/01/16 21:09] – [반교환자] admin
+++ 물리:2차원_이징_모형 [2025/06/12 17:00] (current) – [자유에너지] admin
@@ Line 282: / Line 282: @@
 =====자유에너지=====
 다시 작용을 적어보면
+$$S = \frac{1}{2\pi} \int d^2 x (\varphi \bar{\partial} \varphi + \bar{\varphi} \partial \bar{\varphi} + im\bar{\varphi} \varphi)$$
+이며, 여기에서 미분연산자는 반대칭행렬로 나타낼 수 있음에 유의할 것. 예를 들어 간격 $\Delta$로 $N$개의 입자가 늘어서 있는 길이 $L=N\Delta$의 1차원 계를 생각한다면
 \begin{eqnarray}
-S &=& \frac{1}{2} \int d^2 x (\varphi \partial \bar{\varphi} + \bar{\varphi} \partial \bar{\varphi} + im\bar{\varphi} \varphi)\\
+&&\int dx \left( \frac{1}{2}\varphi \partial_1 \varphi + \frac{1}{2}\bar{\varphi} \partial_1 \bar{\varphi} + im\bar{\varphi} \varphi \right)
-&=& \frac{1}{2} \int d^2 x \begin{pmatrix} \varphi & \bar{\varphi} \end{pmatrix}
+\approx \sum_{j=0}^{N-1} \frac{1}{2} \varphi_j \left( \frac{\varphi_{j+1} - \varphi_{j-1}}{2\Delta} \right) + \frac{1}{2} \bar{\varphi}_j \left( \frac{\bar{\varphi}_{j+1} - \bar{\varphi}_{j-1}}{2\Delta} \right) + im \bar{\varphi}_j \varphi_j\\
-\begin{pmatrix}
+&=& \begin{pmatrix} \varphi_0 & \varphi_1 & \cdots & \varphi_{N-1} & \bar{\varphi}_0 & \bar{\varphi}_1 & \cdots & \bar{\varphi}_{N-1} \end{pmatrix}
-\bar{\partial} & -im/2 \\
+\left( \begin{array}{cccccc|cccccc}
-im/2 & \partial
+& \frac{1}{4\Delta} & 0 & \cdots & 0 & -\frac{1}{4\Delta} & -\frac{im}{2} & 0 & 0 & 0 & \cdots & 0\\
-\end{pmatrix}
+-\frac{1}{4\Delta} & 0 & \frac{1}{4\Delta} & \cdots & 0 & 0 & 0 & -\frac{im}{2} & 0 & 0 & \cdots & 0\\
-\begin{pmatrix} \varphi \\ \bar{\varphi} \end{pmatrix}\\
+\vdots & & & \ddots & & & \vdots & & & & \ddots & \vdots \\
-&\sim&
+\frac{1}{4\Delta} & 0 & 0 & \cdots & -\frac{1}{4\Delta} & 0 & 0 & 0 & 0 & 0 & \cdots & -\frac{im}{2}\\\hline
-\frac{1}{2} \sum_x \begin{pmatrix} \varphi & \bar{\varphi} \end{pmatrix}
+\frac{im}{2} & 0 & 0 & 0 & \cdots & 0 & 0 & \frac{1}{4\Delta} & 0 & \cdots & 0 & -\frac{1}{4\Delta}\\
-\begin{pmatrix}
+ & \frac{im}{2} & 0 & 0 & \cdots & 0 & -\frac{1}{4\Delta} & 0 & \frac{1}{4\Delta} & \cdots & 0 & 0\\
-\bar{\partial} & -im/2 \\
+\vdots & & & \ddots & & \vdots & \vdots & & & & \ddots & \vdots \\
-im/2 & \partial
+& 0 & 0 & 0 & \cdots & -\frac{im}{2} & \frac{1}{4\Delta} & 0 & 0 & \cdots & -\frac{1}{4\Delta} & 0
-\end{pmatrix}
+\end{array}\right)
-\begin{pmatrix} \varphi \\ \bar{\varphi} \end{pmatrix}\\
+\begin{pmatrix} \varphi_0 \\ \varphi_1 \\ \vdots \\ \varphi_{N-1} \\ \bar{\varphi}_0 \\ \bar{\varphi}_1 \\ \vdots \\ \bar{\varphi}_{N-1} \end{pmatrix}\\
-&=&
+&=& -\frac{1}{2} \sum_{\alpha=0}^{2N-1} \sum_{\beta=0}^{2N-1} \tilde{\varphi}_\alpha \Lambda_{\alpha\beta} \tilde{\varphi}_\beta.
-\frac{1}{2}
-\begin{pmatrix} \varphi_{x_1} & \bar{\varphi}_{x_1} & \varphi_{x_2} & \bar{\varphi}_{x_2} & \cdots \end{pmatrix}
-\begin{pmatrix}
-\bar{\partial} & -im/2 & 0 & 0 & \cdots \\
-im/2 & \partial & 0 & 0 & \cdots\\
-& 0 & \bar{\partial} & -im/2 & \cdots\\
- & 0 & im/2 & \partial & \cdots\\
-\vdots & \vdots & \vdots & \vdots & \ddots
-\end{pmatrix}
-\begin{pmatrix} \varphi_{x_1} \\ \bar{\varphi}_{x_1} \\ \varphi_{x_2} \\ \bar{\varphi}_{x_2} \\ \vdots \end{pmatrix}\\
-&\rightarrow& \frac{1}{2} \Psi(\varphi) \Lambda \Psi(\varphi)
 \end{eqnarray}
-이며 이것을 지수함수 위에 올려 적분하는 분배함수는
+여기에서는 중심미분을 사용했지만 앞에서와 일관되게 $\partial_1 \varphi \approx (\varphi_{j+1}-\varphi_j)/\Delta$를 사용해도 마찬가지이다. 편의상 주기적 경계조건을 가정했다.
-$$Z = \int D\varphi D\bar{\varphi} \exp(-S) = \prod_i \text{Pfaff} \Lambda_i.$$
-그러므로 여기에 로그를 취하면
+이것을 지수함수 위에 올려 적분하는 분배함수는
+$$Z = \int D\tilde{\varphi} \exp(-S) = \text{Pfaff}\Lambda = \prod_{n=0}^{N-1} \left( \frac{1}{\Delta^2} \sin^2 k_n \Delta + m^2 \right)^{1/2} \xrightarrow[\Delta \to 0]{} \prod_{n=0}^{N-1} \left( k_n^2 + m^2 \right)^{1/2}$$
+차원 문제로 돌아와서 $k^2=k_x^2 +k_y^2$으로 일반화하고 $Z$에 로그를 취하면
 \begin{eqnarray}
--\beta F &=& \ln Z = \sum_i \ln \text{Pfaff} \Lambda_i\\
+-\beta f &=& \frac{1}{L^2} \ln Z
-&\approx& \int d^2 p \ln (p^2 + m^2)\\
+\propto \frac{1}{2} \int \frac{d^2 k}{(2\pi)^2} \ln (k^2 + m^2)
-&=& \int dp ~2\pi p \ln (p^2 + m^2)\\
+= \frac{1}{8\pi^2} \int dk ~2\pi k \ln (k^2 + m^2)\\
-&=& \pi [(p^2+m^2) \ln(p^2+m^2) - p^2].
+&=& \frac{1}{8\pi} [(k^2+m^2) \ln(k^2+m^2) - k^2]
+= \frac{1}{8\pi} m^2 \ln(m^2) + k^2 \ln(m^2) + O(k^4).
 \end{eqnarray}
-$p \to 0$에서 $-\beta F \propto m^2 \ln m^2$이며 $m = 4(K-K_c)$이므로 $m$으로의 미분은 $K$로의 미분과 대응된다. 즉 $\frac{\partial^2 F}{\partial K^2} \propto \ln m^2$이 되어 비열이 임계점에서 로그 발산을 보인다.
+장파장 영역($k \to 0$)에서 $-\beta f \propto m^2 \ln m^2$이며 $m = 4(K-K_c)$이므로 $m$으로의 미분은 $K$로의 미분과 대응된다. 즉 $\frac{\partial^2 f}{\partial K^2} \propto -\ln m^2$이 되어 비열이 임계점에서 로그 발산을 보인다.
+=====함께 보기=====
+[[물리:이징 장론]]
 ======참고문헌======
-  * Robert Savit, //Duality in field theory and statistical systems//, Rev. Mod. Phys. 52, 453 (1980)
+  * Robert Savit, //Duality in field theory and statistical systems//, Rev. Mod. Phys. 52, 453 (1980).
   * //Ising Field Theory// by A. Zamolodchikov, https://www.weizmann.ac.il/complex/falkovich/courses
   * V. N. Plechko, J. Phys. Studies, 1, 554 (1997).
+  * https://gandhiviswanathan.wordpress.com/2018/10/31/exact-solution-of-the-2d-ising-model-via-grassmann-variables/
+  * J.M Carmona, A. Di Giacomoa, and B. Lucini, //A disorder analysis of the Ising model//, Phys. Lett. B, 485, 126 (2000).
+  * Massimo D’Elia and Luca Tagliacozzo, //Direct numerical computation of disorder parameters//, Phys. Rev. D 74, 114510 (2006).