물리:2차원_이징_모형

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물리:2차원_이징_모형 [2022/01/16 21:09] – [반교환자] admin물리:2차원_이징_모형 [2023/09/05 15:46] (current) – external edit 127.0.0.1
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 =====자유에너지===== =====자유에너지=====
 다시 작용을 적어보면 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} \begin{eqnarray}
-&=& \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 +0 & \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 & -\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 & \cdots & 0 & 0 & \frac{1}{4\Delta} & & \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 & 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 & 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} \end{eqnarray}
-며 이것을 지수함수 위에 올려 적분하는 분배함수는 +여기에서는 중심미분을 사용했지만 앞에서와 일관되게 $\partial_1 \varphi \approx (\varphi_{j+1}-\varphi_j)/\Delta$를 사용해도 마찬가지다. 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 
 +$$Z = \int D\tilde{\varphi} \exp(-S) = \text{Pfaff}\Lambda = \prod_{n=0}^{N-1} \left( \frac{1}{\Delta^2} \sin^2 k_n \Delta + m^2 \right)^{1/2} \xrightarrow[\Delta \to 0]{} \prod_{n=0}^{N-1} \left( k_n^2 + m^2 \right)^{1/2}$$ 
 + 
 +2차원 문제로 돌아와서 $k^2=k_x^2 +k_y^2$으로 일반화하고 $Z$에 로그를 취하면
 \begin{eqnarray} \begin{eqnarray}
--\beta &=& \ln Z \sum_i \ln \text{Pfaff} \Lambda_i\\ +-\beta &=& \frac{1}{L^2} \ln Z 
-&\approx& \int d^2 \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 \ln (p^2 + m^2)\\ +\frac{1}{8\pi^2} \int dk ~2\pi \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} \end{eqnarray}
-$\to 0$에서 $-\beta \propto m^2 \ln m^2$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($\to 0$)에서 $-\beta \propto m^2 \ln m^2$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.
  
  
 ======참고문헌====== ======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   * //Ising Field Theory// by A. Zamolodchikov, https://www.weizmann.ac.il/complex/falkovich/courses
   * V. N. Plechko, J. Phys. Studies, 1, 554 (1997).   * 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).
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