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물리:2차원_이징_모형 [2022/01/16 21:09] – [반교환자] admin | 물리:2차원_이징_모형 [2022/01/17 23:19] – [자유에너지] admin | ||
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다시 작용을 적어보면 | 다시 작용을 적어보면 | ||
\begin{eqnarray} | \begin{eqnarray} | ||
- | S &=& \frac{1}{2} \int d^2 x (\varphi | + | S &=& \frac{1}{2\pi} \int d^2 x (\varphi \bar{\partial} \varphi |
- | &=& \frac{1}{2} \int d^2 x \begin{pmatrix} \varphi & \bar{\varphi} \end{pmatrix} | + | &=& \frac{1}{2\pi} \int d^2 x \begin{pmatrix} \varphi & \bar{\varphi} \end{pmatrix} |
\begin{pmatrix} | \begin{pmatrix} | ||
\bar{\partial} & -im/2 \\ | \bar{\partial} & -im/2 \\ | ||
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\end{pmatrix} | \end{pmatrix} | ||
\begin{pmatrix} \varphi \\ \bar{\varphi} \end{pmatrix}\\ | \begin{pmatrix} \varphi \\ \bar{\varphi} \end{pmatrix}\\ | ||
- | & | ||
- | \frac{1}{2} \sum_x \begin{pmatrix} \varphi & \bar{\varphi} \end{pmatrix} | ||
- | \begin{pmatrix} | ||
- | \bar{\partial} & -im/2 \\ | ||
- | im/2 & \partial | ||
- | \end{pmatrix} | ||
- | \begin{pmatrix} \varphi \\ \bar{\varphi} \end{pmatrix}\\ | ||
- | &=& | ||
- | \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 & 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}\\ | ||
- | & | ||
\end{eqnarray} | \end{eqnarray} | ||
이며 이것을 지수함수 위에 올려 적분하는 분배함수는 | 이며 이것을 지수함수 위에 올려 적분하는 분배함수는 | ||
- | $$Z = \int D\varphi D\bar{\varphi} \exp(-S) = \prod_i \text{Pfaff} \Lambda_i.$$ | + | $$Z = \int D\varphi D\bar{\varphi} \exp(iS) = \prod_i \text{Pfaff} \Lambda_i.$$ |
- | 그러므로 여기에 로그를 취하면 | + | 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} | ||
+ | && | ||
+ | \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} \varphi_0 & \varphi_1 & \cdots & \varphi_{N-1} & \bar{\varphi}_0 & \bar{\varphi}_1 & \cdots & \bar{\varphi}_{N-1} \end{pmatrix} | ||
+ | \left( \begin{array}{cccccc|cccccc} | ||
+ | 0 & \frac{1}{4\Delta} & 0 & \cdots & 0 & -\frac{1}{4\Delta} & -\frac{im}{2} & 0 & 0 & 0 & \cdots & 0\\ | ||
+ | -\frac{1}{4\Delta} & 0 & \frac{1}{4\Delta} & \cdots & 0 & 0 & 0 & -\frac{im}{2} & 0 & 0 & \cdots & 0\\ | ||
+ | \vdots & & & \ddots & & & \vdots & & & & \ddots & \vdots \\ | ||
+ | \frac{1}{4\Delta} & 0 & 0 & \cdots & -\frac{1}{4\Delta} & 0 & 0 & 0 & 0 & 0 & \cdots & -\frac{im}{2}\\\hline | ||
+ | \frac{im}{2} & 0 & 0 & 0 & \cdots & 0 & 0 & \frac{1}{4\Delta} & 0 & \cdots & 0 & -\frac{1}{4\Delta}\\ | ||
+ | 0 & \frac{im}{2} & 0 & 0 & \cdots & 0 & -\frac{1}{4\Delta} & 0 & \frac{1}{4\Delta} & \cdots & 0 & 0\\ | ||
+ | \vdots & & & \ddots & & \vdots & \vdots & & & & \ddots & \vdots \\ | ||
+ | 0 & 0 & 0 & 0 & \cdots & -\frac{im}{2} & \frac{1}{4\Delta} & 0 & 0 & \cdots & -\frac{1}{4\Delta} & 0 | ||
+ | \end{array}\right) | ||
+ | \begin{pmatrix} \varphi_0 \\ \varphi_1 \\ \vdots \\ \varphi_{N-1} \\ \bar{\varphi}_0 \\ \bar{\varphi}_1 \\ \vdots \\ \bar{\varphi}_{N-1} \end{pmatrix}. | ||
+ | \end{eqnarray} | ||
+ | 여기에서는 중심미분을 사용했지만 앞에서와 일관되게 $\partial_1 \varphi \approx (\varphi_{j+1}-\varphi_j)/ | ||
+ | |||
+ | 2차원 문제로 돌아와서 $Z$에 로그를 취하면 | ||
\begin{eqnarray} | \begin{eqnarray} | ||
- | -\beta | + | -\beta |
- | & | + | \propto |
- | &=& \int dp ~2\pi p \ln (p^2 + m^2)\\ | + | = \frac{1}{8\pi^2} |
- | &=& \pi [(p^2+m^2) \ln(p^2+m^2) - p^2]. | + | & |
+ | = \frac{1}{8\pi} m^2 \ln(m^2) + p^2 \ln(m^2) + O(p^4). | ||
\end{eqnarray} | \end{eqnarray} | ||
- | $p \to 0$에서 $-\beta | + | 장파장 영역($p \to 0$)에서 $-\beta |