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--- 수학:허바드-스트라토노비치_변환 [2026/02/28 15:25] – [개요] admin
+++ 수학:허바드-스트라토노비치_변환 [2026/04/14 13:02] (current) – [다변수 함수] admin
@@ Line 9: / Line 9: @@
 $$e^{-ax^2/2} = \sqrt{\frac{aN}{2\pi}} \int_{-\infty}^\infty dm~ e^{-Nam^2/2 - i\sqrt{N}amx}.$$
-======다변수함수에서의 응용======
+======1차원======
+=====실수 변수=====
+$$-\frac{a}{2}x^2+ bx = -\frac{a}{2}\left(x-\frac{b}{a}\right)^2 + \frac{b^2}{2a}$$
+이므로
+\begin{eqnarray*}
+\int_{-\infty}^{\infty} e^{-(a/2)x^2+bx} dx &=& e^{b^2/(2a)} \int_{-\infty}^{\infty} e^{-(a/2)[x-(b/a)]^2} dx\\
+&=& e^{b^2/(2a)} \int_{-\infty-b/a}^{\infty-b/a} e^{-(a/2)y^2} dy\\
+&=& \sqrt{\frac{2\pi}{a}} e^{b^2/(2a)}.
+\end{eqnarray*}
+=====연산자=====
+\begin{eqnarray*}
+\hat{R} &=& \sqrt{\frac{a}{2\pi}} \int_{-\infty}^{\infty} \exp\left(-\frac{a}{2}x^2 \hat{1} + \hat{B}x\right) dx\\
+&=& \sum_{n=0}^{\infty} \frac{{\hat{B}}^n}{n!} \sqrt{\frac{a}{2\pi}} \int_{-\infty}^{\infty} x^n \exp\left( -\frac{a}{2} x^2 \right) dx.
+\end{eqnarray*}
+여기에서 $n$이 홀수인 경우는 적분에 기여하지 않으며, $n=2m$일 때
+$$\int_{-\infty}^{\infty} x^{2m} \exp\left( -\frac{a}{2} x^2 \right) dx = \left( \frac{2}{a} \right)^{m+\frac12} \Gamma\left( m +\frac12 \right)$$
+그리고 르장드르 이중 공식(Legendre duplication formula)에 의해
+$$\Gamma\left(m +\frac12 \right) = 2^{1-2m} \sqrt{\pi} \frac{\Gamma(2m)}{\Gamma(m)}$$
+이므로
+$$\hat{R} = \sum_{m=0}^{\infty} \left( \frac{\hat{B}^2}{2a} \right)^m \frac{1}{m!} = \exp\left( \frac{1}{2a} \hat{B}^2 \right).$$
+======이중 적분======
+=====복소 적분변수=====
+$dz^\ast dz = 2i~dx~dy$임을 이용한다. $\left[ \hat{B}, \hat{B}^\dagger \right]=0$이라면
+\begin{eqnarray*}
+\hat{R} &=& \frac{a}{2\pi i} \int_{\mathbb{C}} dz^\ast dz \exp\left( -z^\ast a z + \hat{B} z^\ast + \hat{B}^\dagger z \right)\\
+&=& \frac{2ia}{2\pi i} \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy \exp\left[ -ax^2 - ay^2 + \hat{B}(x-iy) + \hat{B}^\dagger (x+iy) \right]\\
+&=& \frac{a}{\pi} \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy \exp\left[ -ax^2 - ay^2 + x\left( \hat{B}^\dagger + \hat{B} \right) + iy \left( \hat{B}^\dagger  - \hat{B} \right) \right]\\
+&=& \frac{a}{\pi} \int_{-\infty}^{\infty} dx \exp\left[ -ax^2 + x\left( \hat{B}^\dagger + \hat{B} \right) \right] \int_{-\infty}^{\infty} dy \exp\left[ - ay^2 + iy \left( \hat{B}^\dagger  - \hat{B} \right) \right]\\
+&=& \exp\left[ \frac{1}{4a} \left( \hat{B}^\dagger + \hat{B} \right)^2 \right] \exp\left[ -\frac{1}{4a} \left( \hat{B}^\dagger - \hat{B} \right)^2 \right]\\
+&=& \exp\left( \hat{B}^\dagger \hat{B} /a \right).
+\end{eqnarray*}
+$\hat{B}$가 일반적인 연산자라면
+$$\exp\left(\hat{B}^2/a\right) = \frac{a}{2\pi i}\int dz^\ast dz \exp\left[ -z^\ast a z + \hat{B} (z^\ast + z) \right].$$
+=====그라스만 변수=====
+[[수학:그라스만_대수|그라스만 변수]]의 경우
+\begin{eqnarray*}
+&&\frac{1}{a} \int d\eta^\ast \int d\eta \exp\left( -\eta^\ast a \eta + \hat{B}^\dagger \eta + \eta^\ast \hat{B} \right)\\
+&=& \frac{1}{a} \int d\eta^\ast \int d\eta \left[ 1-\eta^\ast a \eta + \hat{B}^\dagger \eta + \eta^\ast \hat{B} + \frac12 \left( \hat{B}^\dagger \eta \eta^\ast \hat{B} + \eta^\ast \hat{B} \hat{B}^\dagger \eta \right) \right]\\
+&=& \left[ 1 + \frac{1}{2a} \left(\hat{B}^\dagger \hat{B} - \hat{B} \hat{B}^\dagger \right) \right]\\
+&=& \exp \left[ \frac{1}{2a} \left( \hat{B}^\dagger \hat{B} - \hat{B} \hat{B}^\dagger \right) \right].
+\end{eqnarray*}
+만일 $\hat{B}$와 $\hat{B}^\dagger$도 [[수학:그라스만_대수|그라스만 변수]] $\beta$와 $\beta^\ast$로 치환하면
+$$\exp\left( \beta^\ast \beta a \right) = \frac{1}{a} \int d\eta^\ast \int d\eta \exp \left( -\eta^\ast a \eta + \beta^\ast \eta + \eta^\ast \beta \right).$$
+======다변수 함수======
+$\mathbb{A}$가 $n\times n$ 대칭행렬로서 양의 정부호(positive definite) 행렬이라고 하자. $\mathbf{x}$와 $\mathbf{b}$는 $n$차원의 실수 열벡터들이다. 유도는 [[수학:윅의_정리|다차원 가우스 함수의 적분]]을 참조.
+$$\exp\left( \frac12 \mathbf{b}^\intercal \mathbb{A}^{-1} \mathbf{b} \right) = \sqrt{\frac{\det \mathbb{A}}{(2\pi)^n}} \int d^n x \exp\left( -\frac12 \mathbf{x}^\intercal \mathbb{A} \mathbf{x} + \mathbf{b}^\intercal \mathbf{x} \right).$$
 ======함께 보기======
-[[:물리:평균장 이론]]
+  * [[:물리:평균장 이론]]
+  * [[:물리:무작위장 이징 모형]]
-[[:물리:무작위장 이징 모형]]
+======참고문헌======
+  * Krzysztof Byczuk and Paweł Jakubczyk, //Generalized Gaussian integrals with application to the Hubbard–Stratonovich transformation//, [[https://doi.org/10.1119/5.0141045|Am. J. Phys. 91, 840 (2023)]].