Differences
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| Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
| 물리:평균장_이론 [2016/08/17 11:21] – [참고문헌] admin | 물리:평균장_이론 [2018/05/16 09:52] – admin | ||
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| Line 44: | Line 44: | ||
| $$e^{ax^2/ | $$e^{ax^2/ | ||
| 이므로 $a=\beta J$과 $x = \frac{1}{\sqrt{N}} \sum_i S_i$를 대입하면 | 이므로 $a=\beta J$과 $x = \frac{1}{\sqrt{N}} \sum_i S_i$를 대입하면 | ||
| - | $$e^{\frac{\beta J}{2} \frac{1}{N} \left( \sum_i S_i \right)^2} = \frac{\beta J N}{2\pi} \int_{-\infty}^\infty dm~ e^{-N\beta J m^2/2 + \beta J m \sum_i S_i$$ | + | $$e^{\frac{\beta J}{2} \frac{1}{N} \left( \sum_i S_i \right)^2} = \frac{\beta J N}{2\pi} \int_{-\infty}^\infty dm~ e^{-N\beta J m^2/2 + \beta J m \sum_i S_i}$$ |
| 이다. 여기에 대각합을 걸면 $\sum_i S_i$에만 걸리므로, | 이다. 여기에 대각합을 걸면 $\sum_i S_i$에만 걸리므로, | ||
| $$Z = \mbox{Tr} \sqrt{\frac{\beta J N}{2\pi}} \int_{-\infty}^\infty dm~ e^{-N\beta Jm^2/2 + \beta (Jm+h) \sum_i S_i}$$ | $$Z = \mbox{Tr} \sqrt{\frac{\beta J N}{2\pi}} \int_{-\infty}^\infty dm~ e^{-N\beta Jm^2/2 + \beta (Jm+h) \sum_i S_i}$$ | ||