2025-09-16
2025-09-16
Reflecting
Thermal and Statisical Physics I Homework
1.
\[\langle (\Delta m)^3 \rangle = \langle (m - \langle m \rangle)^3 \rangle\] \[= \langle m^3 - 3m^2\langle m \rangle + 3 m \langle m \rangle^2 - \langle m \rangle^3 \rangle\] \[= \langle m^3 \rangle - 3 \langle m^2 \rangle \langle m \rangle + 3 \langle m \rangle \langle m \rangle^2 - \langle m \rangle^3\] \[= \langle m^3 \rangle - 3 \langle m^2 \rangle \langle m \rangle + 2 \langle m \rangle^3\]What is \(\langle m^3 \rangle\)?
\[\langle m^3 \rangle = \frac{1}{i^3}\frac{d^3 \Phi(\theta)}{d\theta^3}\mid _{\theta=0}\] \[= i \frac{d^3}{d\theta^3}(pe^{i\theta} + qe^{-i\theta})^N\mid _{\theta=0}\]Let, \(f(\theta) = pe^{i\theta}+qe^{-i\theta}\)
\[= i \frac{d^2}{d\theta^2} Nf^{N-1}f'\mid _{\theta=0}\] \[= i\frac{d}{d\theta} (N(N-1)f^{N-2}f'^2 + Nf^{N-1}f'')\mid _{\theta=0}\] \[i[N(N-1)(N-2)f^{N-3}f'^3 + N(N-1)f^{N-2}2f'f'' + N(N-1)f^{N-2}f'f'' + Nf^{N-1}f''']_{\theta=0}\] \[f(0) = (p+q) = 1\] \[f'(\theta) = ipe^{i\theta} - iqe^{-i\theta}\] \[f'(0) = i(p-q)\] \[f''(\theta) = i^2 pe^{i\theta} + i^2qe^{-i\theta}\] \[f''(0) = -1\] \[f'''(\theta) = i^3pe^{i\theta} - i^3qe^{-i\theta}\] \[f'''(0) = i^3(p-q) = -i(p-q)\]Thus,
\[= i[N(N-1)(N-2)(i^3)(p-q)^3 + 2N(N-1)i(p-q)(-1)\] \[+ N(N-1)i(p-q)(-1) +N(-i)(p-q)]\] \[= N(N-1)(N-2)(p-q)^3 + 2N(N-1)(p-q) + N(N-1)(p-q) + N(p-q)\] \[= N(N-1)(N-2)(p-q)^3 +3N(N-1)(p-q) + N(p-q)\] \[= N(N-1)(N-2)(p-q)^3 + N(3N - 2)(p-q)\]Thus,
\[\langle (\Delta m)^3 \rangle = N(N-1)(N-2)(p-q)^3 + N(3N - 2)(p-q)\] \[-3[N(N-1)(p-q)^2 + N] \cdot N(p-q) + 2(N(p-q))^3\]\((p-q)^3\) 계수:
\[N(N-1)(N-2) - 3N^2(N-1) + 2N^3\] \[= (N^2-N)(N-2)-3N^3+3N^2 + 2N^3\] \[= N^3 - 2N^2 - N^2 + 2N - 3N^3 + 3N^2 + 2N^3\] \[=2N\]\((p-q)\) 계수:
\[N(3N-2) - 3N^2\] \[= 3N^2 - 2N - 3N^2\] \[= -2N\]Thus,
\[\langle (\Delta m)^3 \rangle = 2N(p-q)^3 - 2N(p-q)\] \[= - 2N(p-q)(1 - (p-q)^2)\] \[= -2N(p-q)((p+q)^2 - (p-q)^2)\] \[= -2N(p-q)(4pq)\] \[= -8Npq(p-q)\]2.
(가)
\[\Phi(0) = \langle 1 \rangle = 1\] \[\langle m \rangle = \frac{1}{i} \Phi'(\theta)\mid _{\theta=0} = \frac{1}{i}\Phi'(0)\] \[\langle m \rangle_{c} = \frac{1}{i} \frac{\Phi'(\theta)}{\Phi(\theta)}\mid _{\theta=0} = \frac{1}{i} \frac{\Phi'(0)}{\Phi(0)} = \frac{1}{i} \Phi'(0)\] \[\therefore ~~ \langle m \rangle = \langle m \rangle_{c}\]\[\langle m^2 \rangle = \frac{1}{i^2} \Phi''(0)\] \[\langle m^2 \rangle_{c} = \frac{1}{i^2} \frac{d}{d\theta}\frac{\Phi'(\theta)}{\Phi(\theta)}\mid _{\theta=0}\] \[= \frac{1}{i^2} \left( \frac{\Phi''(\theta)}{\Phi(\theta)} - \frac{\Phi'(\theta)^2}{\Phi(\theta)^2} \right)\mid _{\theta=0}\] \[=\frac{1}{i^2}(\Phi''(0) - \Phi'(0)^2)\] \[= \langle m^2 \rangle - \langle m \rangle^2\] \[\therefore ~~ \langle m^2 \rangle = \langle m^2 \rangle - \langle m \rangle^2\]
\[\langle m^3 \rangle = \frac{1}{i^3}\Phi'''(0)\] \[\langle m^3 \rangle_{c} = \frac{1}{i^3} \frac{d}{d\theta}\left( \Phi''\Phi^{-1} - \Phi'^2 \Phi^{-2}\right)\mid _{\theta=0}\] \[= \frac{1}{i^3} \left( \Phi'''\Phi^{-1} - \Phi''\Phi^{-2}\Phi' - 2\Phi'\Phi''\Phi^{-2} + 2\Phi'^3\Phi^{-3} \right)\mid _{\theta=0}\] \[= \langle m^3 \rangle - \langle m^2 \rangle \langle m \rangle - 2 \langle m \rangle \langle m^2 \rangle + 2\langle m \rangle^3\] \[= \langle m^3 \rangle - 3\langle m^2 \rangle \langle m \rangle + 2\langle m\rangle^3\]
\[\langle m^4 \rangle = \frac{1}{i^4} \Phi''''(0)\] \[\langle m^4 \rangle_{c} = \frac{1}{i^4} \frac{d}{d\theta} \left( \Phi'''\Phi^{-1} - \Phi''\Phi^{-2}\Phi' - 2\Phi'\Phi''\Phi^{-2} + 2\Phi'^3\Phi^{-3} \right)\mid _{\theta=0}\] \[= \frac{1}{i^4}(\Phi'''' \Phi^{-1} - \Phi''' \Phi^{-2} \Phi' - \Phi''' \Phi^{-2} \Phi' + 2 \Phi'' \Phi^{-3} \Phi'^2 - \Phi'' \Phi^{-2} \Phi''\] \[- 2\Phi''^2 \Phi^{-2} - 2 \Phi' \Phi''' \Phi^{-2} + 4\Phi'^2 \Phi'' \Phi^{-3} + 6\Phi'^{2} \Phi^{-3}\Phi'' - 6 \Phi'^4 \Phi^{-4})\mid _{\theta= 0}\] \[=\langle m^4 \rangle - \langle m^3 \rangle\langle m \rangle - \langle m^3 \rangle\langle m \rangle + 2\langle m^2 \rangle\langle m \rangle^2 - \langle m^2 \rangle^2\] \[- 2 \langle m^2 \rangle^2 - 2 \langle m \rangle \langle m^3 \rangle + 4 \langle m \rangle^2 \langle m^2 \rangle + 6\langle m \rangle^2 \langle m^2 \rangle - 6 \langle m \rangle^4\] \[= \langle m^4 \rangle - 4\langle m^3 \rangle\langle m \rangle + 12\langle m^2 \rangle\langle m \rangle^2 - 3\langle m^2 \rangle^2 - 6\langle m \rangle^4\]
(나)
\(k=1\):
\[\langle m \rangle_{c} = \langle (\Delta m ) \rangle\] \[= \langle m - \langle m \rangle \rangle\] \[= \langle m \rangle - \langle m \rangle = 0\]\(k=2\):
\[\langle (\Delta m)^2 \rangle = \langle (m - \langle m \rangle)^2 \rangle\] \[= \langle m^2 - 2 m \langle m \rangle + \langle m \rangle^2 \rangle\] \[= \langle m^2 \rangle - 2 \langle m \rangle ^2 + \langle m \rangle^2\] \[= \langle m^2 \rangle - \langle m \rangle^2\]\(k=3\):
\[\langle (\Delta m)^3 \rangle = \langle (m - \langle m \rangle)^3 \rangle\] \[= \langle m^3 - 3m^2 \langle m \rangle + 3m \langle m \rangle^2 - \langle m \rangle^3 \rangle\] \[= \langle m^3 \rangle - 3 \langle m^2 \rangle \langle m \rangle + 3 \langle m \rangle \langle m \rangle^2 - \langle m \rangle^3\] \[= \langle m^3 \rangle - 3 \langle m^2 \rangle \langle m \rangle+ 2 \langle m \rangle^3\]\(k=4\):
\[\langle ( \Delta m)^4 \rangle = \langle (m - \langle m \rangle)^4 \rangle\] \[= \langle m^4 - 4m^3 \langle m \rangle + 6m^2\langle m \rangle^2 - 4m\langle m \rangle^3 + \langle m \rangle^4 \rangle\] \[= \langle m^4 \rangle - 4 \langle m^3 \rangle \langle m \rangle + 6 \langle m^2 \rangle^2 - 4 \langle m \rangle \langle m \rangle^3 + \langle m \rangle^4\]\(k=4\)일때 성립하지 않는다.
(다)
\(k=1\):
\[\langle m \rangle = \langle m \rangle_{c}\]\(k=2\):
\[\langle m^2 \rangle = \langle m^2\rangle_{c} + \langle m \rangle_{c}^2\]\(k=3\):
\[\langle m^3 \rangle = \langle m^3 \rangle_{c} + 3 \langle m^2 \rangle \langle m \rangle - 2 \langle m \rangle^3\] \[= \langle m^3 \rangle_{c} + 3 (\langle m^2 \rangle_{c} + \langle m \rangle^2_{c})\langle m \rangle_{c} - 2 \langle m \rangle^3_{c}\] \[= \langle m^3 \rangle_{c} + 3 \langle m^2 \rangle_{c} \langle m \rangle_{c} + \langle m \rangle_{c}^3\]\(k=4\):
\[\langle m^4 \rangle = \langle m^4 \rangle_{c} + 4 \langle m^3 \rangle \langle m \rangle + 3 \langle m^2 \rangle^2 - 12\langle m^2 \rangle \langle m \rangle^2 + 6 \langle m \rangle^4\] \[= \langle m^4 \rangle_{c} + 4(\langle m^3 \rangle_{c} + 3 \langle m^2 \rangle_{c} \langle m \rangle_{c} + \langle m \rangle_{c}^3) \langle m\rangle_{c} + 3(\langle m^2\rangle_{c} + \langle m \rangle_{c}^2)^2\] \[- 12(\langle m^2\rangle_{c} + \langle m \rangle_{c}^2)\langle m \rangle_{c}^2 + 6 \langle m \rangle_{c}^4\] \[= \langle m^4 \rangle_{c} + 4 \langle m^3 \rangle_{c} \langle m \rangle_{c} + 12 \langle m^2 \rangle_{c} \langle m \rangle_{c}^2 + 4\langle m \rangle_{c}^4 + 3\langle m^2 \rangle_{c}^2 +6 \langle m^2 \rangle_{c} \langle m \rangle_{c}^2 + 3\langle m \rangle_{c}^4\] \[- 12 \langle m^2 \rangle_{c} \langle m \rangle_{c}^2 - 12 \langle m \rangle^4_{c} + 6 \langle m\rangle_{c}^4\] \[= \langle m^4 \rangle_{c} + 4\langle m^3 \rangle_{c} \langle m \rangle_{c} +6 \langle m^2 \rangle_{c} \langle m \rangle_{c}^2 + 3 \langle m^2 \rangle^2_{c} + \langle m \rangle^4_{c}\]도형의 개수 = 제곱 수 도형이 점선으로 묶여있음 = 평균기호 내부에서 제곱됨 도형 앞 숫자 = 계수
3.
(가)
\[P(n) = \frac{N!}{n!(N-n)!} p^n (1-p)^{N-n}\] \[= \frac{N!}{n!(N-n)!} \left( \frac{\lambda}{N} \right)^n \left( 1 - \frac{\lambda}{N} \right)^{N-n}\] \[\frac{N! \lambda^n}{n!(N-n)! N^n} \left( 1-\frac{\lambda}{N} \right)^N \left( 1-\frac{\lambda}{N} \right)^{-n}\] \[\implies\ln P(n) = \ln N! - \ln n! - \ln(N-n)! +n\ln p + (N-n)\ln(1-p)\] \[\simeq N\ln N - N - n\ln n + n - (N-n)\ln(N-n)+N-n + n\ln p + (N-n)\ln(1-p)\] \[= (N-n)\ln N + n\ln N - n\ln n - (N-n)\ln(N-n) + n\ln p + (N-n)\ln(1-p)\] \[= -n(\ln n - \ln N - \ln p) - (N-n)(\ln(N-n) - \ln N - \ln(1-p))\] \[= -n \ln \left( \frac{n}{Np} \right) - (N-n) \ln \left( \frac{N-n}{N(1-p)} \right)\] \[\lim_{ N \to \infty }\left( 1+ \frac{x}{N} \right)^N = e^x\] \[e^{-\lambda} = \left( 1- \frac{\lambda}{N} \right)^N\] \[= \left( 1- \frac{Np}{N} \right)^N\] \[= (1-p)^N\] \[= (1-p)^{N-n} (1-p)^n\] \[\implies (1-p)^{N-n} = e^{-\lambda}(1-p)^{-n}\] \[P(n) = \frac{N!}{n!(N-n)!} p^n (1-p)^{-n} e^{-\lambda}\] \[= \frac{N!}{n!(N-n)!} \left( \frac{p}{1-p} \right)^n e^{-\lambda}\] \[\frac{N!}{(N-n)!} \left( \frac{p}{1-p} \right)^n = \lambda^n = (Np)^n\] \[\frac{N!}{(N-n)!} \left( \frac{1}{1-p} \right)^n = N^n\] \[\implies \ln N! - \ln(N-n)! + n \ln \left( \frac{1}{1-p} \right) = n\ln N\] \[\implies N\ln N - N - (N-n)\ln(N-n) + N-n + n\ln\left( \frac{1}{1-p} \right) = n\ln N\] \[\implies (N-n)\ln N + n\ln N - (N-n)\ln(N-n) - n + n\ln\left( \frac{1}{1-p} \right) = n\ln N\] \[\implies -(N-n) \left( \ln \frac{N-n}{N} \right) - n\left( 1 - \ln\left( \frac{1}{1-p} \right) \right) = 0\] \[\lambda^n = (Np)^n = N^n p^n\](나)
(다)
(라)
4.
Thanks:
무기력증을 내게 선물해줘서 감사하다. 이것이 꼭 부정적인걸까? 부정적이다, 긍정적이다 이것은 왜 부정적인지? 왜 긍정적인지? 이유도 알지 못하면서 원숭이처럼 받아들이고 있는 부분이다.
Emotion and Improve:
의욕없고, 지친다. 왤까? 나는 오늘 해야할 것, 해야할 과제, 해야할 공부 들을 어떻게할지 미리 생각한다. 그런 생각을 하고 나면, 그 Task를 통해서 생산되는 스트레스가 감정 기억으로 연관되어 미리 받아버린다. ‘아 해야하는데’를 계속 생각하다 보면, 시작하기도 전에 지쳐버린다. 만약 내가 ‘아 해야하는데’를 계속 생각하지 않았다면, 과연 내가 지쳐버렸을까? 아무것도 하지 않아도 지치는 것은 당연한 것이었다. 그것을 함으로써 받는 스트레스를 미리 받아버렸기 때문이다. 그렇다면, 해야할 것을 미리 생각하지 말자. 그리고 개인적으로 힘들다고 느끼는 상태가 올 때마다 나에겐 큰 기회다. 이런 인사이트, 깨달음, 성찰과 같은 내용들이 내게 선물처럼 하나씩 주어지기 때문이다.