# 第一性原理热统

Posted:   June 08, 2018

Status:   Archived

 Tags :   Statistical-mechanics First-Principle

 Categories :   Physics-study-notes

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# 系统

==用量子力学和经典力学描述系统的差别, 虽然根本上说求别在于运动方程是薛定谔方程还是哈密顿方程. 但是因为我们可以仅用能量和粒子数来描述一个系统, 这两种描述的差别在于这两者的推论: 刘维尔定理是否成立,能量是不是连续分布, 粒子是不是全同的.???==

## 相空间数密度 $\rho$ 的性质

$\rho$ 为相空间点 $(p,q)$ 处的点密度.

$N=\int _ V{\tilde{\rho}{(p,q)}d\Gamma}$

$\rho$ 为归一化的密度

$\rho=\int _ V\frac{\tilde{\rho}}{N}d\Gamma$

## 刘维尔定理

$\frac{\partial}{\partial t}\int _ \omega \rho d \Gamma = -\int _ \sigma \vec{v}\cdot \hat{n} d \sigma$

\begin{align*} \vec{\nabla} \cdot \vec{v}&=(\frac{\partial}{\partial q _ i}\hat{q} _ i,\frac{\partial}{\partial p _ i}\hat{p} _ i)\cdot(\dot{q} _ i \hat{q} _ i,\dot{p} _ i \hat{p} _ i)\\ &=\frac{\partial}{\partial q _ i}\dot{q} _ i {\hat{q} _ i}^2 +\frac{\partial}{\partial p _ i}\dot{p} _ i {\hat{p} _ i}^2+\frac{\partial}{\partial q _ i}\dot{p} _ i \hat{q} _ i \hat{p} _ i +\frac{\partial}{\partial p _ i}\dot{q} _ i \hat{q} _ i \hat{p} _ i\\ &=\frac{\partial}{\partial q _ i}\dot{q} _ i+\frac{\partial}{\partial p _ i}\dot{p} _ i \end{align*}

\begin{align*} \vec{\nabla} \cdot \vec{v}&=\frac{\partial}{\partial q _ i}\dot{q} _ i+\frac{\partial}{\partial p _ i}\dot{p} _ i\\ &=\frac{\partial}{\partial q _ i} \frac{\partial H}{\partial q _ i}+\frac{\partial}{\partial p _ i}\left(-\frac{\partial H}{\partial q _ i}\right)\\ &=0 \end{align*}

\begin{align*} \frac{\partial}{\partial t}\int\omega \rho d \Gamma &= -\int _ \sigma \rho\vec{v}\cdot \hat{n} d \sigma\\ &=-\int _ \omega \mathbf{\nabla}\cdot \left(\rho \vec{v}\right) d \Gamma\\ &=\int _ \omega{\left(\left(\vec{\nabla}\rho\right)\cdot\vec{v}+\rho\left(\vec{\nabla}\cdot\vec{v}\right)\right)d \Gamma }\\ &=\int _ \omega{\left(\vec{\nabla}\rho\right)\cdot\vec{v}d \Gamma }\\ \end{align*}

\begin{align} \frac{\partial}{\partial t}\rho - \left(\vec{\nabla}\rho\right)\cdot\vec{v}&=\frac{\partial}{\partial t}\rho+\frac{\partial\rho}{\partial q _ i}\dot{q} _ i+\frac{\partial\rho}{\partial p _ i}\dot{p} _ i \notag\\ &=\frac{d}{dt}\rho \notag\\ &=0 \end{align}

## 对系统的观测

1. 对同一个系统连续地测量 $O$ 的值. 测量次数很大时, 观测值的加权平均就是最后的观测量.

​ 假设测量对系统的影响任意小 $\leftrightarrow$ 总共只有一次测量, 但假设测量发生在不同的时间, 测量值的平均

2. 对许多个”一样的”系统进行测量. 测量系统数目很大时, 观测值的加权平均就是最后的观测量.

$\mathbb{P}(p,q)d\Gamma =\rho(p,q)d \Gamma$

==也就是说, $\rho$ 包含了系统的全部信息==

## 系统的宏观量计算方法

graph LR
A(微观运动方程)-->B(微观态空间)
B--> C(态空间体积)
C-->|对能量微分|D(能态密度)
D-->E(ρ)
E-->F(观测量)


# 系统的分类与对应的 $\rho$ , 宏观量的计算

## 系统的分类

### 孤立系

$\rho(p,q)=\rho(\varepsilon _ 0)\delta(E-\varepsilon _ 0)$

## 孤立系

### 孤立系的 $\rho$

$\rho(p,q)=\rho(\varepsilon _ 0)\delta(E-\varepsilon _ 0)$

### 孤立系的观测值计算

#### $\ln{\rho}$ 的可加性

$N _ S=\int{\rho _ S \delta(E _ S-\varepsilon _ 0) d \Gamma _ S}\notag\\ N _ A=\int{\rho _ A \delta(E _ A-E _ 1) d \Gamma _ A}\notag\\ N _ B=\int{\rho _ B \delta(E _ B-E _ 2) d \Gamma _ B}\notag\\$

\begin{align*} N _ S(\varepsilon _ 0)&=\int{ dE _ a N _ A(E _ a) N _ B(\varepsilon _ 0-E _ a)}\\ &=\int{ dE \left(\int{\rho _ A \delta(E _ A-E _ a) d \Gamma _ A}\int{\rho _ B \delta(E _ B-E _ a-\varepsilon _ 0) d \Gamma _ B}\right)}\\ &=\int{\rho _ A\rho _ B \left(\int{dE _ A \delta(E _ A-E _ a)\delta(E _ B-E _ a-\varepsilon _ 0)}\right) d\Gamma _ A d\Gamma _ B }\\ &=\int{\rho _ A\rho _ B \delta(E _ B-E _ A-\varepsilon _ 0) d\Gamma _ A d\Gamma _ B }\\ &=\int{\rho _ S \delta(E _ S-\varepsilon _ 0) d \Gamma _ S} \end{align*}

\begin{align} E _ A+E _ B=E _ S\notag\\ d\Gamma _ A\cdot d\Gamma _ B = d\Gamma _ S \label{classicalDGammaProduct} \end{align}

$\rho _ A\rho _ B=\rho _ S \notag$

$\ln{\rho _ A}+\ln{\rho _ B}=\ln{\rho _ S}$

#### 概率密度的表达式

\begin{align} \int\delta(E-\varepsilon _ 0)d\Gamma&=\int\delta(E(p,q)-\varepsilon _ 0)[dp][dq] \notag\\ &= \frac{1}{\left. \frac{dE}{d\Gamma} \right| _ {E=\varepsilon _ 0}} \label{deltaE-E0}\\ \end{align}

\begin{align*} \int \mathbb{P}d\Gamma &=\int\rho(p,q)d \Gamma\\ & =\int\rho(\varepsilon _ 0)\delta(E-\varepsilon _ 0)d\Gamma\\ &= \rho(\varepsilon _ 0)\int\delta(E-\varepsilon _ 0)d\Gamma\\ & = \rho(\varepsilon _ 0) \cdot \frac{1}{\left. \frac{dE}{d\Gamma} \right| _ {E=\varepsilon _ 0}} \\ \end{align*}

$\rho(\varepsilon _ 0) \cdot \frac{1}{\left. \frac{dE}{d\Gamma} \right| _ {E=\varepsilon _ 0}} =1 \notag$

$\rho(p,q)= \left. \frac{dE}{d\Gamma} \right| _ {E=\varepsilon _ 0} \delta(E-\varepsilon _ 0) = \frac{1}{\left. \frac{d\Gamma}{dE} \right| _ {E=\varepsilon _ 0}} \delta(E-\varepsilon _ 0) \notag$

#### 概率密度用能态密度表达

$\begin{gather} \Omega(E)=\frac{d\Gamma}{dE}\\ \mathrm{Volume\ in\ phase\ space}=\int d\Gamma=\int \Omega(E)dE \end{gather}$

$\rho(E)=\frac{1}{\Omega(\varepsilon _ 0)} \delta(E-\varepsilon _ 0)\\$

#### 系统宏观量用能态密度表达

\begin{align} \langle O\rangle&=\int _ {-\infty}^{ {-\infty}}{O(p,q)\mathbb{P}(p,q)dpdq}\notag\\ &=\int _ {-\infty}^{ {-\infty}}O(p,q)\rho(p,q)d \Gamma \notag\\ &=\int _ {-\infty}^{ {-\infty}}O(p,q)\frac{1}{\Omega(\varepsilon _ 0)} \delta(E-\varepsilon _ 0)d\Gamma \notag\\ &=\frac{1}{\Omega(\varepsilon _ 0)} \int _ {-\infty}^{ {-\infty}}O(p,q)\delta(E-\varepsilon _ 0)d\Gamma \\ \end{align}

### 一些特殊的宏观量

#### 能量

==能量和粒子数是仅有的确定系统微观状态的物理量==,

\begin{align} \langle H\rangle &=\frac{1}{\Omega(\varepsilon _ 0)} \int _ {-\infty}^{ {-\infty}}H(p,q)\delta(E-\varepsilon _ 0)d\Gamma \notag\\ & =\int _ {-\infty}^{ {-\infty}}\varepsilon _ 0 \frac{1}{\Omega(\varepsilon _ 0)} d\Gamma \notag\\ & = \varepsilon _ 0\int _ {-\infty}^{ {-\infty}} \frac{1}{\Omega(\varepsilon _ 0)} d\Gamma \notag\\ & = \varepsilon _ 0 \end{align}

### 孤立系的平衡

\begin{align*} 1&=\underbrace{\int\cdots\int} _ {n _ s}\rho _ sd\Gamma _ s\\ &=\underbrace{\int\cdots\int} _ {n _ a}\underbrace{\int\cdots\int} _ {n _ b}\rho _ s d\Gamma _ ad\Gamma _ b\\ &=\underbrace{\int\cdots\int} _ {n _ a}\left(\underbrace{\int\cdots\int} _ {n _ b}\rho _ s d\Gamma _ b\right)d\Gamma _ a\\ 1&=\underbrace{\int\cdots\int} _ {n _ a}\rho _ ad\Gamma _ a \end{align*}

\begin{align} \rho _ a^{\mathrm{equilibrium}}&=\underbrace{\int\cdots\int} _ {n _ b}\rho _ s d\Gamma _ b \notag\\ &=\int\rho _ s d\Gamma _ b\notag\\ &=\int\frac{1}{\Omega _ s(E _ s)}\delta(E _ s-\varepsilon _ s)d\Gamma _ b\notag\\ &=\int\frac{1}{\Omega _ s(E _ s)}\delta(E _ b+E _ a-\varepsilon _ s)d\Gamma _ b\notag\\ &=\frac{1}{\Omega _ s(E _ s)}\int\delta(E _ b-(\varepsilon _ s-E _ a))d\Gamma _ b\notag\\ &=\frac{1}{\Omega _ s(E _ s)}\Omega _ b(\varepsilon _ s-E _ a)\notag\\ \label{rhoaequilibrium} \end{align}

$\rho _ a^{\mathrm{isolated}}=\frac{1}{\Omega _ a(E _ a)}\delta(E _ a-\varepsilon _ a) \label{rhoaisolated}$

$\delta(E _ a-\varepsilon _ a)=\frac{\Omega _ a(E _ a)\Omega _ b(\varepsilon _ s-E _ a)}{\Omega _ s(E _ s)}$

$\delta(E _ b-\varepsilon _ b)=\frac{\Omega _ b(E _ b)\Omega _ a(\varepsilon _ s-E _ b)}{\Omega _ s(E _ s)}$

$\delta(E _ a-\varepsilon _ a)=\frac{\Omega _ a(E _ a)\Omega _ b(E _ b)}{\Omega _ s(E _ s)}=\delta(E _ b-\varepsilon _ b)$

$\frac{1}{\Omega _ a(\varepsilon _ a)}\int\delta(E _ a-\varepsilon _ a)d\Gamma _ a=1 \notag\\ \int\delta(E _ a-\varepsilon _ a)d\Gamma _ a=\Omega _ a(\varepsilon _ a) \notag\\$

\begin{align*} \quad \delta(E _ a-\varepsilon _ a) &=\frac{d\Omega _ a(\varepsilon _ a)}{d\Gamma _ a}\\ &=\frac{d\Omega _ a(\varepsilon _ a)}{d \varepsilon _ a}\frac{d \varepsilon _ a}{d\Gamma _ a}\\ &=\frac{d\Omega _ a(\varepsilon _ a)}{d \varepsilon _ a} {\left.\frac{d E _ a}{d\Gamma _ a}\right|} _ {E _ a=\varepsilon _ a}\\ &=\frac{d\Omega _ a(\varepsilon _ a)}{d \varepsilon _ a}\frac{1}{\Omega(\varepsilon _ a)}\\ &=\left.\frac{d \ln{(\Omega(E _ a))}}{d E _ a}\right| _ {E _ a=\varepsilon _ a} \end{align*}

$\left.\frac{d}{d E _ a}\ln{(\Omega(E _ a))}\right| _ {E _ a=\varepsilon _ a}=\left.\frac{d}{d E _ b}\ln{(\Omega(E _ b))}\right| _ {E _ b=\varepsilon _ b}$

$S(E)=\ln(\Omega(E))$

$S' _ a(E _ a)=S' _ b(E _ b)=\frac{1}{T}$

## 封闭系

### 封闭系的 $\rho$

\begin{align*} 1&=\int\rho _ cd\Gamma _ c\\ 1&=\int \rho _ sd\Gamma _ s\\ &=\iint \rho _ s d \Gamma _ rd\Gamma _ c\\ &=\int \left(\int\rho _ sd\Gamma _ r\right)d\Gamma _ c \end{align*}

\begin{align*} \rho _ c&= \int \rho _ sd\Gamma _ r\\ &=\int\frac{1}{\Omega _ s(E _ s)}\delta(E _ s-(E _ r+E _ c)) d\Gamma _ r\\ &=\int\frac{1}{\Omega _ s(\varepsilon _ s)}\delta(\varepsilon _ s-(E _ r+E _ c))d\Gamma _ r\\ &=\int\frac{1}{\Omega _ s(\varepsilon _ s)}\delta(E _ r-(\varepsilon _ s-E _ c))d\Gamma _ r, \quad (\delta(x)=\delta(-x))\\ &=\frac{\Omega _ r(\varepsilon _ s-E _ c)}{\Omega _ s(\varepsilon _ s)}\\ \end{align*}

#### 封闭系的熵

\begin{align*} \rho _ c&=\frac{\Omega _ r(\varepsilon _ s-E _ c)}{\Omega _ s(\varepsilon _ s)}\\ &=\frac{\Omega _ r(\varepsilon _ s-E _ c)}{\Omega _ b(\varepsilon _ s)}\cdot\frac{\Omega _ b(\varepsilon _ s)}{\Omega _ s(\varepsilon _ s)}\\ &=e^{S _ b(\varepsilon _ s-E _ c)-S _ b(\varepsilon _ s)}\cdot e^{S _ b(\varepsilon _ s)-S _ s(\varepsilon _ s)}\\ &=e^{\frac{S _ b(\varepsilon _ s-E _ c)-S _ b(\varepsilon _ s)}{(-E _ c)}\cdot (-E _ c)}\cdot e^{S _ b(\varepsilon _ s)-S _ s(\varepsilon _ s)}\\ \xrightarrow{E _ c\ll\varepsilon _ s}&=e^{\left.\frac{S _ b(E)}{d E}\right| _ {E=\varepsilon _ s}\cdot (-E _ c)}\cdot e^{S _ b(\varepsilon _ s)-S _ s(\varepsilon _ s)}\\ &=e^{-\frac{1}{k _ 0 T _ r}\cdot E _ c}\cdot e^{S _ b(\varepsilon _ s)-S _ s(\varepsilon _ s)} \end{align*}

$\rho _ c(E)=e^{-\psi -\frac{E}{k _ 0 T}}, \quad \psi=-\left(S _ b(\varepsilon _ s)-S _ s(\varepsilon _ s)\right)$

\begin{align*} 1&=\int\mathbb{P}(p,q)d\Gamma\\ &=\int\rho _ c(E(p,q))d\Gamma\\ &=\int e^{-\psi -\frac{E}{k _ 0 T}} d\Gamma\\ \end{align*}

\begin{align*} e^{\psi}=\int e^{-\frac{E}{k _ 0 T}} d\Gamma\\ \end{align*}

\begin{align} Z&=e^{\psi}=\int e^{-\frac{E}{k _ 0 T}} d\Gamma \\ \rho _ c&=\frac{1}{Z}e^{-\frac{E}{k _ 0 T}}=\frac{e^{-\frac{E}{k _ 0 T}}}{\int e^{-\frac{E}{k _ 0 T}} d\Gamma} \end{align}

\begin{align} \quad \psi &=-\left(S _ b(\varepsilon _ s)-S _ s(\varepsilon _ s)\right)\\ &=-\left(S _ b(\varepsilon _ s)-(S _ b(\varepsilon _ b)+S _ c(\varepsilon _ c))\right)\\ &=-\left((S _ b(\varepsilon _ s)-S _ b(\varepsilon _ b))-S _ c(\varepsilon _ c)\right)\\ &=-\left(\frac{S _ b(\varepsilon _ b+\varepsilon _ c)-S _ b(\varepsilon _ b)}{\varepsilon _ c}\varepsilon _ c-S _ c(\varepsilon _ c)\right)\\ &=-\left(\left.\frac{\partial S _ b(E)}{\partial E}\right| _ {E=\varepsilon _ b}\varepsilon _ c-S _ c(\varepsilon _ c)\right)\\ &=-\left(\frac{\varepsilon _ c}{k _ 0 T}-S _ c(\varepsilon _ c)\right)\\ \end{align}

\begin{align} \psi=-\frac{1}{ {k _ 0 T}}\left(E-T\cdot S(E)\right)\\ \end{align}

#### 封闭系的划分

\begin{align*} \rho^{closed} _ a&=\frac{1}{Z _ a}e^{-\frac{E _ a}{k _ 0T _ a}}\\ \rho^{closed} _ b&=\frac{1}{Z _ b}e^{-\frac{E _ b}{k _ 0T _ b}}\\ \end{align*}

\begin{align*} 1 &= \int\rho^{equilibrium} _ ad\Gamma _ a\\ &=\iint \rho _ s d\Gamma _ a d\Gamma _ b\\ \Rightarrow \rho^{equilibrium} _ a&=\int \rho _ s d\Gamma _ b\\ &=\int \frac{1}{Z _ s}e^{-\frac{E _ s}{k _ 0T _ s}} d\Gamma _ b\\ \end{align*}

==能量的可加性始终成立???==, $E _ a+E _ b=E _ s$ .得到

\begin{align*} \rho^{closed} _ a&= \rho^{equilibrium} _ a\\ \frac{1}{Z _ a}e^{-\frac{E _ a}{k _ 0T _ a}}&=\int \frac{1}{Z _ s}e^{-\frac{E _ s}{k _ 0T _ s}} d\Gamma _ b\\ \frac{1}{Z _ a}e^{-\frac{E _ a}{k _ 0T}}&=\frac{1}{Z _ s}\int e^{-\frac{E _ a}{k _ 0T}-\frac{E _ b}{k _ 0T}}d\Gamma _ b\\ &=\frac{1}{Z _ s} e^{-\frac{E _ a}{k _ 0T}} \int e^{-\frac{E _ b}{k _ 0T}}d\Gamma _ b\\ &=\frac{1}{Z _ s} e^{-\frac{E _ a}{k _ 0T}}\cdot Z _ b\cdot \int \frac{1}{Z _ b}e^{-\frac{E _ b}{k _ 0T}}d\Gamma _ b\\ &=\frac{Z _ b}{Z _ s} e^{-\frac{E _ a}{k _ 0T}} \end{align*}

\begin{align} Z _ aZ _ b&=Z _ s\\ \ln{Z _ a}+\ln{Z _ b}&=\ln{Z _ s} \end{align}

#### 封闭系的稳态

1. 将封闭系S划分为两个封闭子系A和B, 如论如何划分, 这两个封闭子系都好像彼此隔绝.
2. 整个封闭系S表现得像一个孤立系

\begin{align*} E _ a&=F _ a+TS _ a\\ dE _ a&=dF _ a+S _ adT _ a\\ &=(\frac{\partial F}{\partial T}dT+\frac{\partial F}{\partial q}dq)\\ &=\frac{\partial E _ a}{\partial q _ a}dq _ a \end{align*}

$dE _ b=\frac{\partial E}{\partial q _ b}d q _ b$

\begin{align*} dE _ a+dE _ b&=dE _ s=0\\ \end{align*}

$dE _ a=-dE _ b$

$\frac{\partial S _ 1 }{\partial E _ 1}=\frac{\partial S _ 2}{\partial E _ 2}$

\begin{align} \frac{\partial H _ a}{\partial q _ a}dq _ a&=-\frac{\partial H _ b}{\partial q _ b}dq _ b\\ P _ adV _ a&=-P _ bdV _ b& \end{align}

$P _ a=P _ b$

### 封闭系的观测量

#### 封闭系的能量

\begin{align} \langle E\rangle &=\int H(p,q) \mathbb{P}(p,q) d\Gamma\notag\\ & =\int H(p,q) \rho _ cd\Gamma\notag\\ &=\int E\frac{1}{Z}e^{-\frac{E}{k _ 0 T}}\notag\\ &=\frac{1}{Z}\int E e^{-\frac{E}{k _ 0 T} } d\Gamma\\ \end{align}

\begin{align*} \frac{\partial Z}{\partial (\frac{1}{k _ 0T})}&=\frac{\partial }{\partial (\frac{1}{k _ 0T})}{\int e^{-\frac{E}{k _ 0 T} } d\Gamma}\\ \xrightarrow{偏导与积分变量不同可交换}&=\int \frac{\partial }{\partial (\frac{1}{k _ 0T})} e^{-\frac{E}{k _ 0 T} } d\Gamma\\ &=\int -E e^{-\frac{E}{k _ 0 T} } d\Gamma\\ &=-\int E e^{-\frac{E}{k _ 0 T} } d\Gamma\\ \end{align*}

\begin{align} \langle E\rangle &=\frac{1}{Z}\int E e^{-\frac{E}{k _ 0 T} } d\Gamma \notag\\ &=-\frac{1}{Z}\frac{\partial Z}{\partial (\frac{1}{k _ 0T})}\notag \\ &=-\frac{\partial}{\partial (\frac{1}{k _ 0T})} \ln Z\\ &=-\frac{\partial}{\partial (\frac{1}{k _ 0T})} \psi\\ \end{align}

## 开放系

### 开放系的 $\rho$

\begin{align} S(E)&=\ln{\Omega(E)} \notag\\ &=\ln\left(V^N \left(\frac{2e\pi m}{D}\right)^{\frac{ND}{2}}\left(\frac{E}{N}\right)^{\frac{ND}{2}} 2\left(\pi ND\right)^{\frac{1}{2}} \right)\\ \end{align}

\begin{align*} 1&=\underbrace{\int\cdots\int} _ {n _ s}\rho _ {s}(s _ o,E _ o)d\Gamma _ {s,n _ s}\\ &=\underbrace{\int\cdots\int} _ {n _ o}\underbrace{\int\cdots\int} _ {n _ r}\rho _ {s}(s _ o,E _ o) d\Gamma _ {r,n _ r}d\Gamma _ {o,n _ o}\\ &=\underbrace{\int\cdots\int} _ {n _ o}\left(\underbrace{\int\cdots\int} _ {n _ r}\rho _ {s}(s _ o,E _ o) d\Gamma _ {r,n _ r}\right)d\Gamma _ {o,n _ o}\\ 1&=\underbrace{\int\cdots\int} _ {n _ o}\rho _ {o}(n _ o,E _ o)d\Gamma _ {o,n _ o} \end{align*}

\begin{align*} \rho _ {o,n _ o}^{equlibrium}&=\underbrace{\int\cdots\int} _ {n _ r}\rho _ {s}(s _ o,E _ o) d\Gamma _ {r,n _ r}\\ &=\int \rho _ {s}(s _ o,E _ o) d\Gamma _ {r,n _ r}\\ &=\int \frac{1}{\Omega _ {s}(n _ s,E _ s)}\delta(E _ s-\varepsilon _ s)d\Gamma _ {r,n _ r}\\ &=\int \frac{1}{\Omega _ {s}(n _ s,E _ s)}\delta(E _ o+E _ r-\varepsilon _ s)d\Gamma _ {r,n _ r}\\ &=\frac{\Omega _ {r}(n _ r,\varepsilon _ s-E _ o)}{\Omega _ {s}(n _ s,\varepsilon _ s)}\\ &=\frac{\Omega _ {r}(n _ r,\varepsilon _ s-E _ o)}{\Omega _ {s}(n _ s,\varepsilon _ s)}\\ \end{align*}

### 开放系的稳定

graph LR
subgraph 条件
A[T相等]-.->|附加|B[P相等]
B-.->|附加|C[μ相等]
end
subgraph 平衡
A-->|得到|D[孤立系平衡]
B-->|得到|E[封闭系平衡]
D-->|作为基础|E
C-->|得到|F[开放系平衡]
E-->|作为基础|F
end



#### 开放系的熵

\begin{align*} \rho^{equlibrium} _ {o,n _ o}&=\frac{\Omega _ {r}(n _ r,\varepsilon _ s-E _ o)}{\Omega _ {s}(n _ s,\varepsilon _ s)}\\ &=\frac{\Omega _ {r}(n _ r,\varepsilon _ s-E _ o)}{\Omega _ {r}(n _ s,\varepsilon _ s-E _ o)}\cdot\frac{\Omega _ {r}(n _ s,\varepsilon _ s-E _ o)}{\Omega _ {r}(n _ s,\varepsilon _ s)}\cdot\frac{\Omega _ {r}(n _ s,\varepsilon _ s)}{\Omega _ {s}(n _ s,\varepsilon _ s)}\\ &=e^{S _ r(n _ r,\varepsilon _ s-E _ o)-S _ r(n _ s,\varepsilon _ s-E _ o)}e^{S _ r(n _ s,\varepsilon _ s-E _ o)-S _ r(n _ s,\varepsilon _ s)}e^{S _ r(n _ s,\varepsilon _ s)-S _ s(n _ s,\varepsilon _ s)}\\ &=e^{\frac{S _ r(n _ s-n _ o,\varepsilon _ s-E _ o)-S _ r(n _ s,\varepsilon _ s-E _ o)}{-n _ o}(-n _ o)}e^{\frac{S _ r(n _ s,\varepsilon _ s-E _ o)-S _ r(n _ s,\varepsilon _ s)}{-E _ o}(-E _ o)}e^{S _ r(n _ s,\varepsilon _ s)-S _ s(n _ s,\varepsilon _ s)}\\ &=e^{\left.\frac{\partial S _ r}{\partial n}\right| _ {n=n _ s}(-n _ o)}e^{\left.\frac{\partial S _ r}{\partial E}\right| _ {E=\varepsilon _ r}(-E _ o)}e^{S _ r(n _ s,\varepsilon _ s)-S _ s(n _ s,\varepsilon _ s)}\\ &=e^{\left.\frac{\partial S _ r}{\partial E}\frac{\partial E}{\partial n}\right| _ {n=n _ s}(-n _ o)}e^{\left.\frac{\partial S _ r}{\partial E}\right| _ {E=\varepsilon _ r}(-E _ o)}e^{S _ r(n _ s,\varepsilon _ s)-S _ s(n _ s,\varepsilon _ s)}\\ &=e^{-\frac{1}{k _ 0T}\frac{\partial E}{\partial n}\cdot n _ o}e^{-\frac{1}{k _ 0T}E _ o}e^{S _ r(n _ s,\varepsilon _ s)-S _ s(n _ s,\varepsilon _ s)} \end{align*}

$\rho=-\zeta+\frac{\mu}{k _ 0T}N-\frac{1}{k _ 0T}E,\quad \zeta=-(S _ r(n _ s,\varepsilon _ s)-S _ s(n _ s,\varepsilon _ s))$

$Z=e^{\zeta}=\int e^{-\frac{1}{k _ 0T}\frac{\partial E}{\partial n}\cdot n _ o}e^{-\frac{1}{k _ 0T}E _ o}d\Gamma\\ \rho=\frac{1}{Z}e^{-\frac{1}{k _ 0T}\frac{\partial E}{\partial n}\cdot n _ o}e^{-\frac{1}{k _ 0T}E _ o}=\frac{e^{-\frac{1}{k _ 0T}\frac{\partial E}{\partial n}\cdot n _ o}e^{-\frac{1}{k _ 0T}E _ o}}{\int e^{-\frac{1}{k _ 0T}\frac{\partial E}{\partial n}\cdot n _ o}e^{-\frac{1}{k _ 0T}E _ o}d\Gamma}$

\begin{align*} \zeta&=-(S _ r(n _ s,\varepsilon _ s)-S _ s(n _ s,\varepsilon _ s))\\ &=-\bigg(\Big(S _ r(n _ s,\varepsilon _ s)-S _ r(n _ r,\varepsilon _ r)\Big)-S _ o(n _ o,\varepsilon _ o)\bigg)\\ &=-\Bigg(\bigg(\Big(S _ r(n _ s,\varepsilon _ s)-S _ r(n _ s,\varepsilon _ r)\Big)+\Big(S _ r(n _ s,\varepsilon _ r)-S _ r(n _ r,\varepsilon _ r)\Big)\bigg)-S _ o(n _ o,\varepsilon _ o)\Bigg)\\ &=-\left(\frac{S _ r(n _ s,\varepsilon _ s)-S _ r(n _ s,\varepsilon _ r)}{\varepsilon _ o}\varepsilon _ o+\frac{S _ r(n _ s,\varepsilon _ r)-S _ r(n _ r,\varepsilon _ r)}{n _ o}n _ o-S _ o(n _ o,\varepsilon _ o)\right)\\ &=-\left(\frac{1}{k _ 0T}\varepsilon _ o+\frac{\partial S}{\partial E}\frac{\partial E}{\partial n}-S _ o(n _ o,\varepsilon _ o)\right)\\ &=-\left(\frac{1}{k _ 0T}\varepsilon _ o+\frac{1}{k _ 0T}\frac{\partial E}{\partial n}-S _ o(n _ o,\varepsilon _ o)\right)\\ \end{align*}

$\mu=\frac{\partial E}{\partial n}$

\begin{align*} \zeta&=-\left(\frac{1}{k _ 0T}\varepsilon _ o+\frac{1}{k _ 0T}\frac{\partial E}{\partial n}-k _ 0S _ o(n _ o,\varepsilon _ o)\right)\\ &=-\frac{1}{k _ 0T}\left(\varepsilon _ o+\mu-S _ oT\right) \end{align*}

#### 开放系的划分

$\rho _ a^{uncoupled}=\frac{1}{Z _ a}e^{-\frac{1}{k _ 0T}\mu _ a\cdot n _ a}e^{-\frac{1}{k _ 0T}E _ a}\\ \rho _ b^{uncoupled}=\frac{1}{Z _ b}e^{-\frac{1}{k _ 0T}\mu _ b\cdot n _ b}e^{-\frac{1}{k _ 0T}E _ b}$

\begin{align*} 1 &= \int\rho^{equilibrium} _ ad\Gamma _ a\\ &=\iint \rho _ s d\Gamma _ a d\Gamma _ b\\ \Rightarrow \rho^{equilibrium} _ a&=\int \rho _ s d\Gamma _ b\\ &=\int \frac{1}{Z _ s}e^{-\frac{1}{k _ 0T}\mu _ s\cdot n _ s}e^{-\frac{1}{k _ 0T}E _ s} d\Gamma _ b\\ \end{align*}

$E _ a+E _ b=E _ s$ , $T _ a=T _ b$ , $\mu _ a=\mu _ b=\mu _ s$ 得到

\begin{align*} \rho^{uncoupled} _ a&= \rho^{equilibrium} _ a\\ \frac{1}{Z _ a}e^{-\frac{1}{k _ 0T}\mu _ a\cdot n _ a}e^{-\frac{1}{k _ 0T}E _ a} &= \int \frac{1}{Z _ s}e^{-\frac{1}{k _ 0T}\mu _ s\cdot n _ s}e^{-\frac{1}{k _ 0T}E _ s} d\Gamma _ b\\ &= \int \frac{1}{Z _ s}e^{-\frac{1}{k _ 0T}\mu _ s\cdot (n _ a+n _ b)}e^{-\frac{1}{k _ 0T}(E _ a+E _ b)} d\Gamma _ b\\ &=\frac{1}{Z _ s}e^{-\frac{1}{k _ 0T}\mu _ s n _ a}e^{-\frac{1}{k _ 0T}E _ a} \int e^{-\frac{1}{k _ 0T}\mu _ s\cdot n _ b }e^{-\frac{1}{k _ 0T} E _ b } d\Gamma _ b\\ &=\frac{Z _ b}{Z _ s}e^{-\frac{1}{k _ 0T}\mu _ s n _ a}e^{-\frac{1}{k _ 0T}E _ a} \end{align*}

\begin{align} Z _ aZ _ b&=Z _ s\\ \ln{Z _ a}+\ln{Z _ b}&=\ln{Z _ s} \end{align}

#### 开放系的稳态

1. 将开放系S划分为两个开放子系A和B, 如论如何划分, 这两个开放子系都好像彼此隔绝.
2. 整个开放系S表现得像一个孤立系, 也像一个封闭系

???

## 插曲: 以N个无相互作用的粒子体系推导系统的平衡参量

### 自由单粒子系统

\begin{align} d\vec{p}&=\mathscr{A} _ {D-1}p^{D-1}dp \notag \\ \quad p &= {\left(2mE\right)}^{1/2} \label{singleParticleEP} \\ dp &= {\left(2m\right)}^{1/2}E^{-1/2}dE \notag \end{align}

\begin{align*} \Psi &= \int d\Gamma\\ & = \iint d\vec{q} d\vec{p}\\ &= \int V d\vec{p}\\ &= \int V \mathscr{A} _ {D-1}p^{D-1}dp \\\ &= \int V \mathscr{A} _ {D-1} {\left(2mE\right)}^{\frac{D-1}{2}} {\left(2m\right)}^{\frac{1}{2}}E^{-\frac{1}{2}}dE \\ &= \int V \mathscr{A} _ {D-1}{\left(2m\right)}^{\frac{D}{2}}E^{\frac{D-2}{2}}dE \end{align*}

$\frac{d \Gamma}{d E} =\Omega(E) = V \mathscr{A} _ {D-1}{\left(2m\right)}^{\frac{D}{2}}E^{\frac{D-2}{2}}$

### 经典宏观系统

$\vec{\mathfrak{p}}=(\vec{p} _ 1,\vec{p} _ 2,\cdots,\vec{p} _ N)$

$E=\sum _ i \frac{\vec{p} _ i^2}{2m}= \frac{\vec{\mathfrak{p}}^2}{2m}$

\begin{align} d\vec{\mathfrak{p}}&=d\vec{p} _ 1\cdot d\vec{p} _ 2\cdot\cdots\cdot d\vec{p} _ N \notag \qquad 这里仿照d\vec{p}=d(p _ x,p _ y,p _ z)=dp _ xdp _ ydp _ z \notag \\ &=\mathscr{A} _ {ND-1}\mathfrak{p}^{ND-1}d\mathfrak{p}, \\ \mathfrak{p}&=\left( \sum _ i {\vec{p} _ i^2}\right)^{\frac{1}{2}}\notag \\ &=\left( 2m E\right)^{\frac{1}{2}}\\ d\mathfrak{p}&= {\left(2m\right)}^{1/2}E^{-1/2}dE \end{align}

\begin{align*} \Psi &= \int d\Gamma\\ & = \iint [d\vec{q}] [d\vec{p}]\\ &=\left(\iint\cdots\int d\vec{q} _ 1d\vec{q} _ 2\cdots d\vec{q} _ N\right) \left(\iint\cdots\int d\vec{p} _ 1d\vec{p} _ 2\cdots d\vec{p} _ N\right)\\ &=\left((\int d\vec{q} _ 1)(\int d\vec{q} _ 2)\cdots (\int d\vec{q} _ N)\right) \int d\vec{\mathfrak{p}}\\ &\\ &= V\cdot V\cdots V \int \mathscr{A} _ {ND-1}\mathfrak{p}^{ND-1}d\mathfrak{p} \\\ &= \int V^N \mathscr{A} _ {ND-1} {\left(2mE\right)}^{\frac{ND-1}{2}} {\left(2m\right)}^{\frac{1}{2}}E^{-\frac{1}{2}}dE \\ &= \int V^N \mathscr{A} _ {ND-1}{\left(2m\right)}^{\frac{ND}{2}}E^{\frac{ND-2}{2}}dE \end{align*}

$d\Gamma=V^N \mathscr{A} _ {ND-1}{\left(2m\right)}^{\frac{ND}{2}}E^{\frac{ND-2}{2}}dE$

$\Omega(E)=\frac{d\Gamma}{dE} =V^N \mathscr{A} _ {ND-1}{\left(2m\right)}^{\frac{ND}{2}}E^{\frac{ND-2}{2}} \notag$

$\Gamma$ 函数的性质有[^ $\Gamma$ 函数的性质] 利用Gamma函数的性质

$\Gamma(x+1)\approx \sqrt{2\pi x}x^{x}e^{-x}\\ \therefore \Gamma(x)=\frac{\Gamma(x+1)}{x}\approx \sqrt{2\pi x}x^{x-1}e^{-x}$

\begin{align} \Omega(E) &=V^N\mathscr{A} _ {ND-1}\left({2mE}\right)^{DN/2} \notag\\ &=V^N\cdot\left(\frac{2e\pi}{ND}\right)^{ {\frac{ND}{2}}}2\left(\pi ND\right)^{\frac{1}{2}}\cdot \left(2mE\right)^{DN/2}\notag\\ &=V^N \left(\frac{2e\pi m}{D}\right)^{\frac{ND}{2}}\left(\frac{E}{N}\right)^{\frac{ND}{2}} 2\left(\pi ND\right)^{\frac{1}{2}} \end{align}

\begin{align*} \mathrm{Volume\ in\ phase\ space}&=\int \phantom{\rho}d\Gamma=\int \phantom{\rho}\Omega(E)dE g\\ &\qquad\downarrow biased\ with\ \rho \\ \mathrm{Number\ of\ points\ in\ phase\ space }&=\int\rho d\Gamma=\int \rho\Omega(E)dE \\ \end{align*}

### 理想气体的平衡与能态方程

\begin{align} \Omega(E) &=VN\mathscr{A} _ {ND-1}\left({2mE}\right){DN/2} \notag\\ &=VN\cdot\left(\frac{2e\pi}{ND}\right)^{ {\frac{ND}{2}}}2\left(\pi ND\right)^{\frac{1}{2}}\cdot \left(2mE\right)^{DN/2}\notag\\ &=V^N \left(\frac{2e\pi m}{D}\right)^{\frac{ND}{2}}\left(\frac{E}{N}\right)^{\frac{ND}{2}} 2\left(\pi ND\right)^{\frac{1}{2}} \end{align}

\begin{align} S(E)&=\ln{\Omega(E)} \notag\\ &=\ln\left(V^N \left(\frac{2e\pi m}{D}\right)^{\frac{ND}{2}}\left(\frac{E}{N}\right)^{\frac{ND}{2}} 2\left(\pi ND\right)^{\frac{1}{2}} \right)\\ \end{align}

\begin{align*} \frac{1}{T}&=\frac{dS(E)}{dE} \\ &=k _ 0\frac{1}{\Omega(E)}\frac{d\Omega(E)}{dE} \\ &=\frac{k _ 0}{V^N \left(\frac{2e\pi m}{D}\right)^{\frac{ND}{2}}\left(\frac{E}{N}\right)^{\frac{ND}{2}} 2\left(\pi ND\right)^{\frac{1}{2}} }\cdot V^N \left(\frac{2e\pi m}{D}\right)^{\frac{ND}{2}}\left(\frac{1}{N}\right)^{\frac{ND}{2}} 2\left(\pi ND\right)^{\frac{1}{2}} \frac{dE^{\frac{ND}{2}}}{dE} \\ &=k _ 0\frac{ND}{2}E^{-1} \end{align*}

$E=\frac{1}{2}NDk _ 0T$

# 遇到的问题

## 吉布斯佯谬

\begin{align} S(E)&=\ln{\Omega(E)} \notag\\ &=\ln\left(V^N \left(\frac{2e\pi m}{D}\right)^{\frac{ND}{2}}\left(\frac{E}{N}\right)^{\frac{ND}{2}} 2\left(\pi ND\right)^{\frac{1}{2}} \right)\\ \end{align}

# 量子描述

## 量子系统的相空间概率密度

1. 在量子力学中系统的共轭物理量(如坐标和动量)不可以同时确定, 也就不能定义像空间内的一个”点”. 如何来描述一个系统?
2. 在量子力学中一个系统可能处于叠加态, 测量本身有不确定性.

$\hat{\rho}=\sum _ {i}p _ i\left| \psi _ i \right\rangle \left\langle \psi _ i \right|$

\begin{align*} \left\langle O \right\rangle &=\sum _ ip _ i \left\langle O \right\rangle _ i\\ &=\sum _ i p _ i \left\langle \psi _ i \right| \hat{O}\left| \psi _ i \right\rangle \\ &=\sum _ i p _ i \left\langle \psi _ i \right| \mathbf{1} \hat{O}\left| \psi _ i \right\rangle\\ &=\sum _ i p _ i \left\langle \psi _ i \right| \left(\sum _ j \left| \psi _ j \right\rangle \left\langle \psi _ j \right| \right) \hat{O}\left| \psi _ i \right\rangle\\ &=\sum _ i\sum _ j p _ i \left\langle \psi _ i \right| \left| \psi _ j \right\rangle \left\langle \psi _ j \right| \hat{O}\left| \psi _ i \right\rangle\\ &=\sum _ i \left\langle \psi _ i \right| \left(\sum _ j p _ i\left| \psi _ j \right\rangle \left\langle \psi _ j \right| \right) \hat{O}\left| \psi _ i \right\rangle\\ &=\sum _ i \left\langle \psi _ i \right|\hat{\rho} \hat{O}\left| \psi _ i \right\rangle\\ &=\operatorname{Tr}{\hat{\rho} \hat{O}} \end{align*}

## 量子系统的刘维尔定理

$i\hbar\frac{\partial}{\partial t}\left| \psi \right\rangle =\hat{H} \left| \psi \right\rangle$

\begin{align*} \frac{d}{d t}\hat\rho &=\frac{\partial \left| \psi \right\rangle }{\partial t} \left\langle \psi \right|+ \left| \psi \right\rangle\frac{\partial \left\langle \psi \right| }{\partial t} \\ &=\frac{\hat{H}}{i\hbar} \left| \psi \right\rangle \left\langle \psi \right| + \left| \psi \right\rangle \left\langle \psi \right| \frac{\hat{H}}{-i\hbar} \\ &=\frac{1}{i\hbar}[\hat{H},\hat{\rho}] \end{align*}

\begin{align*} \frac{d}{d t}\hat\rho &=\frac{\hat{H}}{i\hbar} \left| \psi \right\rangle \left\langle \psi \right| + \left| \psi \right\rangle \left\langle \psi \right| \frac{\hat{H}}{-i\hbar} \\ &=\frac{E}{i\hbar} \left| \psi \right\rangle \left\langle \psi \right| + \left| \psi \right\rangle \left\langle \psi \right| \frac{E}{-i\hbar} \end{align*}

## 量子系统的能级

==我的问题是, 为什么可以直接把能量除一个hbar就代表相空间的疏密?==

# 附录

## $\delta$ 函数的性质

$\int^{+\infty} _ {-\infty}\delta(ax)dx =\frac{1}{\left| a\right|}\int^{+\infty} _ {-\infty}\delta(y)dy \label{deltaax}$

$\int^{+\infty} _ {-\infty}\delta(ax)dx = \begin{cases} \int^{+\infty} _ {-\infty}\delta(y)\frac{dy}{a}= \frac{1}{a}\int^{+\infty} _ {-\infty}\delta(y) dy ,\quad (a>0) \\\\ \int^{\color{Red}-\infty} _ {\color{Red}+\infty}\delta(y)\frac{dy}{a} =-\frac{1}{a}\int^{\color{Red}+\infty} _ {\color{Red}-\infty}\delta(y) dy ,\quad (a<0) \end{cases} \notag$

$\delta(f(x))=\sum _ {i}\frac{\delta(x-a _ i)}{\lvert f'(a _ i)\rvert}, \qquad \mathrm{where}\ f(a _ i)=0$

\begin{align*} \int^{+\infty} _ {-\infty}g(x)\delta\left(f(x)\right)dx &=\sum _ {i}{\int^{a _ i+\varepsilon} _ {a _ i-\varepsilon}g(x)\delta\left(f(x)\right)dx}\\ & \qquad\qquad\uparrow 只在f(x)=0时 \delta 函数才不为零, 取f(x)=0的邻域计算 \\ \notag\xrightarrow{Taylor\ expension} &= \sum _ {i}{\int^{a _ i+\varepsilon} _ {a _ i-\varepsilon}g(x)\delta\left(f(a _ i)+f'(a _ i)x + \mathcal{O}(x^2)\right)dx}\\ \xrightarrow{\varepsilon\rightarrow 0}&=\sum _ {i}\int^{a _ i+\varepsilon} _ {a _ i-\varepsilon}g(x)\delta\left(f'(a _ i)x \right)dx\\ \xrightarrow{\mathrm{use\ eqn. \ \ref{deltaax}}}&= \sum _ {i}\frac{g(a _ i)}{\lvert f'(a _ i)\rvert} \end{align*}

\begin{align*} \sum _ {i}\frac{g(a _ i)}{\lvert f'(a _ i)\rvert}&=\int^{+\infty} _ {-\infty}\sum _ {i}\frac{\delta(x-a _ i)}{\lvert f'(a _ i)\rvert}dx\\ &=\int^{+\infty} _ {-\infty}g(x)\delta\left(f(x)\right)dx \end{align*}

\begin{align} \delta(f(x))&=\sum _ {i}\frac{\delta(x-a _ i)}{\lvert f'(a _ i)\rvert}, \qquad \mathrm{where}\ f(a _ i)=0 \notag\\ \int _ {-\infty}^{+\infty}\delta(f(x))&=\int^{+\infty} _ {-\infty}1\cdot\delta\left(f(x)\right)dx =\sum _ {i}\frac{1}{\lvert f'(a _ i)\rvert}, \qquad \mathrm{where}\ f(a _ i)=0 \label{integralofdeltafx} \end{align}

## 斯特灵公式

$\Gamma(z)=\left(z-\frac{1}{2}\right)\ln(z)-z+\frac{\ln 2 \pi}{2}$

$\ln(N!)=\sum _ {x=1}^N{\ln{x}}\approx\int _ 1^N\ln{x}dx=N\ln{N}-N+1$

[^\Gamma函数的性质]12

# 参考文献

1. Pathria statistical mechanisc

2. Liu Zhao. thermodynamics