真实メモ

• 原文: https://zhuanlan.zhihu.com/p/66341173
• 发布日期: 2019-05-20
• 分类: 数学方法 / 物理数学工具箱

#这 nm 我草稿本, 别进来乱翻. #

• 光子さん, 待ちなさーい.
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1. $\mathrm{SU}(\mathrm{n})$ 的 Young Diagram

下面说的 Young 图均指 {\rm{SU}(n) 的 Young 图.

基本定义:

• 最一般的 Young 图如上所示, 方框里填上数字后叫 Young 表, 但我还叫 Young 图.
• Young 图由一行行格子拼成, 上图共 $a$ 行, 最多只画到第 $n-1$ 行.
• 若将各行的格子数分别记为 {Y_1},{Y_2}, \cdots ,{Y_{n - 1}.
• Young 图只需满足左对齐行不增即可, 即必须满足 {Y_1} \geqslant {Y_2} \geqslant \cdots \geqslant {Y_{n - 1}.

1. \left[ {\begin{array}{*{20}{c} {\Xi _{\rm{c} \\ {\Xi {'_{\rm{c} \end{array} \right] = R\left[ {\begin{array}{*{20}{c} {\Xi _{\rm{c}^{\bar 3} \\ {\Xi _{\rm{c}^6} \end{array} \right]:

{\Xi }_{\text{c} 的价夸克为 $\text{csq}$, 其中 $\text{s,q}$ 在味道 \text{SU}{\left( 3 \right)}_{\text{f} 近似对称性下被视为全同粒子.

全同费米子体系要求 $\text{s,q}$ 系统呈反对称态, 其量子态 \left| {\text{ }\!\!\psi\!\!\text{ }_{\text{sq} \right\rangle 由四部分构成:

\left| {\text{ }\!\!\psi\!\!\text{ }_{\text{sq} \right\rangle =\left| \text{color} \right\rangle \otimes \left| \text{flavor} \right\rangle \otimes \left| \text{spin} \right\rangle \otimes \left| \text{spatial} \right\rangle .

• 其中颜色部分是 trivial 的, 一正一反即可, 为 \text{SU}{\left( 3 \right)}_{\text{c}:3\otimes \bar{3}={1}_{\text{a}\oplus {8}_{\operatorname{m} 中的反对称单态.
• 味道为 \text{SU}{\left( 3 \right)}_{\text{f}:3\otimes 3={\bar{3}_{\text{a}\oplus {6}_{\text{s} 二选一.
• 自旋显然是经典的 \text{SU}\left( 2 \right):2\otimes 2={1}_{\text{a}\oplus {3}_{\text{s}.
• 而轨道不激发为 $l=0$ 即对称的.
• 所以要想保证整体反对称的话 $|\text{flavor}⟩,|\text{spin}⟩$ 就必须同为对称或反对称.
• 由此以 $|\text{flavor}⟩$ 未标记将两种构型分别记为 \Xi _{\text{c}^{\bar{3},\Xi _{\text{c}^{6}.
• 而物理可观测的质量本征态 {\Xi }_{\text{c},\Xi {'}_{\text{c} 为上述二者的混合.

sym: 对称. asym: 反对称. msym: 混合对称, 交换其中俩对称另外俩则反对称. 对称性和表示的维度并非一一对应, 虽然很多时候这么认为都不会出错.

SU(2):

2 \otimes 2 = {1_{\rm{asym} \oplus {3_{\rm{sym}.

2 \otimes 2 \otimes 2 = {2_{\rm{msym} \oplus {2_{\rm{msym} \oplus {4_{\rm{sym}.

SU(3):

3 \otimes 3 = {\bar 3}_{\rm{asym} \oplus {6_{\rm{sym}?

$3\otimes \bar{3}=1\oplus 8.$

3 \otimes 3 \otimes 3 = {1_{\rm{asym} \oplus {8_{\rm{msym} \oplus {8_{\rm{msym} \oplus {10_{\rm{sym}.

• \bar{\psi }(x)\psi (x)\to \bar{\psi }({x}_{\mathcal{P})\psi ({x}_{\mathcal{P}), 此即宇称为 $+$ 的标量流.
• \bar{\psi }(x){\gamma }^{5}\psi (x)\to -\bar{\psi }({x}_{\mathcal{P}){\gamma }^{5}\psi ({x}_{\mathcal{P}), 此即宇称为 $-$ 的赝标流.
• \bar{\psi }(x){\gamma }^{\mu }\psi (x)\to {\mathcal{P}^{\mu }_{\nu }\bar{\psi }({x}_{\mathcal{P}){\gamma }^{\nu }\psi ({x}_{\mathcal{P}), 此即宇称为 $-$ 的矢量流.

准确来说上面这个流算符的宇称应该记作 {(-)}^{\mu }\equiv {\rm{tr}{\left| {\gamma }^{\mu } \right|}^{2}/4. 这里记作 $-$ 是因为它和 Lorentz 矢量的宇称相同, 而三矢量的宇称就是 $-$ 的.

• \bar{\psi }(x){\gamma }^{\mu }{\gamma }^{5}\psi (x)\to -{\mathcal{P}^{\mu }_{\nu }\bar{\psi }({x}_{\mathcal{P}){\gamma }^{\nu }{\gamma }^{5}\psi ({x}_{\mathcal{P}), 此即宇称为 $+$ 的轴矢流.
• \bar{\psi }(x){\sigma }^{\mu \nu }\psi (x)\to {\mathcal{P}^{\mu }_{\rho }{\mathcal{P}^{\nu }_{\sigma }\bar{\psi }({x}_{\mathcal{P}){\sigma }^{\rho \sigma }\psi ({x}_{\mathcal{P}), 此即混合宇称的张量流.

反对称的 ${\sigma }^{\mu \nu }$ 只有 ${\sigma }^{0i},{\sigma }^{ij}$ 两类分量, 二者的宇称分别为 $-,+.$ 往 $\Gamma$ 里补个 ${\gamma }^5$ 上去也就只是把两类分量的宇称对调一下, 是等价的.

A Lie group is a group which is also a manifold that satisfying the following two conditions:

(a) 群乘映射 $G\times G\to G$ is $C^{\infty }$

(b) 求逆映射 $G\to G$ is $C^{\infty }$

开头第一句中的 manifold 指的从来都是实流形, 哪怕是幺正群对应的流形. 因为这句话太洗脑了, 就不翻译了.

Symmetric Part:

$T_{\left(ab\right)}:=\frac{1}{2}\left(T_{ab}+T_{ba}\right),T_{\left(abc\right)}:=\frac{1}{3!}\left(T_{abc}+T_{acb}+T_{bac}+T_{bca}+T_{cab}+T_{cba}\right),\cdot \cdot \cdot$

Antisymmetric Part:

$T_{\left[ab\right]}:=\frac{1}{2}\left(T_{ab}-T_{ba}\right),T_{\left[abc\right]}:=\frac{1}{3!}\left(T_{abc}-T_{acb}-T_{bac}+T_{bca}+T_{cab}-T_{cba}\right),\cdot \cdot \cdot$

Claims:

$\left(a\right).T_{abc}=T_{\left(abc\right)}\Rightarrow T_{abc}=T_{acb}=T_{bac}=T_{bca}=T_{cab}=T_{cba}$

$\left(b\right).T_{abc}=T_{\left[abc\right]}\Rightarrow T_{abc}=-T_{acb}=-T_{bac}=T_{bca}=T_{cab}=-T_{cba}$

$\left(c\right).T_{\left[abc\right]}{S}^{abc}=T_{\left[abc\right]}{S}^{\left[abc\right]}=T_{abc}{S}^{\left[abc\right]}$

$\left(d\right).T_{\left[a\left[bc\right]\right]}=T_{\left[abc\right]}$

$\left(e\right).T_{\left[\left(ab\right)c\right]}=0$

$\left(f\right).T_{\left[abc\right]}{S}^{\left(abc\right)}=0$

\left( g \right).\left\{ \begin{align} & {T}_{abc}={T}_{\left( abc \right)}\Rightarrow {T}_{\left[ abc \right]}=0 \\ & {T}_{abc}={T}_{\left[ abc \right]}\Rightarrow {T}_{\left( abc \right)}=0 \\ \end{align} \right.

General Linear Group:

未注明 $\mathbb{R},\mathbb{C}$ 一般默认实空间.

$\text{GL}\left(m,\mathbb{R}\right):=\left\{T:V_{real}\to V_{real}|T可逆\right\}$

为非连通流形, 实矢量空间 $V_{real}$ 为 $m$ 维.

$\text{GL}\left(m,\mathbb{C}\right):=\left\{T:V_{complex}\to V_{complex}|T可逆\right\}$

为连通流形, 复矢量空间 $V_{complex}$ 为 $m$ 维.

Orthogonal Group:

\text{O}\left( m \right):=\left\{ {Z}^{a}_{b}\in {\mathcal{T}_{V}\left( 1,1 \right)|\ \ {g}_{cd}={g}_{ab}{Z}^{a}_{c}{Z}^{b}_{d} \right\}\subset \text{GL}\left( \text{m},\mathbb{R} \right)

这里的条件称为保度规条件, 正交群保的度规 ${g}_{ab}$ 是正定度规[[1]](#ref_1). 显然有 $Z^T=Z^{-1}$ , $\left|Z\right|=\pm 1$ 说明是一个非连通流形.

等价定义: \text{O}\left( m \right):=\left\{ Z|Z\in \text{GL}\left( m,\mathbb{R} \right),{Z}^{\text{T}Z=I \right\}

其 lie algebra $\mathcal{O}\left(m\right)=\left\{m\times m实阵A|A^T=-A\right\}$

$\text{O}\left( 2 \right):=\left\{ \left. Z\left( \alpha \right)=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right],\ {Z}'\left( {\alpha }'} \right)=\left[ \begin{matrix} \cos {\alpha }' & \sin {\alpha }' \\ \sin {\alpha }' & -\cos {\alpha }' \\ \end{matrix} \right]\ \right|\ \alpha \in \left[ 0,2\pi \right) \right\}$ 默认笛卡尔坐标系下的二维欧氏空间, $Z,{Z}'$分别为转动与反射后转动.$\text{O}\left(2\right)$ 作为 manifold 是两个圆周.

Indefinite Orthogonal Group：

度规在一组标准正交基表达如下, 其中 $m^{\prime }$ 为 $-1$ 的个数, $m^{\prime \prime }$ 为 $1$ 的个数.

$g_{ab}=\left[\begin{array}{cccccc}-1& & & & & \\ & \ddots & & & & \\ & & -1& & & \\ & & & 1& & \\ & & & & \ddots & \\ & & & & & 1\end{array}\right]$

\text{O}\left( {m}',{m}'' \right):=\left\{ \Lambda _{\ \ b}^{a}\in {\mathcal{T}_{V}\left( 1,1 \right)|\ \ {g}_{cd}={g}_{ab}\Lambda _{\ \ c}^{a}\Lambda _{\ \ d}^{b} \right\}

与正交群的要求类似, 也是全体保度规的 tensor. 显然 $\text{O}\left(0,m\right)\equiv \text{O}\left(m\right)$ 即正交群. 不难发现 $g_{cd}=g_{ab}{\Lambda }_c^a{\Lambda }_d^b\Rightarrow g_{\mu \nu }=g_{\mu \nu }{\Lambda }_{\sigma }^{\mu }{\Lambda }_{\rho }^{\nu }={\left({\Lambda }^T\right)}_{\sigma }^{\mu }{g}_{\mu \nu }{\Lambda }_{\rho }^{\nu }\Rightarrow g={\Lambda }^{T}g\Lambda$

Lorentz group:

$\text{L}\equiv \text{O}\left(1,3\right)$ 群元 $\Lambda$ 满足要求 $\eta ={\Lambda }^T\eta \Lambda$

洛伦兹度规 ${\eta }_{ab}$ 就是只有一个 $-1$ 的 $g_{ab}$ .

Unitary Group:

$\text{U}\left(m\right):=\left\{U\in GL\left(m,\mathbb{C}\right)|U^{†}U=I\right\}$

这里的条件称为保(复)内积条件. 显然有 $U^{†}=U^{-1}$ , $\left|U\right|=e^{i\phi }\phi \in \mathbb{R}$ Unitary group 都是连通 manifold.

其 lie algebra$\mathcal{U}\left(m\right)=\left\{m\times m复阵A|A^{†}=-A\right\}$

其维度: $\dim \text{U}\left(m\right)=\dim \mathcal{U}\left(m\right)=m^2$

Special Group:

$\text{S}\text{L}\left(m\right):=\left\{T\in GL\left(m\right)|\left|T\right|=1\right\}\subset \text{G}\text{L}\left(m\right)$

---

$\text{S}\text{O}\left(m\right):=\left\{Z\in O\left(m\right)|\left|Z\right|=1\right\}\subset \text{O}\left(m\right)$

其 lie algebra $\mathcal{S}\mathcal{O}\left(m\right)=\left\{m\times m实阵A|A^T=-A\right\}=\mathcal{O}\left(m\right)$

$\text{S}\text{O}\left(2\right):=\left\{I,Z\left(\alpha \right)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\right\}$

$\text{O}\left(2\right)$ 作为 manifold 是两个圆周, 其中一个包含恒等元的是他的子群 $\text{SO}\left(2\right)$ .

$\text{S}\text{O}\left(3\right):=\left\{绕某过原点轴转动某个角度的所有映射\right\}$

其 manifold 是个半径为 $\pi$ 的对径认同的三维实心球. 补充: 球内点到球心距离大小表征旋转的角度大小, 与球心连线表征转轴.

---

$\text{S}\text{U}\left(m\right):=\left\{U\in U\left(m\right)|\left|U\right|=1\right\}\subset \text{U}\left(m\right)$

其 lie algebra$\mathcal{S}\mathcal{U}\left(m\right)=\left\{m\times m复阵A|A^{†}=-A,trA=0\right\}\subset \mathcal{U}\left(m\right)$

与 $O\left(m\right)$ 和 $SO\left(m\right)$ 对应的 $\mathcal{S}\mathcal{O}\left(m\right)=\mathcal{O}\left(m\right)$ 情况不一样, 这里 $\mathcal{S}\mathcal{U}\left(m\right)\ne \mathcal{U}\left(m\right)$.

其维度: $\dim \text{S}\text{U}\left(m\right)=\dim \mathcal{S}\mathcal{U}\left(m\right)=m^2-1$

由 $\dim \mathcal{U}\left(m\right)=m^2$ 与结论" $\mathcal{S}\mathcal{U}\left(m\right)$ 内元素 $A$ 对角元是纯虚数且 $\text{t}\text{r}A=0$ " 可知 $\dim \mathcal{S}\mathcal{U}\left(m\right)=m^2-1$ 是一个很显然的结论.

$\mathcal{S}\mathcal{U}\left(2\right)$ : 其基矢为泡利矩阵 $\left\{{\tau }_1=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],{\tau }_2=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right],{\tau }_3=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\right\}$

一般为了结构常数好看, 实际上会选择 $\left\{E_i=-\frac{i}{2}{\tau }_i|i=1,2,3\right\}$ 作为基矢. 这样选择基矢会得到满足具有轮换对称性的表达式 $\left[ {E}_{1},{E}_{2} \right]={E}_{3}$ .

结构张量(Structure Tensors)与结构常数(Structure Constants):

A. 结构张量(Structure Tensors):

给定李代数有与之对应的李括号 $\left[\cdot ,\cdot \right]$ , 其本质是一个李代数上的双线性映射 $\mathcal{V}\times \mathcal{V}\to \mathcal{V}$ .[2]

所以本质上来说李括号就是一个 $(1,2)$ 型的张量 $C_{bc}^a$ :

即 ${\left[v,u\right]}^c=C_{ab}^c{v}^a{u}^b,\forall v^a,u^b\in \mathcal{V}$ .

此时我们就称这个张量 $C_{ab}^c$ 为此李代数的结构张量, 其满足两条性质:[3]

$\left(a\right).C_{ab}^c=-C_{ba}^c\left(b\right).C_{a[b}^c{C}_{de]}^a=T_{\left[bde\right]}^c=0$

B. 结构常数(Structure Constants):

李代数就是李群的恒等元的切空间, 就是说它本质上是一个线性空间, 那么对一个线性空间我们总可以选择一组基矢量 $\left\{{\left(e_{\mu }\right)}^a\right\}$ 来刻画这个空间.

我们将结构张量作用在任意两个基矢上可得:

${\left[e_{\mu },e_v\right]}^c=C_{ab}^c{\left(e_{\mu }\right)}^a{\left(e_{\nu }\right)}^b=C_{\mu \nu }^c=C_{\mu \nu }^d{\left(e^{\sigma }\right)}_d{\left(e_{\sigma }\right)}^c=C_{\mu \nu }^{\sigma }{\left(e_{\sigma }\right)}^c$

而这里的 $C_{\mu \nu }^{\sigma }$ 就被称作此李代数的结构常数, 其自然满足性质:

$\left(a\right).C_{\mu \nu }^{\sigma }=-C_{\nu \mu }^{\sigma }\left(b\right).C_{\mu [\nu }^{\sigma }{C}_{\rho \tau ]}^{\mu }=T_{\left[\nu \rho \tau \right]}^{\sigma }=0$

显然, 结构常数是依赖于基底的选取的.

理论上给出一组满足上面两个关系的数, 就可以找到一个李群对应的李代数以其为结构常数.

普通导数算符(Ordinary Derivative Operaters)

符号定义:

{\partial }_{a}T_{\ \ c}^{b}:={\left( d{x}^{\mu } \right)}_{a}{\left( \frac{\partial }{\partial {x}^{\nu } \right)}^{b}{\left( d{x}^{\sigma } \right)}_{c}{\partial }_{\mu }T_{\ \ \sigma }^{\nu }={\left( d{x}^{\mu } \right)}_{a}{\left( \frac{\partial }{\partial {x}^{\nu } \right)}^{b}{\left( d{x}^{\sigma } \right)}_{c}\frac{\partial T_{\ \ \sigma }^{\nu }{\partial {x}^{\mu }

\left\{ \begin{align} & {\partial }_{a}{u}^{a}={\left( d{x}^{\mu } \right)}_{a}{\left( \frac{\partial }{\partial {x}^{\nu } \right)}^{a}{\partial }_{\mu }{u}^{\nu }=\frac{\partial {x}^{\mu }{\partial {x}^{\nu }{\partial }_{\mu }{u}^{\nu }={\partial }_{\nu }{u}^{\nu }=\frac{\partial {u}^{\nu }{\partial {x}^{\nu } \\ & {\partial }_{a}{u}^{b}={\left( d{x}^{\mu } \right)}_{a}{\left( \frac{\partial }{\partial {x}^{\nu } \right)}^{b}{\partial }_{\mu }{u}^{\nu }={\left( d{x}^{\mu } \right)}_{a}{\left( \frac{\partial }{\partial {x}^{\nu } \right)}^{b}\frac{\partial {u}^{\nu }{\partial {x}^{\mu } \\ \end{align} \right. 其中上下标 $\mu$ 也是包含求和操作的, 例: {\partial }_{a}f={\left( d{x}^{\mu } \right)}_{a}{\partial }_{\mu }f={\left( d{x}^{\mu } \right)}_{a}\frac{\partial f}{\partial {x}^{\mu }

矢量场对易子的定义: ${\left[v,u\right]}^a=v^b{\nabla }_b{u}^a-u^b{\nabla }_b{v}^a$

守恒流: ${\partial }_a{J}^a=0\Rightarrow {\partial }_0{J}^0+{\partial }_i{J}^i=\frac{\partial \rho }{\partial t}+\nabla \cdot \vec{J}=0$

Others:

$\left|e^A\right|=e^{trA}$

闭子集的定义是补集为开集的子集.

✦General Linear Groups:

$GL\left(m\right):=$ $m$ 维实矢量空间 $V$ 上全体可逆线性映射 $T$ 的集合.

非连通流形

✦Orthogonal Groups:

O\left( m \right):=\left\{ Z_{\ \ b}^{a}\in {\mathcal{T}_{V}\left( 1,1 \right)|\ \ {g}_{cd}={g}_{ab}Z_{\ \ c}^{a}Z_{\ \ d}^{b} \right\}\subset GL\left( m \right)

显然有: $Z^T=Z^{-1}$ , $\left|Z\right|=\pm 1$ (说明非连通流形)

$O\left(2\right):=$包括恒等映射 $I$ 与 \left\{ \begin{align} &逆时针转动 Z\left( \alpha \right)=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ &反射后转动 {Z}'\left( {\alpha }'} \right)=\left[ \begin{matrix} \cos {\alpha }' & \sin {\alpha }' \\ \sin {\alpha }' & -\cos {\alpha }' \\ \end{matrix} \right] \\ \end{align} \right.

默认笛卡尔坐标系下的二维欧氏空间.

✦Indefinite orthogonal group：

度规在一组标准正交基表达如下, 其中 $m^{\prime }$ 为 $-1$ 的个数, $m^{\prime \prime }$ 为 $1$ 的个数.

$g_{ab}=\left[\begin{array}{cccccc}-1& & & & & \\ & \cdot \cdot \cdot & & & & \\ & & -1& & & \\ & & & 1& & \\ & & & & \cdot \cdot \cdot & \\ & & & & & 1\end{array}\right]$

O\left( {m}',{m}'' \right):=\left\{ \Lambda _{\ \ b}^{a}\in {\mathcal{T}_{V}\left( 1,1 \right)|\ \ {g}_{cd}={g}_{ab}\Lambda _{\ \ c}^{a}\Lambda _{\ \ d}^{b} \right\}

与正交群的要求类似, 也是全体保度规的 tensor. 显然 $O\left(0,m\right)\equiv O\left(m\right)$ 即正交群. 不难发现 $g_{cd}=g_{ab}{\Lambda }_c^a{\Lambda }_d^b\Rightarrow g_{\mu \nu }=g_{\mu \nu }{\Lambda }_{\sigma }^{\mu }{\Lambda }_{\rho }^{\nu }={\left({\Lambda }^T\right)}_{\sigma }^{\mu }{g}_{\mu \nu }{\Lambda }_{\rho }^{\nu }\Rightarrow g={\Lambda }^{T}g\Lambda$

上式的特例Lorentz group:$L\equiv O\left(1,3\right)$ 群元 $\Lambda$ 满足要求 $\eta ={\Lambda }^T\eta \Lambda$

洛伦兹度规 ${\eta }_{ab}$ 就是只有一个 $-1$ 的 $g_{ab}$

✦Special Groups:

$SL\left(m\right):=\left\{T\in GL\left(m\right)|\left|T\right|=1\right\}\subset GL\left(m\right)$

$SO\left(m\right):=\left\{Z\in O\left(m\right)|\left|Z\right|=1\right\}\subset O\left(m\right)$

其 manifold 是俩圆, 其中一个包含恒等元是他的子群 $SO\left(2\right)$ .

$SO\left(2\right):\left\{I,Z\left(\alpha \right)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\right\}$

其 manifold 是个圆.

$SO\left(3\right):\left\{绕某过原点轴转动某个角度的所有映射\right\}$

其 manifold 是个半径为 $\pi$ 的对径认同的三维实心球. 补充: 球内点到球心距离大小表征旋转的角度大小, 与球心连线表征转轴.

$F_M\left(k,l\right)$ 指的是流形 $M$ [4]上的 $\left(k,l\right)$ 型张量场的集合, F for field.

$T_V\left(k,l\right)$ 指的向量空间 $V$ [5]上的 $\left(k,l\right)$ 型张量的集合, T for tensor.

\begin{align} & T_{\ \ \ \nu }^{'\mu }=T_{\ \ \ b}^{a}{\left( d{x}'}^{\mu } \right)}_{a}{\left( \frac{\partial }{\partial {x}'}^{\nu } \right)}^{b}=T_{\ \ b}^{a}\frac{\partial {x}'}^{\mu }{\partial {x}^{\rho }{\left( d{x}^{\rho } \right)}_{a}\frac{\partial {x}^{\sigma }{\partial {x}'}^{\nu }{\left( \frac{\partial }{\partial {x}^{\sigma } \right)}^{b}=\frac{\partial {x}'}^{\mu }{\partial {x}^{\rho }\frac{\partial {x}^{\sigma }{\partial {x}'}^{\nu }T_{\ \ \sigma }^{\rho } \\ & {\left( d{x}'}^{\mu } \right)}_{a}={\left( d{x}'}^{\mu } \right)}_{b}{\left( d{x}^{\rho } \right)}_{a}{\left( \frac{\partial }{\partial {x}^{\rho } \right)}^{b}=\frac{\partial {x}'}^{\mu }{\partial {x}^{\rho }{\left( d{x}^{\rho } \right)}_{a} \\ & {\left( \frac{\partial }{\partial {x}'}^{\nu } \right)}^{b}={\left( \frac{\partial }{\partial {x}'}^{\nu } \right)}^{a}{\left( d{x}^{\sigma } \right)}_{a}{\left( \frac{\partial }{\partial {x}^{\sigma } \right)}^{b}=\frac{\partial {x}^{\sigma }{\partial {x}'}^{\nu }{\left( \frac{\partial }{\partial {x}^{\sigma } \right)}^{b} \\ \end{align}

高量:

证明 $e^{A}B{e}^{-A}=\sum _{n=0}\frac{1}{n!}\left[A^{\left(n\right)},B\right]$

构造 $f\left(x\right)=e^{Ax}B{e}^{-Ax}$

$\frac{d}{dx}f\left(x\right)=e^{Ax}AB{e}^{-Ax}-e^{Ax}BA{e}^{-Ax}=e^{Ax}\left(AB-BA\right)e^{-Ax}=e^{Ax}\left[A,B\right]e^{-Ax}$

\frac{d}^{2}{d{x}^{2}f\left( x \right)={e}^{Ax}A\left[ A,B \right]{e}^{-Ax}-{e}^{Ax}\left[ A,B \right]A{e}^{-Ax}={e}^{Ax}\left[ A,\left[ A,B \right] \right]{e}^{-Ax}

$\cdot \cdot \cdot$

\frac{d}^{n}{d{x}^{n}f\left( x \right)={e}^{Ax}\left[ {A}^{\left( n \right)},B \right]{e}^{-Ax}

上面可以使用数学归纳法证明.

{e}^{Ax}B{e}^{-Ax}=f\left( x \right)=\sum\limits_{n=0}{\left. \frac{1}{n!}\frac{d}^{n}{d{x}^{n}f\left( x \right) \right|}_{x=0}{x}^{n}=\sum\limits_{n=0}{\frac{1}{n!}\left[ {A}^{\left( n \right)},B \right]{x}^{n}

令上式 $x=1$ 即得证.

喀兴林书上二次量子化部分定义的对称或反对称基矢为:

${|n;{\lambda }_a,{\lambda }_b,\cdot \cdot \cdot ,{\lambda }_z⟩}^{\prime }=\frac{1}{n!}\sum _s{P}_s{\epsilon }_s^p{|{\lambda }_a⟩}_1{|{\lambda }_b⟩}_2\cdot \cdot \cdot {|{\lambda }_z⟩}_n$

我则认为应该定义做:

\left| n;{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle =\frac{1}{\sqrt{n!{n}_{a}!{n}_{b}!\cdot \cdot \cdot {n}_{z}!}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\left| {\lambda }_{a} \right\rangle }_{1}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot {\left| {\lambda }_{z} \right\rangle }_{n}

二者对应关系为:

{\left| n;{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle }^{\prime }=\sqrt{\frac{n}_{a}!{n}_{b}!\cdot \cdot \cdot {n}_{z}!}{n!}\left| n;{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle

这样一来就会直接改变产生湮灭算符的定义:

咯书为: $a^{†}\left({\lambda }_i\right){|n;{\lambda }_a,{\lambda }_b,\cdot \cdot \cdot ,{\lambda }_z⟩}^{\prime }=\sqrt{n+1}{|n+1;{\lambda }_i,{\lambda }_a,{\lambda }_b,\cdot \cdot \cdot ,{\lambda }_z⟩}^{\prime }$

利用我的定义的矢量就要改为: {a}^{\dagger }\left( {\lambda }_{i} \right)\sqrt{\frac{n}_{a}!{n}_{b}!\cdot \cdot \cdot {n}_{z}!}{n!}\left| n;{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle =\sqrt{n+1}\left[ \begin{align} & {\delta }_{ia}\sqrt{\frac{\left( {n}_{a}+1 \right)!{n}_{b}!\cdot \cdot \cdot {n}_{z}!}{\left( n+1 \right)!}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ & +{\delta }_{ib}\sqrt{\frac{n}_{a}!\left( {n}_{b}+1 \right)!\cdot \cdot \cdot {n}_{z}!}{\left( n+1 \right)!}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ & +\cdot \cdot \cdot \\ & +{\delta }_{iz}\sqrt{\frac{n}_{a}!{n}_{b}!\cdot \cdot \cdot \left( {n}_{z}+1 \right)!}{\left( n+1 \right)!}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ \end{align} \right]

化简后并不复杂 {a}^{\dagger }\left( {\lambda }_{i} \right)\left| n;{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle =\left[ \begin{align} & {\delta }_{ia}\sqrt{n}_{a}+1}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ & +{\delta }_{ib}\sqrt{n}_{b}+1}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ & +\cdot \cdot \cdot \\ & +{\delta }_{iz}\sqrt{n}_{z}+1}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ \end{align} \right]

但其实还有一种情况没考虑, 考虑上就是:

$i\notin \left\{a,b,\cdot \cdot \cdot ,z\right\}$ 时:

$a^{†}\left({\lambda }_i\right)|n;{\lambda }_a,{\lambda }_b,\cdot \cdot \cdot ,{\lambda }_z⟩=|n+1;{\lambda }_i,{\lambda }_a,{\lambda }_b,\cdot \cdot \cdot ,{\lambda }_z⟩$

$i\in \left\{a,b,\cdot \cdot \cdot ,z\right\}$ 时:

{a}^{\dagger }\left( {\lambda }_{i} \right)\left| n;{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle =\left[ \begin{align} & \ \ \ \ \ {\delta }_{ia}\sqrt{n}_{a}+1}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ & +{\delta }_{ib}\sqrt{n}_{b}+1}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ & +\cdot \cdot \cdot \\ & +{\delta }_{iz}\sqrt{n}_{z}+1}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle \\ \end{align} \right]

或者写作 {a}^{\dagger }\left( {\lambda }_{i} \right)\left| n;{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle =\sqrt{n}_{i}+1}\left| n+1;{\lambda }_{i},{\lambda }_{a},{\lambda }_{b},\cdot \cdot \cdot ,{\lambda }_{z} \right\rangle

量子力学:

双电子体系的自旋耦合:

单电子自旋算符记作 \vec{S}={S}_{x}{\vec{e}_{x}+{S}_{y}{\vec{e}_{y}+{S}_{z}{\vec{e}_{z}

其中自旋 $z$ 方向分量算符 $S_z$ 对应本征值 $\frac{\hslash }{2},-\frac{\hslash }{2}$ 的本征态分别记作 $|\uparrow ⟩,|\downarrow ⟩$

现在定义算符 $\vec{\sigma }=\frac{2}{\hslash }\vec{S}$ 即 $S_i=\frac{\hslash }{2}{\sigma }_i$

不难证明在 $S_z$ 表象(或 ${\sigma }_z$ 表象)下, 相关算符与本征态的矩阵表达如下:

${\sigma }_x=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],{\sigma }_y=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right],{\sigma }_z=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right],|\uparrow ⟩=\left[\begin{array}{c}1\\ 0\end{array}\right],|\downarrow ⟩=\left[\begin{array}{c}0\\ 1\end{array}\right]$

通过矩阵运算可以得到下列关系:

{\sigma }_{x}\left| \uparrow \right\rangle =\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 1 \\ 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\ 1 \\ \end{matrix} \right]=\left| \downarrow \right\rangle \ ,\ {\sigma }_{x}\left| \downarrow \right\rangle =\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{align} & 0 \\ & 1 \\ \end{align} \right]=\left[ \begin{matrix} 1 \\ 0 \\ \end{matrix} \right]=\left| \uparrow \right\rangle

${\sigma }_y|\uparrow ⟩=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=i\left[\begin{array}{c}0\\ 1\end{array}\right]=i|\downarrow ⟩,{\sigma }_y|\downarrow ⟩=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=-i\left[\begin{array}{c}1\\ 0\end{array}\right]=-i|\uparrow ⟩$

双电子体系的对称基矢量有如下四种:

出于方便起见, 我们永远将电子 $1$ 的算符与态放在张量积的左边, 即 $|\uparrow ⟩|\downarrow ⟩={|\uparrow ⟩}_1\otimes {|\downarrow ⟩}_2$

三个对称三重态 \left\{ \begin{align} & \left| {\varphi }_{1} \right\rangle =\left| \uparrow \right\rangle \left| \uparrow \right\rangle \\ & \left| {\varphi }_{2} \right\rangle =\left| \downarrow \right\rangle \left| \downarrow \right\rangle \\ & \left| {\varphi }_{3} \right\rangle =\frac{1}{\sqrt{2}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle +\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right) \\ \end{align} \right.

与一个反对称单重态 \left| {\varphi }_{4} \right\rangle =\frac{1}{\sqrt{2}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle -\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right)

容易证明它们两两正交, 全部归一, 双电子自旋是四维体系故不可能存在更多基矢.

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对于哈密顿量为 \vec{H}=\alpha {\vec{S}_{1}\cdot {\vec{S}_{2}=\alpha \left( {S}_{1x}{S}_{2x}+{S}_{1y}{S}_{2y}+{S}_{1z}{S}_{2z} \right) 的体系

我们可以通过关系 {\left( {\vec{S}_{1}+{\vec{S}_{2} \right)}^{2}=S_{1}^{2}+S_{2}^{2}+2{\vec{S}_{1}\cdot {\vec{S}_{2} 得到 {\vec{S}_{1}\cdot {\vec{S}_{2} 的表达式:

{\vec{S}_{1}\cdot {\vec{S}_{2}=\frac{1}{2}\left[ {\left( {\vec{S}_{1}+{\vec{S}_{2} \right)}^{2}-S_{1}^{2}-S_{2}^{2} \right]=\frac{1}{2}\left[ {\left( {\vec{S}_{1}+{\vec{S}_{2} \right)}^{2}-\frac{1}{2}\left( \frac{3}{2} \right){\hbar }^{2}-\frac{1}{2}\left( \frac{3}{2} \right){\hbar }^{2} \right]=\frac{1}{2}\left[ {\left( {\vec{S}_{1}+{\vec{S}_{2} \right)}^{2}-\left( \frac{3}{2} \right){\hbar }^{2} \right]

显然 {S}^{2}={\left( {\vec{S}_{1}+{\vec{S}_{2} \right)}^{2} 的本征值有 $0,2{\hslash }^2$ 所以分立的能谱有两条谱线:

E=\left\{ \begin{align} & {E}_{t}=\frac{1}{2}\left[ 2-\frac{3}{2} \right]{\hbar }^{2}=\frac{1}{4}{\hbar }^{2}\alpha \ \ \ \ \ {S}^{2}=2{\hbar }^{2} \\ & {E}_{s}=\frac{1}{2}\left[ 0-\frac{3}{2} \right]{\hbar }^{2}=-\frac{3}{4}{\hbar }^{2}\alpha \ \ \ {S}^{2}=0 \\ \end{align} \right. 其中 t for triplet, s for singlet.

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为了进一步理解, 可以直接证明四个对称或反对称的基矢都是 {\vec{S}_{1}\cdot {\vec{S}_{2} 的本征矢:

\begin{align} & {\vec{S}_{1}\cdot {\vec{S}_{2}\left| {\varphi }_{1} \right\rangle =\left( {S}_{1x}{S}_{2x}+{S}_{1y}{S}_{2y}+{S}_{1z}{S}_{2z} \right)\left| \uparrow \right\rangle \left| \uparrow \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={S}_{1x}{S}_{2x}\left| \uparrow \right\rangle \left| \uparrow \right\rangle +{S}_{1y}{S}_{2y}\left| \uparrow \right\rangle \left| \uparrow \right\rangle +{S}_{1z}{S}_{2z}\left| \uparrow \right\rangle \left| \uparrow \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\hbar }^{2}{4}\left| \downarrow \right\rangle \left| \downarrow \right\rangle -\frac{\hbar }^{2}{4}\left| \downarrow \right\rangle \left| \downarrow \right\rangle +\frac{\hbar }^{2}{4}\left| \uparrow \right\rangle \left| \uparrow \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\hbar }^{2}{4}\left| {\varphi }_{1} \right\rangle \\ \end{align} \begin{align} & {\vec{S}_{1}\cdot {\vec{S}_{2}\left| {\varphi }_{2} \right\rangle =\left( {S}_{1x}{S}_{2x}+{S}_{1y}{S}_{2y}+{S}_{1z}{S}_{2z} \right)\left| \downarrow \right\rangle \left| \downarrow \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\hbar }^{2}{4}\left| \uparrow \right\rangle \left| \uparrow \right\rangle -\frac{\hbar }^{2}{4}\left| \uparrow \right\rangle \left| \uparrow \right\rangle +\frac{\hbar }^{2}{4}\left| \downarrow \right\rangle \left| \downarrow \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\hbar }^{2}{4}\left| {\varphi }_{2} \right\rangle \\ \end{align} \begin{align} & {\vec{S}_{1}\cdot {\vec{S}_{2}\left| {\varphi }_{3} \right\rangle =\left( {S}_{1x}{S}_{2x}+{S}_{1y}{S}_{2y}+{S}_{1z}{S}_{2z} \right)\frac{1}{\sqrt{2}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle +\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\sqrt{2}\left[ {S}_{1x}{S}_{2x}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle +\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right)+{S}_{1y}{S}_{2y}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle +\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right)+{S}_{1z}{S}_{2z}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle +\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right) \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\sqrt{2}\left[ \frac{\hbar }^{2}{4}\left( \left| \downarrow \right\rangle \left| \uparrow \right\rangle +\left| \uparrow \right\rangle \left| \downarrow \right\rangle \right)+\frac{\hbar }^{2}{4}\left( \left| \downarrow \right\rangle \left| \uparrow \right\rangle +\left| \uparrow \right\rangle \left| \downarrow \right\rangle \right)-\frac{\hbar }^{2}{4}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle +\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right) \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\hbar }^{2}{4}\left| {\varphi }_{3} \right\rangle \\ \end{align} \begin{align} & {\vec{S}_{1}\cdot {\vec{S}_{2}\left| {\varphi }_{4} \right\rangle =\left( {S}_{1x}{S}_{2x}+{S}_{1y}{S}_{2y}+{S}_{1z}{S}_{2z} \right)\frac{1}{\sqrt{2}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle -\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\sqrt{2}\left[ {S}_{1x}{S}_{2x}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle -\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right)+{S}_{1y}{S}_{2y}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle -\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right)+{S}_{1z}{S}_{2z}\left( \left| \uparrow \right\rangle \left| \downarrow \right\rangle -\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right) \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\sqrt{2}\left[ \frac{\hbar }^{2}{4}\left( \left| \downarrow \right\rangle \left| \uparrow \right\rangle -\left| \uparrow \right\rangle \left| \downarrow \right\rangle \right)+\frac{\hbar }^{2}{4}\left( \left| \downarrow \right\rangle \left| \uparrow \right\rangle -\left| \uparrow \right\rangle \left| \downarrow \right\rangle \right)+\frac{\hbar }^{2}{4}\left( -\left| \uparrow \right\rangle \left| \downarrow \right\rangle +\left| \downarrow \right\rangle \left| \uparrow \right\rangle \right) \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{3{\hbar }^{2}{4}\left| {\varphi }_{4} \right\rangle \\ \end{align}

显然他们也是算符 {S}^{\text{2}=\frac{3}{2}{\hbar }^{2}+2{\vec{S}_{1}\cdot {\vec{S}_{2} 的本征矢.

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概率流密度与动量的联系:

\begin{align} & \int{\vec{j}\cdot d\vec{r}=\frac{\hbar }{2mi}\int{\left( {\psi }^{*}\nabla \psi -\nabla {\psi }^{*}\psi \right)\cdot d\vec{r} \\ & \ \ \ \ \ \ \ \ \ \ =\frac{1}{2m}\int{\left( \left\langle \psi | {\vec{r} \right\rangle \left\langle {\vec{r} \right|P\left| \psi \right\rangle +\left\langle \psi \right|P\left| {\vec{r} \right\rangle \left\langle {\vec{r} | \psi \right\rangle \right)\cdot d\vec{r} \\ & \ \ \ \ \ \ \ \ \ \ =\frac{1}{2m}\left[ \int{\left\langle \psi | {\vec{r} \right\rangle \left\langle {\vec{r} \right|P\left| \psi \right\rangle \cdot d\vec{r}+\int{\left\langle \psi \right|P\left| {\vec{r} \right\rangle \left\langle {\vec{r} | \psi \right\rangle \cdot d\vec{r} \right] \\ & \ \ \ \ \ \ \ \ \ \ =\frac{1}{2m}\left[ \left\langle \psi \right|P\left| \psi \right\rangle +\left\langle \psi \right|P\left| \psi \right\rangle \right]=\frac{\left\langle P \right\rangle }{m} \\ \end{align}

其中第二步可看作 \left\langle {\vec{r} \right|P\left| \psi \right\rangle =\frac{\hbar }{i}\nabla \left\langle {\vec{r} | \psi \right\rangle \Rightarrow \left\langle \psi \right|{P}^{\dagger }\left| {\vec{r} \right\rangle =-\frac{\hbar }{i}\nabla \left\langle \psi | {\vec{r} \right\rangle

可以严格推导 \left\langle \psi \right|{P}^{\dagger }\left| {\vec{r} \right\rangle =\int{\left\langle \psi | {\vec{r}'} \right\rangle \left\langle {\vec{r}'} \right|{P}^{\dagger }\left| {\vec{r} \right\rangle d{\vec{r}'} \begin{align} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\psi }^{*}\left( {\vec{r}'} \right)\frac{\hbar }{i}\nabla \delta \left( \vec{r}-{\vec{r}' \right)d{\vec{r}'} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\psi }^{*}\left( {\vec{r}'} \right)\frac{\hbar }{i}d\delta \left( \vec{r}-{\vec{r}' \right)} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{\hbar }{i}\int{\delta \left( \vec{r}-{\vec{r}' \right)\frac{d{\psi }^{*}\left( {\vec{r}'} \right)}{d{\vec{r}'}d{\vec{r}'} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{\hbar }{i}\int{\delta \left( \vec{r}-{\vec{r}' \right)\nabla {\psi }^{*}\left( {\vec{r}'} \right)d{\vec{r}'} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{\hbar }{i}\nabla \left\langle \psi | {\vec{r} \right\rangle \\ \end{align}

热力学与统计物理:

\left\{ \begin{align} & \frac{\omega }_{l}\mp {a}_{l}{a}_{l}\text{=}{e}^{\alpha +\beta {\varepsilon }_{l} \\ & \frac{\omega }_{l}\mp {a}_{l}{\omega }_{l}\text{=}\frac{1}{1\pm {e}^{-\alpha -\beta {\varepsilon }_{l} \\ \end{align} \right.

{a}_{l}=\frac{\omega }_{l}{e}^{\alpha +\beta {\varepsilon }_{l}\pm 1}\Rightarrow {\omega }_{l}\mp {a}_{l}={\omega }_{l}\left( \frac{e}^{\alpha +\beta {\varepsilon }_{l}{e}^{\alpha +\beta {\varepsilon }_{l}\pm 1} \right)={e}^{\alpha +\beta {\varepsilon }_{l}{a}_{l}

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在用巨正则分布计算物理量的时候通常都要先算出 $\bar{N}$ 来, 因为其他物理量求完总会有 $\bar{N}$, 再者就是需要 $\bar{N}\text{=}{e}^{-\alpha }Z$ 的关系来确定 $\alpha$ 的值.有时$\alpha$是从大热源确定的, 比如吸附现象的大热源: 气体环境.

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正则系综理论:\left\{ \begin{align} & {\rho }_{s}=\frac{1}{Z}{e}^{-\beta {E}_{s}\ \ \ \ Z=\sum\limits_{s}{e}^{-\beta {E}_{s} \\ & \bar{E}=\sum\limits_{s}{E}_{s}{\rho }_{s}=-\frac{\partial \ln z}{\partial \beta } \\ & {\bar{Y}_{i}=\sum\limits_{s}{\frac{\partial {E}_{s}{\partial {y}_{i}{\rho }_{s}=-\frac{1}{\beta }\frac{\partial \ln Z}{\partial {y}_{i} \\ & S=k\left[ \ln Z-\beta \frac{\partial \ln Z}{\partial \beta } \right] \\ \end{align} \right.

玻尔兹曼分布: \left\{ \begin{align} & {a}_{l}={\omega }_{l}{e}^{-\alpha -\beta {\varepsilon }_{l}\ \ \ \ {Z}_{1}=\sum\limits_{l}{\omega }_{l}{e}^{-\beta {\varepsilon }_{l} \\ & N=\sum\limits_{l}{a}_{l}={e}^{-\alpha }{Z}_{1} \\ & E=\sum\limits_{l}{\varepsilon }_{l}{a}_{l}=-N\frac{\partial \ln {Z}_{1}{\partial \beta } \\ & {Y}_{i}=\sum\limits_{l}{\frac{\partial {\varepsilon }_{l}{\partial {y}_{i}{a}_{l}=-\frac{N}{\beta }\frac{\partial \ln {Z}_{1}{\partial {y}_{i} \\ & S=Nk\left[ \ln {Z}_{1}-\beta \frac{\partial \ln {Z}_{1}{\partial \beta } \right]-\left( k\ln N! \right) \\ \end{align} \right.

巨正则系综理论: \left\{ \begin{align} & {\rho }_{Ns}=\frac{1}{\Xi }{e}^{-\alpha N-\beta {E}_{s}\ \ \ \ \ \Xi =\sum\limits_{N}{\sum\limits_{s}{e}^{-\alpha N-\beta {E}_{s} \\ & \bar{N}=\sum\limits_{N}{\sum\limits_{s}{N{\rho }_{Ns}=-\frac{\partial \ln \Xi }{\partial \alpha } \\ & \bar{E}=\sum\limits_{N}{\sum\limits_{s}{E}_{s}{\rho }_{Ns}=-\frac{\partial \ln \Xi }{\partial \beta } \\ & {\bar{Y}_{i}=\sum\limits_{N}{\sum\limits_{s}{\frac{\partial {E}_{s}{\partial {y}_{i}{\rho }_{Ns}\text{=}-\frac{1}{\beta }\frac{\partial \ln \Xi }{\partial {y}_{i} \\ & S=k\left[ \ln \Xi -\alpha \frac{\partial \ln \Xi }{\partial \alpha }-\beta \frac{\ln \Xi }{\partial \beta } \right] \\ \end{align} \right.

费米/玻色分布: \left\{ \begin{align} & {a}_{l}=\frac{\omega }_{l}{e}^{\alpha +\beta {\varepsilon }_{l}\pm 1}\ \ \ \ \Xi =\prod\limits_{l}{\left( 1\pm {e}^{-\alpha -\beta {\varepsilon }_{l} \right)}^{\pm {\omega }_{l} \\ & \ln \Xi =\sum\limits_{l}{\pm {\omega }_{l}\ln \left( 1\pm {e}^{-\alpha -\beta {\varepsilon }_{l} \right)} \\ & N\text{=}\sum\limits_{l}{a}_{l}=\sum\limits_{l}{\frac{\omega }_{l}{e}^{-\alpha -\beta {\varepsilon }_{l}{1\pm {e}^{-\alpha -\beta {\varepsilon }_{l}=-\frac{\partial \ln \Xi }{\partial \alpha } \\ & E=\sum\limits_{l}{\varepsilon }_{l}{a}_{l}=\sum\limits_{l}{\varepsilon }_{l}\frac{\omega }_{l}{e}^{-\alpha -\beta {\varepsilon }_{l}{1\pm {e}^{-\alpha -\beta {\varepsilon }_{l}=-\frac{\partial \ln \Xi }{\partial \beta } \\ & {Y}_{i}=\sum\limits_{l}{\frac{\partial {\varepsilon }_{l}{\partial {y}_{i}{a}_{l}=-\frac{1}{\beta }\frac{\partial \ln \Xi }{\partial {y}_{i} \\ & S=k\left[ \ln \Xi -\alpha \frac{\partial \ln \Xi }{\partial \alpha }-\beta \frac{\partial \ln \Xi }{\partial \beta } \right] \\ \end{align} \right.

固体物理:

晶格振动:

格子波矢数目=原胞数 $N$ 波矢数=\frac{V}{\left( \text{2}\pi \right)}^{\text{3}\int_{第一布里渊区}{d\vec{q}=\frac{N{\Omega }_{原胞体积}{\left( \text{2}\pi \right)}^{\text{3}\Omega _{倒格原胞体积}^{*}=N

色散关系 \omega =\omega \left( {\vec{q} \right) 数=声学支数+光学支数=一个原胞内的自由度数 $3s$

振动模式 $\omega$ 数=波矢数x色散关系数=固体自由度数 $3Ns$

参考

• ^标准正交基下为单位阵
• ^这个式子的意思是从李代数里给定两个元素就可以确定得到第三个元素的映射.
• ^(b)可以用雅可比恒等式证明.
• ^这就是整个 manifold 上的场了
• ^manifold 上一点的切空间