量子力学基础公式(初版)

文章目录
  1. 1. 可能会常用的数学公式
  2. 2. Schrödinger方程
  3. 3. 常见势场
  4. 4. 常见定理
  5. 5. 常见对易关系
  6. 6. 一维谐振子升降算符
  7. 7. 角动量算符
  8. 8. 自旋角动量算符
  9. 9. 占有数算符
  10. 10. 近似方法

  前言:这篇博客是由于我没活整所以决定写的,意在总结一下做量子力学习题可能会用到的一些常用公式(某人:做题蛆闹麻了【你已经被我预判到了】),至于之后应该会结合各类教材以及个人的理解出个《简明量子力学教程》的系列博客,正如Feynman所言"What I cannot create, I do not understand"。当然会不会真的写就看后人智慧了(笑)

 此外,个人整理的公式不一定全,也不一定有逻辑顺序,我是想到什么就写什么的,没有按常见的章节分类,完全按我一时兴起的分类,所以这只是初版,日后可能会做些许改动与增补。



可能会常用的数学公式

  1. 三角函数:

sinx=eixeix2isinx = \frac{e^{ix}-e^{-ix}}{2i}

cosx=eix+eix2cosx = \frac{e^{ix}+e^{-ix}}{2}

sinαcosβ=12[sin(α+β)+sin(αβ)]sin \alpha cos \beta = \frac{1}{2}[sin(\alpha + \beta)+sin(\alpha - \beta)]

cosαsinβ=12[sin(α+β)sin(αβ)]cos \alpha sin \beta = \frac{1}{2}[sin(\alpha + \beta)-sin(\alpha - \beta)]

cosαcosβ=12[cos(α+β)+cos(αβ)]cos \alpha cos \beta = \frac{1}{2}[cos(\alpha + \beta)+cos(\alpha - \beta)]

sinαsinβ=12[cos(α+β)sin(αβ)]sin \alpha sin \beta = -\frac{1}{2}[cos(\alpha + \beta)-sin(\alpha - \beta)]

  1. 积分公式:

+eax2dx=πa\int_{-\infty}^{+\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}

+x2eax2dx=12aπa\int_{-\infty}^{+\infty}x^2 e^{-ax^2}dx=\frac{1}{2a} \sqrt{\frac{\pi}{a}}

0+xneαxdx=n!αn+1\int_{0}^{+\infty}x^{n}e^{-\alpha x}dx=\frac{n!}{\alpha ^{n+1}}

0+x2neβx2dx=(2n1)!!2nπβ2n+1\int_{0}^{+\infty}x^{2n}e^{-\beta x^2}dx=\frac{(2n-1)!!}{2^n} \sqrt{\frac{\pi}{\beta ^{2n+1}}}

0π2sinnxdx=0π2cosnxdx={n1nn3n2...231nn1nn3n2...3412π2n\int_{0}^{\frac{\pi}{2}}sin ^n x dx=\int_{0}^{\frac{\pi}{2}}cos ^n x dx= \begin{cases} \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdot ... \frac{2}{3} \cdot 1 \qquad n为奇数\\ \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdot ... \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \qquad n为偶数 \end{cases}

  1. 泰勒展开:

1+x=n(1)n1(2n3)!!2n!!xn=1+12x18x2+116x3...\sqrt{1+x} = \sum_{n}(-1)^{n-1}\frac{(2n-3)!!}{2n!!}x^n =1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - ...

ex=n1n!xn=1+x+12!x2+13!x3+...e^x = \sum_{n}\frac{1}{n!}x^n = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + ...

sinx=n(1)n(2n+1)!x2n+1=x13!x3+15!x5...sinx = \sum_{n}\frac{(-1)^n}{(2n+1)!}x^{2n+1}= x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - ...

cosx=n(1)n(2n)!x2n=112!x2+14!x4...cosx = \sum_{n}\frac{(-1)^n}{(2n)!}x^{2n}= 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - ...

ln(1+x)=n(1)nn+1xn+1=x12x2+13x3...ln(1+x)=\sum_{n}\frac{(-1)^n}{n+1}x^{n+1}= x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - ...

11x=nxn=1+x+x2+...\frac{1}{1-x}=\sum_{n}x^n = 1 + x + x^2 + ...

11+x=n(1)nxn=1x+x2...\frac{1}{1+x}=\sum_{n}(-1)^n x^n =1 - x + x^2 - ...



Schrödinger方程

  1. 含时Schrödinger方程:

itΨ(r,t)=H^Ψ(r,t)i\hbar \frac{\partial}{\partial t}\Psi (r,t)=\hat{H}\Psi (r,t)\\

其中:H^=22μ2+V(r,t)\hat{H}=-\frac{\hbar ^2}{2\mu}\nabla ^2 + V(r,t)

  1. 定态Schrödinger方程:

H^Ψ(r)=EΨ(r)\hat{H}\Psi (r)=E\Psi (r)

  1. 时间演化算符:

U^=eiEt/\hat{U}=e^{-iEt/ \hbar}

  1. 概率流密度:

J=i2μ[ψψψψ]J=-\frac{i\hbar}{2\mu}[\psi ^* \nabla \psi - \psi \nabla \psi ^*]

常见势场

  1. 一维无限深势阱(三维作简单乘积即可):

V(x)={00<x<aelseV(x)= \begin{cases} 0 \qquad 0 < x < a \\ \infty \qquad else \\ \end{cases}

其定态能量与波函数为:

En=n2π222μa2n=1,2,3...E_n = \frac{n^2 \pi ^2 \hbar ^2}{2 \mu a^2} \qquad n=1,2,3...

ψn(x)={2asinnπxa0<x<a0else\psi _n (x)= \begin{cases} \sqrt{\frac{2}{a}}sin \frac{n \pi x}{a} \qquad 0 < x < a \\ 0 \qquad else\\ \end{cases}


  1. 一维谐振子势 $V(x)=\frac{1}{2}\mu \omega ^2 x^2 $ 的定态能量与波函数为(三维同理作简单乘积即可):

En=(n+12)ωn=0,1,2,3...E_n = (n + \frac{1}{2})\hbar \omega \qquad n=0,1,2,3...

\ket{n}=N_n e^{-\alpha ^2 x^2 /2}H_n (\alpha x)

其中

α=μω,Nn=απ1/22nn!n=0,1,2,3...\alpha = \sqrt{\frac{\mu \omega}{\hbar}} , N_n =\sqrt{\frac{\alpha}{\pi ^{1/2}2^n n!}} \qquad n=0,1,2,3...

例:

\ket{0} = (\frac{\mu \omega}{\pi \hbar})^{\frac{1}{4}}e^{-\frac{\mu \omega}{2\hbar}x^2}
  1. 一维 $\delta $ 势场 $V(x) = A \delta (x-a) $ ,其一阶导数不满足连续条件,存在阶跃条件:

ψ(a+)ψ(a)=2μA2ψ(a)\psi ^{'}(a^{+})- \psi ^{'}(a^{-})=\frac{2\mu A}{\hbar ^2}\psi (a)


  1. 平面转子能量与波函数:

Em=m222μr2=m222IE_m = \frac{m^2 \hbar ^2}{2\mu r^2} = \frac{m^2 \hbar^2}{2I}

ψm(φ)=12πeimφ\psi _m (\varphi)=\sqrt{\frac{1}{2\pi}}e^{im \varphi}


  1. 中心力场(A.T.Field!)中定态波函数可表示为:

ψ(r)=R(r)Ylm(θφ)=u(r)rYlm(θφ)\psi (r) = R(r) Y_{lm}(\theta \varphi)=\frac{u(r)}{r}Y_{lm}(\theta \varphi)

其中径向部分 u(r)u(r) 满足方程:

[22μd2dr2+l(l+1)22μr2+V(r)]u(r)=Eu(r)[-\frac{\hbar ^2}{2\mu}\frac{d^2}{dr^2} + \frac{l(l+1)\hbar ^2}{2\mu r^2} + V(r)]u(r) = Eu(r)

角向部分的前几个球谐函数为:

Y00=14π,Y11=18πsinθeiφ,Y11=38πsinθeiφ,Y10=34πcosθY_{00}=\frac{1}{\sqrt{4\pi}},Y_{11}=-\sqrt{\frac{1}{8\pi}}sin \theta e^{i \varphi},Y_{1-1}=\sqrt{\frac{3}{8\pi}}sin \theta e^{-i \varphi},Y_{10}=\sqrt{\frac{3}{4\pi}}cos \theta


  1. 电荷为q的粒子在电磁场中的 H^\hat{H} 为:

H^=12μ[p^qcA(r,t)]2+qΦ(r,t)\hat{H}=\frac{1}{2\mu}[\hat{p}- \frac{q}{c}A(r,t)]^2 + q\varPhi (r,t)


  1. 类氢势场 V(r)=Ze2rV(r)=-\frac{Ze^2}{r} 的定态能量(就写个基态,后面懒得写了自己找吧)与定态波函数为:

En=Z2e22an2E_n = -\frac{Z^2 e^2}{2an^2}

ψ=Z3πa3eZra\psi = \sqrt{\frac{Z^3}{\pi a^3}}e^{-\frac{Zr}{a}}

其中a为玻尔半径 a=2μe2a=\frac{\hbar ^2}{\mu e^2}


常见定理

  1. Hellmann–Feynman定理:
\frac{\partial E_n}{\partial \lambda}=\bra{\psi _n} \frac{\partial \hat{H}}{\partial \lambda} \ket{\psi _n}
  1. Virial定理:

2<T>=<rV(r)>2<T>=<r \cdot \nabla V(r)>


  1. Ehrenfest定理(即Heisenberg绘景中取期望):

ddt<H^>=1i<[A^,H^]>+<A^t>\frac{d}{dt}<\hat{H}>=\frac{1}{i\hbar}<[\hat{A},\hat{H}]> + <\frac{\partial \hat{A}}{\partial t}>


常见对易关系

  1. B-H恒等式:

eA^B^eA^=B^+[A^,B^]+12![A^,[A^,B^]]+13![A^[A^,[A^,B^]]]+...e^{\hat{A}}\hat{B}e^{-\hat{A}}=\hat{B}+[\hat{A},\hat{B}]+\frac{1}{2!}[\hat{A},[\hat{A},\hat{B}]]+\frac{1}{3!}[\hat{A}[\hat{A},[\hat{A},\hat{B}]]]+...


  1. 常见对易

[f(x),p^]=idf(x)dx[f(x),\hat{p}]=i \hbar \frac{d f(x)}{dx}

[x,f(p^)]=idf(p)dp[x,f(\hat{p})]=i \hbar \frac{d f(p)}{dp}

[Lα^,Lβ^]=iεα,β,γLγ^[\hat{L_{\alpha}},\hat{L_{\beta}}]=i\hbar \varepsilon _{\alpha , \beta , \gamma} \hat{L_{\gamma}}

[Sα^,Sβ^]=iεα,β,γSγ^[\hat{S_{\alpha}},\hat{S_{\beta}}]=i\hbar \varepsilon _{\alpha , \beta , \gamma} \hat{S_{\gamma}}

[σα^,σβ^]=2iεα,β,γσγ^[\hat{\sigma _{\alpha}},\hat{\sigma _{\beta}}]=2i \varepsilon _{\alpha , \beta , \gamma} \hat{\sigma _{\gamma}}

[Lα^,pβ^]=iεα,β,γpγ^[\hat{L_{\alpha}},\hat{p_{\beta}}]=i\hbar \varepsilon _{\alpha , \beta , \gamma} \hat{p_{\gamma}}

[p^,H^]=iV(x)x[\hat{p},\hat{H}]=-i\hbar \frac{\partial V(x)}{\partial x}

[x^,H^]=ip^μ[\hat{x},\hat{H}]=-i\hbar \frac{\hat{p}}{\mu}

[rα,pα^]=i[r_{\alpha},\hat{p_{\alpha}}]=i\hbar


一维谐振子升降算符

  1. 降算符与升算符定义:

a^=12μω(mωx^+ip^),a^=12μω(mωx^ip^)\hat{a}=\frac{1}{2\mu \omega \hbar}(m \omega \hat{x} + i\hat{p}) , \hat{a^{\dagger}}=\frac{1}{2\mu \omega \hbar}(m \omega \hat{x} - i\hat{p})



  1. 对易关系:

[a,a]=1[a,a^{\dagger}]=1



  1. 升降作用:
a^{\dagger}\ket{n}=\sqrt{n+1}\ket{n+1} a\ket{n}=\sqrt{n}\ket{n-1}

  1. Hamiltonian可表示为:

H^=(N^+12)ω,N^=aa\hat{H}=(\hat{N}+\frac{1}{2})\hbar \omega ,其中\hat{N}=a^{\dagger}a为厄米算符



  1. 递推关系(其中 α=μω\alpha = \sqrt{\frac{\mu \omega}{\hbar}} ):
x \ket{n} = \frac{1}{\alpha} (\sqrt{\frac{n}{2}} \ket{n-1} + \sqrt{\frac{n+1}{2}} \ket{n+1} ) \frac{d}{dx} \ket{n} = \frac{1}{\alpha} (\sqrt{\frac{n}{2}} \ket{n-1} - \sqrt{\frac{n+1}{2}} \ket{n+1} ) x^2 \ket{n} = \frac{1}{\alpha ^2}[\frac{\sqrt{n(n-1)}}{2}\ket{n-2} + (n+\frac{1}{2})\ket{n} + \frac{\sqrt{(n+1)(n+2)}}{2}\ket{n+2}] \frac{d^2}{dx^2} \ket{n} = \frac{\alpha ^2}{2}[\sqrt{n(n-1)}\ket{n-2} - (2n+1)\ket{n} + \sqrt{(n+1)(n+2)}\ket{n+2}]

角动量算符

  1. 角动量升降算符定义与衍生:

L±^=Lx^+iLy^\hat{L_{\pm}} = \hat{L_x} + i \hat{L_y}

Lx^=12(L+^+L^),Ly^=12i(L+^L^)\hat{L_x}=\frac{1}{2}(\hat{L_+} + \hat{L_-}),\hat{L_y}=\frac{1}{2i}(\hat{L_+} - \hat{L_-})

L2^=L^L+^+Lz^+Lz2^\hat{L^2}=\hat{L_-} \hat{L_+} + \hbar \hat{L_z} + \hat{L_z^2}


  1. 角动量算符与球谐函数的关系:

L2^Ylm=l(l+1)2Ylm\hat{L^2}Y_{lm}= l(l+1)\hbar ^2 Y_{lm}

Lz^Ylm=mYlm\hat{L_z}Y_{lm}=m \hbar Y_{lm}

L±^Ylm=l(l+1)m(m±1)Ylm±1\hat{L_{\pm}}Y_{lm}=\hbar \sqrt{l(l+1)-m(m \pm 1)} Y{lm \pm 1}

cosθYlm=αYl+1m+βYl1mcos\theta Y_{lm}=\alpha Y_{l+1m} + \beta Y_{l-1m}


  1. 角动量算符的矩阵表示(在l=1的 {L2,Lz}\{L^2,L_z\} 表象下):

Lx^=2(010101010)Ly^=2(0i0i0i0i0)Lz^=(100000001)\hat{L_x}=\frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix} \hat{L_y}=\frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \\ \end{pmatrix} \hat{L_z}=\hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}


  1. 角动量算符的本征值与本征矢:

lx=,ψ+=12(121)lx=0,ψ0=12(101)lx=,ψ=12(121)l_x =\hbar,\psi _+ =\frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{2} \\ 1\\ \end{pmatrix} l_x =0,\psi _0 =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ -1\\ \end{pmatrix} l_x =-\hbar,\psi _- =\frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt{2} \\ 1\\ \end{pmatrix}

ly=,ψ+=12(12i1)ly=0,ψ0=12(101)ly=,ψ=12(12i1)l_y =\hbar,\psi _+ =\frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{2i} \\ -1\\ \end{pmatrix} l_y =0,\psi _0 =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1\\ \end{pmatrix} l_y =-\hbar,\psi _- =\frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt{2i} \\ 1\\ \end{pmatrix}

lz=,ψ+=(100)lz=0,ψ0=(010)lz=,ψ=(001)l_z =\hbar,\psi _+ = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} l_z =0,\psi _0 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix} l_z =-\hbar,\psi _- = \begin{pmatrix} 0 \\ 0 \\ 1\\ \end{pmatrix}


自旋角动量算符

  1. 电子的自旋算符矩阵表示( SzS_z 表象下):

Sx^=2(0110)Sy^=2(0ii0)Sz^=2(1001)\hat{S_x} = \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \hat{S_y} = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix} \hat{S_z} = \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}


  1. 泡利矩阵( σz\sigma _z 表象下):

σx=(0110)σy=(0ii0)σz=(1001)\sigma _x= \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \sigma _y= \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix} \sigma _z= \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}

满足广义Euler公式:

eiασj^=cosα+iσj^sinα,σj2^=1e^{i\alpha \hat{\sigma _j}}=cos \alpha +i\hat{\sigma _j}sin\alpha , \hat{\sigma _j^2}=1

满足关系:

[σα^,σβ^]=2iεα,β,γσγ^[\hat{\sigma _{\alpha}},\hat{\sigma _{\beta}}]=2i \varepsilon _{\alpha , \beta , \gamma} \hat{\sigma _{\gamma}}

σα^σβ^=σβ^σα^=iεαβγσγ^\hat{\sigma _{\alpha}}\hat{\sigma _{\beta}}=-\hat{\sigma _{\beta}}\hat{\sigma _{\alpha}}=i\varepsilon _{\alpha \beta \gamma}\hat{\sigma _{\gamma}}


  1. 自旋角动量的本征值与本征矢:

sz=2,ψ+=(10)sz=2,ψ=(01)s_z = \frac{\hbar}{2} , \psi _+ = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} s_z = -\frac{\hbar}{2} , \psi _- = \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix}

sx=2,φ+=12(11)sx=2,φ=12(11)s_x = \frac{\hbar}{2} , \varphi _+ = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} s_x = -\frac{\hbar}{2} , \varphi _- =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}

sy=2,ϕ+=12(1i)sy=2,ϕ=12(1i)s_y = \frac{\hbar}{2} , \phi _+ = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \\ \end{pmatrix} s_y = -\frac{\hbar}{2} , \phi _- =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i \\ \end{pmatrix}


  1. 自旋投影算符的矩阵形式:

Sn^=2(cosθsinθeiφsinθeiφcosθ)\hat{S_n}=\frac{\hbar}{2} \begin{pmatrix} cos\theta & sin\theta e^{-i \varphi} \\ sin\theta e^{i \varphi} & -cos\theta \\ \end{pmatrix}


  1. 自旋投影算符的本征值与本征矢:

sn=2,ψ+=(cosθ2eiφ/2sinθ2eiφ/2)sn=2,ψ=(sinθ2eiφ/2cosθ2eiφ/2)s_n = \frac{\hbar}{2},\psi _+ = \begin{pmatrix} cos\frac{\theta}{2}e^{-i \varphi /2} \\ sin\frac{\theta}{2}e^{i \varphi /2} \\ \end{pmatrix} s_n = -\frac{\hbar}{2},\psi _- = \begin{pmatrix} -sin\frac{\theta}{2}e^{-i \varphi /2} \\ cos\frac{\theta}{2}e^{i \varphi /2} \\ \end{pmatrix}


  1. 轨道/自旋角动量耦合:

J2^=L2^+S2^+2L^S^=L2^+342+2L^S^\hat{J^2}=\hat{L^2} + \hat{S^2} + 2\hat{L}\hat{S} =\hat{L^2} + \frac{3}{4}\hbar ^2 + 2\hat{L} \cdot \hat{S}

L^S^=12(J2^L2^342)\hat{L} \cdot \hat{S} =\frac{1}{2}(\hat{J^2}-\hat{L^2}-\frac{3}{4}\hbar ^2)


  1. 两个自旋1/2耦合为三重态与单态:
\ket{11}=\ket{\uparrow \uparrow} \ket{1-1}=\ket{\downarrow \downarrow} \ket{10}=\frac{1}{\sqrt{2}}(\ket{\uparrow \downarrow} + \ket{\downarrow \uparrow} ) \ket{00}=\frac{1}{\sqrt{2}}(\ket{\uparrow \downarrow} - \ket{\downarrow \uparrow} )

占有数算符

  1. 对玻色子体系:

[ai,aj]=δij,[ai,aj]=[ai,aj]=0,ij=1,2...[a_i,a_j^{\dagger}]=\delta _{ij},[a_i,a_j]=[a_i^{\dagger},a_j^{\dagger}]=0 ,i、j=1,2...

由对易关系可知Ni^=aiai\hat{N_i}=a_i^{\dagger}a_i的本征值ni=0,1,2...n_i =0,1,2...

a_i \ket{n_i}=\sqrt{n_i}\ket{n_i -1} a_i^{\dagger} \ket{n_i}=\sqrt{n_i +1}\ket{n_i +1} a_i \ket{n_1 n_2 ... n_i ...}=\sqrt{n_i}\ket{n_1 n_2 ... n_i -1 ...} a_i^{\dagger} \ket{n_1 n_2 ... n_i ...}=\sqrt{n_i +1}\ket{n_1 n_2 ... n_i +1 ...}

  1. 对费米子体系:

{ai,aj}=aiaj+ajai=δij,{ai,aj}={ai,aj}=0,ij=1,2...\{a_i,a_j^{\dagger}\}=a_i a_j^{\dagger}+a_j^{\dagger}a_i=\delta _{ij},\{a_i,a_j\}=\{a_i^{\dagger},a_j^{\dagger}\}=0,i、j=1,2...

由反对易关系可知Ni^=aiai\hat{N_i}=a_i^{\dagger}a_i的本征值ni=0,1n_i =0,1

a_i \ket{n_i}=\sqrt{n_i}\ket{n_i -1} a_i^{\dagger} \ket{n_i}=\sqrt{1-n_i}\ket{n_i +1} a_i \ket{n_1 n_2 ... n_i ...}=(-1)^m \sqrt{n_i} \ket{n_1 n_2 ... n_i -1 ...} a_i^{\dagger} \ket{n_1 n_2 ... n_i ...}=(-1)^m \sqrt{1-n_i}\ket{n_1 n_2 ... n_i +1 ...}

m=α=1i1nαm=\sum_{\alpha = 1}^{i-1}n_{\alpha}



近似方法

  1. 定态非简并微扰论(能量至三级修正,波函数至一级修正):
E_n^{(1)}=\bra{\psi _n^{(0)}}H^{'}\ket{\psi _n^{(0)}}=H_{nn}^{'}

En(2)=mnHmn2En(0)Em(0)E_n^{(2)}=\sum_{m \ne n} \frac{|H_{mn}^{'}|^2}{E_n^{(0)} - E_m^{(0)}}

En(3)=m,mnHnmHmmHmn(En(0)Em(0))(En(0)Em(0))mnHnnHmn2(En(0)Em(0))2E_n^{(3)}=\sum_{m,m^{'}\ne n}\frac{H_{nm}^{'}H_{mm^{'}}^{'}H_{m^{'}n}^{'}}{(E_n^{(0)} - E_m^{(0)})(E_n^{(0)} - E_{m^{'}}^{(0)})} - \sum_{m \ne n}\frac{H_{nn}^{'}|H_{mn}^{'}|^2}{(E_n^{(0)} - E_m^{(0)})^2}

ψn(1)=mnHmnEn(0)Em(0)ψm(0)\psi _n^{(1)}=\sum_{m \ne n}\frac{H_{mn}^{'}}{E_n^{(0)} - E_m^{(0)}}\psi _m^{(0)}


  1. 定态简并微扰论:写出H’在简并能级下的本征子空间的矩阵元后解一级近似方程就行了,二级近似方程懒得写,自己搜去吧。

  1. 变分法:将试探波函数代入计算,并令其偏导为0,带回求基态的近似能量与近似波函数
E(\alpha)=\bra{\psi}\hat{H}\ket{\psi},\frac{\partial E(\alpha)}{\partial \alpha}
  1. 含时微扰:粒子跃迁的概率为:

Wkm(t)=120tHmkeiωmktdt2W_{k\rarr m}(t)=\frac{1}{\hbar ^2}|\int_0^t H_{mk}^{'}e^{i \omega _{mk}t}dt|^2

其中

H_{mk}^{'}(t)=\bra{m}\hat{H^{'}}\ket{k},\omega _{mk}=\frac{E_m -E_k}{\hbar}
  1. 黄金规则公式:

w=2πHmk(E)2ρ(E)w=\frac{2\pi}{\hbar}|H_{mk}^{'}(E)|^2\rho (E)

其中

H_{mk}^{'}(E)=\bra{m}\hat{H^{'}}\ket{k}
  1. 强度为I的连续光照原子发生跃迁的概率速率(电偶极近似):

wkm=4π2e232I(ωmk)rmk2w_{k \rarr m}=\frac{4\pi ^2 e^2}{3\hbar ^2}I(|\omega _{mk}|)|r_{mk}|^2

其中

rmk2=xmk2+ymk2+zmk2|r_{mk}|^2=|x_{mk}|^2 + |y_{mk}|^2 + |z_{mk}|^2

x_{mk}=\bra{m}x\ket{k},y_{mk}=\bra{m}y\ket{k},z_{mk}=\bra{m}z\ket{k}

电偶极跃迁选择定则:

Δl=±1,Δm=0,±1\Delta l=\pm 1 ,\Delta m=0,\pm 1


  1. 原子的自发跃迁速率为:

Akm=4e2ωkm33c3rmk2A_{k \rarr m}=\frac{4 e^2\omega _{km}^3}{3\hbar c^3}|r_{mk}|^2