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
三角函数:
s i n x = e i x − e − i x 2 i sinx = \frac{e^{ix}-e^{-ix}}{2i}
s i n x = 2 i e i x − e − i x
c o s x = e i x + e − i x 2 cosx = \frac{e^{ix}+e^{-ix}}{2}
c o s x = 2 e i x + e − i x
s i n α c o s β = 1 2 [ s i n ( α + β ) + s i n ( α − β ) ] sin \alpha cos \beta = \frac{1}{2}[sin(\alpha + \beta)+sin(\alpha - \beta)]
s i n α c o s β = 2 1 [ s i n ( α + β ) + s i n ( α − β ) ]
c o s α s i n β = 1 2 [ s i n ( α + β ) − s i n ( α − β ) ] cos \alpha sin \beta = \frac{1}{2}[sin(\alpha + \beta)-sin(\alpha - \beta)]
c o s α s i n β = 2 1 [ s i n ( α + β ) − s i n ( α − β ) ]
c o s α c o s β = 1 2 [ c o s ( α + β ) + c o s ( α − β ) ] cos \alpha cos \beta = \frac{1}{2}[cos(\alpha + \beta)+cos(\alpha - \beta)]
c o s α c o s β = 2 1 [ c o s ( α + β ) + c o s ( α − β ) ]
s i n α s i n β = − 1 2 [ c o s ( α + β ) − s i n ( α − β ) ] sin \alpha sin \beta = -\frac{1}{2}[cos(\alpha + \beta)-sin(\alpha - \beta)]
s i n α s i n β = − 2 1 [ c o s ( α + β ) − s i n ( α − β ) ]
积分公式:
∫ − ∞ + ∞ e − a x 2 d x = π a \int_{-\infty}^{+\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}
∫ − ∞ + ∞ e − a x 2 d x = a π
∫ − ∞ + ∞ x 2 e − a x 2 d x = 1 2 a π a \int_{-\infty}^{+\infty}x^2 e^{-ax^2}dx=\frac{1}{2a} \sqrt{\frac{\pi}{a}}
∫ − ∞ + ∞ x 2 e − a x 2 d x = 2 a 1 a π
∫ 0 + ∞ x n e − α x d x = n ! α n + 1 \int_{0}^{+\infty}x^{n}e^{-\alpha x}dx=\frac{n!}{\alpha ^{n+1}}
∫ 0 + ∞ x n e − α x d x = α n + 1 n !
∫ 0 + ∞ x 2 n e − β x 2 d x = ( 2 n − 1 ) ! ! 2 n π β 2 n + 1 \int_{0}^{+\infty}x^{2n}e^{-\beta x^2}dx=\frac{(2n-1)!!}{2^n} \sqrt{\frac{\pi}{\beta ^{2n+1}}}
∫ 0 + ∞ x 2 n e − β x 2 d x = 2 n ( 2 n − 1 ) ! ! β 2 n + 1 π
∫ 0 π 2 s i n n x d x = ∫ 0 π 2 c o s n x d x = { n − 1 n ⋅ n − 3 n − 2 ⋅ . . . 2 3 ⋅ 1 n 为 奇 数 n − 1 n ⋅ n − 3 n − 2 ⋅ . . . 3 4 ⋅ 1 2 ⋅ π 2 n 为 偶 数 \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}
∫ 0 2 π s i n n x d x = ∫ 0 2 π c o s n x d x = { n n − 1 ⋅ n − 2 n − 3 ⋅ . . . 3 2 ⋅ 1 n 为 奇 数 n n − 1 ⋅ n − 2 n − 3 ⋅ . . . 4 3 ⋅ 2 1 ⋅ 2 π n 为 偶 数
泰勒展开:
1 + x = ∑ n ( − 1 ) n − 1 ( 2 n − 3 ) ! ! 2 n ! ! x n = 1 + 1 2 x − 1 8 x 2 + 1 16 x 3 − . . . \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 - ...
1 + x = n ∑ ( − 1 ) n − 1 2 n ! ! ( 2 n − 3 ) ! ! x n = 1 + 2 1 x − 8 1 x 2 + 1 6 1 x 3 − . . .
e x = ∑ n 1 n ! x n = 1 + x + 1 2 ! x 2 + 1 3 ! x 3 + . . . e^x = \sum_{n}\frac{1}{n!}x^n = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + ...
e x = n ∑ n ! 1 x n = 1 + x + 2 ! 1 x 2 + 3 ! 1 x 3 + . . .
s i n x = ∑ n ( − 1 ) n ( 2 n + 1 ) ! x 2 n + 1 = x − 1 3 ! x 3 + 1 5 ! x 5 − . . . sinx = \sum_{n}\frac{(-1)^n}{(2n+1)!}x^{2n+1}= x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - ...
s i n x = n ∑ ( 2 n + 1 ) ! ( − 1 ) n x 2 n + 1 = x − 3 ! 1 x 3 + 5 ! 1 x 5 − . . .
c o s x = ∑ n ( − 1 ) n ( 2 n ) ! x 2 n = 1 − 1 2 ! x 2 + 1 4 ! x 4 − . . . cosx = \sum_{n}\frac{(-1)^n}{(2n)!}x^{2n}= 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - ...
c o s x = n ∑ ( 2 n ) ! ( − 1 ) n x 2 n = 1 − 2 ! 1 x 2 + 4 ! 1 x 4 − . . .
l n ( 1 + x ) = ∑ n ( − 1 ) n n + 1 x n + 1 = x − 1 2 x 2 + 1 3 x 3 − . . . ln(1+x)=\sum_{n}\frac{(-1)^n}{n+1}x^{n+1}= x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - ...
l n ( 1 + x ) = n ∑ n + 1 ( − 1 ) n x n + 1 = x − 2 1 x 2 + 3 1 x 3 − . . .
1 1 − x = ∑ n x n = 1 + x + x 2 + . . . \frac{1}{1-x}=\sum_{n}x^n = 1 + x + x^2 + ...
1 − x 1 = n ∑ x n = 1 + x + x 2 + . . .
1 1 + x = ∑ n ( − 1 ) n x n = 1 − x + x 2 − . . . \frac{1}{1+x}=\sum_{n}(-1)^n x^n =1 - x + x^2 - ...
1 + x 1 = n ∑ ( − 1 ) n x n = 1 − x + x 2 − . . .
Schrödinger方程
含时Schrödinger方程:
i ℏ ∂ ∂ t Ψ ( r , t ) = H ^ Ψ ( r , t ) i\hbar \frac{\partial}{\partial t}\Psi (r,t)=\hat{H}\Psi (r,t)\\
i ℏ ∂ t ∂ Ψ ( r , t ) = H ^ Ψ ( r , t )
其中:H ^ = − ℏ 2 2 μ ∇ 2 + V ( r , t ) \hat{H}=-\frac{\hbar ^2}{2\mu}\nabla ^2 + V(r,t) H ^ = − 2 μ ℏ 2 ∇ 2 + V ( r , t )
定态Schrödinger方程:
H ^ Ψ ( r ) = E Ψ ( r ) \hat{H}\Psi (r)=E\Psi (r)
H ^ Ψ ( r ) = E Ψ ( r )
时间演化算符:
U ^ = e − i E t / ℏ \hat{U}=e^{-iEt/ \hbar}
U ^ = e − i E t / ℏ
概率流密度:
J = − i ℏ 2 μ [ ψ ∗ ∇ ψ − ψ ∇ ψ ∗ ] J=-\frac{i\hbar}{2\mu}[\psi ^* \nabla \psi - \psi \nabla \psi ^*]
J = − 2 μ i ℏ [ ψ ∗ ∇ ψ − ψ ∇ ψ ∗ ]
常见势场
一维无限深势阱(三维作简单乘积即可):
V ( x ) = { 0 0 < x < a ∞ e l s e V(x)=
\begin{cases}
0 \qquad 0 < x < a \\
\infty \qquad else \\
\end{cases}
V ( x ) = { 0 0 < x < a ∞ e l s e
其定态能量与波函数为:
E n = n 2 π 2 ℏ 2 2 μ a 2 n = 1 , 2 , 3... E_n = \frac{n^2 \pi ^2 \hbar ^2}{2 \mu a^2} \qquad n=1,2,3...
E n = 2 μ a 2 n 2 π 2 ℏ 2 n = 1 , 2 , 3 . . .
ψ n ( x ) = { 2 a s i n n π x a 0 < x < a 0 e l s e \psi _n (x)=
\begin{cases}
\sqrt{\frac{2}{a}}sin \frac{n \pi x}{a} \qquad 0 < x < a \\
0 \qquad else\\
\end{cases}
ψ n ( x ) = { a 2 s i n a n π x 0 < x < a 0 e l s e
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):
E n = ( n + 1 2 ) ℏ ω n = 0 , 1 , 2 , 3... E_n = (n + \frac{1}{2})\hbar \omega \qquad n=0,1,2,3...
E n = ( n + 2 1 ) ℏ ω n = 0 , 1 , 2 , 3 . . .
\ket{n}=N_n e^{-\alpha ^2 x^2 /2}H_n (\alpha x)
其中
α = μ ω ℏ , N n = α π 1 / 2 2 n n ! 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...
α = ℏ μ ω , N n = π 1 / 2 2 n n ! α n = 0 , 1 , 2 , 3 . . .
例:
\ket{0} = (\frac{\mu \omega}{\pi \hbar})^{\frac{1}{4}}e^{-\frac{\mu \omega}{2\hbar}x^2}
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 − ) = 2 μ A ℏ 2 ψ ( a ) \psi ^{'}(a^{+})- \psi ^{'}(a^{-})=\frac{2\mu A}{\hbar ^2}\psi (a)
ψ ′ ( a + ) − ψ ′ ( a − ) = ℏ 2 2 μ A ψ ( a )
平面转子能量与波函数:
E m = m 2 ℏ 2 2 μ r 2 = m 2 ℏ 2 2 I E_m = \frac{m^2 \hbar ^2}{2\mu r^2} = \frac{m^2 \hbar^2}{2I}
E m = 2 μ r 2 m 2 ℏ 2 = 2 I m 2 ℏ 2
ψ m ( φ ) = 1 2 π e i m φ \psi _m (\varphi)=\sqrt{\frac{1}{2\pi}}e^{im \varphi}
ψ m ( φ ) = 2 π 1 e i m φ
中心力场(A.T.Field!)中定态波函数可表示为:
ψ ( r ) = R ( r ) Y l m ( θ φ ) = u ( r ) r Y l m ( θ φ ) \psi (r) = R(r) Y_{lm}(\theta \varphi)=\frac{u(r)}{r}Y_{lm}(\theta \varphi)
ψ ( r ) = R ( r ) Y l m ( θ φ ) = r u ( r ) Y l m ( θ φ )
其中径向部分 u ( r ) u(r) u ( r ) 满足方程:
[ − ℏ 2 2 μ d 2 d r 2 + l ( l + 1 ) ℏ 2 2 μ r 2 + V ( r ) ] u ( r ) = E u ( 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)
[ − 2 μ ℏ 2 d r 2 d 2 + 2 μ r 2 l ( l + 1 ) ℏ 2 + V ( r ) ] u ( r ) = E u ( r )
角向部分的前几个球谐函数为:
Y 00 = 1 4 π , Y 11 = − 1 8 π s i n θ e i φ , Y 1 − 1 = 3 8 π s i n θ e − i φ , Y 10 = 3 4 π c o s θ Y_{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
Y 0 0 = 4 π 1 , Y 1 1 = − 8 π 1 s i n θ e i φ , Y 1 − 1 = 8 π 3 s i n θ e − i φ , Y 1 0 = 4 π 3 c o s θ
电荷为q的粒子在电磁场中的 H ^ \hat{H} H ^ 为:
H ^ = 1 2 μ [ p ^ − q c A ( r , t ) ] 2 + q Φ ( r , t ) \hat{H}=\frac{1}{2\mu}[\hat{p}- \frac{q}{c}A(r,t)]^2 + q\varPhi (r,t)
H ^ = 2 μ 1 [ p ^ − c q A ( r , t ) ] 2 + q Φ ( r , t )
类氢势场 V ( r ) = − Z e 2 r V(r)=-\frac{Ze^2}{r} V ( r ) = − r Z e 2 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:
E n = − Z 2 e 2 2 a n 2 E_n = -\frac{Z^2 e^2}{2an^2}
E n = − 2 a n 2 Z 2 e 2
ψ = Z 3 π a 3 e − Z r a \psi = \sqrt{\frac{Z^3}{\pi a^3}}e^{-\frac{Zr}{a}}
ψ = π a 3 Z 3 e − a Z r
其中a为玻尔半径 a = ℏ 2 μ e 2 a=\frac{\hbar ^2}{\mu e^2} a = μ e 2 ℏ 2
常见定理
Hellmann–Feynman定理:
\frac{\partial E_n}{\partial \lambda}=\bra{\psi _n} \frac{\partial \hat{H}}{\partial \lambda} \ket{\psi _n}
Virial定理:
2 < T > = < r ⋅ ∇ V ( r ) > 2<T>=<r \cdot \nabla V(r)>
2 < T > = < r ⋅ ∇ V ( r ) >
Ehrenfest定理(即Heisenberg绘景中取期望):
d d t < H ^ > = 1 i ℏ < [ A ^ , H ^ ] > + < ∂ A ^ ∂ t > \frac{d}{dt}<\hat{H}>=\frac{1}{i\hbar}<[\hat{A},\hat{H}]> + <\frac{\partial \hat{A}}{\partial t}>
d t d < H ^ > = i ℏ 1 < [ A ^ , H ^ ] > + < ∂ t ∂ A ^ >
常见对易关系
B-H恒等式:
e A ^ B ^ e − A ^ = B ^ + [ A ^ , B ^ ] + 1 2 ! [ A ^ , [ A ^ , B ^ ] ] + 1 3 ! [ 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}]]]+...
e A ^ B ^ e − A ^ = B ^ + [ A ^ , B ^ ] + 2 ! 1 [ A ^ , [ A ^ , B ^ ] ] + 3 ! 1 [ A ^ [ A ^ , [ A ^ , B ^ ] ] ] + . . .
常见对易
[ f ( x ) , p ^ ] = i ℏ d f ( x ) d x [f(x),\hat{p}]=i \hbar \frac{d f(x)}{dx}
[ f ( x ) , p ^ ] = i ℏ d x d f ( x )
[ x , f ( p ^ ) ] = i ℏ d f ( p ) d p [x,f(\hat{p})]=i \hbar \frac{d f(p)}{dp}
[ x , f ( p ^ ) ] = i ℏ d p d f ( p )
[ L α ^ , L β ^ ] = i ℏ ε α , β , γ L γ ^ [\hat{L_{\alpha}},\hat{L_{\beta}}]=i\hbar \varepsilon _{\alpha , \beta , \gamma} \hat{L_{\gamma}}
[ L α ^ , L β ^ ] = i ℏ ε α , β , γ L γ ^
[ S α ^ , S β ^ ] = i ℏ ε α , β , γ S γ ^ [\hat{S_{\alpha}},\hat{S_{\beta}}]=i\hbar \varepsilon _{\alpha , \beta , \gamma} \hat{S_{\gamma}}
[ S α ^ , S β ^ ] = i ℏ ε α , β , γ S γ ^
[ σ α ^ , σ β ^ ] = 2 i ε α , β , γ σ γ ^ [\hat{\sigma _{\alpha}},\hat{\sigma _{\beta}}]=2i \varepsilon _{\alpha , \beta , \gamma} \hat{\sigma _{\gamma}}
[ σ α ^ , σ β ^ ] = 2 i ε α , β , γ σ γ ^
[ L α ^ , p β ^ ] = i ℏ ε α , β , γ p γ ^ [\hat{L_{\alpha}},\hat{p_{\beta}}]=i\hbar \varepsilon _{\alpha , \beta , \gamma} \hat{p_{\gamma}}
[ L α ^ , p β ^ ] = i ℏ ε α , β , γ p γ ^
[ p ^ , H ^ ] = − i ℏ ∂ V ( x ) ∂ x [\hat{p},\hat{H}]=-i\hbar \frac{\partial V(x)}{\partial x}
[ p ^ , H ^ ] = − i ℏ ∂ x ∂ V ( x )
[ x ^ , H ^ ] = − i ℏ p ^ μ [\hat{x},\hat{H}]=-i\hbar \frac{\hat{p}}{\mu}
[ x ^ , H ^ ] = − i ℏ μ p ^
[ r α , p α ^ ] = i ℏ [r_{\alpha},\hat{p_{\alpha}}]=i\hbar
[ r α , p α ^ ] = i ℏ
一维谐振子升降算符
降算符与升算符定义:
a ^ = 1 2 μ ω ℏ ( m ω x ^ + i p ^ ) , a † ^ = 1 2 μ ω ℏ ( m ω x ^ − i p ^ ) \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})
a ^ = 2 μ ω ℏ 1 ( m ω x ^ + i p ^ ) , a † ^ = 2 μ ω ℏ 1 ( m ω x ^ − i p ^ )
对易关系:
[ a , a † ] = 1 [a,a^{\dagger}]=1
[ a , a † ] = 1
升降作用:
a^{\dagger}\ket{n}=\sqrt{n+1}\ket{n+1}
a\ket{n}=\sqrt{n}\ket{n-1}
Hamiltonian可表示为:
H ^ = ( N ^ + 1 2 ) ℏ ω , 其 中 N ^ = a † a 为 厄 米 算 符 \hat{H}=(\hat{N}+\frac{1}{2})\hbar \omega ,其中\hat{N}=a^{\dagger}a为厄米算符
H ^ = ( N ^ + 2 1 ) ℏ ω , 其 中 N ^ = a † a 为 厄 米 算 符
递推关系(其中 α = μ ω ℏ \alpha = \sqrt{\frac{\mu \omega}{\hbar}} α = ℏ μ ω ):
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
角动量升降算符定义与衍生:
L ± ^ = L x ^ + i L y ^ \hat{L_{\pm}} = \hat{L_x} + i \hat{L_y}
L ± ^ = L x ^ + i L y ^
L x ^ = 1 2 ( L + ^ + L − ^ ) , L y ^ = 1 2 i ( L + ^ − L − ^ ) \hat{L_x}=\frac{1}{2}(\hat{L_+} + \hat{L_-}),\hat{L_y}=\frac{1}{2i}(\hat{L_+} - \hat{L_-})
L x ^ = 2 1 ( L + ^ + L − ^ ) , L y ^ = 2 i 1 ( L + ^ − L − ^ )
L 2 ^ = L − ^ L + ^ + ℏ L z ^ + L z 2 ^ \hat{L^2}=\hat{L_-} \hat{L_+} + \hbar \hat{L_z} + \hat{L_z^2}
L 2 ^ = L − ^ L + ^ + ℏ L z ^ + L z 2 ^
角动量算符与球谐函数的关系:
L 2 ^ Y l m = l ( l + 1 ) ℏ 2 Y l m \hat{L^2}Y_{lm}= l(l+1)\hbar ^2 Y_{lm}
L 2 ^ Y l m = l ( l + 1 ) ℏ 2 Y l m
L z ^ Y l m = m ℏ Y l m \hat{L_z}Y_{lm}=m \hbar Y_{lm}
L z ^ Y l m = m ℏ Y l m
L ± ^ Y l m = ℏ l ( l + 1 ) − m ( m ± 1 ) Y l m ± 1 \hat{L_{\pm}}Y_{lm}=\hbar \sqrt{l(l+1)-m(m \pm 1)} Y{lm \pm 1}
L ± ^ Y l m = ℏ l ( l + 1 ) − m ( m ± 1 ) Y l m ± 1
c o s θ Y l m = α Y l + 1 m + β Y l − 1 m cos\theta Y_{lm}=\alpha Y_{l+1m} + \beta Y_{l-1m}
c o s θ Y l m = α Y l + 1 m + β Y l − 1 m
角动量算符的矩阵表示(在l=1的 { L 2 , L z } \{L^2,L_z\} { L 2 , L z } 表象下):
L x ^ = ℏ 2 ( 0 1 0 1 0 1 0 1 0 ) L y ^ = ℏ 2 ( 0 − i 0 i 0 − i 0 i 0 ) L z ^ = ℏ ( 1 0 0 0 0 0 0 0 − 1 ) \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}
L x ^ = 2 ℏ ⎝ ⎛ 0 1 0 1 0 1 0 1 0 ⎠ ⎞ L y ^ = 2 ℏ ⎝ ⎛ 0 i 0 − i 0 i 0 − i 0 ⎠ ⎞ L z ^ = ℏ ⎝ ⎛ 1 0 0 0 0 0 0 0 − 1 ⎠ ⎞
角动量算符的本征值与本征矢:
l x = ℏ , ψ + = 1 2 ( 1 2 1 ) l x = 0 , ψ 0 = 1 2 ( 1 0 − 1 ) l x = − ℏ , ψ − = 1 2 ( 1 − 2 1 ) 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}
l x = ℏ , ψ + = 2 1 ⎝ ⎛ 1 2 1 ⎠ ⎞ l x = 0 , ψ 0 = 2 1 ⎝ ⎛ 1 0 − 1 ⎠ ⎞ l x = − ℏ , ψ − = 2 1 ⎝ ⎛ 1 − 2 1 ⎠ ⎞
l y = ℏ , ψ + = 1 2 ( 1 2 i − 1 ) l y = 0 , ψ 0 = 1 2 ( 1 0 1 ) l y = − ℏ , ψ − = 1 2 ( 1 − 2 i 1 ) 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}
l y = ℏ , ψ + = 2 1 ⎝ ⎛ 1 2 i − 1 ⎠ ⎞ l y = 0 , ψ 0 = 2 1 ⎝ ⎛ 1 0 1 ⎠ ⎞ l y = − ℏ , ψ − = 2 1 ⎝ ⎛ 1 − 2 i 1 ⎠ ⎞
l z = ℏ , ψ + = ( 1 0 0 ) l z = 0 , ψ 0 = ( 0 1 0 ) l z = − ℏ , ψ − = ( 0 0 1 ) 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}
l z = ℏ , ψ + = ⎝ ⎛ 1 0 0 ⎠ ⎞ l z = 0 , ψ 0 = ⎝ ⎛ 0 1 0 ⎠ ⎞ l z = − ℏ , ψ − = ⎝ ⎛ 0 0 1 ⎠ ⎞
自旋角动量算符
电子的自旋算符矩阵表示( S z S_z S z 表象下):
S x ^ = ℏ 2 ( 0 1 1 0 ) S y ^ = ℏ 2 ( 0 − i i 0 ) S z ^ = ℏ 2 ( 1 0 0 − 1 ) \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}
S x ^ = 2 ℏ ( 0 1 1 0 ) S y ^ = 2 ℏ ( 0 i − i 0 ) S z ^ = 2 ℏ ( 1 0 0 − 1 )
泡利矩阵( σ z \sigma _z σ z 表象下):
σ x = ( 0 1 1 0 ) σ y = ( 0 − i i 0 ) σ z = ( 1 0 0 − 1 ) \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}
σ x = ( 0 1 1 0 ) σ y = ( 0 i − i 0 ) σ z = ( 1 0 0 − 1 )
满足广义Euler公式:
e i α σ j ^ = c o s α + i σ j ^ s i n α , σ j 2 ^ = 1 e^{i\alpha \hat{\sigma _j}}=cos \alpha +i\hat{\sigma _j}sin\alpha , \hat{\sigma _j^2}=1
e i α σ j ^ = c o s α + i σ j ^ s i n α , σ j 2 ^ = 1
满足关系:
[ σ α ^ , σ β ^ ] = 2 i ε α , β , γ σ γ ^ [\hat{\sigma _{\alpha}},\hat{\sigma _{\beta}}]=2i \varepsilon _{\alpha , \beta , \gamma} \hat{\sigma _{\gamma}}
[ σ α ^ , σ β ^ ] = 2 i ε α , β , γ σ γ ^
σ α ^ σ β ^ = − σ β ^ σ α ^ = i ε α β γ σ γ ^ \hat{\sigma _{\alpha}}\hat{\sigma _{\beta}}=-\hat{\sigma _{\beta}}\hat{\sigma _{\alpha}}=i\varepsilon _{\alpha \beta \gamma}\hat{\sigma _{\gamma}}
σ α ^ σ β ^ = − σ β ^ σ α ^ = i ε α β γ σ γ ^
自旋角动量的本征值与本征矢:
s z = ℏ 2 , ψ + = ( 1 0 ) s z = − ℏ 2 , ψ − = ( 0 1 ) s_z = \frac{\hbar}{2} , \psi _+ =
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
s_z = -\frac{\hbar}{2} , \psi _- =
\begin{pmatrix}
0 \\
1 \\
\end{pmatrix}
s z = 2 ℏ , ψ + = ( 1 0 ) s z = − 2 ℏ , ψ − = ( 0 1 )
s x = ℏ 2 , φ + = 1 2 ( 1 1 ) s x = − ℏ 2 , φ − = 1 2 ( 1 − 1 ) 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}
s x = 2 ℏ , φ + = 2 1 ( 1 1 ) s x = − 2 ℏ , φ − = 2 1 ( 1 − 1 )
s y = ℏ 2 , ϕ + = 1 2 ( 1 i ) s y = − ℏ 2 , ϕ − = 1 2 ( 1 − i ) 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}
s y = 2 ℏ , ϕ + = 2 1 ( 1 i ) s y = − 2 ℏ , ϕ − = 2 1 ( 1 − i )
自旋投影算符的矩阵形式:
S n ^ = ℏ 2 ( c o s θ s i n θ e − i φ s i n θ e i φ − c o s θ ) \hat{S_n}=\frac{\hbar}{2}
\begin{pmatrix}
cos\theta & sin\theta e^{-i \varphi} \\
sin\theta e^{i \varphi} & -cos\theta \\
\end{pmatrix}
S n ^ = 2 ℏ ( c o s θ s i n θ e i φ s i n θ e − i φ − c o s θ )
自旋投影算符的本征值与本征矢:
s n = ℏ 2 , ψ + = ( c o s θ 2 e − i φ / 2 s i n θ 2 e i φ / 2 ) s n = − ℏ 2 , ψ − = ( − s i n θ 2 e − i φ / 2 c o s θ 2 e i φ / 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}
s n = 2 ℏ , ψ + = ( c o s 2 θ e − i φ / 2 s i n 2 θ e i φ / 2 ) s n = − 2 ℏ , ψ − = ( − s i n 2 θ e − i φ / 2 c o s 2 θ e i φ / 2 )
轨道/自旋角动量耦合:
J 2 ^ = L 2 ^ + S 2 ^ + 2 L ^ S ^ = L 2 ^ + 3 4 ℏ 2 + 2 L ^ ⋅ 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}
J 2 ^ = L 2 ^ + S 2 ^ + 2 L ^ S ^ = L 2 ^ + 4 3 ℏ 2 + 2 L ^ ⋅ S ^
L ^ ⋅ S ^ = 1 2 ( J 2 ^ − L 2 ^ − 3 4 ℏ 2 ) \hat{L} \cdot \hat{S} =\frac{1}{2}(\hat{J^2}-\hat{L^2}-\frac{3}{4}\hbar ^2)
L ^ ⋅ S ^ = 2 1 ( J 2 ^ − L 2 ^ − 4 3 ℏ 2 )
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
对玻色子体系:
[ a i , a j † ] = δ i j , [ a i , a j ] = [ a i † , a j † ] = 0 , i 、 j = 1 , 2... [a_i,a_j^{\dagger}]=\delta _{ij},[a_i,a_j]=[a_i^{\dagger},a_j^{\dagger}]=0 ,i、j=1,2...
[ a i , a j † ] = δ i j , [ a i , a j ] = [ a i † , a j † ] = 0 , i 、 j = 1 , 2 . . .
由对易关系可知N i ^ = a i † a i \hat{N_i}=a_i^{\dagger}a_i N i ^ = a i † a i 的本征值n i = 0 , 1 , 2... n_i =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 ...}
对费米子体系:
{ a i , a j † } = a i a j † + a j † a i = δ i j , { a i , a j } = { a i † , a j † } = 0 , i 、 j = 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...
{ a i , a j † } = a i a j † + a j † a i = δ i j , { a i , a j } = { a i † , a j † } = 0 , i 、 j = 1 , 2 . . .
由反对易关系可知N i ^ = a i † a i \hat{N_i}=a_i^{\dagger}a_i N i ^ = a i † a i 的本征值n i = 0 , 1 n_i =0,1 n 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 = 1 i − 1 n α m=\sum_{\alpha = 1}^{i-1}n_{\alpha}
m = α = 1 ∑ i − 1 n α
近似方法
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}^{'}
E n ( 2 ) = ∑ m ≠ n ∣ H m n ′ ∣ 2 E n ( 0 ) − E m ( 0 ) E_n^{(2)}=\sum_{m \ne n} \frac{|H_{mn}^{'}|^2}{E_n^{(0)} - E_m^{(0)}}
E n ( 2 ) = m = n ∑ E n ( 0 ) − E m ( 0 ) ∣ H m n ′ ∣ 2
E n ( 3 ) = ∑ m , m ′ ≠ n H n m ′ H m m ′ ′ H m ′ n ′ ( E n ( 0 ) − E m ( 0 ) ) ( E n ( 0 ) − E m ′ ( 0 ) ) − ∑ m ≠ n H n n ′ ∣ H m n ′ ∣ 2 ( E n ( 0 ) − E m ( 0 ) ) 2 E_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}
E n ( 3 ) = m , m ′ = n ∑ ( E n ( 0 ) − E m ( 0 ) ) ( E n ( 0 ) − E m ′ ( 0 ) ) H n m ′ H m m ′ ′ H m ′ n ′ − m = n ∑ ( E n ( 0 ) − E m ( 0 ) ) 2 H n n ′ ∣ H m n ′ ∣ 2
ψ n ( 1 ) = ∑ m ≠ n H m n ′ E n ( 0 ) − E m ( 0 ) ψ m ( 0 ) \psi _n^{(1)}=\sum_{m \ne n}\frac{H_{mn}^{'}}{E_n^{(0)} - E_m^{(0)}}\psi _m^{(0)}
ψ n ( 1 ) = m = n ∑ E n ( 0 ) − E m ( 0 ) H m n ′ ψ m ( 0 )
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。
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}
含时微扰:粒子跃迁的概率为:
W k → m ( t ) = 1 ℏ 2 ∣ ∫ 0 t H m k ′ e i ω m k t d t ∣ 2 W_{k\rarr m}(t)=\frac{1}{\hbar ^2}|\int_0^t H_{mk}^{'}e^{i \omega _{mk}t}dt|^2
W k → m ( t ) = ℏ 2 1 ∣ ∫ 0 t H m k ′ e i ω m k t d t ∣ 2
其中
H_{mk}^{'}(t)=\bra{m}\hat{H^{'}}\ket{k},\omega _{mk}=\frac{E_m -E_k}{\hbar}
黄金规则公式:
w = 2 π ℏ ∣ H m k ′ ( E ) ∣ 2 ρ ( E ) w=\frac{2\pi}{\hbar}|H_{mk}^{'}(E)|^2\rho (E)
w = ℏ 2 π ∣ H m k ′ ( E ) ∣ 2 ρ ( E )
其中
H_{mk}^{'}(E)=\bra{m}\hat{H^{'}}\ket{k}
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):
w k → m = 4 π 2 e 2 3 ℏ 2 I ( ∣ ω m k ∣ ) ∣ r m k ∣ 2 w_{k \rarr m}=\frac{4\pi ^2 e^2}{3\hbar ^2}I(|\omega _{mk}|)|r_{mk}|^2
w k → m = 3 ℏ 2 4 π 2 e 2 I ( ∣ ω m k ∣ ) ∣ r m k ∣ 2
其中
∣ r m k ∣ 2 = ∣ x m k ∣ 2 + ∣ y m k ∣ 2 + ∣ z m k ∣ 2 |r_{mk}|^2=|x_{mk}|^2 + |y_{mk}|^2 + |z_{mk}|^2
∣ r m k ∣ 2 = ∣ x m k ∣ 2 + ∣ y m k ∣ 2 + ∣ z m k ∣ 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:
Δ l = ± 1 , Δ m = 0 , ± 1 \Delta l=\pm 1 ,\Delta m=0,\pm 1
Δ l = ± 1 , Δ m = 0 , ± 1
原子的自发跃迁速率为:
A k → m = 4 e 2 ω k m 3 3 ℏ c 3 ∣ r m k ∣ 2 A_{k \rarr m}=\frac{4 e^2\omega _{km}^3}{3\hbar c^3}|r_{mk}|^2
A k → m = 3 ℏ c 3 4 e 2 ω k m 3 ∣ r m k ∣ 2