量子力学第四弹——角动量

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相对于位置和动量等力学量，在量子力学中角动量的内容丰富的多，因此值得进行专门的、深入的探讨。

注：从本讲起省略算符头上的帽子，并假设读者自知。

轨道角动量$\hat L$

Definition

$$\mathbf{L= x\times p}$$

$$L_\alpha=\epsilon_{\alpha\beta\gamma}\mathbf{x_\beta p_\gamma}$$

Commutation Relations

1. $[L_\alpha,L_\beta]=i\hbar \epsilon_{\alpha\beta\gamma}L_\gamma$
2. $[\mathbf L^2,L_\alpha]\equiv 0\Rightarrow Common \,Eigenstate$ $\Rightarrow Degeneracy$

Eigenvalue Problems

$$\begin{align*}\mathbf L^2|l,m_l\rangle=&l(l+1)\hbar^2|l,m_l\rangle,\quad l=0,1,\dots\\L_z|l,m_l\rangle=&m_l\hbar|l,m_l\rangle ,\quad m_l=-l,-l+1,\dots,+l\end{align*}$$

自旋角动量$\hat S$

Definition

$$[S_\alpha,S_\beta]=i\hbar S_\gamma$$

$$\Rightarrow[\mathbf S^2,S_\alpha]\equiv 0$$

Eigenvalue Problems

$$\begin{align*}\mathbf S^2|s,m_s\rangle=&s(s+1)\hbar^2|s,m_s\rangle,\quad s=\frac 12\\S_z|s,m_s\rangle=&m_s\hbar|s,m_s\rangle ,\quad m_s=-\frac 12,+\frac 12\end{align*}$$

In this course, only electron, a Fermion, is considerd, which means we assume

s
$s$ equals



12

$\frac 12$.

$$\begin{align*}\mathbf S^2|s,m_s\rangle=&\frac 34\hbar^2|s,m_s\rangle\\S_z|s,m_s\rangle=&\pm\frac 12\hbar|s,m_s\rangle\end{align*}$$

Orthonormal & Completeness

Let $|m_s=+\dfrac12\rangle=|1\rangle, |m_s=-\dfrac12\rangle=|0\rangle$

$$\begin{cases}\langle i|j\rangle=\delta_{ij}\\\sum_{i=0}^1|i\rangle\langle i|=\mathbb I(unit \,operator)\end{cases}$$

升降算符

$\bf J$总角动量

Definition

$$J_\pm=J_x\pm iJ_y$$

Properties

1. ⎧⎩⎨⎪⎪J±|j,mj⟩=C|j,mj±1⟩, where C=ℏ(j±mj+1)(j∓mj)−−−−−−−−−−−−−−−−−√J+|j,+j⟩≡0J−|j,−j⟩≡0

   $$\begin{cases}J_\pm|j,m_j\rangle=C|j,m_j\pm 1\rangle, \text{ where } C=\hbar\sqrt{(j\pm m_j+1)(j\mp m_j)}\\J_+|j,+j\rangle\equiv0\\J_-|j,-j\rangle\equiv0\end{cases}$$
   (to be prooved )

2. $[J_\pm,\mathbf J^2]\equiv 0$
   

   proof
   

   [J±,J2]===[Jx±iJy,J2][Jx,J2]±i[Jy,J2]0

   $$\begin{align*}[J_\pm,\mathbf J^2]=&[J_x\pm iJ_y,\mathbf J^2]\\ =&[J_x,\mathbf J^2]\pm i[J_y,\mathbf J^2]\\ =&0\end{align*}$$

3. $[J_\pm,J_z] = \mp \hbar J_\pm$
   

   proof
   

   [J±,Jz]=====[Jx±iJy,Jz][Jx,Jz]±i[Jy,Jz]−iℏJy±i⋅iℏJx∓ℏ(Jx±iJy)∓ℏJ±

   $$\begin{align*}[J_\pm,J_z]=&[J_x\pm iJ_y,J_z]\\ =&[J_x,J_z]\pm i[J_y,J_z]\\ =&-i\hbar J_y\pm i\cdot i\hbar J_x\\ =&\mp\hbar(J_x\pm iJ_y)\\ =&\mp\hbar J_\pm\end{align*}$$

4. $[J_\pm,J_\mp]=\pm 2\hbar J_z$
   

   proof
   

   [J±,J∓]====[Jx±iJy,Jx∓iJy][Jx,Jx]+[Jy,Jy]∓i([Jx,Jy]−[Jy,Jx])0+0±2ℏJz±2ℏJz

   $$\begin{align*}[J_\pm,J_\mp]=&[J_x\pm iJ_y,J_x\mp iJ_y]\\=&[J_x,J_x]+[J_y,J_y]\mp i([J_x,J_y]-[J_y,J_x])\\=&0+0\pm 2\hbar J_z\\=&\pm 2\hbar J_z\end{align*}$$

5. $J_\pm J_\mp=\mathbf J^2-J_z^2\pm\hbar J_z$
   

   proof
   

   J±J∓====(Jx±iJy)(Jx∓iJy)J2x+J2y∓i(JxJy−JyJx)J2x+J2y±ℏJzJ2−J2z±ℏJz

   $$\begin{align*}J_\pm J_\mp=&(J_x\pm iJ_y)(J_x\mp iJ_y)\\=&J_x^2+J_y^2\mp i(J_xJ_y-J_yJ_x)\\=&J_x^2+J_y^2\pm \hbar J_z\\=&\mathbf J^2-J_z^2\pm \hbar J_z\end{align*}$$

泡利矩阵

Definition

$S_\alpha=\dfrac {\hbar}2\sigma_\alpha$ so that $\sigma_\alpha^2=1$.

Properties

1. $[\sigma_\alpha,\sigma_\beta]=2i\epsilon_{\alpha\beta\gamma}\sigma_\gamma$
   

   proof
   

   [σα,σβ]=====[2ℏSα,2ℏSβ]4ℏ2[Sα,Sβ]4ℏ2⋅iℏϵαβγSγ2iϵαβγ⋅2ℏSγ2iϵαβγσγ

   $$\begin{align*}[\sigma_\alpha,\sigma_\beta]=&\left[\frac 2\hbar S _\alpha,\frac 2\hbar S_\beta\right]\\=&\frac 4{\hbar^2} [S_\alpha,S_\beta]\\=&\frac 4{\hbar^2}\cdot i\hbar\epsilon_{\alpha\beta\gamma}S_\gamma\\=&2i\epsilon_{\alpha\beta\gamma}\cdot \frac 2\hbar S_\gamma\\=&2i\epsilon_{\alpha\beta\gamma}\sigma_\gamma\end{align*}$$

2. $[\sigma_\alpha,\sigma_\beta]_+=\sigma_\alpha\sigma_\beta+\sigma_\beta\sigma_\alpha\equiv 0(\alpha\ne\beta)$
   

   proof
   根据prop 1
   

   σασβ−σβσα=2iϵαβγσγ

   $$\begin{align*}\sigma_\alpha\sigma_\beta-\sigma_\beta\sigma_\alpha=2i\epsilon_{\alpha\beta\gamma}\sigma_\gamma\end{align*}$$
   
   两边左乘
   
   σα
   $\sigma_\alpha$
   
   
   

   σβ−σασβσα=2iϵαβγσασγ

   $$\sigma_\beta-\sigma_\alpha\sigma_\beta\sigma_\alpha=2i\epsilon_{\alpha\beta\gamma}\sigma_\alpha\sigma_\gamma$$
   
   两边右乘
   
   σα
   $\sigma_\alpha$
   
   
   

   σασβσα−σβ=2iϵαβγσγσα

   $$\sigma_\alpha\sigma_\beta\sigma_\alpha-\sigma_\beta=2i\epsilon_{\alpha\beta\gamma}\sigma_\gamma\sigma_\alpha$$
   
   两式相加
   
   
   

   4iϵαβγ(σασγ+σγσα)=0

   $$4i\epsilon_{\alpha\beta\gamma}(\sigma_\alpha\sigma_\gamma+\sigma_\gamma\sigma_\alpha)= 0$$
   
   因此
   
   
   

   [σα,σβ]+=σασβ+σβσα≡0

   $$[\sigma_\alpha,\sigma_\beta]_+=\sigma_\alpha\sigma_\beta+\sigma_\beta\sigma_\alpha\equiv 0$$ (provided
   
   α≠β
   $\alpha\ne\beta$)
   
   if
   
   α=β
   $\alpha=\beta$,
   
   
   

   [σα,σβ]+=σασβ+σβσα=2

   $$[\sigma_\alpha,\sigma_\beta]_+=\sigma_\alpha\sigma_\beta+\sigma_\beta\sigma_\alpha=2$$
   
   To sum up
   
   
   

   [σα,σβ]+=σασβ+σβσα=2δαβ

   $$[\sigma_\alpha,\sigma_\beta]_+=\sigma_\alpha\sigma_\beta+\sigma_\beta\sigma_\alpha=2\delta_{\alpha\beta}$$

3. $\sigma_\alpha\sigma_\beta=\delta_{\alpha\beta}+i\epsilon_{\alpha\beta\gamma}\sigma_\gamma$
   

   proof
   

   σασβ−σβσα=σασβ+σβσα=2iϵαβγσγ2iϵαβγσγ+2δαβ

   $$\begin{align*}\sigma_\alpha\sigma_\beta-\sigma_\beta\sigma_\alpha=&2i\epsilon_{\alpha\beta\gamma}\sigma_\gamma\\ \sigma_\alpha\sigma_\beta+\sigma_\beta\sigma_\alpha=&2i\epsilon_{\alpha\beta\gamma}\sigma_\gamma+2\delta_{\alpha\beta}\end{align*}$$
   
   Therefore
   
   
   

   σασβ=δαβ+iϵαβγσγ

   $$\sigma_\alpha\sigma_\beta=\delta_{\alpha\beta}+i\epsilon_{\alpha\beta\gamma}\sigma_\gamma$$

4. DEF: $\sigma_\pm=\dfrac{\sigma_x\pm i\sigma_y}2$
5. DEF: $\sigma_+=|1\rangle\langle0|,\sigma_-=|0\rangle\langle1|$
6. ⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪σx=(0110)σy=(0i−i0)σz=(100−1)

   $$\begin{cases}\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}\\\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}\\\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\end{cases}$$

proof
$\sigma_+=|1\rangle\langle0|=\begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}0&1\end{pmatrix}=\begin{pmatrix}0&1\\0&0\end{pmatrix}$
$\sigma_-=|0\rangle\langle1|=\begin{pmatrix}0\\1\end{pmatrix}\begin{pmatrix}1&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}$
$\sigma_x=\sigma_++\sigma_-=\begin{pmatrix}0&1\\1&0\end{pmatrix}$
$\sigma_y=-i(\sigma_+-\sigma_-)=\begin{pmatrix}0&i\\-i&0\end{pmatrix}$

$\begin{cases}\sigma_z|1\rangle=+1|1\rangle\\\sigma_z|0\rangle=-1|0\rangle\end{cases}$
$\begin{align*}\sigma_z=&\mathbb I\sigma_z\mathbb I\\=&(|0\rangle\langle0|+|1\rangle\langle1|)\sigma_z(|0\rangle\langle0|+|1\rangle\langle1|)\\=&|1\rangle\langle1|-|0\rangle\langle0|\\=&\begin{pmatrix}1&0\\0&0\end{pmatrix}-\begin{pmatrix}0&0\\0&1\end{pmatrix}\\=&\begin{pmatrix}1&0\\0&-1\end{pmatrix}\end{align*}$

总角动量$\hat J$

Definition

$\mathbf{J=L+S}$

Properties

1. $[J_\alpha,J_\beta]=i\hbar \epsilon_{\alpha\beta\gamma}L_\gamma$
   $[J_\alpha,S_\beta]\equiv0$
2. $[\mathbf J^2,J_\alpha]\equiv 0$
   
   ⇒{
   J2|j,mj⟩=Jz|j,mj⟩=j(j+1)ℏ2|j,mj⟩,j=0,1,…mjℏ|j,mj⟩,mj=−j,−j+1,…,+j

   $$\Rightarrow\begin{cases}\mathbf J^2|j,m_j\rangle=&j(j+1)\hbar^2|j,m_j\rangle,\quad j=0,1,\dots\\J_z|j,m_j\rangle=&m_j\hbar|j,m_j\rangle ,\quad m_j=-j,-j+1,\dots,+j\end{cases}$$

3. $[J_{\pm},\mathbf J^2]\equiv0$
4. $[J_\pm,J_z]=\mp\hbar J_\pm$
5. $[J_\pm,J_\mp]=\pm 2\hbar J_z$
6. $J_\pm J_\mp=\mathbf J^2-J_z^2\pm\hbar J_z$

角动量的本征值问题

用纯代数的方法，借助升降算符的性质，求解角动量的本征值问题，虽然没有物理图像，但是过程简洁。

$$[\mathbf J^2,J_\alpha]=0$$

$$\Rightarrow\begin{cases}\mathbf J^2|\lambda,m\rangle=\lambda\hbar^2|\lambda,m\rangle\\J_z|\lambda,m\rangle=m\hbar|\lambda,m\rangle \end{cases}$$ (

λ,m
$\lambda,m$ unknown)

$$[J_\pm,J_z]=\mp\hbar J_\pm$$

$$\Rightarrow J_zJ_\pm=J_\pm J_z\pm\hbar J_\pm$$

$$\Rightarrow\begin{cases}\mathbf J^2[J_\pm|\lambda,m\rangle] =J_\pm\mathbf J^2|\lambda,m\rangle=\lambda\hbar^2[J_\pm|\lambda,m\rangle]\\J_z[J_\pm|\lambda,m\rangle]=(J_\pm J_z\pm\hbar J_\pm)|\lambda,m\rangle=(m\pm1)\hbar[J_\pm|\lambda,m\rangle]\end{cases}$$

which means

$$J_\pm|\lambda,m\rangle=C|\lambda,m\pm1 \rangle$$

However

$$\mathbf J^2\geqslant J_z^2\Rightarrow m^2\leqslant \lambda$$

Let

m0
$m_0$ &

m0+N
$m_0+N$ be the extremums

Thus

$$\begin{cases}J_+|\lambda,m_0\rangle\equiv0\\J_-|\lambda,m_0+N\rangle\equiv0\end{cases}$$

As

$$\begin{align*}\langle\lambda,m|J_\pm J_\mp|\lambda,m\rangle=&\langle\lambda,m|\mathbf J^2-J_z^2\pm\hbar J_z|\lambda,m\rangle\\=&(\lambda^2-m^2\pm m)\hbar^2\end{align*}$$

We can get

$$\begin{cases}\lambda^2-m_0^2+ m_0=0\\\lambda^2-(m_0+N)^2- (m_0+N)=0\end{cases}$$



⇒
$\Rightarrow$

$$\begin{cases}m_0=-\dfrac N2\\\lambda=\dfrac N2\left(\dfrac N2+1\right)\end{cases}$$

Let

j=N2,N=0,1,2,...
$j=\dfrac N2, N=0,1,2,...$

$$\begin{cases}\mathbf J^2|j,m_j\rangle=j(j+1)\hbar^2|j,m_j\rangle,\quad j=0,\dfrac 12,1,\dfrac 32,\dots\\J_z|j,m_j\rangle=m_j\hbar|j,m_j\rangle ,\quad m_j=-j,-j+1,\dots,+j\end{cases}$$

角动量的耦合

Two electron’s coupling

$\bf S=S_1+S_2$

1. Commutation Relations
   $[S_{1\alpha},S_1^2]\equiv 0,[S_{2\alpha},S_2^2]\equiv 0,[S_{\alpha},S^2]\equiv 0$
   $[S_{1\alpha},S_{1\beta}]=i\hbar \epsilon_{\alpha\beta\gamma} S_{1\gamma},[S_{2\alpha},S_{2\beta}]=i\hbar \epsilon_{\alpha\beta\gamma }S_{2\gamma},[S_{\alpha},S_{\beta}]=i\hbar \epsilon_{\alpha\beta\gamma }S_{\gamma}$
   $[S_{1\alpha},S_{2\beta}]=0$
2. Eigenvalue Equations

Tricks

1. ⎛⎝⎜123⎞⎠⎟⨂(10100)=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜101002020030300⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟

   $$\left(\matrix {1\\2\\3}\right)\bigotimes \left(\matrix {10\\100}\right)=\left(\matrix {10\\100\\20\\200\\30\\300}\right)$$
   低维矢量到高维矢量。

2.

Thanks to Prof. Guo Hong