사용자:종잇조각/연습장

1.

12.5.3

(1)

Since $J_{x}$, $J_{y}$, and $J_{z}$ are Hermitian, $J_{x}=J_{x}^{\dagger }$, $J_{y}=J_{y}^{\dagger }$, and $J_{z}=J_{z}^{\dagger }$, all of their eigenvalues are real, and there is a set of eigenvectors that forms a complete basis of the Hilbert space.
For any Hermitian operator, say ${\hat {A}}$, let its eigenbasis, $\left\{|a_{1}\rangle ,|a_{2}\rangle ,\cdots \right\}$. Then for any $|\psi \rangle$,

$\langle \psi |{\hat {A}}|\psi \rangle =\sum _{i,j}\langle \psi |a_{i}\rangle \langle a_{i}|{\hat {A}}|a_{j}\rangle \langle a_{j}|\psi \rangle =\sum _{i,j}a_{j}\langle a_{i}|\psi \rangle \langle a_{i}|a_{j}\rangle \langle a_{j}|\psi \rangle =\sum _{i,j}a_{j}\delta _{ij}\langle a_{i}|\psi \rangle \langle a_{j}|\psi \rangle =\sum _{i}a_{i}\left|\langle a_{i}|\psi \rangle \right|^{2}\in \mathbb {R}$

${\begin{aligned}\langle J_{x}\rangle &=\langle jm|J_{x}|jm\rangle =\langle jm|-{i \over \hbar }\left[J_{y},J_{z}\right]|jm\rangle =-{i \over \hbar }\left(\langle jm|J_{y}J_{z}|jm\rangle -\langle jm|J_{z}J_{y}|jm\rangle \right)\\&=-{i \over \hbar }\left(m\hbar \langle jm|J_{y}|jm\rangle -\langle jm|J_{y}^{\dagger }J_{z}^{\dagger }|jm\rangle ^{\dagger }\right)=-{i \over \hbar }\left(m\hbar \langle jm|J_{y}|jm\rangle -\langle jm|J_{y}J_{z}|jm\rangle ^{\dagger }\right)\\&=-{i \over \hbar }\left(m\hbar \langle jm|J_{y}|jm\rangle -m\hbar \langle jm|J_{y}|jm\rangle ^{\dagger }\right)=m\ {\text{Im}}\left\{\langle jm|J_{y}|jm\rangle \right\}=0\left(\because \langle jm|J_{y}|jm\rangle \in \mathbb {R} \right)\end{aligned}}$
${\begin{aligned}\langle J_{y}\rangle &=\langle jm|J_{y}|jm\rangle =\langle jm|-{i \over \hbar }\left[J_{z},J_{x}\right]|jm\rangle =-{i \over \hbar }\left(\langle jm|J_{z}J_{x}|jm\rangle -\langle jm|J_{x}J_{z}|jm\rangle \right)\\&={i \over \hbar }\left(m\hbar \langle jm|J_{x}|jm\rangle -\langle jm|J_{x}^{\dagger }J_{z}^{\dagger }|jm\rangle ^{\dagger }\right)={i \over \hbar }\left(m\hbar \langle jm|J_{x}|jm\rangle -\langle jm|J_{x}J_{z}|jm\rangle ^{\dagger }\right)\\&={i \over \hbar }\left(m\hbar \langle jm|J_{x}|jm\rangle -m\hbar \langle jm|J_{x}|jm\rangle ^{\dagger }\right)=-m\ {\text{Im}}\left\{\langle jm|J_{x}|jm\rangle \right\}=0\left(\because \langle jm|J_{x}|jm\rangle \in \mathbb {R} \right)\end{aligned}}$

(2)

$\langle J_{x}^{2}\rangle +\langle J_{y}^{2}\rangle =\langle J_{x}^{2}+J_{y}^{2}\rangle =\langle J^{2}-J_{z}^{2}\rangle =\langle jm|J^{2}-J_{z}^{2}|jm\rangle =j\left(j+1\right)\hbar ^{2}-m^{2}\hbar ^{2}$

By symmetry, $\langle J_{x}^{2}\rangle =\langle J_{y}^{2}\rangle$.

$\therefore \langle J_{x}^{2}\rangle =\langle J_{y}^{2}\rangle ={1 \over 2}\hbar ^{2}\left[j\left(j+1\right)-m^{2}\right]$

(3)

$\left(\Delta J_{x}\right)^{2}=\langle J_{x}^{2}\rangle -\langle J_{x}\rangle ^{2}={1 \over 2}\hbar ^{2}\left[j\left(j+1\right)-m^{2}\right]$ and $\left(\Delta J_{y}\right)^{2}=\langle J_{y}^{2}\rangle -\langle J_{y}\rangle ^{2}={1 \over 2}\hbar ^{2}\left[j\left(j+1\right)-m^{2}\right]$.
Since $J_{x}={{J_{+}+J_{-}} \over 2}$ and $J_{y}={{J_{+}-J_{-}} \over {2i}}$,
${\begin{aligned}J_{x}|jm\rangle &={1 \over 2}\left(J_{+}|jm\rangle +J_{-}|jm\rangle \right)={1 \over 2}\left(J_{+}|jm\rangle +J_{-}|jm\rangle \right)\\&={1 \over 2}\left[\hbar \left\{\left(j-m\right)\left(j+m+1\right)\right\}^{1/2}|j,m+1\rangle +\hbar \left\{\left(j+m\right)\left(j-m+1\right)\right\}^{1/2}|j,m-1\rangle \right]\\J_{y}|jm\rangle &={1 \over 2i}\left(J_{+}|jm\rangle -J_{-}|jm\rangle \right)={1 \over 2i}\left(J_{+}|jm\rangle -J_{-}|jm\rangle \right)\\&={1 \over 2i}\left[\hbar \left\{\left(j-m\right)\left(j+m+1\right)\right\}^{1/2}|j,m+1\rangle -\hbar \left\{\left(j+m\right)\left(j-m+1\right)\right\}^{1/2}|j,m-1\rangle \right]\end{aligned}}$

$\langle j,m+1|j,m+1\rangle =\langle j,m-1|j,m-1\rangle =1,\langle j,m-1|j,m+1\rangle =\langle j,m+1|j,m-1\rangle =0\Rightarrow$

${\begin{aligned}\langle jm|J_{x}J_{y}|jm\rangle &=\left(J_{x}^{\dagger }|jm\rangle \right)^{\dagger }J_{y}|jm\rangle =\left(J_{x}|jm\rangle \right)^{\dagger }J_{y}|jm\rangle \\&={1 \over 2}\left[\hbar \left\{\left(j-m\right)\left(j+m+1\right)\right\}^{1/2}\langle j,m+1|+\hbar \left\{\left(j+m\right)\left(j-m+1\right)\right\}^{1/2}\langle j,m-1|\right]\times \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {1 \over 2i}\left[\hbar \left\{\left(j-m\right)\left(j+m+1\right)\right\}^{1/2}|j,m+1\rangle -\hbar \left\{\left(j+m\right)\left(j-m+1\right)\right\}^{1/2}|j,m-1\rangle \right]\\&={{\hbar ^{2}} \over {4i}}\left\{\left(j-m\right)\left(j+m+1\right)-\left(j+m\right)\left(j-m+1\right)\right\}\\&=-{{m\hbar ^{2}} \over {2i}}\end{aligned}}$

${\begin{aligned}\langle J_{x}\rangle ^{2}\langle J_{y}\rangle ^{2}-\left|\langle jm|J_{x}J_{y}|jm\rangle \right|^{2}&={\hbar ^{4} \over 4}\left[j\left(j+1\right)-m^{2}\right]^{2}-{{m^{2}\hbar ^{4}} \over {4}}={\hbar ^{4} \over 4}\left\{j\left(j+1\right)-m^{2}+m\right\}\left\{j\left(j+1\right)-m^{2}-m\right\}\\&={\hbar ^{4} \over 4}\left(j+m\right)\left(j-m+1\right)\left(j+m+1\right)\left(j-m\right)={\hbar ^{4} \over 4}\left(j^{2}-m^{2}\right)\left\{\left(j+1\right)^{2}-m^{2}\right\}\end{aligned}}$

$j\geq 0$ and $j\geq m\geq -j$. So $\left(j+1\right)^{2}\geq j^{2}\geq m^{2}$

$\therefore \langle J_{x}\rangle ^{2}\langle J_{y}\rangle ^{2}\geq \left|\langle jm|J_{x}J_{y}|jm\rangle \right|^{2}$

(4)

When $m=\pm j$, $j^{2}-m^{2}=0$. Then $\langle J_{x}\rangle ^{2}\langle J_{y}\rangle ^{2}-\left|\langle jm|J_{x}J_{y}|jm\rangle \right|^{2}=0$

$\therefore \langle J_{x}\rangle ^{2}\langle J_{y}\rangle ^{2}=\left|\langle j,\pm j|J_{x}J_{y}|j,\pm j\rangle \right|^{2}$

12.5.10

$-\hbar ^{2}\left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{1 \over {\sin ^{2}\theta }}{\partial ^{2} \over {\partial \phi ^{2}}}\right)\psi _{\alpha \beta }(\theta ,\phi )=\alpha \psi _{\alpha \beta }(\theta ,\phi )$

$\Rightarrow -\hbar ^{2}\left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{1 \over {\sin ^{2}\theta }}{\partial ^{2} \over {\partial \phi ^{2}}}\right)\left(P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }\right)=\alpha P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }$

$\Rightarrow -{{\hbar ^{2}} \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }+{{\beta ^{2}\hbar ^{2}} \over {\sin ^{2}\theta }}P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }=\alpha P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }$

$\Rightarrow {1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}P_{\alpha }^{\beta }(\theta )-{{\beta ^{2}} \over {\sin ^{2}\theta }}P_{\alpha }^{\beta }(\theta )=-{\alpha \over {\hbar ^{2}}}P_{\alpha }^{\beta }(\theta )$

$\Rightarrow \left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{\alpha \over {\hbar ^{2}}}-{{\beta ^{2}} \over {\sin ^{2}\theta }}\right)P_{\alpha }^{\beta }(\theta )=0$

$\therefore \left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{\alpha \over {\hbar ^{2}}}-{{m^{2}} \over {\sin ^{2}\theta }}\right)P_{\alpha }^{m}(\theta )=0$

${d \over {d\theta }}P_{\alpha }^{m}={{du} \over {d\theta }}{d \over {du}}P_{\alpha }^{m}=-\sin \theta {d \over {du}}P_{\alpha }^{m}$

${\begin{aligned}\Rightarrow {1 \over {\sin \theta }}{d \over {d\theta }}\sin \theta {d \over {d\theta }}P_{\alpha }^{m}&=-{1 \over {\sin \theta }}{d \over {d\theta }}\sin ^{2}\theta {d \over {du}}P_{\alpha }^{m}=-{1 \over {\sin \theta }}{d \over {d\theta }}\left\{\left(1-u^{2}\right){d \over {du}}P_{\alpha }^{m}\right\}\\&=-{1 \over {\sin \theta }}{{du} \over {d\theta }}{d \over {du}}\left\{\left(1-u^{2}\right){d \over {du}}P_{\alpha }^{m}\right\}=\left(1-u^{2}\right){d^{2} \over {du^{2}}}P_{\alpha }^{m}-2u{d \over {du}}P_{\alpha }^{m}\end{aligned}}$

For $m=0$,

$\left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{\alpha \over {\hbar ^{2}}}\right)P_{\alpha }^{0}(\theta )=\left(1-u^{2}\right){d^{2} \over {du^{2}}}P_{\alpha }^{0}-2u{d \over {du}}P_{\alpha }^{0}+{\alpha \over {\hbar ^{2}}}P_{\alpha }^{0}=0$

$\Rightarrow \left(1-u^{2}\right)\sum _{n=2}^{\infty }C_{n}n\left(n-1\right)u^{n-2}-2u\sum _{n=1}^{\infty }C_{n}nu^{n-1}+{\alpha \over {\hbar ^{2}}}\sum _{n=0}^{\infty }C_{n}u^{n}=0$

$\Rightarrow \sum _{n=0}^{\infty }C_{n+2}\left(n+2\right)\left(n+1\right)u^{n}-\sum _{n=2}^{\infty }C_{n}n\left(n-1\right)u^{n}-2\sum _{n=1}^{\infty }C_{n}nu^{n}+{\alpha \over {\hbar ^{2}}}\sum _{n=0}^{\infty }C_{n}u^{n}=0$

$\Rightarrow \sum _{n=0}^{\infty }\left\{C_{n+2}\left(n+2\right)\left(n+1\right)-C_{n}n\left(n-1\right)-2C_{n}n+{\alpha \over {\hbar ^{2}}}C_{n}\right\}u^{n}=0\left(\because n=1\Rightarrow n\left(n-1\right)=0,n=0\Rightarrow n\left(n-1\right)=n=0\right)$

$\Rightarrow C_{n+2}\left(n+2\right)\left(n+1\right)=\left(n^{2}+n-{\alpha \over {\hbar ^{2}}}\right)C_{n}$

$\therefore \lim _{n\to \infty }{C_{n+2} \over C_{n}}=\lim _{n\to \infty }{n^{2}+n-\alpha /\hbar ^{2} \over \left(n+2\right)\left(n+1\right)}=1$

Let $a_{0}=1$, and $a_{n+1}={4n^{2}+2n-\alpha /\hbar ^{2} \over {\left(2n+2\right)\left(2n+1\right)}}a_{n}$.

$\sum a_{n}$ diverges if for any $N\in \mathbb {N}$, there is $n>N$ such that $a_{n}\neq 0$.

${\begin{aligned}C_{0}C_{1}>0&\Rightarrow \left|P_{\alpha }^{0}(0)\right|\geq \sum \left|C_{0}\right|a_{n}\\C_{0}C_{1}<0&\Rightarrow \left|P_{\alpha }^{0}(\pi )\right|\geq \sum \left|C_{1}\right|a_{n}\\C_{0}=0&\Rightarrow \left|P_{\alpha }^{0}(\pi )\right|\geq \sum \left|C_{1}\right|a_{n}\\C_{1}=0&\Rightarrow \left|P_{\alpha }^{0}(0)\right|\geq \sum \left|C_{0}\right|a_{n}\\\end{aligned}}$

Since $C_{0}$ and $C_{1}$ cannot both be zero at the same time, one of $P_{\alpha }^{0}(0)$ or $P_{\alpha }^{0}(\pi )$ diverges unless the series terminates.

${\begin{aligned}{\alpha \over {\hbar ^{2}}}=l\left(l+1\right)&\Rightarrow C_{n+2}\left(n+2\right)\left(n+1\right)=\left(n^{2}+n-l^{2}-l\right)C_{n}\\&\Rightarrow C_{n+2}\left(n+2\right)\left(n+1\right)=\left(n+l+1\right)\left(n-l\right)C_{n}\\&\Rightarrow {\text{for all k}}\in \mathbb {N} ,C_{l+2k}=0\end{aligned}}$

If $l$ is even and $C_{1}=0$ , $P_{\alpha }^{0}$ is even. If $l$ is odd and $C_{0}=0$ , $P_{\alpha }^{0}$ is odd. For other values of $C_{0}$ or $C_{1}$ the series diverges and lacks any physical meaning.

For $l=0$, $C_{2k-1}=C_{2k}=0$ for all $k\in \mathbb {N} \Rightarrow P_{0}=1$.

For $l=1$, $C_{2k-2}=C_{2k+1}=0$ for all $k\in \mathbb {N} \Rightarrow P_{1}=u$.

For $l=2$, $C_{2k+2}=C_{2k-1}=0$ for all $k\in \mathbb {N}$ and $2C_{2}=-6C_{0}\Rightarrow P_{2}=3u^{2}-1$.

They are equal to $Y_{0}^{0}=\left(4\pi \right)^{{\text{-}}1/2},Y_{1}^{0}={\sqrt {3 \over {4\pi }}}u,Y_{2}^{0}={\sqrt {5 \over {16\pi }}}\left(3u^{2}-1\right)$ if we ignore the scaling factors.

12.6.9

The eigenvalue equation on $U$ is $\left[-{\hbar ^{2} \over {2\mu }}{d^{2} \over {dr^{2}}}+V(r)+{l\left(l+1\right)\hbar ^{2} \over {2\mu r^{2}}}\right]U=EU$

Since $l=0$, ${\begin{cases}-{\hbar ^{2} \over {2\mu }}{d^{2} \over {dr^{2}}}U=EU&r<r_{0}\\-{\hbar ^{2} \over {2\mu }}{d^{2} \over {dr^{2}}}U=\left(E-V_{0}\right)U&r>r_{0}\end{cases}}$

Let $k'={{\sqrt {2\mu E}} \over \hbar }$, $\kappa ={{\sqrt {2\mu \left(V_{0}-E\right)}} \over \hbar }$, and $U_{E}(r)={\begin{cases}U_{1}&r<r_{0}\\U_{2}&r>r_{0}\end{cases}}$.

The general solutions of each differential equations are $U_{1}=Ae^{ik'r}+Be^{-ik'r}$ and $U_{2}=Ce^{\kappa r}+De^{-\kappa r}$.

Boundary conditions:

(1) $U\to 0{\text{ as }}r\to 0$

$A+B=0$

(2) $U\to 0{\text{ as }}r\to \infty$

$C=0$

(3) continuity

$Ae^{ik'r_{0}}-Ae^{-ik'r_{0}}=De^{-\kappa r_{0}}$

(4) continuity of the derivative

$ik'Ae^{ik'r_{0}}+ik'Ae^{-ik'r_{0}}=-\kappa De^{-\kappa r_{0}}$

${\begin{aligned}ik'Ae^{ik'r_{0}}+ik'Ae^{-ik'r_{0}}=-\kappa \left(Ae^{ik'r_{0}}-Ae^{-ik'r_{0}}\right)&\Rightarrow ik'e^{ik'r_{0}}+ik'e^{-ik'r_{0}}=-\kappa \left(e^{ik'r_{0}}-e^{-ik'r_{0}}\right)\\&\Rightarrow 2ik'\cos \left(k'r_{0}\right)=-2i\kappa \sin \left(k'r_{0}\right)\end{aligned}}$

$\therefore k'/\kappa =-\tan \left(k'r_{0}\right)$

Since $k'^{2}+\kappa ^{2}={{2\mu V_{0}} \over \hbar ^{2}}$, if we take $\alpha =k'r_{0}$ and $\beta =\kappa r_{0}$, then $\alpha ^{2}+\beta ^{2}={{2\mu V_{0}{r_{0}}^{2}} \over \hbar ^{2}}$ and $\beta =-\alpha \cot \alpha$.

The minimum value of $\alpha ^{2}+\beta ^{2}$ on the first quadrant of the graph $\beta =-\alpha \cot \alpha$ is $\pi ^{2} \over 4$.

$\alpha ^{2}+\beta ^{2}={{2\mu V_{0}{r_{0}}^{2}} \over \hbar ^{2}}={\pi ^{2} \over 4}\Rightarrow V_{0}={{\pi ^{2}\hbar ^{2}} \over {8\mu {r_{0}}^{2}}}$

$\therefore$ If $V_{0}<{{\pi ^{2}\hbar ^{2}} \over {8\mu {r_{0}}^{2}}}$ there are no bound states.

12.6.11

(1)

$v'=\left(l+1\right)y^{l}\sum _{n=0}^{\infty }C_{n}y^{n}+y^{l+1}\sum _{n=1}^{\infty }nC_{n}y^{n-1}=y^{l}\sum _{n=0}^{\infty }\left(n+l+1\right)C_{n}y^{n}$

$v''=ly^{l-1}\sum _{n=0}^{\infty }\left(n+l+1\right)C_{n}y^{n}+y^{l}\sum _{n=1}^{\infty }n\left(n+l+1\right)C_{n}y^{n-1}=y^{l-1}\sum _{n=0}^{\infty }\left(n+l\right)\left(n+l+1\right)C_{n}y^{n}$

${\begin{aligned}v''&-2yv'+\left[2\lambda -1-{{l\left(l+1\right)} \over y^{2}}\right]v=0\\&\Rightarrow y^{l-1}\sum _{n=0}^{\infty }\left(n+l\right)\left(n+l+1\right)C_{n}y^{n}-2y^{l+1}\sum _{n=0}^{\infty }\left(n+l+1\right)C_{n}y^{n}+\left(2\lambda -1\right)y^{l+1}\sum _{n=0}^{\infty }C_{n}y^{n}-l\left(l+1\right)y^{l-1}\sum _{n=0}^{\infty }C_{n}y^{n}=0\\&\Rightarrow \sum _{n=0}^{\infty }\left(n+l\right)\left(n+l+1\right)C_{n}y^{n}-2\sum _{n=0}^{\infty }\left(n+l+1\right)C_{n}y^{n+2}+\left(2\lambda -1\right)\sum _{n=0}^{\infty }C_{n}y^{n+2}-l\left(l+1\right)\sum _{n=0}^{\infty }C_{n}y^{n}=0\\&\Rightarrow \left(l+1\right)\left(l+2\right)C_{1}y-l\left(l+1\right)C_{1}y+\sum _{n=2}^{\infty }\left(n+l\right)\left(n+l+1\right)C_{n}-2\left(n+l+1\right)C_{n-2}y^{n}+\left(2\lambda -1\right)C_{n-2}y^{n}-l\left(l+1\right)C_{n}y^{n}=0\\&\Rightarrow 2\left(l+1\right)C_{1}y+\sum _{n=2}^{\infty }\left\{n\left(n+2l+1\right)C_{n}-\left(2n+2l-2\lambda +3\right)C_{n-2}\right\}y^{n}=0\\&\Rightarrow C_{1}=0,\ n\left(n+2l+1\right)C_{n}=\left(2n+2l-2\lambda +3\right)C_{n-2}\\&\Rightarrow {\text{for any }}k\in \mathbb {N} ,C_{2k-1}=0\end{aligned}}$

If $C_{0}=0$, then for every $n$, $C_{n}=0$, which leads to a nonphysical zero wave function.

$\therefore C_{0}\neq 0$

For $y=1$, $v=\sum _{n=0}^{\infty }C_{n}=\sum _{k=0}^{\infty }C_{2k}$.

Unless the series terminates, ${\text{as }}k\to \infty ,{C_{2k+2} \over C_{2k}}\to {1 \over {k+1}}\Rightarrow v\sim y^{l+1}e^{y^{2}/2}$, which goes to infinity as r increases.

This cannot be a physical solution, so the series must terminate at certain $k\Rightarrow 4k+2l-2\lambda +3=0$

$\therefore E=\left(2k+l+3/2\right)\hbar \omega$

(2)

For each $n$, the allowed $l$ values are $n,n-2,\cdots ,1{\text{ or }}0$.

For each $l$, there are $2l+1$-fold degeneracy, so the total degeneracy is $\left(2n+1\right)+\left(2n-3\right)+\cdots +3{\text{ or }}1$.

If $n=2k$, $\left(4k+1\right)+\left(4k-3\right)+\cdots +1={{\left(4k+2\right)\left(k+1\right)} \over 2}={{\left(n+1\right)\left(n+2\right)} \over 2}$.

If $n=2k-1$, $\left(4k-1\right)+\left(4k-5\right)+\cdots +3={{\left(4k+2\right)k} \over 2}={{\left(n+1\right)\left(n+2\right)} \over 2}$.

$\therefore \psi _{n}$ has ${{\left(n+1\right)\left(n+2\right)} \over 2}$-fold degeneracy

Also $Y_{l}^{m}$ are odd if $l$ is odd and even if $l$ is even.

$\therefore$ parity of $\psi _{n}$ is $\left({\text{-}}1\right)^{n}$

The results are equal to the results obtained from the Cartesian solutions.

(3)

$n=2k+l=0\Rightarrow l=k=m=0$

Then $v=C_{0}y$ because $C_{k}=0$ for all $k\in \mathbb {N}$.

To normalize $U_{00}$, $\int _{0}^{\infty }\left|U_{00}\right|^{2}dr={C_{0}}^{2}\int _{0}^{\infty }y^{2}e^{-y^{2}}\left({\hbar \over {\mu \omega }}\right)^{1/2}dy={C_{0}}^{2}\left({\hbar \over {\mu \omega }}\right)^{1/2}{{\sqrt {\pi }} \over 4}=1\Rightarrow U_{00}=2\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}ye^{-y^{2}/2}$.

$\because \int _{0}^{\infty }x^{2}e^{-x^{2}}=\left[-{x \over 2}e^{-x^{2}}\right]_{0}^{\infty }+\int _{0}^{\infty }{1 \over 2}e^{-x^{2}}={{\sqrt {\pi }} \over 4}$

$\psi _{000}=U_{00}/rY_{0}^{0}=2\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}\left({\mu \omega \over {4\pi \hbar }}\right)^{1/2}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)=\left({{\mu \omega } \over {\pi \hbar }}\right)^{3/4}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)$

$n=2k+l=1\Rightarrow l=1,k=0,m={\text{-}}1,0,1$

Then $v=C_{0}y^{2}$ because $C_{k}=0$ for all $k\in \mathbb {N}$.

To normalize $U_{01}$, $\int _{0}^{\infty }\left|U_{01}\right|^{2}dr={C_{0}}^{2}\int _{0}^{\infty }y^{4}e^{-y^{2}}\left({\hbar \over {\mu \omega }}\right)^{1/2}dy={C_{0}}^{2}\left({\hbar \over {\mu \omega }}\right)^{1/2}{{3{\sqrt {\pi }}} \over 8}=1\Rightarrow U_{01}={\sqrt {8 \over 3}}\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}y^{2}e^{-y^{2}/2}$.

$\because \int _{0}^{\infty }x^{4}e^{-x^{2}}=\left[-{x^{3} \over 2}e^{-x^{2}}\right]_{0}^{\infty }+\int _{0}^{\infty }{3x^{2} \over 2}e^{-x^{2}}=\left[-{3x \over 4}e^{-x^{2}}\right]_{0}^{\infty }+\int _{0}^{\infty }{3 \over 4}e^{-x^{2}}={{3{\sqrt {\pi }}} \over 8}$

$\psi _{11\pm 1}=U_{01}/rY_{1}^{\pm 1}=\mp {\sqrt {8 \over 3}}{\sqrt {3 \over {8\pi }}}\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}\left({\mu \omega \over {\hbar }}\right)r\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\sin \theta e^{\pm \phi },\psi _{110}=U_{01}/rY_{1}^{0}={\sqrt {8 \over 3}}{\sqrt {3 \over {4\pi }}}\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}\left({\mu \omega \over {\hbar }}\right)r\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\cos \theta ,$

$\Rightarrow \psi _{11\pm 1}=\mp {\sqrt {\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}r\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\sin \theta e^{\pm i\phi },\psi _{110}={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}r\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\cos \theta$

Cartesian solutions for $n=0,1$ are:

$\psi '_{000}=\psi _{0}(x)\psi _{0}(y)\psi _{0}(z)={A_{0}}^{3}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)=\left({{\mu \omega } \over {\pi \hbar }}\right)^{3/4}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)$

$\psi '_{100}=\psi _{1}(x)\psi _{0}(y)\psi _{0}(z)={A_{0}}^{2}A_{1}\left({\mu \omega \over {\hbar }}\right)^{1/2}x\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}x\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)$

By the same logic,

$\psi '_{010}={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}y\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)$

$\psi '_{001}={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}z\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)$

So $\psi _{000}=\psi '_{000}$ and

$\psi _{11\pm 1}=\mp {\sqrt {\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)r\sin \theta \left(\cos \phi \pm i\sin \phi \right)=\mp {\sqrt {\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\left(x\pm iy\right)=\mp {1 \over {\sqrt {2}}}\psi '_{100}-{i \over {\sqrt {2}}}\psi '_{010}$

$\psi _{110}={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}z\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)=\psi '_{001}$

$\therefore \psi _{000}=\psi '_{000},\psi _{111}=-{1 \over {\sqrt {2}}}\psi '_{100}-{i \over {\sqrt {2}}}\psi '_{010},\psi _{110}=\psi '_{001},\psi _{11{\text{-}}1}={1 \over {\sqrt {2}}}\psi '_{100}-{i \over {\sqrt {2}}}\psi '_{010}$

13.1.5

$\langle {\dot {\Omega }}\rangle =-{i \over \hbar }\langle \left[\Omega ,H\right]\rangle$

$\left[\Omega ,H\right]=\left[XP_{X}+YP_{Y}+ZP_{Z},H\right]$

$\left[XP_{X},H\right]={1 \over {2m}}\left[XP_{X},P_{X}^{2}+P_{Y}^{2}+P_{Z}^{2}\right]-e^{2}\left[XP_{X},{1 \over R}\right]$

${\begin{aligned}\left[XP_{X},P_{X}^{2}+P_{Y}^{2}+P_{Z}^{2}\right]&=\left[XP_{X},P_{X}^{2}\right]=XP_{X}^{3}-P_{X}^{2}XP_{X}\\&=XP_{X}^{3}-P_{X}XP_{X}^{2}+P_{X}XP_{X}^{2}-P_{X}^{2}XP_{X}=\left[X,P_{X}\right]P_{X}^{2}+P_{X}\left[X,P_{X}\right]P_{X}=2i\hbar P_{X}^{2}\end{aligned}}$

${i \over \hbar }\langle x',y',z'|\left[XP_{X},{1 \over R}\right]|x,y,z\rangle =x'\delta '(x-x')\delta (y-y')\delta (z-z'){1 \over {\sqrt {x^{2}+y^{2}+z^{2}}}}-{1 \over {\sqrt {x'^{2}+y'^{2}+z'^{2}}}}x'\delta '(x-x')\delta (y-y')\delta (z-z')$

${\begin{aligned}{i \over \hbar }\int \langle n,l,m&|x',y',z'\rangle \langle x',y',z'|\left[XP_{X},{1 \over R}\right]|x,y,z\rangle \langle x,y,z|n,l,m\rangle dxdydzdx'dy'dz'\\&=\int \psi _{nlm}^{*}x{\partial \over {\partial x}}{1 \over {\sqrt {x^{2}+y^{2}+z^{2}}}}\psi _{nlm}-\psi _{nlm}^{*}{x \over {\sqrt {x^{2}+y^{2}+z^{2}}}}{\partial \over {\partial x}}\psi _{nlm}dxdydz\\&=\int \psi _{nlm}^{*}x{\partial \over {\partial x}}\left({1 \over r}\right)\psi _{nlm}dxdydz\\&=\int -\psi _{nlm}^{*}{x^{2} \over r^{3}}\psi _{nlm}dxdydz=-\langle {x^{2} \over r^{3}}\rangle \end{aligned}}$

$\langle {\dot {\Omega }}\rangle =-{i \over \hbar }\langle \left[\Omega ,H\right]\rangle ={1 \over m}\langle P^{2}\rangle -e^{2}\langle {1 \over r}\rangle =2\langle T\rangle +\langle V\rangle =0\left(\because \langle {x^{2} \over r^{3}}\rangle +\langle {y^{2} \over r^{3}}\rangle +\langle {z^{2} \over r^{3}}\rangle =\langle {1 \over r}\rangle \right)$

$\therefore \langle T\rangle =-{1 \over 2}\langle V\rangle$

13.2.1

(1)

$\mathbf {n} ={{\mathbf {p} \times \mathbf {l} } \over m}-{e^{2} \over r}\mathbf {r} ={{\mathbf {p} \times \left(\mathbf {r} \times \mathbf {p} \right)} \over m}-{e^{2} \over r}\mathbf {r} ={{\mathbf {r} p^{2}-\mathbf {p} \left(\mathbf {p} \cdot \mathbf {r} \right)} \over m}-{e^{2} \over r}\mathbf {r} =\left({p^{2} \over m}-{e^{2} \over r}\right)\mathbf {r} -{\mathbf {p} \cdot \mathbf {r} \over m}\mathbf {p}$

(2)

By energy conservation, ${p^{2} \over {2m}}-{e^{2} \over r}=E$.

Let the angle between $\mathbf {p}$ and $\mathbf {r}$ be $\theta$. i.e. $\mathbf {p} \cdot \mathbf {r} =pr\cos \theta$

$\mathbf {n} =\left(2E+{e^{2} \over r}\right)\mathbf {r} -{\mathbf {p} \cdot \mathbf {r} \over m}\mathbf {p} =\left(2E+{e^{2} \over r}\right)\mathbf {r} -\left({\dot {\mathbf {r} }}\cdot \mathbf {r} \right)\mathbf {p} =\left(2E+{e^{2} \over r}\right)\mathbf {r} -{1 \over 2}{d \over {dt}}\left(\mathbf {r} \cdot \mathbf {r} \right)\mathbf {p} =\left(2E+{e^{2} \over r}\right)\mathbf {r} -{1 \over 2}{d \over {dt}}\left(r^{2}\right)\mathbf {p}$

When the particle arrives at the maximum or minimum distance from the origin, ${d \over {dt}}\left(r^{2}\right)=0$.

So $\mathbf {n} =\mathbf {r} _{\text{max}}\left(2E+{e^{2} \over r_{\text{max}}}\right)=\mathbf {r} _{\text{min}}\left(2E+{e^{2} \over r_{\text{min}}}\right)$ and $\mathbf {n} \parallel \mathbf {r} _{\text{max}}\parallel \mathbf {r} _{\text{min}}$.

If $\mathbf {n}$, $\mathbf {r} _{\text{max}}$, and $\mathbf {r} _{\text{min}}$ are all pointing to the same direction, $\mathbf {r} _{\text{max}}=r_{\text{max}}{\hat {r}},\mathbf {r} _{\text{min}}=r_{\text{min}}{\hat {r}}$.

$\mathbf {n} =r_{\text{max}}\left(2E+{e^{2} \over r_{\text{max}}}\right){\hat {r}}=\left(2Er_{\text{max}}+e^{2}\right){\hat {r}}=r_{\text{min}}\left(2E+{e^{2} \over r_{\text{min}}}\right)=\left(2Er_{\text{min}}+e^{2}\right){\hat {r}}\Rightarrow 2Er_{\text{max}}+e^{2}=2Er_{\text{min}}+e^{2}\Rightarrow r_{\text{max}}=r_{\text{min}}$

This does not hold for general cases, so $\mathbf {r} _{\text{max}}=-r_{\text{max}}{\hat {r}}$.

$\mathbf {n} =-r_{\text{max}}\left(2E+{e^{2} \over r_{\text{max}}}\right){\hat {r}}=-\left(2Er_{\text{max}}+e^{2}\right){\hat {r}}=r_{\text{min}}\left(2E+{e^{2} \over r_{\text{min}}}\right)=\left(2Er_{\text{min}}+e^{2}\right){\hat {r}}\Rightarrow -2Er_{\text{max}}-e^{2}=2Er_{\text{min}}+e^{2}\Rightarrow r_{\text{max}}+r_{\text{min}}=-{e^{2} \over E}$

$2E+{e^{2} \over r_{\text{min}}}\geq 0$, so $\mathbf {r} _{\text{max}}$ and $\mathbf {r} _{\text{min}}$ are antiparallel to each other.

For circular orbits, $r_{\text{max}}+r_{\text{min}}=2r=-{e^{2} \over E}\Rightarrow 2E+{e^{2} \over r}=0$.

$\therefore \mathbf {n} =0$

2.

(1)

$\left[{\hat {X}},{\hat {P}}_{x}\right]=\left[{\hat {Y}},{\hat {P}}_{y}\right]=i\hbar ,\ \left[{\hat {X}},{\hat {Y}}\right]=\left[{\hat {P}}_{x},{\hat {P}}_{y}\right]=0\Rightarrow$,

${\begin{aligned}\left[{\hat {Q}}_{1},{\hat {P}}_{1}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {P}}_{x}\right]-\left[{\hat {P}}_{y},{\hat {Y}}\right]+{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {P}}_{x}\right]-{\hbar \over a^{2}}\left[{\hat {X}},{\hat {Y}}\right]\right)\\&=i\hbar \\\left[{\hat {Q}}_{2},{\hat {P}}_{2}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {P}}_{x}\right]-\left[{\hat {P}}_{y},{\hat {Y}}\right]-{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {P}}_{x}\right]+{\hbar \over a^{2}}\left[{\hat {X}},{\hat {Y}}\right]\right)\\&=i\hbar \\\left[{\hat {Q}}_{1},{\hat {Q}}_{2}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right)-{1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {X}}\right]-{a^{4} \over \hbar ^{2}}\left[{\hat {P}}_{y},{\hat {P}}_{y}\right]-{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {X}}\right]+{a^{2} \over \hbar }\left[{\hat {X}},{\hat {P}}_{y}\right]\right)\\&=0\\\left[{\hat {P}}_{1},{\hat {P}}_{2}\right]&={1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right)\\&={1 \over 2}\left(\left[{\hat {P}}_{x},{\hat {P}}_{x}\right]-{\hbar ^{2} \over a^{4}}\left[{\hat {Y}},{\hat {Y}}\right]-{\hbar \over a^{2}}\left[{\hat {Y}},{\hat {P}}_{x}\right]+{\hbar \over a^{2}}\left[{\hat {P}}_{x},{\hat {Y}}\right]\right)\\&=0\\\left[{\hat {Q}}_{1},{\hat {P}}_{2}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {P}}_{x}\right]+\left[{\hat {P}}_{y},{\hat {Y}}\right]+{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {P}}_{x}\right]+{\hbar \over a^{2}}\left[{\hat {X}},{\hat {Y}}\right]\right)\\&=0\\\left[{\hat {Q}}_{2},{\hat {P}}_{1}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {P}}_{x}\right]+\left[{\hat {P}}_{y},{\hat {Y}}\right]-{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {P}}_{x}\right]-{\hbar \over a^{2}}\left[{\hat {X}},{\hat {Y}}\right]\right)\\&=0\\\end{aligned}}$

(2)

$\left[{\hat {X}},{\hat {P}}_{y}\right]=\left[{\hat {Y}},{\hat {P}}_{x}\right]=0\Rightarrow {\hat {Q_{1}}}^{2}-{\hat {Q_{2}}}^{2}={a^{2} \over \hbar }{\hat {X}}{\hat {P}}_{y},\ {\hat {P_{1}}}^{2}-{\hat {P_{2}}}^{2}=-{\hbar \over a^{2}}{\hat {Y}}{\hat {P}}_{x}$

$\left({a^{2} \over 2\hbar }{\hat {P_{1}}}^{2}+{\hbar \over 2a^{2}}{\hat {Q_{1}}}^{2}\right)-\left({a^{2} \over 2\hbar }{\hat {P_{2}}}^{2}+{\hbar \over 2a^{2}}{\hat {Q_{2}}}^{2}\right)={a^{2} \over 2\hbar }\left({\hat {P_{1}}}^{2}-{\hat {P_{2}}}^{2}\right)+{\hbar \over 2a^{2}}\left({\hat {Q_{1}}}^{2}-{\hat {Q_{2}}}^{2}\right)={1 \over 2}\left({\hat {X}}{\hat {P}}_{y}-{\hat {Y}}{\hat {P}}_{x}\right)={\hat {L}}_{z}$

(3)

Take ${\hat {\alpha }}={\sqrt {1 \over 2a^{2}}}{\hat {Q}}_{1}+i{\sqrt {a^{2} \over 2\hbar ^{2}}}{\hat {P}}_{1}$, ${\hat {\beta }}={\sqrt {1 \over 2a^{2}}}{\hat {Q}}_{2}+i{\sqrt {a^{2} \over 2\hbar ^{2}}}{\hat {P}}_{2}$, and ${{\hat {L}}_{z} \over \hbar }={\hat {l}}_{z}$.
Then ${\hat {\alpha }}^{\dagger }={\sqrt {1 \over 2a^{2}}}{\hat {Q}}_{1}-i{\sqrt {a^{2} \over 2\hbar ^{2}}}{\hat {P}}_{1}$ and ${\hat {\beta }}^{\dagger }={\sqrt {1 \over 2a^{2}}}{\hat {Q}}_{2}-i{\sqrt {a^{2} \over 2\hbar ^{2}}}{\hat {P}}_{2}$, because ${\hat {P}}_{1}$, ${\hat {P}}_{2}$, ${\hat {Q}}_{1}$, and ${\hat {Q}}_{2}$ are all hermitian operators.

${\begin{aligned}{\hat {\alpha }}^{\dagger }{\hat {\alpha }}&={1 \over 2a^{2}}{\hat {Q_{1}}}^{2}+{a^{2} \over 2\hbar ^{2}}{\hat {P_{1}}}^{2}+{i \over {2\hbar }}\left[{\hat {Q}}_{1},{\hat {P}}_{1}\right]=\left({a^{2} \over 2\hbar ^{2}}{\hat {P_{1}}}^{2}+{1 \over 2a^{2}}{\hat {Q_{1}}}^{2}\right)-{1 \over 2}\\{\hat {\beta }}^{\dagger }{\hat {\beta }}&={1 \over 2a^{2}}{\hat {Q_{2}}}^{2}+{a^{2} \over 2\hbar ^{2}}{\hat {P_{2}}}^{2}+{i \over {2\hbar }}\left[{\hat {Q}}_{2},{\hat {P}}_{2}\right]=\left({a^{2} \over 2\hbar ^{2}}{\hat {P_{2}}}^{2}+{1 \over 2a^{2}}{\hat {Q_{2}}}^{2}\right)-{1 \over 2}\end{aligned}}\Rightarrow {\hat {l}}_{z}={\hat {\alpha }}^{\dagger }{\hat {\alpha }}-{\hat {\beta }}^{\dagger }{\hat {\beta }}$

Take $h_{\alpha }={\hat {\alpha }}^{\dagger }{\hat {\alpha }}+{1 \over 2}$ and $h_{\beta }={\hat {\beta }}^{\dagger }{\hat {\beta }}+{1 \over 2}$. Then ${\hat {l}}_{z}=h_{\alpha }-h_{\beta }$

$\left[{\hat {Q}}_{1},{\hat {P}}_{2}\right]=\left[{\hat {Q}}_{2},{\hat {P}}_{1}\right]=0$, so $\left[{\hat {\alpha }},{\hat {\beta }}\right]=0$. Also, $\left[{\hat {\alpha }},{\hat {\alpha }}^{\dagger }\right]={\hat {\alpha }}{\hat {\alpha }}^{\dagger }-{\hat {\alpha }}^{\dagger }{\hat {\alpha }}=1$ and $\left[{\hat {\beta }},{\hat {\beta }}^{\dagger }\right]={\hat {\beta }}{\hat {\beta }}^{\dagger }-{\hat {\beta }}^{\dagger }{\hat {\beta }}=1$.

Other commutation relations are:
$\left[{\hat {\alpha }},h_{\alpha }\right]={\hat {\alpha }}{\hat {\alpha }}^{\dagger }{\hat {\alpha }}-{\hat {\alpha }}^{\dagger }{\hat {\alpha }}^{2}=\left[{\hat {\alpha }},{\hat {\alpha }}^{\dagger }\right]{\hat {\alpha }}={\hat {\alpha }}$, $\left[{\hat {\alpha }}^{\dagger },h_{\alpha }\right]={\hat {\alpha ^{\dagger }}}^{2}{\hat {\alpha }}-{\hat {\alpha }}^{\dagger }{\hat {\alpha }}{\hat {\alpha }}^{\dagger }=-{\hat {\alpha }}^{\dagger }\left[{\hat {\alpha }},{\hat {\alpha }}^{\dagger }\right]=-{\hat {\alpha }}^{\dagger }$, $\left[{\hat {\beta }},h_{\beta }\right]=-{\hat {\beta }}$, and $\left[{\hat {\beta }}^{\dagger },h_{\beta }\right]={\hat {\beta }}^{\dagger }$.

By the same logic with the harmonic oscillators, $E_{\alpha }=n_{\alpha }+{1 \over 2}$ and $E_{\beta }=n_{\beta }+{1 \over 2}$. Since ${\hat {\alpha }}$ and ${\hat {\beta }}$ commute, an eigenstate of one operator is an eigenstate of another operator.

$l=E_{\alpha }-E_{\beta }=n_{\alpha }-n_{\beta }\in \mathbb {N}$

$\therefore L_{z}=m\hbar$ where $m\in \mathbb {N}$

3.

(1)

$\psi (\mathbf {r} )=A'e^{-\alpha r}Y_{0}^{0}$

Since the spherical harmonics are orthonormal to each other, the wave function only has the $|0,0\rangle$ component.

$\therefore P_{lm}={\begin{cases}1&l=m=0\\0&otherwise\end{cases}}$

(2)

${\begin{aligned}\psi (\mathbf {r} )&=A\left(r\sin \theta \cos \phi +r\sin \theta \sin \phi +2r\cos \theta \right)e^{-\alpha r}\\&=A\left(\sin \theta {{e^{i\phi }+e^{-i\phi }} \over 2}-i\sin \theta {{e^{i\phi }-e^{-i\phi }} \over 2}+\cos \theta \right)re^{-\alpha r}\\&=A\left\{{1 \over 2}\left(1-i\right)\sin \theta e^{i\phi }+{1 \over 2}\left(1+i\right)\sin \theta e^{-i\phi }+\cos \theta \right\}re^{-\alpha r}\\&=A\left\{-{\sqrt {2\pi \over 3}}\left(1-i\right)Y_{1}^{1}+{\sqrt {2\pi \over 3}}\left(1+i\right)Y_{1}^{{\text{-}}1}+{\sqrt {4\pi \over 3}}Y_{1}^{0}\right\}re^{-\alpha r}\end{aligned}}$

$\left|{\sqrt {2\pi \over 3}}\left(1-i\right)\right|^{2}=\left|{\sqrt {2\pi \over 3}}\left(1+i\right)\right|^{2}=\left|{\sqrt {4\pi \over 3}}\right|^{2}={4\pi \over 3}$

Since the spherical harmonics are orthonormal to each other, the probability of observing $|1,1\rangle$, $|1,{\text{-}}1\rangle$, and $|1,0\rangle$ are same and others are impossible.

$\therefore P_{lm}={\begin{cases}{1 \over 3}&(l,m)=(1,{\text{-}}1),\ (1,0),\ or\ (1,1)\\0&otherwise\\\end{cases}}$

(3)

${\begin{aligned}\psi (\mathbf {r} )&=A\left(\sin ^{2}\theta \sin \phi \cos \phi +\sin \theta \cos \theta \sin \phi +\sin \theta \cos \theta \cos \phi \right)r^{2}e^{-\alpha r^{2}}\\&=A\left(\sin ^{2}\theta {{e^{2i\phi }-e^{-2i\phi }} \over 4i}+\sin \theta \cos \theta {{e^{i\phi }-e^{-i\phi }} \over 2i}+\sin \theta \cos \theta {{e^{i\phi }+e^{-i\phi }} \over 2}\right)r^{2}e^{-\alpha r^{2}}\\&=A\left\{{1 \over 4i}\sin ^{2}\theta e^{2i\phi }-{1 \over 4i}\sin ^{2}\theta e^{-2i\phi }+{1 \over 2}\left(1-i\right)\sin \theta \cos \theta e^{i\phi }+{1 \over 2}\left(1+i\right)e^{-i\phi }\sin \theta \cos \theta \right\}r^{2}e^{-\alpha r^{2}}\\&=A\left\{-i{\sqrt {2\pi \over 15}}Y_{2}^{2}+i{\sqrt {2\pi \over 15}}Y_{2}^{{\text{-}}2}-\left(1-i\right){\sqrt {2\pi \over 15}}Y_{2}^{1}+\left(1+i\right){\sqrt {2\pi \over 15}}Y_{2}^{{\text{-}}1}\right\}r^{2}e^{-\alpha r^{2}}\end{aligned}}$

$\left|-i{\sqrt {2\pi \over 15}}\right|^{2}=\left|i{\sqrt {2\pi \over 15}}\right|^{2}={2\pi \over 15}$ and $\left|-\left(1-i\right){\sqrt {2\pi \over 15}}\right|^{2}=\left|\left(1+i\right){\sqrt {2\pi \over 15}}\right|^{2}={8\pi \over 15}$

The spherical harmonics are orthonormal to each other, so $P_{2{\text{-}}2}:P_{2{\text{-}}1}:P_{21}:P_{22}=1:4:4:1$ and $P_{2{\text{-}}2}+P_{2{\text{-}}1}+P_{21}+P_{22}=1$.

$\therefore P_{lm}={\begin{cases}{1 \over 10}&(l,m)=(2,{\text{-}}2)\ or\ (2,2)\\{2 \over 5}&(l,m)=(2,{\text{-}}1)\ or\ (2,1)\\0&otherwise\end{cases}}$

4.

Since $\psi (\mathbf {r} )$ is a energy eigenstate, it satisfies the T.I.S.E.
$\left\{-{\hbar ^{2} \over {2\mu }}\left({1 \over r^{2}}{\partial \over {\partial r}}r^{2}{\partial \over {\partial r}}+{1 \over {r^{2}\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{1 \over {r^{2}\sin ^{2}\theta }}{\partial ^{2} \over {\partial \phi ^{2}}}\right)+V(r)\right\}\psi (\mathbf {r} )=E\psi (\mathbf {r} )$

$\psi (\mathbf {r} )$ is independent of $\theta$ and $\phi$.
${\begin{aligned}\left(-{\hbar ^{2} \over {2\mu }}{1 \over r^{2}}{\partial \over {\partial r}}r^{2}{\partial \over {\partial r}}+V(\mathbf {r} )\right)Ae^{-\alpha r}&=EAe^{-\alpha r}\\{{\alpha \hbar ^{2}} \over {2\mu }}{1 \over r^{2}}{\partial \over \partial r}\left(r^{2}e^{-\alpha r}\right)+V\left(r\right)e^{-\alpha r}&=Ee^{-\alpha r}\\{{\alpha \hbar ^{2}} \over {2\mu }}{1 \over r^{2}}\left(2r-\alpha r^{2}\right)e^{-\alpha r}+V(r)e^{-\alpha r}&=Ee^{-\alpha r}\\{{\alpha \hbar ^{2}} \over {2\mu }}{1 \over r^{2}}\left(2r-\alpha r^{2}\right)+V(r)&=E\\{{\alpha \hbar ^{2}} \over {\mu r}}-{{\alpha ^{2}\hbar ^{2}} \over {2\mu }}+V(r)&=E\end{aligned}}$

$V(r)=-{{\alpha \hbar ^{2}} \over {\mu r}}+E+{{\alpha ^{2}\hbar ^{2}} \over {2\mu }}\Rightarrow \lim _{r\to \infty }V(r)=E+{{\alpha ^{2}\hbar ^{2}} \over {2\mu }}=0$

$\therefore V(r)=-{{\alpha \hbar ^{2}} \over {\mu r}}$ and $E=-{{\alpha ^{2}\hbar ^{2}} \over {2\mu }}$

5.

(1)

For ground state, $n=1$. Then the only possible $(l,m)=(0,0)$.
So the corresponding eigenstate is $\psi _{1,0,0}(\mathbf {r} )=BR_{10}Y_{0}^{0}=Be^{-r/{a_{0}}}$.
To normalize $\psi _{1,0,0}(\mathbf {r} )$,
${\begin{aligned}\langle \psi _{1,0,0}|\psi _{1,0,0}\rangle &=\int _{0}^{\infty }\int _{0}^{\pi }\int _{0}^{2\pi }\left|\psi _{1,0,0}(r,\theta ,\phi )\right|^{2}r^{2}\sin \theta d\phi d\theta dr\\&=\int _{0}^{\infty }\int _{0}^{\pi }\int _{0}^{2\pi }B^{2}e^{-2r/{a_{0}}}r^{2}\sin \theta d\phi d\theta dr\\&=2\pi B^{2}\left(\left[-{{a_{0}} \over 2}r^{2}e^{-2r/{a_{0}}}\right]_{0}^{\infty }+a_{0}\int _{0}^{\infty }re^{-2r/{a_{0}}}dr\right)\\&=2\pi B^{2}\left(\left[-{{a_{0}}^{2} \over 2}re^{-2r/{a_{0}}}\right]_{0}^{\infty }+{a_{0}}^{2}\int _{0}^{\infty }e^{-2r/{a_{0}}}dr\right)\\&=2\pi B^{2}\left[-{{a_{0}}^{3} \over 2}re^{-2r/{a_{0}}}\right]_{0}^{\infty }=\pi {a_{0}}^{3}B^{2}=1\Rightarrow B=\left({1 \over {\pi {a_{0}}^{3}}}\right)^{1/2}\end{aligned}}$
By the same logic, $A=\left({1 \over {\pi {a}^{3}}}\right)^{1/2}$.

${\begin{aligned}\langle \psi _{1,0,0}|\psi (\mathbf {r} )\rangle &=\int _{0}^{\infty }\int _{0}^{\pi }\int _{0}^{2\pi }\psi _{1,0,0}^{*}(r,\theta ,\phi )\psi (r,\theta ,\phi )r^{2}\sin \theta d\phi d\theta dr\\&={1 \over \pi }\int _{0}^{\infty }\int _{0}^{\pi }\int _{0}^{2\pi }\left(aa_{0}\right)^{-3/2}r^{2}\exp \left\{-\left({1 \over a}+{1 \over a_{0}}\right)r\right\}\sin \theta d\phi d\theta dr\\&=4\left(aa_{0}\right)^{-3/2}\int _{0}^{\infty }r^{2}\exp \left\{-\left({{a+a_{0}} \over {aa_{0}}}\right)r\right\}dr\\&=4\left(aa_{0}\right)^{-3/2}\int _{0}^{\infty }\left({aa_{0} \over {a+a_{0}}}\right)^{2}t^{2}e^{-t}\left({aa_{0} \over {a+a_{0}}}\right)dt\\&=4\left({{\sqrt {aa_{0}}} \over {a+a_{0}}}\right)^{3}\int _{0}^{\infty }t^{2}e^{-t}dt\\&=8\left({{\sqrt {aa_{0}}} \over {a+a_{0}}}\right)^{3}\end{aligned}}$.

cf.) $\int _{0}^{\infty }x^{2}e^{-x}dx=-\left[x^{2}e^{-x}\right]_{0}^{\infty }+2\int _{0}^{\infty }xe^{-x}dx=-\left[xe^{-x}\right]_{0}^{\infty }+2\int _{0}^{\infty }e^{-x}dx=2$

$\therefore P=\left\|\langle \psi _{1,0,0}|\psi (\mathbf {r} )\rangle \right\|^{2}=64\left({{\sqrt {aa_{0}}} \over {a+a_{0}}}\right)^{6}$

(2)

Reduced mass of $^{3}{\text{H}}$ is ${{\left(m_{p}+2m_{n}\right)m_{e}} \over {m_{p}+2m_{n}+m_{e}}}=9.10773026\times 10^{{\text{-}}31}\,{\text{kg}}$ so its bohr radius is $a_{1}=5.29273186\times 10^{{\text{-}}11}\,{\text{m}}$.

Reduced mass of $^{3}{\text{He}}^{+}$ is ${{\left(2m_{p}+m_{n}\right)m_{e}} \over {2m_{p}+m_{n}+m_{e}}}=9.10772950\times 10^{{\text{-}}31}\,{\text{kg}}$ so its bohr radius is $a_{2}=5.29273230\times 10^{{\text{-}}11}\,{\text{m}}$.

From the result of (1), $1-P=5.18\times 10^{-15}$.

6.

${\begin{aligned}H={\sqrt {\mathbf {p} ^{2}c^{2}+\mu ^{2}c^{4}}}+V(\mathbf {r} )&\Rightarrow \left(H-V(\mathbf {r} )\right)^{2}=\mathbf {p} ^{2}c^{2}+\mu ^{2}c^{4}\\&\Rightarrow \left(H-V(\mathbf {r} )\right)^{2}|\psi (\mathbf {r} )\rangle =\left(\mathbf {p} ^{2}c^{2}+\mu ^{2}c^{4}\right)|\psi (\mathbf {r} )\rangle \\\end{aligned}}$

The potential is rotationally invariant. The Hamiltonian operator is only the function of ${\hat {\mathbf {R} }}^{2}$ and ${\hat {\mathbf {P} }}^{2}$.

So the Hamiltonian and Angular momentum operators mutually commute, which means that $|l,m\rangle$ are eigenstates. $\Rightarrow \psi _{E}(\mathbf {r} )=R_{El}Y_{l}^{m}$

${\begin{aligned}\left(E-V(r)\right)^{2}R_{El}Y_{l}^{m}&=-\hbar ^{2}c^{2}\nabla ^{2}R_{El}Y_{l}^{m}+\mu ^{2}c^{4}R_{El}Y_{l}^{m}=-\hbar ^{2}c^{2}\left[{1 \over r^{2}}{\partial \over {\partial r}}\left(r^{2}{\partial \over {\partial r}}\right)-{1 \over {\hbar ^{2}r^{2}}}{\hat {\mathbf {L} }}^{2}\right]R_{El}Y_{l}^{m}+\mu ^{2}c^{4}R_{El}Y_{l}^{m}\\&=-\hbar ^{2}c^{2}\left[{1 \over r^{2}}{\partial \over {\partial r}}\left(r^{2}{\partial \over {\partial r}}\right)-{{l\left(l+1\right)} \over r^{2}}\right]R_{El}Y_{l}^{m}+\mu ^{2}c^{4}R_{El}Y_{l}^{m}\end{aligned}}$

Dividing each side with $Y_{l}^{m}$ we get,

${\begin{aligned}\left(E-V(r)\right)^{2}R_{El}&=-\hbar ^{2}c^{2}\left[{1 \over r^{2}}{d \over {dr}}\left(r^{2}{d \over {dr}}\right)-{{l\left(l+1\right)} \over r^{2}}\right]R_{El}+\mu ^{2}c^{4}R_{El}\end{aligned}}$

Let $U_{El}=rR_{El}$.

$\left(E-V(r)\right)^{2}U_{El}=\left(E+{e^{2} \over {4\pi \epsilon _{0}r}}\right)^{2}U_{El}=\left[-\hbar ^{2}c^{2}{d^{2} \over {dr^{2}}}+{{l\left(l+1\right)\hbar ^{2}c^{2}} \over {r^{2}}}\right]U_{El}+\mu ^{2}c^{4}U_{El}$

$\left[{d^{2} \over {dr^{2}}}-{{l\left(l+1\right)} \over r^{2}}+{1 \over {\hbar ^{2}c^{2}}}\left(E+{e^{2} \over {4\pi \epsilon _{0}r}}\right)^{2}-{{\mu ^{2}c^{2}} \over \hbar ^{2}}\right]U_{El}=0$

$\left[{d^{2} \over {dr^{2}}}-{{l\left(l+1\right)} \over r^{2}}+\left({e^{2} \over {4\pi \epsilon _{0}\hbar cr}}\right)^{2}+{1 \over {\hbar ^{2}c^{2}}}\left(E^{2}+{e^{2}E \over {2\pi \epsilon _{0}r}}-\mu ^{2}c^{4}\right)\right]U_{El}=0$

$\left[{d^{2} \over {dr^{2}}}-\left\{l\left(l+1\right)-\left({e^{2} \over {4\pi \epsilon _{0}\hbar c}}\right)^{2}\right\}{1 \over r^{2}}+{1 \over {\hbar ^{2}c^{2}}}\left(E^{2}+{e^{2}E \over {2\pi \epsilon _{0}r}}-\mu ^{2}c^{4}\right)\right]U_{El}=0$

Take $\kappa ^{2}={1 \over {\hbar ^{2}c^{2}}}\left(\mu ^{2}c^{4}-E^{2}\right)$ and $l\left(l+1\right)-\left({e^{2} \over {4\pi \epsilon _{0}\hbar c}}\right)^{2}=l'\left(l'+1\right)$.

$\left[{d^{2} \over {dr^{2}}}-{{l'\left(l'+1\right)} \over r^{2}}-\kappa ^{2}+{e^{2}E \over {2\pi \epsilon _{0}\hbar ^{2}c^{2}r}}\right]U_{El}=0$

As $r\to \infty$ the equation becomes ${d^{2} \over {dr^{2}}}U_{El}-\kappa ^{2}U_{El}=0$.

The solution to this equation is $Ae^{-\kappa r}$. Take $\rho =\kappa r$ and $U_{El}=e^{-\rho }v_{El}$.

$\left[\kappa ^{2}{d^{2} \over {d\rho ^{2}}}-\kappa ^{2}{{l'\left(l'+1\right)} \over \rho ^{2}}-\kappa ^{2}+\kappa {e^{2}E \over {2\pi \epsilon _{0}\hbar ^{2}c^{2}\rho }}\right]v_{El}e^{-\rho }=0$

$\left[{d^{2} \over {d\rho ^{2}}}-{{l'\left(l'+1\right)} \over \rho ^{2}}-1+{e^{2}E \over {2\pi \epsilon _{0}\hbar ^{2}c^{2}\kappa \rho }}\right]v_{El}e^{-\rho }=0$

${d^{2}v_{El} \over {d\rho ^{2}}}e^{-\rho }-2{dv_{El} \over {d\rho }}e^{-\rho }+v_{El}e^{-\rho }+\left[-{{l'\left(l'+1\right)} \over \rho ^{2}}-1+{e^{2}E \over {2\pi \epsilon _{0}\hbar ^{2}c^{2}\kappa \rho }}\right]v_{El}e^{-\rho }=0$

${d^{2}v_{El} \over {d\rho ^{2}}}-2{dv_{El} \over {d\rho }}+v_{El}+\left[-{{l'\left(l'+1\right)} \over \rho ^{2}}-1+{e^{2}E \over {2\pi \epsilon _{0}\hbar ^{2}c^{2}\kappa \rho }}\right]v_{El}=0$

${d^{2}v_{El} \over {d\rho ^{2}}}-2{dv_{El} \over {d\rho }}-\left[{{l'\left(l'+1\right)} \over \rho ^{2}}-{e^{2}E \over {2\pi \epsilon _{0}\hbar ^{2}c^{2}\kappa \rho }}\right]v_{El}=0$

Take $\lambda ={E \over {2\pi \epsilon _{0}\hbar ^{2}c^{2}\kappa }}$.

${d^{2}v_{El} \over {d\rho ^{2}}}-2{dv_{El} \over {d\rho }}-\left[{{l'\left(l'+1\right)} \over \rho ^{2}}-{e^{2}\lambda \over \rho }\right]v_{El}=0$

This equation is same as (13.1.8). $\Rightarrow e^{2}\lambda =2\left(k+l'+1\right)$.

$\left({e^{2} \over {2\pi \epsilon _{0}\hbar ^{2}c^{2}}}\right)^{2}E^{2}=4\left(k+l'+1\right)^{2}\kappa ^{2}$

$\left({e^{2} \over {4\pi \epsilon _{0}\hbar c}}\right)^{2}E^{2}=\left(k+l'+1\right)^{2}\left(\mu ^{2}c^{4}-E^{2}\right)$

Let ${e^{2} \over {4\pi \epsilon _{0}\hbar c}}=\alpha$.

$\therefore E_{k}^{2}={\mu ^{2}c^{4}\left(k+l'+1\right)^{2} \over {\alpha ^{2}+\left(k+l'+1\right)^{2}}}$ where $k\in \mathbb {N}$

$l\left(l+1\right)-\left({e^{2} \over {4\pi \epsilon _{0}\hbar c}}\right)^{2}=l\left(l+1\right)-\alpha ^{2}=l'\left(l'+1\right)\Rightarrow l'^{2}+l'+\alpha ^{2}-l\left(l+1\right)=0\Rightarrow l'=-{1 \over 2}+{\sqrt {l\left(l+1\right)-\alpha ^{2}+{1 \over 4}}}$

$\alpha \approx {1 \over 137}$, so $l'$ always exists and is real.