14.2 Quantum dynamics

中文

The discussion of Section 14.1 refers only to quantum statics – the essential framework in which quantum descriptions are to be set. The dynamical evolution of quantum systems is determined by a hermitian operator $H$, possibly but not usually a function oftime $H=H(t)$ such that the time development of any state $|\psi (t)⟩$ of the system is given by Schrödinger’s equation

$$i\hslash \frac{\mathrm{d}}{\mathrm{d}t}|\psi ⟩=H|\psi ⟩.\text{\hspace{1em}(14.13)}$$

The operator $H$ is known as the Hamiltonian or energy operator. Equation (14.13) guar antees that all inner products are preserved for, taking the adjoint gives

$$-i\hslash \frac{\mathrm{d}}{\mathrm{d}t}⟨\psi |=⟨\psi |H^{\ast }=⟨\psi |H$$

and for any pair of states $|\psi ⟩$ and $|\varphi ⟩$ ,

$$\begin{array}{c}i\hslash \frac{\mathrm{d}}{\mathrm{d}t}⟨\psi |\varphi ⟩=i\hslash ((\frac{\mathrm{d}}{\mathrm{d}t}⟨\psi |)|\varphi ⟩+⟨\psi |(\frac{\mathrm{d}}{\mathrm{d}t}|\varphi ⟩))\\ =-⟨\psi |H|\varphi ⟩+⟨\psi |H|\varphi ⟩=0.\end{array}$$

In particular the normalization $\Vert \psi (t)\Vert =\Vert \varphi (t)\Vert =1$ is preserved by Schrödinger’s equa tion. Since

$$⟨\psi (t)|\varphi (t)⟩=⟨\psi (0)|\varphi (0)⟩$$

for all pairs of states, there exists a unitary operator $U$($t$) such that

$$|\psi (t)⟩=U(t)|\psi (0)⟩.\text{\hspace{1em}(14.14)}$$

If $H$ is independent of $t$ then

$$U(t)={\mathrm{e}}^{(-i/\hslash )Ht}\text{\hspace{1em}(14.15)}$$

where the exponential function can be defined as in the comments prior to Theorem 13.26 at the end of Chapter 13. If $H$ is a complete hermitian operator $\begin{array}{r}H={\sum }_n{\lambda }_n|{\psi }_n⟩⟨{\psi }_n|\end{array}$ then

$${\mathrm{e}}^{(-i/\hslash )Ht}=\sum _n{\mathrm{e}}^{(-i/\hslash )\lambda t}|{\psi }_n⟩⟨{\psi }_n|$$

and for a self-adjoint operator

$$H={\int }_{-\infty }^{\infty }\lambda \mathrm{d}{E}_{\lambda }⟹{\mathrm{e}}^{(-i/\hslash )Ht}={\int }_{-\infty }^{\infty }{\mathrm{e}}^{(-i/\hslash )\lambda t}\mathrm{d}{E}_{\lambda }.$$

To prove Eq. (14.15) substitute Eq. (14.14) in Schrödinger’s equation

$$i\hslash \frac{\mathrm{d}U}{\mathrm{d}t}|\psi (0)⟩=HU|\psi (0)⟩,$$

and since $|\psi (0)⟩$ is an arbitrary initial state vector,

$$i\hslash \frac{\mathrm{d}U}{\mathrm{d}t}=HU.$$

Setting $U(t)={\mathsf{e}}^{(-i/\hslash )Ht}V(t)$ (always possible since the operato ${\mathrm{e}}^{(-i/\hslash )Ht}$ is invertible with inverse ${\mathsf{e}}^{(i/\hslash )Ht})$ we obtain

$${\mathrm{e}}^{(-i/\hslash )Ht}HV(t)+i\hslash {\mathrm{e}}^{(-i/\hslash )Ht}\frac{\mathrm{d}V(t)}{\mathrm{d}t}=H{\mathrm{e}}^{(-i/\hslash )Ht}V(t).$$

As $H$ and ${\mathrm{e}}^{(-i/\hslash )Ht}$ commute it follows that $V(t)=const.=V(0)=I$ since $U(0)=I$ on setting $t=0$ in Eq. (14.14).

The Heisenberg picture

The above description of the evolution of a quantum mechanical system is called the Schrödinger picture. There is an equivalent version called the Heisenberg picture in which the states are treated as constant, but observables undergo a dynamic evolution. The idea is to perform a unitary transformation on $\mathcal{H},$ simultaneously on states and operators:

$$\begin{array}{cc}|\psi ⟩\mapsto |{\psi }^{\prime }⟩=U^{\ast }|\psi ⟩& (|\psi ⟩=U|{\psi }^{\prime }⟩)\\ A\mapsto A^{\prime }=U^{\ast }AU,& \end{array}$$

where $U$ is given by Eq. (14.15). This transformation has the effect of bringing every solution of Schrödinger’s equation to rest, for $\mathrm{if}|\psi (t)$ is a solution of $\mathrm{Eq}$ . Eq. (14.13) then

$$|{\psi }^{\prime }⟩=U^{\ast }|\psi (t)⟩=U^{\ast }U|\psi (0)⟩=|\psi (0)⟩⟹\frac{\mathrm{d}}{\mathrm{d}t}|{\psi }^{\prime }(t)⟩=0.$$

It preserves all matrix elements, and in particular all expectation values:

$$⟨A^{\prime }{⟩}_{{\psi }^{\prime }}=⟨{\psi }^{\prime }|A^{\prime }|{\psi }^{\prime }⟩=⟨\psi |U{U}^{\ast }AU{U}^{\ast }|\psi ⟩=⟨\psi |A|\psi ⟩=⟨A{⟩}_{\psi }.$$

Thus the states and observables are physically equivalent in the two pictures.

We derive a dynamical equation for the Heisenberg operator $A^{\prime }:$

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}{A}^{\prime }& =\frac{\mathrm{d}}{\mathrm{d}t}(U^{\ast }AU)\\ & =\frac{\mathrm{d}{U}^{\ast }}{\mathrm{d}t}AU+U^{\ast }\frac{\mathrm{d}A}{\mathrm{d}t}U+U^{\ast }A\frac{\mathrm{d}U}{\mathrm{d}t}\\ & =\frac{1}{-i\hslash }{U}^{\ast }HAU+U^{\ast }\frac{\mathrm{d}A}{\mathrm{d}t}U+\frac{1}{i\hslash }{U}^{\ast }AHU\end{array}$$

since, by Eq. (14.15),

$$\frac{\mathrm{d}U}{\mathrm{d}t}=\frac{1}{i\hslash }HU,\frac{\mathrm{d}{U}^{\ast }}{\mathrm{d}t}=\frac{1}{-i\hslash }{U}^{\ast }{H}^{\ast }=\frac{1}{-i\hslash }{U}^{\ast }H.$$

Hence

$$\frac{\mathrm{d}}{\mathrm{d}t}{A}^{\prime }=\frac{1}{i\hslash }[A^{\prime },H^{\prime }]+\frac{\partial A^{\prime }}{\partial t}\text{\hspace{1em}(14.16)}$$

where

$$\frac{\partial A^{\prime }}{\partial t}=U^{\ast }\frac{\mathrm{d}A}{\mathrm{d}t}U={\left(\frac{\mathrm{d}A}{\mathrm{d}t}\right)}^{\prime }.$$

The motivation for this identification is the following: if $|{\psi }_i⟩$ is a rest basis of $\mathcal{H}$ in the Schrödinger picture, so that $\mathrm{d}|{\psi }_i⟩/\mathrm{d}t=0$ , and if $|{\psi }_i^{\prime }⟩=U^{\ast }|{\psi }_i⟩$ is the ‘moving basis obtained from it, then

$$⟨{\psi }_i^{\prime }|A^{\prime }|{\psi }_j^{\prime }⟩=⟨{\psi }_i|A|{\psi }_j⟩$$

and

$$⟨{\psi }_i^{\prime }|\frac{\partial A^{\prime }}{\partial t}|{\psi }_j^{\prime }⟩=⟨{\psi }_i|U\frac{\partial A^{\prime }}{\partial t}{U}^{\ast }|{\psi }_j⟩=⟨{\psi }_i|\frac{\mathrm{d}A}{\mathrm{d}t}|{\psi }_j⟩=\frac{\mathrm{d}}{\mathrm{d}t}⟨{\psi }_i|A|{\psi }_j⟩.$$

Thus the matrix elements of $\partial A^{\prime }/\partial t$ measure the explicit time rate of change of the matrix elements of the operator $A$ in the Schrödinger representation.

If $A$ is an operator having no explicit time dependence, so that $\partial A^{\prime }/\partial t=0$ , then $A^{\prime }$ is a constant of the motion if and only if it commutes with the Hamiltonian, $[A^{\prime },H^{\prime }]=0$

$$\frac{\mathrm{d}{A}^{\prime }}{\mathrm{d}t}=0⟺[A^{\prime }{H}^{\prime }]=[A,H]=0.$$

In particular, since every operator commutes with itself, the Hamiltonian $H^{\prime }$ is a constant of the motion if and only if it is time independent, $\partial H^{\prime }/\partial t=0$

Example 14.4

For an electron of charge $e$, mass $m$ and spin $\frac{1}{2}$ , notation as in Example 14.1, the Hamiltonian in a magnetic field $B$ is given by

$$H=\frac{-e\hslash }{2mc}\mathbf{\sigma }\cdot \mathbf{B}.$$

If $B$ is parallel to the $z$-axis then $H=-(e\hslash /2mc){\sigma }_{z}B$ and setting $|\psi ⟩={\psi }_1(t)|+z⟩+{\psi }_2(t)|-z⟩$ , Schrödinger’s equation Eq. (14.13) can be written as the two differential equation

$$i\hslash \mathrm{d}{\dot{\psi }}_1=-\frac{e\hslash }{2mc}B{\psi }_1,i\hslash \mathrm{d}{\dot{\psi }}_2=\frac{e\hslash }{2mc}B{\psi }_2$$

with solutions

$${\psi }_1(t)={\psi }_{10}{\mathrm{e}}^{i(\omega /2)t},{\psi }_2(t)={\psi }_{20}{\mathrm{e}}^{-i(\omega /2)t},$$

where $\omega =eB/mc$ . Substituting in the expectation values

$$⟨\psi (t)|\sigma |\psi (t)⟩=\left(\begin{array}{c}\sin \theta \cos \varphi (t)\\ \sin \theta \sin \varphi (t)\\ \cos \theta (t)\end{array}\right)$$

results in $\theta (t)={\theta }_0=\mathrm{const}$ . and

$$\cos \varphi (t)=\frac{{\mathrm{e}}^{i({\varphi }_0-\omega t)}+{\mathrm{e}}^{-i({\varphi }_0-\omega t)}}{2}=\cos ({\varphi }_0-\omega t).$$

Hence $\varphi (t)={\varphi }_0-\omega t$ , and the motion is a precession with angular velocity $\omega$ about the direction of the magnetic field.

In the Heisenberg picture, set ${\sigma }_x={\sigma }_x(t)$ , etc., where ${\sigma }_x(0)={\sigma }_1,{\sigma }_y(0)={\sigma }_2,{\sigma }_z(0)={\sigma }_3$ are the Pauli values, Eq. (14.11). From the commutation relations

$$[{\sigma }_1,{\sigma }_2]=2i{\sigma }_3,[{\sigma }_2,{\sigma }_3]=2i{\sigma }_1,[{\sigma }_3,{\sigma }_1]=2i{\sigma }_2\text{\hspace{1em}(14.17)}$$

and ${\sigma }_x=U^{\ast }{\sigma }_{1}U$ , etc., it follows that

$$[{\sigma }_x,{\sigma }_y]=2i{\sigma }_z,\text{ etc. }$$

Heisenberg equations of motion are

$$\begin{array}{l}{\dot{\sigma }}_x=\frac{1}{i\hslash }[{\sigma }_x,H]=\omega {\sigma }_y,\\ {\dot{\sigma }}_y=-\omega {\sigma }_x,\\ {\dot{\sigma }}_z=0.\end{array}$$

Hence ${\ddot{\sigma }}_x=\omega {\dot{\sigma }}_y=-{\omega }^2{\sigma }_x$ and the solution of Heisenberg’s equation is

$${\sigma }_x={\mathsf{Ae}}^{i\omega t}+{\mathsf{Be}}^{-i\omega t},{\sigma }_y=\frac{1}{\omega }{\dot{\sigma }}_x.$$

The $2\times 2$ matrices $A$, $B$ are evaluated by initial values at $t=0$ , resulting in

$$\begin{array}{c}{\sigma }_x(t)=\cos \omega t{\sigma }_1+\sin \omega t{\sigma }_2,\\ {\sigma }_y(t)=-\sin \omega t{\sigma }_1+\cos \omega t{\sigma }_2\\ {\sigma }_z={\sigma }_3=\text{ const }.\end{array}$$

Correspondence with classical mechanics and wave mechanics

For readers familiar with Hamiltonian mechanics (see Section 16.5), the following correspondence can be set up between classical and quantum mechanics:

	Quantum mechanics	Classical mechanics
State space	Hilbert space $\mathcal{H}$	Phase space $\Gamma$
States	Normalized kets $|\psi ⟩\in \mathcal{H}$	Points $(q_i,p_j)\in \Gamma$
Observables	Self-adjoint operators in $\mathcal{H}$ ; multiple values ${\lambda }_{b}i$ in each state $|\psi ⟩$ with probability $P=⟨{\psi }_i|\psi {⟩}^2$	Real functions $f(q_i,p_j)$ on phase space; one value for each state
Commutators	Bracket commutators [$A$, $B$]	Poisson brackets ($f$, $g$)
Dynamics	1. Schrödinger picture $i\hslash \frac{\mathrm{d}}{\mathrm{d}t}|\psi ⟩=H|\psi ⟩$ 2. Heisenberg picture $\dot{A}=\frac{1}{i\hslash }[A,H]+\frac{\partial A}{\partial t}$	1. Hamilton’s equations ${\dot{q}}_i=\frac{\partial H}{\partial p_i},{\dot{p}}_i=-\frac{\partial H}{\partial q_i}$ 2. Poisson bracket form $\dot{f}=(f,H)+\frac{\partial f}{\partial t}$

If $f$ and $g$ are classical observables with quantum mechanical equivalents $F$ and $G$ then, from Heisenberg’s equation of motion, the proposal is that the commutator $[F,G]$ corresponds to the $i\hslash$ times the Poisson bracket,

$$[F,G]⟷i\hslash (f,g)=i\hslash \left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial q_i}\right).$$

For example if $Q_i$ are position operators representing classical variable $q_i$ and $P_i=-i\hslash \partial /\partial q_i$ the momentum operators, then the classical canonical commutation relations imply

$$\begin{array}{l}(q_i,q_j)=0⟹[Q_i,Q_j]=0,\\ (p_i,p_j)=0⟹[P_i,P_j]=0,\\ (q_i,p_j)={\delta }_{ij}⟹[Q_i,P_j]=i\hslash {\delta }_{ij}I.\end{array}$$

Generalizing from the one-dimensional case, we assume is the set of differentiable functions in $L^2({\mathbb{R}}^n)$ such that $x_i\psi (x_1,\dots ,x_n)$ belongs to $L^2({\mathbb{R}}^n)$ for each $x_i$ . The above commutation relations are satisfied by the standard operators:

$$Q_i\psi (x_1,\dots ,x_n)=x_i\psi (x_1,\dots ,x_n),P_i\psi (x_1,\dots ,x_n)=-i\hslash \frac{\partial \psi }{\partial x_i}.$$

For a particle in a potential $V(x,y,z)$ the Hamiltonian is $H={\mathbf{p}}^2/2m+V(x,y,z)$ , which corresponds to the quantum mechanical Schrödinger equation

$$i\hslash \frac{\partial \psi }{\partial t}=-\frac{{\hslash }^2}{2m}{\nabla }^2\psi +V(\mathbf{r})\psi .\text{\hspace{1em}(14.18)}$$

Exercise

Show that the probability density $P(\mathbf{r},t)=\psi \overline{\psi }$ satisfies the conservation equation

$$\frac{\partial P}{\partial t}=-\nabla \mathbf{J}\text{ where }\mathbf{J}=\frac{i\hslash }{2m}(\psi \nabla \overline{\psi }-\overline{\psi }\nabla \psi ).$$

A trial solution of Eq. (14.18) by separation of variables, $\psi =T(t)\varphi (\mathbf{x})$ , results in

$$\psi ={\mathrm{e}}^{-i\omega t}\varphi (\mathbf{x})$$

where $\varphi (\mathbf{x})$ satisfies the time-independent Schrödinger equation

$$H\varphi (\mathbf{x})=-\frac{{\hslash }^2}{2m}{\nabla }^2\varphi (\mathbf{x})+V(\mathbf{r})\varphi (\mathbf{x})=E\varphi (\mathbf{x})$$

where $E$ is given by Planck’s relation, $E=\hslash \omega =h\nu$ . From its classical analogue, the eigenvalue $E$ of the Hamiltonian is interpreted as the energy of the system, and if the Hamiltonian is a complete operator with discrete spectrum $E_n$ then the general solution of the Schrödinger equation is given by

$$\psi (\mathbf{x},t)=\sum _n{c}_n{\varphi }_n(\mathbf{x}){\mathrm{e}}^{-i{E}_{n}t/\hslash }$$

where

$$H{\varphi }_n(\mathbf{x})=E_n{\varphi }_n(\mathbf{x}).$$

Harmonic oscillator

The classical one-dimensional harmonic oscillator has Hamiltonian

$$H_{\mathrm{cl}}=\frac{1}{2m}{p}^2+\frac{k}{2}{q}^2.$$

Its quantum mechanical equivalent should have energy operator

$$H=\frac{1}{2m}{P}^2+\frac{k}{2}{Q}^2\text{\hspace{1em}(14.19)}$$

where

$$P=-i\hslash \frac{\mathrm{d}}{\mathrm{d}x},[Q,P]=i\hslash I.$$

Set

$$A=\frac{1}{\sqrt{\omega \hslash }}(\frac{1}{\sqrt{2m}}P+i\sqrt{\frac{k}{2}}Q)$$

where $\omega =\sqrt{k/m}$ and we find

$$H=\omega \hslash (N+\frac{1}{2}I)\text{\hspace{1em}(14.20)}$$

where $N$ is the self-adjoint operator $N=A{A}^{\ast }=N^{\ast }$

It is not hard to show that

$$[A,A^{\ast }]=A{A}^{\ast }-A^{\ast }A=-I,\text{\hspace{1em}(14.21)}$$

and from the identities in Problem 14.3 it follows that

$$[N,A]=[A{A}^{\ast },A]=A[A^{\ast },A]+[A,A]A^{\ast }=A,\text{\hspace{1em}(14.22)}$$

$$[N,A^{\ast }]=[A,N^{\ast }{]}^{\ast }=-[N,A{]}^{\ast }=-A^{\ast }.\text{\hspace{1em}(14.23)}$$

All eigenvalues of $N$ are non-negative, $n\ge 0,$ , for if $N|{\psi }_n⟩=n|{\psi }_n⟩$ then

$$0\le \Vert A^{\ast }{\psi }_n{\Vert }^2=⟨{\psi }_n|A{A}^{\ast }|{\psi }_n⟩=⟨{\psi }_n|N|{\psi }_n⟩=n⟨{\psi }_n|{\psi }_n⟩.\text{\hspace{1em}(14.24)}$$

Let $n_0\ge 0$ be the lowest eigenvalue. Using Eq. (14.23), the state $A^{\ast }|{\psi }_n⟩$ is an eigenstate of $N$ with eigenvalue $(n-1)$ )

$$N{A}^{\ast }|{\psi }_n⟩=(A^{\ast }N-A^{\ast })|{\psi }_n⟩=(n-1)A^{\ast }|{\psi }_n⟩.$$

Hence $A^{\ast }|{\psi }_{n_0}⟩=0$ , else $n_0-1$ would be an eigenvalue, contradicting $n_0$ being lowest, and setting $n=n_0$ in Eq. (14.24) gives $n_0=0$ . Furthermore, if $n$ is an eigenvalue then $A|{\psi }_n⟩\ne 0$ is an eigenstate with eigenvalue $(n+1)$ for, by Eqs. Eq. (14.22) and Eq. (14.21)

$$NA|{\psi }_n⟩=(AN+A)|{\psi }_n⟩=(n+1)|{\psi }_n⟩,$$ $$\Vert A|{\psi }_n⟩{\Vert }^2=⟨{\psi }_n|A^{\ast }A|{\psi }_n⟩=⟨{\psi }_n|A{A}^{\ast }+I|{\psi }_n⟩=(n+1)⟨{\psi }_n|{\psi }_n⟩>0.$$

The eigenvalues of $N$ are therefore $n=0,1,2,3,...$ . and the eigenvalues of $H$ are $\frac{1}{2}\hslash \omega .\begin{array}{r}\frac{3}{2}\hslash \omega ,\dots ,(n+\frac{1}{2})\hslash \omega ,\end{array}$

Angular momentum

A similar analysis can be used to find the eigenvalues of the angular momentum operators $L_i={ϵ}_{ijk}{Q}_j{P}_k$ . Using the identities in Problems 14.3 and 14.4 it is straightforward to derive the commutation relations of angular momentum

$$[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k\text{\hspace{1em}(14.25)}$$

and

$$[L^2,L_i]=0\text{\hspace{1em}(14.26)}$$

where $L^2=(L_1{)}^2+(L_2{)}^2+(L_3{)}^2$ is the total angular momentum.

Exercise

Prove the identities Eq. (14.25) and Eq. (14.26).

Any set of three operators $L_1,L_2,L_3$ satisfying the commutation relations Eq. (14.25) are said to be angular momentum operators. If they are of the form $L_i={ϵ}_{ijk}{Q}_j{P}_k$ then we term them orbital angular momentum, else they are called spin angular momentum, or a combination thereof. If we set $J_i=L_1/\hslash$ and $J^2=(J_1{)}^2+(J_2{)}^2+(J_3{)}^2$ then

$$[J_i,J_j]=i{ϵ}_{ijk}{J}_k,[J^2,J_i]=0$$

and since $J_3$ and $J^2$ commute there exist, by Theorem 14.2, a common set of eigenvectors $|j^{2}m⟩$ such that

$$J^2|j^{2}m⟩=j^2|j^{2}m⟩,J_3|j^{2}m⟩=m|j^{2}m⟩.$$

Thus

$$⟨j^{2}m|J^2|j^{2}m⟩=\sum _{k=1}^3⟨j^{2}m|(J_k{)}^2|j^{2}m⟩=\sum _{k=1}^3\Vert J_k|j^{2}m⟩{\Vert }^2\ge m^2\Vert |j^{2}m⟩{\Vert }^2.$$

Since the left-hand side is equal to $j^2\Vert |j^{2}m⟩{\Vert }^2$ we have $j^2\ge m^2\ge 0$ , and there is an upper and lower bound to the eigenvalue $m$ for any fixed $j^2$ . Let this upper bound be $l$.

If we set $J_{\pm }=J_1\pm i{J}_2$ , then it is simple to show the identities

$$[J_3,J_{\pm }]=\pm J_{\pm },J_{\pm }{J}_{\mp }=J^2-(J_3{)}^2\pm J_3.\text{\hspace{1em}(14.27)}$$

Exercise

Prove the identities Eq. (14.27).

Hence $J_{\pm }$ are raising and lowering operators for the eigenvalues of $J_3$ ,

$$J_3{J}_{\pm }|j^{2}m⟩=\left(J_{\pm }{J}_3\pm J_{\pm }\right)|j^{2}m⟩=(m\pm 1)J_{\pm }|j^{2}m⟩$$

while they leave the eigenvalue of $J^2$ alone,

$$J^2\left(J_{\pm }|j^{2}m⟩\right)=J_{\pm }{J}^2|j^{2}m⟩=j^2{J}_{\pm }{J}^2|j^{2}m⟩.$$

Since $l$ is the maximum possible value of $m$, we must have $J_{\pm }|j^{2}l⟩=0$ and using the second identity of Eq. (14.27) we have

$$J_{-}{J}_{+}|j^{2}l⟩=(J^2-(J_3{)}^2-J_3)|j^{2}l⟩=(j^2-l^2-l)|j^{2}l⟩=0,$$

whence $j^2=l(l+1)$ . Since for each integer $n,(J_{-}{)}^n|j^{2}l⟩$ is an eigenket of $J_3$ with eigenvalue $(l-n)$ and the eigenvalues of $J_3$ are bounded below, there exists an integer $n$ such that $(J_{-}{)}^n|j^{2}l⟩\ne 0$ and $(J_{-}{)}^{n+1}|j^{2}l⟩=0$ . Using Eq. (14.27) we deduce

$$\begin{array}{c}0=J_{+}{J}_{-}(J_{-}{)}^n|j^{2}l⟩=(J^2-(J_3{)}^2\pm J_3)(J_{-}{)}^n|j^{2}l⟩\\ =(j^2-(l-n{)}^2+(l-n)\left)\right(J_{-}{)}^n|j^{2}l⟩\end{array}$$

so that

$$j^2=(l-n)(l-n-1)=l^2+l$$

from which it follows that $\begin{array}{r}l=\frac{1}{2}n\end{array}$ . Thus the eigenvalues of total angular momentum $L^2$ are of the form $l(l+1){\hslash }^2$ where $l$ has integral or half integral values. The eigenspaces are $(2l+1)$ -degenerate, and the simultaneous eigenstates of $L_3$ have eigenvalues $m$- where $-l\le m\le m$ . For orbital angular momentum it turns out that the value of $l$ is always integral, but spin eigenstates may have all possible eigenvalues, $\begin{array}{r}l=0,\frac{1}{2},1,\frac{3}{2},...\end{array}$ . depending on the particle in question.

Problems

Problem 14.6

In the Heisenberg picture show that the time evolution of the expection value of an operator $A$ is given by

$$\frac{\mathrm{d}}{\mathrm{d}t}⟨A^{\prime }{⟩}_{{\psi }^{\prime }}=\frac{1}{i\hslash }⟨[A^{\prime },H^{\prime }]{⟩}_{{\psi }^{\prime }}+⟨\frac{\partial A^{\prime }}{\partial t}{⟩}_{{\psi }^{\prime }}.$$

Convert this to an equation in the Schrödinger picture for the time evolution of $⟨A{⟩}_{\psi }$

Problem 14.7

For a particle of spin half in a magnetic field with Hamiltonian given in Example 14.4, show that in the Heisenberg picture

$$\begin{array}{l}⟨{\sigma }_x(t){⟩}_{\mathbf{n}}=\sin \theta \cos (\varphi -\omega t),\\ ⟨{\sigma }_y(t){⟩}_{\mathbf{n}}=\sin \theta \sin (\varphi -\omega t),\\ ⟨{\sigma }_z(t){⟩}_{\mathbf{n}}=\cos \theta .\end{array}$$

Problem 14.8

A particle of mass $m$ is confined by an infinite potential barrier to remain within a box $0\le x,y,z\le a_{:}$ , so that the wave function vanishes on the boundary of the box. Show that the energy levels are

$$E=\frac{1}{2m}\frac{{\pi }^2{\hslash }^2}{a^2}\left(n_1^2+n_2^2+n_3^2\right),$$

where $n_1,n_2,n_3$ are positive integers, and calculate the stationary wave functions ${\psi }_E(\mathbf{x})$ . Verify that the lowest energy state is non-degenerate, but the next highest is triply degenerate.

Problem 14.9

For a particle with Hamiltonian

$$H=\frac{{\mathbf{p}}^2}{2m}+V(\mathbf{x})$$

show from the equation of motion in the Heisenberg picture that

$$\frac{\mathrm{d}}{\mathrm{d}t}⟨\mathbf{r}\cdot \mathbf{p}⟩=⟨\frac{{\mathbf{p}}^2}{m}⟩-⟨\mathbf{r}\cdot \nabla V⟩.$$

This is called the Virial theorem. For stationary states, show that

$$2⟨T⟩=⟨\mathbf{r}\cdot \nabla V⟩$$

where $T$ is the kinetic energy. If $V\propto r^n$ this reduces to the classical result $⟨2T+nV⟩=0.$

Problem 14.10

Show that the $n$th normalized eigenstate of the harmonic oscillator is given by

$$|{\psi }_n⟩=\frac{1}{(n!{)}^{1/2}}{A}^n|{\psi }_0⟩.$$

Show from $A^{\ast }{\psi }_0=0$ that

$${\psi }_0=c{\mathrm{e}}^{-\sqrt{km}{x}^2/2\hslash }\text{ where }c={\left(\frac{m\omega }{\pi \hslash }\right)}^{1/4}$$

and the $n$th eigenfunction is

$${\psi }_n(x)=\frac{i^n}{(2^{n}n!{)}^{1/2}}{\left(\frac{m\omega }{\pi \hslash }\right)}^{1/4}{\mathrm{e}}^{-m\omega x^2/2\hslash }{H}_n\left(\sqrt{\frac{m\omega }{\hslash }}x\right),$$

where $H_n(y)$ is the $n$th hermite polynomial (see Example 13.7).

Problem 14.11

For the two-dimensional harmonic oscillator define operators $A_1,A_2$ such that

$$[A_i,A_j]=[A_i^{\ast },A_j^{\ast }]=0,[A_i,A_j^{\ast }]=-{\delta }_{ij},H=\hslash \omega (2N+I)$$

where $i,j=1$ , 2 and $N$ is the number operator

$$N=\frac{1}{2}(A_1{A}_1^{\ast }+A_2{A}_2^{\ast }).$$

Let $J_1,J_2$ and $J_3$ be the operators

$$J_1=\frac{1}{2}(A_2{A}_1^{\ast }+A_1{A}_2^{\ast }),J_2=\frac{1}{2}i(A_2{A}_1^{\ast }-A_1{A}_2^{\ast }),J_3=\frac{1}{2}(A_1{A}_1^{\ast }-A_2{A}_2^{\ast }),$$

and show that:

(a) The $J_i$ satisfy the angular momentum commutation relations $[J_1,J_2]=i{J}_3$ , etc.

(b) $J^2=J_1^2+J_2^2+J_3^2=N(N+1).$

(c) $[J^2,N]=0,[J_3,N]=0.$

From the properties of angular momentum deduce the energy levels and their degeneracies for the two-dimensional harmonic oscillator.

Problem 14.12

Show that the eigenvalues of the three-dimensional harmonic oscillator have the form $\begin{array}{r}(n+\frac{3}{2}.\end{array}$ )-$\omega$ where $n$ is a non-negative integer. Show that the degeneracy of the $n$th eigenvalue $i\frac{1}{2}(n^2+3n\overline{+}2)$ . Find the corresponding eigenfunctions.