14.3 Symmetry transformations

中文

Consider two observers $O$ and $O^{\prime }$ related by a symmetry transformation such as a translation ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{a}.$ , or rotation ${\mathbf{x}}^{\prime }=\mathsf{A}\mathbf{x}$ where ${\mathsf{AA}}^T={\mathsf{I}}_{:}$ , etc. For any state corresponding to a ray $[|\psi ⟩]$ according to $O$ let $O^{\prime }$ assign the ray $[|{\psi }^{\prime }⟩]$ , and for any observable assigned the self-adjoint operator $A$ by $O$ let $O^{\prime }$ assign the operator $A^{\prime }.$ . Since the physical elements are determined by the modulus squared of the matrix elements between states (being the probability of transition between states) and the expectation values of observables, this correspondence is said to be a symmetry transformation if

$$|⟨{\varphi }^{\prime }|{\psi }^{\prime }⟩{|}^2=|⟨\varphi |\psi ⟩{|}^2\text{\hspace{1em}(14.28)}$$

$$⟨A^{\prime }{⟩}_{{\psi }^{\prime }}=⟨{\psi }^{\prime }|A^{\prime }|{\psi }^{\prime }⟩=⟨A{⟩}_{\psi }=⟨\psi |A|\psi ⟩\text{\hspace{1em}(14.29)}$$

for all states $|\psi ⟩$ and observables $A$.

Theorem 14.3 · Wigner’s theorem

(Wigner) A ray correspondence that satisfies Eq. (14.28)for all rays is generated up to aphase by a transformation $U:\mathcal{H}\to \mathcal{H}$ that is either unitary or anti-unitary.

A unitary transformation was defined in Chapter 13 as a linear transformation, $U(|\psi ⟩+\alpha |\varphi ⟩)=U|\psi ⟩+\alpha U|\varphi ⟩$ , which preserves inner products

$$⟨U\psi |U\varphi ⟩=⟨\psi |\varphi ⟩⟺U{U}^{\ast }=U^{\ast }U=I.$$

A transformation $V$ is said to be antilinear if

$$V(|\psi ⟩+\alpha |\varphi ⟩)=V|\psi ⟩+\overline{\alpha }V|\varphi ⟩.$$

The adjoint $V^{\ast }$ is defined by

$$⟨V^{\ast }\psi |\varphi ⟩=\overline{⟨\psi |V\varphi ⟩}=⟨V\varphi |\psi ⟩,$$

in order that it too will be antilinear (we pay no attention to domains here). An operator $V$ is called anti-unitary if it is antilinear and $V^{\ast }V=V{V}^{\ast }=I$ . In this case

$$⟨V\psi |V\varphi ⟩=\overline{⟨\psi |\varphi ⟩}=⟨\varphi |\psi ⟩$$

for all vectors $|\psi ⟩$ and $|\varphi ⟩$ .

Proof outline

We prove Wigner’s theorem in the case of a two-dimensional Hilbert space. The full proof is along similar lines. $\mathrm{If}|e_1⟩,|e_2⟩$ is an orthonorma basis then, up to phases, so is $|e_1^{\prime }⟩,|e_2^{\prime }⟩$ ,

$${\delta }_{ij}=|⟨e_i|e_j⟩{|}^2=|⟨e_i^{\prime }|e_j^{\prime }⟩{|}^2.$$

Let $|\psi ⟩=a_1|e_1⟩+a_2|e_2⟩$ be any unit vector, $⟨\psi \mid \psi ⟩=|a_1{|}^2+|a_2{|}^2=1$ . Set

$$|{\psi }^{\prime }⟩=a_1^{\prime }|e_1^{\prime }⟩+a_2^{\prime }|e_2^{\prime }⟩$$

and we have, from Eq. (14.28),

$$|a_1^{\prime }{|}^2=|⟨e_1^{\prime }|{\psi }^{\prime }⟩{|}^2=|⟨e_1|\psi ⟩{|}^2=|a_1{|}^2,$$

and similarly $|a_2^{\prime }{|}^2=|a_2{|}^2$ . Hence we can set real angles $\alpha ,\theta ,\phi$ , etc. such that

$$a_1=\cos \alpha {\mathrm{e}}^{i\theta },a_1^{\prime }=\cos \alpha {\mathrm{e}}^{i{\theta }^{\prime }},$$ $$a_2=\sin \alpha {\mathrm{e}}^{i\phi },a_2^{\prime }=\sin \alpha {\mathrm{e}}^{i{\phi }^{\prime }}.$$

Let $|{\psi }_1⟩$ and $|{\psi }_2⟩$ be an arbitrary pair of unit vectors,

$$|{\psi }_i⟩=\cos {\alpha }_i{\mathrm{e}}^{i{\theta }_1}|e_1⟩+\sin {\alpha }_i{\mathrm{e}}^{i{\phi }_1}|e_2⟩,$$

then $|⟨{\psi }_1^{\prime }|{\psi }_2^{\prime }⟩{|}^2=|⟨{\psi }_1|{\psi }_2⟩{|}^2$ implies

$$\cos ({\theta }_2^{\prime }-{\phi }_2^{\prime }-{\theta }_1^{\prime }+{\phi }_1^{\prime })=\cos ({\theta }_2-{\phi }_2-{\theta }_1+{\phi }_1).$$

Hence

$${\theta }_2^{\prime }-{\phi }_2^{\prime }-({\theta }_1^{\prime }-{\phi }_1^{\prime })=\pm ({\theta }_2-{\phi }_2-({\theta }_1-{\phi }_1)).\text{\hspace{1em}(14.30)}$$

Define an angle $\delta$ by

$${\theta }_1^{\prime }-{\phi }_1^{\prime }=\delta \pm ({\theta }_1-{\phi }_1),$$

and it follows from Eq. (14.30) that

$${\theta }_2^{\prime }-{\phi }_2^{\prime }=\delta \pm ({\theta }_2-{\phi }_2).$$

Hence for an arbitrary vector $|\psi ⟩$

$${\theta }^{\prime }-{\phi }^{\prime }=\delta \pm (\theta -\phi ).$$

For the + sign this results in the transformation

$$|\psi ⟩\mapsto |{\psi }^{\prime }⟩={\mathrm{e}}^{i({\phi }^{\prime }-\phi )}\left(a_1{\mathrm{e}}^{i\delta }|e_1^{\prime }⟩+a_2|e_2^{\prime }⟩\right)$$

while for the - sign it is

$$|\psi ⟩\mapsto |{\psi }^{\prime }⟩={\mathrm{e}}^{i({\phi }^{\prime }+\phi )}(\overline{a_1}{\mathrm{e}}^{i\delta }|e_1^{\prime }⟩+\overline{a_2}|e_2^{\prime }⟩).$$

These transformations are, up to a phase ${\mathrm{e}}^{if}$ , unitary and anti-unitary respectively. This is Wigner’s theorem. -

Exercise

Show that the phase $f={\phi }^{\prime }\pm \phi$ is independent of the state.

If $U$ is a unitary transformation and $A$ is an observable, then

$$⟨A^{\prime }{⟩}_{{\psi }^{\prime }}=⟨{\psi }^{\prime }|A^{\prime }|{\psi }^{\prime }⟩=⟨\psi |U^{\ast }{\mathrm{e}}^{-if}{A}^{\prime }{\mathrm{e}}^{if}U|\psi ⟩=⟨U^{\ast }{A}^{\prime }U{⟩}_{\psi }$$

and the requirement Eq. (14.29) implies that this holds for arbitrary vectors $|\psi ⟩$ if and only if

$$A^{\prime }=UA{U}^{\ast }.$$

Performing two symmetries $g$ and $h$ in succession results in a symmetry transformation of satisfying

$$U(g)U(h)={\mathrm{e}}^{i\phi }U(gh)$$

where the phase $\phi$ may depend on $g$ and $h$. This is called a projective or ray representation of the group $G$ on the Hilbert space . It is not in general possible to choose the phases such that all ${\mathrm{e}}^{i\phi }=1$ , giving a genuine representation. For a continuous group (see Section 10.8), elements in the component of the identity $G_0$ must be unitary since they are connected continuously with the identity element, which is definitely unitary. Anti-unitary transfor mations can only correspond to group elements in components that are not continuously connected with the identity.

Infinitesimal generators

If $G$ is a Lie group, the elements of which are unitary transformations characterized as in Section 6.5 by a set of real parameters $U=U(a_1,\dots ,a_n)$ such that $U(0,\dots ,0)=I$ , we define the infinitesimal generators by

$$S_j=-i\frac{\partial U}{\partial a_j}{|}_{\mathbf{a}=0}.\text{\hspace{1em}(14.31)}$$

These are self-adjoint operators since $U{U}^{\ast }=I$ implies that

$$0={\frac{\partial U}{\partial a_j}|}_{\mathbf{a}=0}+{\frac{\partial U^{\ast }}{\partial a_j}|}_{\mathbf{a}=0}=i(S_j-S_j^{\ast }).$$

Note that self-adjoint operators do not form a Lie algebra, since their commutator is not in general self-adjoint. However, as seen in Section 6.5, Problem 6.12, the operators $i{S}_j$ do form a Lie algebra,

$$[i{S}_i,i{S}_j]=\sum _{k=1}^n{C}_{ij}^{k}i{S}_k.$$

Exercise

Show that an operator $S$ satisfies $S^{\ast }=-S$ iff it is of the form $S=iA$ where $A$ is self adjoint. Show that the commutator product preserves this property.

Example 14.5

If $S$ is a hermitian operator the set of unitary transformations $U(a)={\mathrm{e}}^{iaS}$ where $-\infty <a<\infty$ is a one-parameter group of unitary transformations,

$$U(a)U^{\ast }(a)={\mathrm{e}}^{iaS}{\mathrm{e}}^{-iaS}=I,U(a)U(b)={\mathrm{e}}^{iaS}{\mathrm{e}}^{ibS}={\mathrm{e}}^{i(a+b)S}=U(a+b).$$

Its infinitesimal generator is

$${-i\frac{\partial }{\partial a}{\mathrm{e}}^{iaS}|}_{a=0}=S.$$

Example 14.6

Let $O$ and $O^{\prime }$ be two observers related by a displacement of the origin through the vector $a$. Let the state vectors be related by

$$|{\psi }^{\prime }⟩=T(\mathbf{a})|\psi ⟩.$$

If $Q$ is the position operator then

$${\mathbf{q}}^{\prime }=⟨\mathbf{Q}{⟩}_{{\psi }^{\prime }}=⟨{\psi }^{\prime }|\mathbf{Q}|{\psi }^{\prime }⟩=\mathbf{q}-\mathbf{a}=⟨\psi |\mathbf{Q}|\psi ⟩-\mathbf{a}⟨\psi |I|\psi ⟩,$$

whence

$$T^{\ast }(\mathbf{a})\mathbf{Q}T(\mathbf{a})=\mathbf{Q}-\mathbf{a}I.\text{\hspace{1em}(14.32)}$$

Taking the partial derivative with respect to $a_i$ at $\mathbf{a}=0$ we find

$$-i{S}_i{Q}_j+i{Q}_j{S}_i=-{\delta }_{ij}\text{ where }{S}_j=-i\frac{\partial T}{\partial a_j}{|}_{\mathbf{a}=0}.$$

Hence $[S_i,Q_j]=-i{\delta }_{ij}$ , and we may expect

$$S_i=\frac{1}{\hslash }{P}_i$$

where $P_i$ are the momentum operators

$$P_i=-i\hslash \frac{\partial }{\partial q_i},[Q_i,P_j]=i\hslash {\delta }_{ij}.$$

This is consistent with ${\mathbf{P}}^{\prime }=\mathbf{P}.$ , since

$$T^{\ast }(\mathbf{a})P_{i}T(\mathbf{a})=P_i⟹[S_j,P_i]\propto [P_j,P_i]=0.$$

To find the translation operators $T(\mathbf{a})$ , we use the group property $T(\mathbf{a})T(\mathbf{b})=T(\mathbf{a}+\mathbf{b})$ and take the derivative with respect to $b_i$ at $\mathbf{b}=0$ 1

$$iT(\mathbf{a})S_i=T(\mathbf{a})\frac{\partial T}{\partial b_i}{|}_{\mathbf{b}=0}=\frac{\partial T}{\partial a_i}(\mathbf{a}).$$

The solution of this operator equation may be written

$$\begin{array}{c}T(\mathbf{a})={\mathrm{e}}^{i{\sum }_i{a}_i{S}_i}={\mathrm{e}}^{i\mathbf{a}\cdot \mathbf{S}}\\ ={\mathrm{e}}^{i\mathbf{a}\cdot \mathbf{P}/\hslash }={\mathrm{e}}^{i{a}_1{P}_1/\hslash }{\mathrm{e}}^{i{a}_2{P}_2/\hslash }{\mathrm{e}}^{i{a}_3{P}_3/\hslash }\end{array}$$

since the $P_i$ commute with each other. It is left as an exercise (Problem 14.14) to verify Eq. (14.32).

Example 14.7

Two observers related by a rotation through an angle $\theta$ about the $z$-axis

$$q_1^{\prime }=q_1\cos \theta +q_2\sin \theta ,q_2^{\prime }=-q_1\sin \theta +q_2\cos \theta ,q_3^{\prime }=q_3$$

are related by a unitary operator $R$($\theta$) such that

$$|{\psi }^{\prime }⟩=R(\theta )|\psi ⟩,R^{\ast }(\theta )R(\theta )=I.$$

In order to arrive at the correct transformation of expectation values we require that

$$R^{\ast }(\theta )Q_{1}R(\theta )=Q_1\cos \theta +Q_2\sin \theta ,\text{\hspace{1em}(14.33)}$$

$$R^{\ast }(\theta )Q_{2}R(\theta )=-Q_1\sin \theta +Q_2\cos \theta ,\text{\hspace{1em}(14.34)}$$

$$R^{\ast }(\theta )Q_{3}R(\theta )=Q_3.\text{\hspace{1em}(14.35)}$$

Setting

$$J=-i\frac{\mathrm{d}R}{\mathrm{d}\theta }{|}_{\theta =0},J^{\ast }=J$$

we find on taking derivatives at $\theta =0$ of Eqs. Eq. (14.33)–Eq. (14.35)

$$[J,Q_1]=i{Q}_2,[J,Q_2]=-i{Q}_1,[J,Q_3]=0.$$

A solution is the $z$-component of angular momentum,

$$J=\frac{1}{\hslash }{L}_3=\frac{1}{\hslash }(Q_1{P}_2-Q_2{P}_1),$$

since $[J,Q_1]=i{Q}_2$ , etc. (see Problem 14.4).

It is again easy to verify the group property $R({\theta }_1)R({\theta }_2)=R({\theta }_1+{\theta }_2)$ and as in the translational example above,

$$iR(\theta )J=\frac{\mathrm{d}R}{\mathrm{d}\theta }⟹R\theta ={\mathrm{e}}^{i\theta J}={\mathrm{e}}^{i\theta L_3/\hslash }.$$

It is again left as an exercise to show that this operator satisfies Eqs. Eq. (14.33)–Eq. (14.35). For a rotation of magnitude $\theta$ about an axis $n$ the rotation operator is

$$R(\mathbf{n},\theta )={\mathrm{e}}^{i\theta \mathbf{n}\cdot \mathbf{L}/\hslash }$$

where $L$ is the angular momentum operator having components $L_i={ϵ}_{ijk}{Q}_j{P}_k$ . Since these operators do not commute, satisfying the commutation relations Eq. (14.25), we have in general

$$R(\mathbf{n},\theta )\ne {\mathrm{e}}^{i{\theta }_1{L}_1/\hslash }{\mathrm{e}}^{i{\theta }_2{L}_2/\hslash }{\mathrm{e}}^{i{\theta }_3{L}_3/\hslash }.$$

Exercise

Show that the transformation of momentum components under a rotation with infinitesimal generator $J$ is

$$[J,P_1]=i{P}_2,[J,P_2]=-i{P}_1,[J,P_3]=0.$$

Example 14.8

Under a time translation $t^{\prime }=t-\tau$ , we have $|{\psi }^{\prime }(t^{\prime })⟩=|\psi (t)⟩$ , so that

$$|{\psi }^{\prime }(t)⟩=|\psi (t+\tau )⟩=T(\tau )|\psi (t)⟩.$$

Hence, by Schrödinger’s equation

$$iS|\psi (t)⟩=\frac{\partial T}{\partial \tau }{|}_{\tau =0}|\psi (t)⟩=\frac{\mathrm{d}}{\mathrm{d}t}|\psi (t)⟩=\frac{1}{i\hslash }H|\psi (t)⟩.$$

The infinitesimal generator of the time translation is essentially the Hamiltonian, $S=-H/\hslash$ . If the Hamiltonian $H$ is time-independent,

$$T(\tau )={\mathrm{e}}^{-iH\tau /\hslash }.\text{\hspace{1em}(14.36)}$$

Conserved quantities

Under a time-dependent unitary transformation

$$|{\psi }^{\prime }⟩=U(t)|\psi ⟩.$$

Schrödinger’s equation Eq. (14.13) results in

$$\begin{array}{r}i\hslash \frac{\mathrm{d}}{\mathrm{d}t}|{\psi }^{\prime }⟩=i\hslash \frac{\partial U}{\partial t}|\psi ⟩+i\hslash U(t)\frac{\mathrm{d}}{\mathrm{d}t}|\psi ⟩\\ =H^{\prime }|{\psi }^{\prime }⟩\end{array}$$

where

$$H^{\prime }=UH{U}^{\ast }+i\hslash \frac{\partial U}{\partial t}{U}^{\ast }.\text{\hspace{1em}(14.37)}$$

Exercise

Show that under an anti-unitary transformation

$$H^{\prime }=-UH{U}^{\ast }+i\hslash \frac{\partial U}{\partial t}{U}^{\ast }.$$

$U$ is called a Hamiltonian symmetry if $H^{\prime }=H$ . Then, multiplying Eq. (14.37) by $U$ on the right gives

$$[U,H]+i\hslash \frac{\partial U}{\partial t}=0.\text{\hspace{1em}(14.38)}$$

If $U$ is independent of time then $U$ commutes with the Hamiltonian, $[U,H]=0$

If $G$ is an $n$-parameter Lie group ofunitary Hamiltonian symmetries $U(t,a_1,a_2,\dots ,a_n)$ , having infinitesimal generators

$$S_i(t)=-i\frac{\partial U}{\partial a_i}{|}_{\mathbf{a}=0},$$

then differentiating Eq. (14.38) with respect to $a_i$ gives

$$[S_i,H]+i\hslash \frac{\partial S_i}{\partial t}=0.\text{\hspace{1em}(14.39)}$$

Any hermitian operator $S(t)$ satisfying this equation is said to be a constant of the motion or conserved quantity, for Schrödinger’s equation implies that its expection values are constant:

$$\begin{array}{rl}& \frac{\mathrm{d}}{\mathrm{d}t}⟨S{⟩}_{\psi }=\frac{\mathrm{d}}{\mathrm{d}t}⟨\psi |S|\psi ⟩=⟨\frac{\mathrm{d}\psi }{\mathrm{d}t}|S|\psi ⟩+⟨\psi |S\frac{\mathrm{d}}{\mathrm{d}t}|\psi ⟩+⟨\psi |\frac{\partial S}{\partial t}|\psi ⟩\\ & =\frac{-1}{i\hslash }⟨\psi |HS|\psi ⟩+\frac{1}{i\hslash }⟨\psi |SH|\psi ⟩+⟨\psi |\frac{\partial S}{\partial t}|\psi ⟩\\ & =\frac{1}{i\hslash }⟨\psi |[S,H]+i\hslash \frac{\partial S}{\partial t}|\psi ⟩=0.\end{array}$$

Exercise

Show that in the Heisenberg picture, this is equivalent to

$$\frac{\mathrm{d}{S}_{\mathrm{H}}}{\mathrm{d}t}=0\text{ where }{S}_{\mathrm{H}}={\mathrm{e}}^{-iHt/\hslash }S{\mathrm{e}}^{iHt/\hslash }.$$

From Examples 14.6 and 14.7 it follows that invariance of the Hamiltonian under trans lations and rotations is equivalent to conservation of momentum and angular momentum respectively. In both cases the infinitesimal generators are time-independent. Ifthe Hamilto nian is invariant under time translations, having generator $S=-H/\hslash$ (see Example 14.8), then Eq. (14.39) reduces to

$$-\frac{1}{\hslash }[H,H]-i\frac{\partial H}{\partial t}=0,$$

which is true if and only if $H$ has no explicit time dependence, $\partial H/\partial t=0$

Discrete symmetries

There are a number of important symmetries of a more discrete nature, illustrated in the following examples.

Example 14.9

Consider a spatial inversion $\mathbf{r}\mapsto {\mathbf{r}}^{\prime }=-\mathbf{r}.$ , which can be thought of as a rotation by $180^{\circ }$ about the $z$-axis followed by a reflection $x^{\prime }=x,y^{\prime }=y,z^{\prime }=-z$ . Let $\Pi$

be the operator on $\mathcal{H}$ induced by such an inversion, satisfying

$${\Pi }^{\ast }{Q}_i\Pi =-Q_i,{\Pi }^{\ast }{P}_i\Pi =-P_i.$$

By Wigner’s theorem,

$${\Pi }^{\ast }[Q_i,P_j]\Pi ={\Pi }^{\ast }i\hslash {\delta }_{ij}\Pi =\{\begin{array}{ll}i\hslash {\delta }_{ij}& \text{ if }\Pi \text{ is unitary },\\ -i\hslash {\delta }_{ij}& \text{ if }\Pi \text{ is anti - unitary }.\end{array}$$

It turns out that ! must be a unitary operator, for

$${\Pi }^{\ast }[Q_i,P_j]\Pi =[{\Pi }^{\ast }{Q}_i\Pi ,{\Pi }^{\ast }{P}_j\Pi ]=[-Q_i,-P_j]=[Q_i,P_j]=i\hslash {\delta }_{ij}.$$

Note also that angular momentum operators are invariant under spatial reflections,

$${\Pi }^{\ast }{L}_i\Pi =L_i\text{ where }\mathbf{L}=\mathbf{Q}\times \mathbf{P}.$$

Since successive reflections result in the identity ${\Pi }^2=I_{:}$ , we have ${\Pi }^{\ast }=\Pi$ . Hence $\Pi$ is a hermitian operator, corresponding to an observable called parity, having eigenvalues $\pm 1$. States of eigenvalue 1, $\Pi |\psi ⟩=|\psi ⟩$ , are said to be of even parity, while those of eigenvalue $-1$ are called odd parity, $\Pi |\psi ⟩=-|\psi ⟩$ . Every state can be decomposed as a sum of an even and an odd parity state,

$$|\psi ⟩=\frac{1}{2}(I+\Pi )|\psi ⟩+\frac{1}{2}(I-\Pi )|\psi ⟩=|{\psi }_{+}⟩+|{\psi }_{-}⟩.$$

Exercise

Show that if $[\Pi ,H]=0$ , the parity of any state is preserved throughout its motion, and eigenstates of $H$ with non-degenerate eigenvalue have definite parity.

Example 14.10

In classical physics, if $q$($t$) is a solution of Newton’s equations then so is the reverse motion ${\mathbf{q}}_{\mathrm{rev}}(t)=\mathbf{q}(t)$ having opposite momentum ${\mathbf{p}}_{\mathrm{rev}}(t)=-\mathbf{p}(-t)$ . If $O^{\prime }$ is an observer having time in the reversed direction $t^{\prime }=-t$ to that of an observer $O$, let the time-reversed states be

$$|{\psi }^{\prime }⟩=\Theta |\psi ⟩,$$

where ” is the time-reversal operator. Since we require

$${\Theta }^{\ast }{Q}_i\Theta =Q_i,{\Theta }^{\ast }{P}_i\Theta =-P_i$$

a similar discussion to that in Example 14.9 gives

$$\begin{array}{r}{\Theta }^{\ast }[Q,P]\Theta ={\Theta }^{\ast }i\hslash I\Theta =\pm i\hslash I\\ =[Q,-P]=-i\hslash I.\end{array}$$

Hence time-reversal ” is an anti-unitary operator.

If the Hamiltonian $H$ is invariant under time reversal, $[H,\Theta ]=0$ , then applying ” to Schrödinger’s equation Eq. (14.13) gives

$$-i\hslash \frac{\mathrm{d}}{\mathrm{d}t}\Theta |\psi (t)⟩=\Theta i\hslash \frac{\mathrm{d}}{\mathrm{d}t}|\psi ⟩=\Theta H|\psi ⟩=H\Theta |\psi (t)⟩.$$

Changing the time variable $t\mathrm{to}-t$

$$i\hslash \frac{\mathrm{d}}{\mathrm{d}t}\Theta |\psi (-t)⟩=H\Theta |\psi (-t)⟩.$$

It follows that $|{\psi }_{\mathrm{rev}}(t)⟩=\Theta |\psi (-t)⟩$ is a solution of Schrödinger’s equation, which may be thought of as the time-reversed solution. In this sense, the dynamics of quantum mechanics is time-reversable, but note that because of the anitilinear nature of the operator $\Theta$, a complex conjugation is required in addition to time inversion. For example in the position representation, if $\psi (\mathbf{r},t)$ is a solution ofSchrödinger’s wave equation Eq. (14.18), then $\psi (\mathbf{r},-t)$ is not in general a solution. However, taking the complex conjugate shows that ${\psi }_{\mathrm{rev}}(t)=\psi (\mathbf{r},-t)$ is a solution of Eq. (14.18),

$$i\hslash \frac{\partial }{\partial t}\overline{\psi (\mathbf{r},-t)}=-\left(\frac{{\hslash }^2}{2m}{\nabla }^2+V(\mathbf{r})\right)\overline{\psi (\mathbf{r},-t)}.$$

Identical particles

Consider a system consisting of $N$ indistinguishable particles. If the Hilbert space of each individual particle is we take the Hilbert space of the entire system to be the tensor product

$${\mathcal{H}}^N=\mathcal{H}\otimes \mathcal{H}\otimes \cdots \otimes \mathcal{H}.$$

As in Chapter 7 this may be regarded as the tensor space spanned by free formal products

$$|{\psi }_1{\psi }_2\dots {\psi }_N⟩\equiv |{\psi }_1⟩|{\psi }_2⟩\dots |{\psi }_N⟩$$

where each $|{\psi }_i⟩\in \mathcal{H}$ , subject to identifications

$$\left(\alpha |{\psi }_1⟩+\beta |\varphi ⟩\right)|{\psi }_2⟩\cdots =\alpha |{\psi }_1⟩|{\psi }_2⟩\cdots +\beta |\varphi ⟩|{\psi }_2⟩\dots ,\text{ etc. }$$

The inner product on ${\mathcal{H}}^N$ is defined by

$$⟨{\psi }_1{\psi }_2\dots {\psi }_N|{\varphi }_1{\varphi }_2\dots {\varphi }_N⟩=⟨{\psi }_1|{\varphi }_1⟩⟨{\psi }_2|{\varphi }_2⟩\dots ⟨{\psi }_N|{\varphi }_N⟩.$$

For each pair $1\le i<j\le N$ define the permutation operator $P_{ij}:{\mathcal{H}}^N\to {\mathcal{H}}^N$ by

$$⟨{\varphi }_1\dots {\varphi }_i\dots {\varphi }_j\dots {\varphi }_N|P_{ij}\psi ⟩=⟨{\varphi }_1\dots {\varphi }_j\dots {\varphi }_i\dots {\varphi }_N|\psi ⟩$$

for all $|{\varphi }_i⟩\in \mathcal{H}$ . This is a linear operator that ‘interchanges particles’ $i$ and $j$,

$$P_{ij}|{\psi }_1\dots {\psi }_i\dots {\psi }_j\dots {\psi }_N⟩=|{\psi }_1\dots {\psi }_j\dots {\psi }_i\dots {\psi }_N⟩.$$

$\mathrm{If}|{\psi }_a⟩(a=1,2,...)$ is an o.n. basis of the Hilbert space then the set of all vectors

$$|{\psi }_{a_1}{\psi }_{a_2}\dots {\psi }_{a_N}⟩\equiv |{\psi }_{a_1}⟩|{\psi }_{a_2}⟩\dots |{\psi }_{a_N}⟩$$

forms an o.n. basis of ${\mathcal{H}}^N$ . Since $P_{i}j$ transforms any such o.n. basis to an o.n. basis it must be a unitary operator. These statements extend to a general permutation $P$, since it can be written as a product of interchanges

$$P=P_{ij}{P}_{kl}\dots$$

As there is no dynamical way ofdetecting an interchange ofidentical particles, the expection values of the Hamiltonian must be invariant under permutations, $⟨H{⟩}_{P\psi }=⟨H{⟩}_{\psi }$ , so that

$$⟨\psi |P^{\ast }HP|\psi ⟩=⟨\psi |H|\psi ⟩$$

for all $|\psi ⟩\in \mathcal{H}$ . Hence $P^{\ast }HP=H$ and as $P$ is unitary, $P{P}^{\ast }=I$

$$[H,P]=0.$$

This is yet another example of a discrete symmetry. In classical mechanics it is taken as given that all particles have an individuality and are in principle distinguishable. It is basic to the philosophy of quantum mechanics, however, that since no physical procedure exists for ‘marking’ identical particles such as a pair of electrons in order to keep track of them, there can be no method even in principle of distinguishing between them.

All interchanges have the property $(P_{ij}{)}^2=I$ , from which they are necessarily hermitian, $P_{ij}=P_{ij}^{\ast }$ . Every interchange therefore corresponds to an observable. It has eigenvalues $ϵ\pm 1$ and, since $P_{ij}$ is a constant of the motion, any eigenstate

$$P_{ij}|\psi ⟩=ϵ|\psi ⟩$$

remains an eigenstate corresponding to the same eigenvalue, for

$$ϵ=⟨\psi |P_{ij}|\psi ⟩=⟨P_{ij}{⟩}_{\psi }=const.$$

Since no physical observable can distinguish between states related by a permutation, a similar argument to that used for the Hamiltonian shows that every observable $A$ commutes with all permutation operators,

$$[A,P]=0.$$

Hence, $\mathrm{if}|\psi$ is a non-degenerate eigenstate of $A,$ , it is an eigenstate of every permutation operator $P_{:}$ ,

$$\begin{array}{rl}A|\psi ⟩=a|\psi ⟩& ⟹AP|\psi ⟩=PA|\psi ⟩=aP|\psi ⟩\\ & ⟹P|\psi ⟩=p|\psi ⟩\end{array}$$

for some factor $p=\pm 1.\mathrm{If},$ as is commonly assumed, every state is representable as a sum of non-degenerate common eigenvectors of a commuting set of complete observables $A$, $B$, $C$, … we must then assume that every physical state $|\psi ⟩$ of the system is a common eigenvector of all permutation operators. In particular, for all interchanges $P_{ij}$

$$P_{ij}|\psi ⟩=p_{ij}|\psi ⟩\text{ where }{p}_{ij}=\pm 1.$$

All $p_{ij}$ are equal for the state $|\psi ⟩$ , since for any pair $k$, $l$

$$P_{ij}|\psi ⟩=P_{ki}{P}_{lj}{P}_{kl}{P}_{lj}{P}_{ki}|\psi ⟩,$$

from which it follows that $p_{ij}=p_{kl}$ since

$$p_{ij}|\psi ⟩=(p_{ki}{)}^2(p_{lj}{)}^2{p}_{kl}|\psi ⟩=p_{kl}|\psi ⟩.$$

Thus for all permutations either $P|\psi ⟩=|\psi ⟩\mathrm{or}P|\psi ⟩=(-1{)}^P|\psi ⟩$ . In the first case, $p_{ij}=1$ the state is said to be symmetrical and the particles are called bosons or to obey Bose– Einstein statistics. If $p_{ij}=-1$ the state is antisymmetrical, the particles are said to be fermions and obey Fermi–Dirac statistics. It turns out that bosons are always particles of integral spin such as photons or mesons, while fermions such as electrons or protons always have half-integral spin. This is known as the spin-statistics theorem, but lies beyond the scope of this book (see, for example, [9]).

The celebrated Pauli exclusion principle asserts that two identical fermions cannot occupy the same state, for if

$$|\psi ⟩=|{\psi }_1⟩\dots |{\psi }_i⟩\dots |{\psi }_j⟩\dots |{\psi }_N⟩$$

and $|{\psi }_i⟩=|{\psi }_j⟩$ then

$$P_{ij}|\psi ⟩=|\psi ⟩=-|\psi ⟩$$

since every state has eigenvalue 1. Hence $|\psi ⟩=0$

Problems

Problem 14.13

If the operator $K$ is complex conjugation with respect to a complete o.n. set,

$$K(\sum _i\alpha |e_i⟩)=\sum _i\overline{{\alpha }_i}|e_i⟩,$$

show that every anti-unitary operator $V$ can be written in the form $V=UK$ , where $U$ is a unitary operator.

Problem 14.14

For any pair of operators $A$ and $B$ show by induction on the coefficients that

$${\mathrm{e}}^{aB}A{\mathrm{e}}^{-aB}=A+a[B,A]+\frac{a^2}{2!}[B,[B,A]]+\frac{a^3}{3!}[B,[B,[B,A]]]+\dots$$

Hence show the relation Eq. (14.32) holds for $T(\mathbf{a})={\mathbf{e}}^{i\mathbf{a}\cdot \mathbf{P}/\hslash }$

Problem 14.15

Using the expansion in Problem 14.14 show that $R(\theta )={\mathbf{e}}^{i\theta L_3/\hslash }$ satisfies Eqs. Eq. (14.33)–Eq. (14.35).

Problem 14.16

Show that the time reversal of angular momentum $\mathbf{L}=\mathbf{Q}\times \mathbf{P}$ is ${\Theta }^{\ast }{L}_i\Theta =-L_i,$ and that the commutation relations $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$ are only preserved if ” is anti-unitary.