2025-05-30
Solutions to Principles of Quantum Mechanics
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目录

Chapter 12 Rotational Invariance and Angular Momentum
12.1 Translations in Two Dimensions
12.2 Rotations in Two Dimensions
12.3 The Eigenvalue Problem of $L_{z}$
12.4 Angular Momentum in Three Dimensions
12.5 The Eigenvalue Problem of $L^{2}$ and $L_{z}$
12.6 Solution of Rotationally Invariant Problems

Chapter 12 Rotational Invariance and Angular Momentum

12.1 Translations in Two Dimensions

Exercise 12.1.1 Verify that a^P\hat{a} \cdot \mathbf{P} is the generator of infinitesimal translations along a by considering the relation

x,yIiδaPψ=ψ(xδax,yδay)\langle x, y| I-\frac{i}{\hbar} \boldsymbol{\delta} a \cdot \mathbf{P}|\psi\rangle=\psi\left(x-\delta a_x, y-\delta a_y\right)

12.2 Rotations in Two Dimensions

Exercise 12.2.1 Provide the steps linking Eq. (12.2.8) to Eq. (12.2.9). [Hint: Recall the derivation of Eq. (11.2.8) from Eq. (11.2.6).]

Exercise 12.2.2 Using these commutation relations (and your keen hindsight) derive Lz=XPyYPxL_z=X P_y-Y P_x. At least show that Eqs. (12.2.16) and (12.2.17) are consistent with Lz=XPyYPxL_z=X P_y-Y P_x.

Exercise 12.2.3 Derive Eq. (12.2.19) by doing a coordinate transformation on Eq. (12.2.10), and also by the direct method mentioned above.

Exercise 12.2.4 Rederive the equivalent of Eq. (12.2.23) keeping terms of order εxεz2\varepsilon_x \varepsilon_z^2. (You may assume εy=0\varepsilon_y=0.) Use this information to rewrite Eq. (12.2.24) to order εxεz2\varepsilon_x \varepsilon_z^2. By equating coefficients of this term deduce the constraint

2LzPxLz+PxLz2+Lz2Px=2Px-2 L_z P_x L_z+P_x L_z^2+L_z^2 P_x=\hbar^2 P_x

This seems to conflict with statement (1) made above, but not really, in view of the identity

2ΛΩΛ+ΩΛ2+Λ2Ω[Λ,[Λ,Ω]]-2 \Lambda \Omega \Lambda+\Omega \Lambda^2+\Lambda^2 \Omega \equiv[\Lambda,[\Lambda, \Omega]]

Using the identify, verify that the new constraint coming from the εxεz2\varepsilon_x \varepsilon_z^2 term is satisfied given the commutation relations between PxP_x, PyP_y, and LzL_z.

12.3 The Eigenvalue Problem of LzL_{z}

Exercise 12.3.1 Provide the steps linking Eq. (12.3.5) to Eq. (12.3.6).

Exercise 12.3.2 Let us try to deduce the restriction on lzl_{z} from another angle. Consider a superposition of two allowed lzl_{z} eigenstates:

ψ(ρ,ϕ)=A(ρ)eiϕlz/+B(ρ)eiϕlz/\psi(\rho, \phi)=A(\rho) \mathrm{e}^{\mathrm{i}\phi l_{z} / \hbar}+B(\rho) \mathrm{e}^{\mathrm{i} \phi l_{z} / \hbar}

By demanding that upon a 2π2 \pi rotation we get the same physical state (not necessarily the same state vector), show that lzlz=ml_z-l_z^{\prime}=m \hbar, where mm is an integer. By arguing on the grounds of symmetry that the allowed values of lzl_z must be symmetric about zero, show that these values are either ,3/2,/2,/2,3/2,\ldots, 3 \hbar / 2, \hbar / 2,-\hbar / 2,-3 \hbar / 2, \ldots or ,2,,0,,2,\ldots, 2 \hbar, \hbar, 0,-\hbar,-2 \hbar, \ldots It is not possible to restrict lzl_z any further this way.

Exercise 12.3.3 A particle is described by a wave function

ψ(ρ,ϕ)=Aeρ2/2Δ2cos2ϕ\psi(\rho, \phi)=A e^{-\rho^2/2 \Delta^2} \cos ^2 \phi

Show (by expressing cos2ϕ\cos^2 \phi in terms of Φm\Phi_m) that

P(lz=0)=2/3P(lz=2)=1/6P(lz=2)=1/6\begin{gathered} P\left(l_z=0\right)=2 / 3 \\ P\left(l_z=2 \hbar\right)=1 / 6 \\ P\left(l_z=-2 \hbar\right)=1 / 6 \end{gathered}

(Hint: Argue that the radial part eρ2/2Δ2e^{-\rho^2 / 2 \Delta^2} is irrelevant here.)

Exercise 12.3.4 A particle is described by a wave function

ψ(ρ,ϕ)=Aeρ2/2Δ2(ρΔcosϕ+sinϕ)\psi(\rho, \phi)=A\mathrm{e}^{-\rho^2 / 2 \Delta^2}\left(\dfrac{\rho}{\Delta} \cos \phi+\sin \phi\right)

Show that

P(lz=)=P(lz=)=12P\left(l_z=\hbar\right)=P\left(l_z=-\hbar\right)=\dfrac{1}{2}

Exercise 12.3.5 Note that the angular momentum seems to generate a repulsive potential in Eq. (12.3.13). Calculate its gradient and identify it as the centrifugal force.

Exercise 12.3.6 Consider a particle of mass μ\mu constrained to move on a circle of radius aa. Show that H=Lz2/2μa2H=L_z^2/2\mu a^2. Solve the eigenvalue problem of HH and interpret the degeneracy.

Exercise 12.3.7 (The Isotropic Oscillator) Consider the Hamiltonian

H=Px2+Py22μ+12μω2(X2+Y2)H=\frac{P_x^2+P_y^2}{2 \mu}+\frac{1}{2} \mu \omega^2\left(X^2+Y^2\right)

(1) Convince yourself [H,Lz]=0\left[H, L_z\right]=0 and reduce the eigenvalue problem of HH to the radial differential equation for REm(ρ)R_{E m}(\rho).

(2) Examine the equation as ρ0\rho \rightarrow 0 and show that

REm(ρ)ρ0ρmR_{E m}(\rho) \xrightarrow[\rho \rightarrow 0]{} \rho^{|m|}

(3) Show likewise that up to powers of ρ\rho

REm(ρ)ρeμωρ2/2R_{E m}(\rho) \xrightarrow[\rho \rightarrow \infty]{} \mathrm{e}^{-\mu \omega \rho^2/2 \hbar}

So assume that REm(ρ)=ρmeμωρ2/2UEm(ρ)R_{E m}(\rho)=\rho^{|m|} \mathrm{e}^{-\mu \omega \rho^2 / 2 \hbar} U_{E m}(\rho).

(4) Switch to dimensionless variables ε=E/ω\varepsilon=E/\hbar\omega, y=(μω/)1/2ρy=(\mu \omega / \hbar)^{1/2}\rho.

(5) Convert the equation for RR into an equation for UU. (I suggest proceeding in two stages: R=ymfR=y^{|m|}f, f=ey2/2Uf=\mathrm{e}^{-y^2/2}U.) You should end up with

U+[(2m+1y)2y]U+(2ε2m2)U=0U^{\prime \prime}+\left[\left(\frac{2|m|+1}{y}\right)-2 y\right] U^{\prime}+(2 \varepsilon-2|m|-2) U=0

(6) Argue that a power series for UU of the form

U(y)=r=0CryrU(y)=\sum_{r=0}^{\infty} C_r y^r

will lead to a two-term recursion relation.

(7) Find the relation between Cr+2C_{r+2} and CrC_r. Argue that the series must terminate at some finite rr if the yy \rightarrow \infty behavior of the solution is to be acceptable. Show ε=r+m+1\varepsilon=r+|m|+1 leads to termination after rr terms. Now argue that rr is necessarily even——i.e, r=2kr=2 k. (Show that if rr is odd, the behavior of RR as ρ0\rho \rightarrow 0 is not ρm\rho^{|m|}.) So finally you must end up with

E=(2k+m+1)ω,k=0,1,2,E=(2 k+|m|+1) \hbar \omega, \quad k=0,1,2, \ldots

Define n=2k+mn=2 k+|m|, so that

En=(n+1)ωE_n=(n+1) \hbar \omega

(8) For a given nn, what are the allowed values of m|m|? Given this information show that for a given nn, the degeneracy is n+1n+1. Compare this to what you found in Cartesian coordinates (Exercise 10.2.2).

(9) Write down all the normalized eigenfunctions corresponding to n=0,1n=0,1.

(10) Argue that the n=0n=0 function must equal the corresponding one found in Cartesian coordinates. Show that the two n=2n=2 solutions are linear combinations of their counterparts in Cartesian coordinates. Verify that the parity of the states is (1)n(-1)^n as you found in Cartesian coordinates.

Exercise 12.3.8 Consider a particle of charge qq in a vector potential

A=B2(yi+xj)\mathbf{A}=\frac{B}{2}(-y \mathbf{i}+x \mathbf{j})

(1) Show that the magnetic field is B=Bk\mathbf{B}=B \mathbf{k}.

(2) Show that a classical particle in this potential will move in circles at an angular frequency ω0=qB/μc\omega_0=q B / \mu c.

(3) Consider the Hamiltonian for the corresponding quantum problem:

H=[Px+qYB/2c]22μ+[PyqXB/2c]22μH=\frac{\left[P_x+q Y B / 2 c\right]^2}{2 \mu}+\frac{\left[P_y-q X B / 2 c\right]^2}{2 \mu}

Show that Q=(cPx+qYB/2)/qBQ=\left(c P_x+q Y B / 2\right) / q B and P=(PyqXB/2c)P=\left(P_y-q X B / 2 c\right) are canonical. Write HH in terms of PP and QQ and show that allowed levels are E=(n+1/2)ω0E=(n+1 / 2) \hbar \omega_0.

(4) Expand HH out in terms of the original variables and show

H=H(ω02,μ)ω02LzH=H\left(\frac{\omega_0}{2}, \mu\right)-\frac{\omega_0}{2} L_z

where H(ω0/2,μ)H\left(\omega_0 / 2, \mu\right) is the Hamiltonian for an isotropic two-dimensional harmonic oscillator of mass μ\mu and frequency ω0/2\omega_0 / 2. Argue that the same basis that diagonalized H(ω0/2,μ)H\left(\omega_0 / 2, \mu\right) will diagonalize HH. By thinking in terms of this basis, show that the allowed levels for HH are E=(k+12m12m+12)ω0E=\left(k+\dfrac{1}{2}|m|-\dfrac{1}{2} m+\dfrac{1}{2}\right) \hbar \omega_0, where kk is any integer and mm is the angular momentum. Convince yourself that you get the same levels from this formula as from the earlier one [E=(n+1/2)ω0]\left[E=(n+1 / 2) \hbar \omega_0\right]. We shall return to this problem in Chapter 21.

12.4 Angular Momentum in Three Dimensions

Exercise 12.4.1 (1) Verify that Eqs. (12.4.9) and Eq. (12.4.8) are equivalent, given the definition of εijk\varepsilon_{i j k}.

(2) Let U1,U2U_1, U_2, and U3U_3 be three energy eigenfunctions of a single particle in some potential. Construct the wave function ψA(x1,x2,x3)\psi_A\left(x_1, x_2, x_3\right) of three fermions in this potential, one of which is in U1U_1, one in U2U_2, and one in U3U_3, using the εijk\varepsilon_{i j k} tensor.

Exercise 12.4.2 (1) Verify Eq. (12.4.2) by first constructing the 3×33 \times 3 matrices corresponding to R(εxi)R\left(\varepsilon_x \mathbf{i}\right) and R(εyj)R\left(\varepsilon_y \mathbf{j}\right), to order ε\varepsilon.

(2) Provide the steps connecting Eqs. (12.4.3) and (12.4.4a).

(3) Verify that LxL_x and LyL_y defined in Eq. (12.4.1) satisfy Eq. (12.4.4a). The proof for other commutators follows by cyclic permutation.

Exercise 12.4.3 We would like to show that θ^L\hat{\theta} \cdot \mathbf{L} generates rotations about the axis parallel to θ^\hat{\theta}. Let δθ\delta\boldsymbol{\theta} be an infinitesimal rotation parallel to θ\boldsymbol{\theta}.

(1) Show that when a vector r\mathbf{r} is rotated by an angle δθ\delta \boldsymbol{\theta}, it changes to r+δθ×r\mathbf{r}+\delta \boldsymbol{\theta} \times \mathbf{r}. (It might help to start with rδθ\mathbf{r} \perp \delta \boldsymbol{\theta} and then generalize.)

(2) We therefore demand that (to first order, as usual)

ψ(r)U[R(δθ)]ψ(rδθ×r)=ψ(r)(δθ×r)ψ\psi(\mathbf{r}) \xrightarrow[{U[R(\delta \boldsymbol{\theta})]}]{ } \psi(\mathbf{r}-\delta \boldsymbol{\theta} \times \mathbf{r})=\psi(\mathbf{r})-(\delta \boldsymbol{\theta} \times \mathbf{r}) \cdot \nabla \psi

Comparing to U[R(δθ)]=I(iδθ/)Lθ^U[R(\delta \boldsymbol{\theta})]=I-(i \delta \theta / \hbar) L_{\hat{\theta}}, show that Lθ^=θ^LL_{\hat{\theta}}=\hat{\theta} \cdot \mathbf{L}.

Exercise 12.4.4 Recall that V\mathbf{V} is a vector operator if its components ViV_i transform as

U[R]ViU[R]=jRijVj(12.4.13)U^{\dagger}[R] V_i U[R]=\sum_j R_{i j} V_j\tag{12.4.13}

(1) For an infinitesimal rotation δθ\delta \boldsymbol{\theta}, show, on the basis of the previous exercise, that

jRijVj=Vi+(δθ×V)i=Vi+jkεijk(δθ)jVk\sum_j R_{i j} V_j=V_i+(\delta \boldsymbol{\theta} \times \mathbf{V})_i=V_i+\sum_j \sum_k \varepsilon_{i j k}(\delta \theta)_j V_k

(2) Feed in U[R]=1(i/)δθLU[R]=1-(i / \hbar) \delta \boldsymbol{\theta} \cdot \mathbf{L} into the left-hand side of Eq. (12.4.13) and deduce that

[Vi,Lj]=ikεijkVk\left[V_i, L_j\right]=i \hbar \sum_k \varepsilon_{i j k} V_k

12.5 The Eigenvalue Problem of L2L^{2} and LzL_{z}

Exercise 12.5.1 Consider a vector field Ψ(x,y)\Psi(x, y) in two dimensions. From Fig. 12.1 it follows that under an infinitesimal rotation εzk\varepsilon_z \mathbf{k},

ψxψx(x,y)=ψx(x+yεz,yxεz)ψy(x+yεz,yxεz)εzψyψy(x,y)=ψx(x+yεz,yxεz)εz+ψy(x+yεz,yxεz)\begin{aligned} & \psi_x \rightarrow \psi_x^{\prime}(x, y)=\psi_x\left(x+y \varepsilon_z, y-x \varepsilon_z\right)-\psi_y\left(x+y \varepsilon_z, y-x \varepsilon_z\right) \varepsilon_z \\ & \psi_y \rightarrow \psi_y^{\prime}(x, y)=\psi_x\left(x+y \varepsilon_z, y-x \varepsilon_z\right) \varepsilon_z+\psi_y\left(x+y \varepsilon_z, y-x \varepsilon_z\right) \end{aligned}

Show that (to order εz\varepsilon_z)

(ψxψy)=((1001)iεz(Lz00Lz)iεz(0ii0))(ψxψy)\begin{pmatrix} \psi_x^{\prime} \\ \psi_y^{\prime} \end{pmatrix}=\left(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}-\frac{\mathrm{i} \varepsilon_z}{\hbar}\begin{pmatrix} L_z & 0 \\ 0 & L_z \end{pmatrix}-\frac{\mathrm{i} \varepsilon_z}{\hbar}\begin{pmatrix} 0 & -\mathrm{i} \hbar \\ i \hbar & 0 \end{pmatrix}\right)\begin{pmatrix} \psi_x \\ \psi_y \end{pmatrix}

so that

Jz=Lz(1)I(2)+I(1)Sz(2)=Lz+Sz\begin{aligned} J_z & =L_z^{(1)} \otimes I^{(2)}+I^{(1)} \otimes S_z^{(2)} \\ & =L_z+S_z \end{aligned}

where I(2)I^{(2)} is a 2×22 \times 2 identity matrix with respect to the vector components, I(1)I^{(1)} is the identity operator with respect to the argument (x,y)(x, y) of Ψ(x,y)\Psi(x, y). This example only illustrates the fact that Jz=Lz+SzJ_z=L_z+S_z if the wave function is not a scalar. An example of half-integral eigenvalues will be provided when we consider spin in a later chapter. (In the present example, SzS_z has eigenvalues ±\pm \hbar.)

Exercise 12.5.2 (1) Verify that the 2×22 \times 2 matrices Jx(1/2)J_x^{(1 / 2)}, Jy(1/2)J_y^{(1 / 2)}, and Jz(1/2)J_z^{(1 / 2)} obey the commutation rule [Jx(1/2),Jy(1/2)]=iJz(1/2)\left[J_x^{(1 / 2)}, J_y^{(1 / 2)}\right]=\mathrm{i} \hbar J_z^{(1 / 2)}.

(2) Do the same for the 3×33 \times 3 matrices Ji(1)J_i^{(1)}.

(3) Construct the 4×44 \times 4 matrices and verify that

[Jx(3/2),Jy(3/2)]=iJz(3/2)\left[J_x^{(3 / 2)}, J_y^{(3 / 2)}\right]=\mathrm{i} \hbar J_z^{(3 / 2)}

Exercise 12.5.3 (1) Show that Jx=Jy=0\left\langle J_x\right\rangle=\left\langle J_y\right\rangle=0 in a state jm|j m\rangle.

(2) Show that in these states

Jx2=Jy2=122[j(j+1)m2]\left\langle J_x^2\right\rangle=\left\langle J_y^2\right\rangle=\frac{1}{2} \hbar^2\left[j(j+1)-m^2\right]

(use symmetry arguments to relate Jx2\left\langle J_x^2\right\rangle to Jy2\left\langle J_y^2\right\rangle).

(3) Check that ΔJxΔJy\Delta J_x \cdot \Delta J_y from part (2) satisfies the inequality imposed by the uncertainty principle [Eq. (9.2.9)].

(4) Show that the uncertainty bound is saturated in the state j,±j|j, \pm j\rangle.

Exercise 12.5.4 (1) Argue that the eigenvalues of Jx(j)J_x^{(j)} and Jy(j)J_y^{(j)} are the same as those of Jz(j)J_z^{(j)}, namely, j,(j1),,(j)j \hbar,(j-1) \hbar, \ldots, (-j\hbar). Generalize the result to θ^J(j)\hat{\theta} \cdot \mathbf{J}^{(j)}.

(2) Show that

(Jj)[J(j1)][J(j2)](J+j)=0(J-j \hbar)[J-(j-1) \hbar][J-(j-2) \hbar] \cdots(J+j \hbar)=0

where Jθ^J(j)J \equiv \hat{\theta} \cdot \mathbf{J}^{(j)}. (Hint: In the case J=JzJ=J_z what happens when both sides are applied to an arbitrary eigenket jm|j m\rangle? What about an arbitrary superpositions of such kets?)

(3) It follows from (2) that J2j+1J^{2 j+1} is a linear combination of J0,J1,,J2jJ^0, J^1, \ldots, J^{2 j}. Argue that the same goes for J2j+kJ^{2 j+k}, k=1,2,k=1,2, \ldots.

Exercise 12.5.5 Using results from the previous exercise and Eq. (12.5.23), show that

(1) D(1/2)[R]=exp(iθ^J(1/2)/)=cos(θ/2)I(1/2)(2i/)sin(θ/2)θ^J(1/2)D^{(1 / 2)}[R]=\exp \left(-\mathrm{i} \hat{\theta} \cdot \mathbf{J}^{(1 / 2)} / \hbar\right)=\cos (\theta / 2) I^{(1 / 2)}-(2\mathrm{i} / \hbar) \sin (\theta / 2) \hat{\theta} \cdot \mathbf{J}^{(1 / 2)}

(2) D(1)[R]=exp(iθxJx(1)/)=(cosθx1)(Jx(1))2isinθx(Jx(1))+I(1)D^{(1)}[R]=\exp \left(-\mathrm{i} \theta_x J_x^{(1)} / \hbar\right)=\left(\cos \theta_x-1\right)\left(\frac{J_x^{(1)}}{\hbar}\right)^2-\mathrm{i} \sin \theta_x\left(\frac{J_x^{(1)}}{\hbar}\right)+I^{(1)}

Exercise 12.5.6 Consider the family of states jj,,jm,,j,j|j j\rangle, \ldots,|j m\rangle, \ldots,|j,-j\rangle. One refers to them as states of the same magnitude but different orientation of angular momentum. If ones takes this remark literally, i.e., in the classical sense, one is led to believe that one may rotate these into each other, as is the case for classical states with these properties. Consider, for instance, the family 1,1,1,0,1,1|1,1\rangle,|1,0\rangle,|1,-1\rangle. It may seem, for example, that the state with zero angular momentum along the zz axis, 1,0|1,0\rangle, may be obtained by rotating 1,1|1,1\rangle by some suitable (12π\frac{1}{2} \pi?) angle about the xx axis. Using D(1)[R(θxi)]D^{(1)}\left[R\left(\theta_x \mathbf{i}\right)\right] from part (2) in the last exercise show that

1,0D(1)[R(θxi)]1,1for anyθx|1,0\rangle \neq D^{(1)}\left[R\left(\theta_x \mathbf{i}\right)\right]|1,1\rangle \quad \text{for any} \theta_x

The error stems from the fact that classical reasoning should be applied to J\langle\mathbf{J}\rangle, which responds to rotations like an ordinary vector, and not directly to jm|j m\rangle, which is a vector in Hilbert space. Verify that J\langle\mathbf{J}\rangle responds to rotations like its classical counterpart, by showing that J\langle\mathbf{J}\rangle in the state D(1)[R(θxi)]1,1D^{(1)}\left[R\left(\theta_x \mathbf{i}\right)\right]|1,1\rangle is [sinθxj+cosθxk]\hbar\left[-\sin \theta_x \mathbf{j}+\cos \theta_x \mathbf{k}\right].

Exercise 12.5.7 (Euler Angles) Rather than parametrize an arbitrary rotation by the angle θ\boldsymbol{\theta}, which describes a single rotation by θ\theta about an axis parallel to θ\boldsymbol{\theta}, we may parametrize it by three angles, γ,β\gamma, \beta, and α\alpha called Euler angles, which define three successive rotations:

U[R(α,β,γ)]=eiαJz/eiβJy/eiγJz/U[R(\alpha, \beta, \gamma)]=\mathrm{e}^{-\mathrm{i} \alpha J_z / \hbar} \mathrm{e}^{-\mathrm{i} \beta J_y / \hbar} \mathrm{e}^{-\mathrm{i} \gamma J_z / \hbar}

(1) Construct D(1)[R(α,β,γ)]D^{(1)}[R(\alpha, \beta, \gamma)] explicitly as a product of three 3×33 \times 3 matrices. (Use the result from Exercise 12.5.5 with JxJyJ_x \rightarrow J_y.)

(2) Let it act on 1,1|1,1\rangle and show that J\langle\mathbf{J}\rangle in the resulting state is

J=(sinβcosαi+sinβsinαj+cosβk)\langle\mathbf{J}\rangle=\hbar(\sin \beta \cos \alpha \mathbf{i}+\sin \beta \sin \alpha \mathbf{j}+\cos \beta \mathbf{k})

(3) Show that for no value of α,β\alpha, \beta, and γ\gamma can one rotate 1,1|1,1\rangle into just 1,0|1,0\rangle.

(4) Show that one can always rotate any 1,m|1, m\rangle into a linear combination that involves 1,m\left|1, m^{\prime}\right\rangle, i.e.,

1,mD(1)[R(α,β,γ)]1,m0\left\langle 1, m^{\prime}\right| D^{(1)}[R(\alpha, \beta, \gamma)]|1, m\rangle \neq 0

for some α,β,γ\alpha, \beta, \gamma and any m,mm, m^{\prime}. \noindent (5) To see that one can occasionally rotate jm|j m\rangle into jm\left|j m^{\prime}\right\rangle, verify that a 180180^{\circ} rotation about the yy axis applied to 1,1|1,1\rangle turns it into 1,1|1,-1\rangle.

Exercise 12.5.8 Verify that

Lxcoordinatebasisi(sinϕθ+cosϕcotθϕ)Lycoordinatebasisi(cosϕθ+sinϕcotθϕ)\begin{aligned} & L_x \xrightarrow[\substack{\text{coordinate}\\ \text{basis}}]{} \mathrm{i} \hbar\left(\sin \phi \dfrac{\partial}{\partial \theta}+\cos \phi \cot \theta \dfrac{\partial}{\partial \phi}\right) \\ & L_y \xrightarrow[\substack{\text{coordinate}\\ \text{basis}}]{} \mathrm{i} \hbar\left(-\cos \phi \dfrac{\partial}{\partial \theta}+\sin \phi \cot \theta \dfrac{\partial}{\partial \phi}\right) \end{aligned}

Exercise 12.5.9 Show that L2L^2 above is Hermitian in the sense

ψ1(L2ψ2)dΩ=[ψ2(L2ψ1)dΩ]\int \psi_1^*\left(L^2 \psi_2\right) \mathrm{d} \Omega=\left[\int \psi_2^*\left(L^2 \psi_1\right) \mathrm{d} \Omega\right]^*

The same goes for LzL_z, which is insensitive to θ\theta and is Hermitian with respect to the ϕ\phi integration.

Exercise 12.5.10 Write the differential equation corresponding to

L2αβ=ααβL^2|\alpha \beta\rangle=\alpha|\alpha \beta\rangle

in the coordinate basis, using the L2L^2 operator given in Eq. (12.5.36). We already know β=m\beta=m \hbar from the analysis of i(/ϕ)-\mathrm{i} \hbar(\partial / \partial \phi). So assume that the simultaneous eigenfunctions have the form

ψαm(θ,ϕ)=Pαm(θ)eimϕ\psi_{\alpha m}(\theta, \phi)=P_\alpha^m(\theta) \mathrm{e}^{\mathrm{i}m \phi}

and show that PαmP_\alpha^m satisfies the equation

(1sinθθsinθθ+α2m2sin2θ)Pαm(θ)=0\left(\frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \sin \theta \frac{\partial}{\partial \theta}+\frac{\alpha}{\hbar^2}-\frac{m^2}{\sin ^2 \theta}\right) P_\alpha^m(\theta)=0

We need to show that

(1) α2=l(l+1)\dfrac{\alpha}{\hbar^2}=l(l+1), l=0,1,2,l=0,1,2,\ldots

(2) ml|m| \leqslant l

We will consider only part (1) and that too for the case m=0m=0. By rewriting the equation in terms of u=cosθu=\cos \theta, show that Pα0P_\alpha^0 satisfies

(1u2)d2Pα0du22udPa0du+(α2)Pα0=0\left(1-u^2\right) \frac{\mathrm{d}^2 P_\alpha^0}{\mathrm{d} u^2}-2 u \frac{\mathrm{d} P_a^0}{\mathrm{d} u}+\left(\frac{\alpha}{\hbar^2}\right) P_\alpha^0=0

Convince yourself that a power series solution

Pα0=n=0CnunP_\alpha^0=\sum_{n=0}^{\infty} C_n u^n

will lead to a two-term recursion relation. Show that (Cn+2/Cn)1\left(C_{n+2} / C_n\right) \rightarrow 1 as nn \rightarrow \infty. Thus the series diverges when u1|u| \rightarrow 1 (θ0\theta \rightarrow 0 or π\pi). Show that if α/2=(l)(l+1)\alpha / \hbar^2=(l)(l+1); l=0,1,2,l=0,1,2,\ldots, the series will terminate and be either an even or odd function of uu. The functions Pα0(u)=Pl(l+1)20(u)Pl0(u)Pl(u)P_\alpha^0(u)=P_{l(l+1) \hbar^2}^0(u) \equiv P_l^0(u) \equiv P_l(u) are just the Legendre polynomials up to a scale factor. Determine P0P_0, P1P_1, and P2P_2 and compare (ignoring overall scales) with the Yl0Y_{l}^{0} functions.

Exercise 12.5.11 Derive Y11Y_1^1 starting from Eq.(12.5.28) and normalize it yourself. [Remember the (1)l(-1)^l factor from Eq. (12.5.32).] Lower it to get Y10Y_1^0 and Y11Y_1^{-1} and compare it with Eq. (12.5.39).

Exercise 12.5.12 Since L2L^2 and LzL_z commute with Π\Pi, they should share a basis with it. Verify that under parity Ylm(1)lYlmY_l^m \rightarrow (-1)^l Y_l^m. (First show that θπθ\theta \rightarrow \pi -\theta, ϕϕ+π\phi \rightarrow \phi+\pi under parity. Prove the result for YllY_l^l. Verify that LL_{-}does not alter the parity, thereby proving the result for all YlmY_l^m.)

Exercise 12.5.13 Consider a particle in a state described by

ψ=N(x+y+2z)eαr\psi=N(x+y+2 z) \mathrm{e}^{-\alpha r}

where NN is a normalization factor.

(1) Show, by rewriting the Y1±1.0Y_1^{ \pm 1.0} functions in terms of x,y,zx, y, z, and rr, that

Y1±1=(34π)1/2x±iy21/2rY10=(34π)1/2zr(12.5.42)\begin{aligned} Y_1^{ \pm 1} & =\mp\left(\frac{3}{4 \pi}\right)^{1 / 2} \frac{x \pm \mathrm{i} y}{2^{1 / 2} r} \\ Y_1^0 & =\left(\frac{3}{4 \pi}\right)^{1 / 2} \frac{z}{r} \end{aligned}\tag{12.5.42}

(2) Using this result, show that for a particle described by ψ\psi above, P(lz=0)=2/3P\left(l_z=0\right)=2 / 3; P(lz=+)=1/6=P(lz=)P\left(l_z=+\hbar\right)=1 / 6=P\left(l_z=-\hbar\right).

Exercise 12.5.14 Consider a rotation θxi\theta_x \mathbf{i}. Under this

xxyycosθxzsinθxzzcosθz+ysinθx\begin{aligned} & x \rightarrow x \\ & y \rightarrow y \cos \theta_x-z \sin \theta_x \\ & z \rightarrow z \cos \theta_z+y \sin \theta_x \end{aligned}

Therefore we must have

ψ(x,y,z)U[R(θxi)]ψR=ψ(x,ycosθxzsinθx,zcosθx+ysinθx)\psi(x, y, z) \xrightarrow[{U\left[R\left(\theta_x \mathbf{i}\right)\right]}]{} \psi_R=\psi\left(x, y \cos \theta_x-z \sin \theta_x, z \cos \theta_x+y \sin \theta_x\right)

Let us verify this prediction for a special case

ψ=Azer2/a2\psi=A z \mathrm{e}^{-r^2 / a^2}

which must go into

ψR=A(zcosθxysinθx)er2/a2\psi_R=A\left(z \cos \theta_x-y \sin \theta_x\right) \mathrm{e}^{-r^2 / a^2}

(1) Expand ψ\psi in terms of Y11,Y10,Y11Y_1^1, Y_1^0, Y_1^{-1}.

\noindent (2) Use the matrix eiθxLx/\mathrm{e}^{-\mathrm{i} \theta_x L_x / \hbar} to find the fate of ψ\psi under this rotation.\footnote{Use Exercise 12.5.15.} Check your result against that anticipated above. [Hint: (1) ψY10\psi \sim Y_1^0, which corresponds to

(010)\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

(2) Use Eq. (12.5.42).]

12.6 Solution of Rotationally Invariant Problems

Exercise 12.6.1 A particle is described by the wave function

ψE(r,θ,ϕ)=Aera0(a0=const)\psi_E(r, \theta, \phi)=A e^{-r a_0} \quad\left(a_0=\text{const}\right)

(1) What is the angular momentum content of the state?

(2) Assuming ψE\psi_E is an eigenstate in a potential that vanishes as rr \rightarrow \infty, find EE. (Match leading terms in Schrödinger's equation.)

(3) Having found EE, consider finite rr and find V(r)V(r).

Exercise 12.6.2 Provide the steps connecting Eq. (12.6.3) and Eq. (12.6.5).

Exercise 12.6.3 Show that Eq. (12.6.7b) follows from Eq. (12.6.7a).

Exercise 12.6.4 (1) Show that

δ3(rr)δ(xx)δ(yy)δ(zz)=1r2sinθδ(rr)δ(θθ)δ(ϕϕ)\delta^3\left(\mathbf{r}-\mathbf{r}^{\prime}\right) \equiv \delta\left(x-x^{\prime}\right) \delta\left(y-y^{\prime}\right) \delta\left(z-z^{\prime}\right)=\frac{1}{r^2 \sin \theta} \delta\left(r-r^{\prime}\right) \delta\left(\theta-\theta^{\prime}\right) \delta\left(\phi-\phi^{\prime}\right)

(consider a test function).

(2) Show that

2(1/r)=4πδ3(r)\nabla^2(1 / r)=-4 \pi \delta^3(\mathbf{r})

[Hint: First show that 2(1/r)=0\nabla^2(1 / r)=0 if r0r \neq 0. To see what happens at r=0r=0, consider a small sphere centered at the origin and use Gauss's law and the identity 2ϕ=ϕ\nabla^2 \phi=\nabla \cdot \nabla \phi]. (Or compare this equation to Poisson's equation in electrostatics 2ϕ=4πρ\nabla^2 \phi=-4 \pi \rho. Here ρ=δ3(r)\rho=\delta^3(\mathbf{r}), which represents a unit point charge at the origin. In this case we know from Coulomb's law that ϕ=1/r\phi=1 / r.)

Exercise 12.6.5 Show that DlD_l is nondegenerate in the space of functions UU that vanish as r0r \rightarrow 0. (Recall the proof of Theorem 15, Section 5.6.) Note that UElU_{El} is nondegenerate even for E>0E>0. This means that EE, ll, and mm, label a state fully in three dimensions.

Exercise 12.6.6 (1) Verify that Eqs. (12.6.21) and (12.6.22) are equivalent to Eq. (12.6.20).

(2) Verify Eq. (12.6.24).

Exercise 12.6.7 Verify that j0j_{0} and j1j_{1} have the limits given by Eq. (12.6.33).

Exercise 12.6.8 Find the energy levels of a particle in a spherical box of radius r0r_{0} in the l=0l=0 sector.

Exercise 12.6.9 Show that the quantization condition for l=0l=0 bound states in a spherical well of depth V0-V_0 and radius r0r_0 is

k/κ=tankr0k^{\prime} / \kappa=-\tan k^{\prime} r_0

where kk^{\prime} is the wave number inside the well and iκ\mathrm{i}\kappa is the complex wave number for the exponential tail outside. Show that there are no bound states for V0<π22/8μr02V_0<\pi^2 \hbar^2 / 8 \mu r_0^2. (Recall Exercise 5.2.6.)

Exercise 12.6.10 Verify Eq. (12.6.41) given that

(1) 11Pl(cosθ)Pl(cosθ)d(cosθ)=[2/(2l+1)]δll\displaystyle\int_{-1}^1 P_l(\cos \theta) P_{l^{\prime}}(\cos \theta) \mathrm{d}(\cos \theta)=[2 /(2 l+1)] \delta_{l l^{\prime}}.

(2) Pl(x)=12ll!dl(x21)dxlP_l(x)=\dfrac{1}{2^{l} l!} \dfrac{\mathrm{d}^l\left(x^2-1\right)^{\prime}}{\mathrm{d} x^l}

(3) 01(1x2)mdx=(2m)!!(2m+1)!!\displaystyle\int_0^1\left(1-x^2\right)^m \mathrm{d} x=\frac{(2 m)!!}{(2 m+1)!!}

[Hint: Consider the limit kr0k r \rightarrow 0 after projecting out ClC_l.]

Exercise 12.6.11 (1) By combining Eqs. (12.6.48) and (12.6.49) derive the two-term recursion relation. Argue that C00C_0 \neq 0 if UU is to have the right properties near y=0y=0. Derive the quantizations condition, Eq. (12.6.50).

(2) Calculate the degeneracy and parity at each nn and compare with Exercise 10.2.3, where the problem was solved in Cartesian coordinates.

(3) Construct the normalized eigenfunction ψnlm\psi_{nlm} for n=0n=0 and 11. Write them as linear combinations of the n=0n=0 and n=1n=1 eigenfunctions obtained in Cartesian coordinates.