# Nondegenerate states in 3-d spherically symmetric systems

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.5.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

In solving the Schrödinger equation for spherically symmetric potentials, we found that we could reduce the problem to the equation

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (1)$

where ${U_{El}\left(r\right)}$ is related to the radial function by

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (2)$

We can write 1 as an eigenvalue equation for the operator ${D_{l}}$ in the form

$\displaystyle D_{l}\left(r\right)U_{El}=EU_{El} \ \ \ \ \ (3)$

with

$\displaystyle D_{l}\left(r\right)\equiv-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}} \ \ \ \ \ (4)$

We can show that, provided ${U_{El}\left(r\right)\rightarrow0}$ as ${r\rightarrow0}$, there are no degenerate eigenstates (that is, any state ${U_{El}}$ that is an eigenstate with energy ${E}$ is unique up to a scaling factor). The proof is similar to that in 1-d quantum mechanics, and goes by contradiction.

We suppose that there are two different functions ${U_{1}}$ and ${U_{2}}$ that satisfy 1 for the same energy ${E}$ (and same angular momentum number ${l}$). We then have

 $\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{1}$ $\displaystyle =$ $\displaystyle EU_{1}\ \ \ \ \ (5)$ $\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{2}$ $\displaystyle =$ $\displaystyle EU_{2} \ \ \ \ \ (6)$

Multiply the first by ${U_{2}}$ and the second by ${U_{1}}$ and subtract to get

$\displaystyle U_{2}U_{1}^{\prime\prime}-U_{1}U_{2}^{\prime\prime}=0 \ \ \ \ \ (7)$

This expression is

$\displaystyle U_{2}U_{1}^{\prime\prime}-U_{1}U_{2}^{\prime\prime}=\frac{d}{dr}\left(U_{2}U_{1}^{\prime}-U_{1}U_{2}^{\prime}\right)=0 \ \ \ \ \ (8)$

which we can integrate to get

$\displaystyle U_{2}U_{1}^{\prime}-U_{1}U_{2}^{\prime}=C \ \ \ \ \ (9)$

for some constant ${C}$. This relation is valid for all ${r}$, so we can choose ${r=0}$ where ${U_{2}\left(0\right)=U_{1}\left(0\right)=0}$, which shows that ${C=0}$. Therefore

$\displaystyle \frac{U_{1}^{\prime}}{U_{1}}=\frac{U_{2}^{\prime}}{U_{2}} \ \ \ \ \ (10)$

Integrating gives us

$\displaystyle \ln U_{1}=\ln U_{2}+K \ \ \ \ \ (11)$

for some other constant ${K}$, so

$\displaystyle U_{1}=e^{K}U_{2} \ \ \ \ \ (12)$

That is, any two eigenfunctions with the same eigenvalue ${E}$ are multiples of each other, so represent the same state, which is nondegenerate.

Note that the derivation didn’t rely on the value of ${U}$ anywhere except at ${r=0}$, so there is no requirement that, for example, ${U\rightarrow0}$ as ${r\rightarrow\infty}$. Also, the derivation is valid whatever the sign of ${E}$.

# Spherically symmetric potentials: hermiticity of the radial function

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.3.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

The Schrödinger equation in 3-d for a potential that depends only on ${r}$ is

$\displaystyle -\frac{\hbar^{2}}{2\mu}\left[\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{r^{2}\sin^{2}\theta}\left(\frac{\partial^{2}\psi}{\partial\phi^{2}}\right)\right]+V\psi=E\psi \ \ \ \ \ (1)$

Eigenfunctions in this equation satisfy

$\displaystyle \psi=R_{Elm}\left(r\right)Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (2)$

where the subscript ${Elm}$ refers to the energy ${E}$ and the angular momentum quantum numbers ${l}$ and ${m}$. ${Y_{l}^{m}}$ is a spherical harmonic and ${R_{Elm}}$ is the radial function which depends on the potential ${V}$. With the substitution

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (3)$

the differential equation reduces to

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (4)$

The quantity in the square brackets is an operator which will call ${D_{l}\left(r\right)}$:

$\displaystyle D_{l}\left(r\right)\equiv-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}} \ \ \ \ \ (5)$

Equation 4 is similar to the 1-d Schrödinger equation except that the variable ${r}$ goes from 0 to ${\infty}$ rather than from ${-\infty}$ to ${\infty}$, and the potential is modified by the ‘centrifugal term’ ${\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}}$. Because ${r}$ begins at 0 rather than ${-\infty}$, the usual boundary conditions on ${U}$ (that it tend to zero at ${\pm\infty}$) must also be modified. We can get the new boundary conditions by imposing the hermiticity condition, which says that

 $\displaystyle \int_{0}^{\infty}U_{1}^*\left(D_{l}U_{2}\right)dr$ $\displaystyle =$ $\displaystyle \left[\int_{0}^{\infty}U_{2}^*\left(D_{l}U_{1}\right)dr\right]^*\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}\left(D_{l}U_{1}\right)^*U_{2}dr \ \ \ \ \ (7)$

The two terms ${V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}}$ in 5 are real and multiplicative, so the hermiticity condition is automatically satisfied for them. For the derivative term, we can use the usual integration by parts.

 $\displaystyle \int_{0}^{\infty}U_{1}^*\left(\frac{d^{2}}{dr^{2}}U_{2}\right)dr$ $\displaystyle =$ $\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}^{\infty}-\int_{0}^{\infty}\frac{dU_{1}^*}{dr}\frac{dU_{2}}{dr}dr\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}^{\infty}-\left.U_{2}\frac{dU_{1}^*}{dr}\right|_{0}^{\infty}+\int_{0}^{\infty}U_{2}\left(\frac{d^{2}}{dr^{2}}U_{1}^*\right)dr \ \ \ \ \ (9)$

If we require

$\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}^{\infty}-\left.U_{2}\frac{dU_{1}^*}{dr}\right|_{0}^{\infty}=0 \ \ \ \ \ (10)$

then we have

 $\displaystyle \int_{0}^{\infty}U_{1}^*\left(\frac{d^{2}}{dr^{2}}U_{2}\right)dr$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}U_{2}\left(\frac{d^{2}}{dr^{2}}U_{1}^*\right)dr\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\int_{0}^{\infty}U_{2}^*\left(\frac{d^{2}}{dr^{2}}U_{1}\right)dr\right]^* \ \ \ \ \ (12)$

and the hermiticity condition 6 is satisfied.

# Spherically symmetric potentials – a simple example

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.1.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

The Schrödinger equation in 3-d for a potential that depends only on ${r}$ is

$\displaystyle -\frac{\hbar^{2}}{2\mu}\left[\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{r^{2}\sin^{2}\theta}\left(\frac{\partial^{2}\psi}{\partial\phi^{2}}\right)\right]+V\psi=E\psi \ \ \ \ \ (1)$

The angular part of the operator on the LHS is essentially the angular momentum operator ${L^{2}}$ (times ${1/2\mu r^{2}}$):

$\displaystyle L^{2}=-\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right] \ \ \ \ \ (2)$

, so we can write this as

$\displaystyle -\frac{\hbar^{2}}{2\mu}\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+V\psi+\frac{L^{2}}{2\mu r^{2}}\psi=E\psi \ \ \ \ \ (3)$

Eigenfunctions in this equation satisfy

$\displaystyle \psi=R_{Elm}\left(r\right)Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (4)$

where the subscript ${Elm}$ refers to the energy ${E}$ and the angular momentum quantum numbers ${l}$ and ${m}$. ${Y_{l}^{m}}$ is a spherical harmonic and ${R_{Elm}}$ is the radial function which depends on the potential ${V}$. The eigenvalues of ${L^{2}}$ are ${l\left(l+1\right)\hbar^{2}}$ so 3 becomes

$\displaystyle -\frac{\hbar^{2}}{2\mu}\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial R_{El}}{\partial r}\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}R_{El}+VR_{El}=ER_{El} \ \ \ \ \ (5)$

We’ve dropped the ${m}$ from ${R_{Elm}}$ since, for a spherically symmetric potential, the radial function is independent of ${m}$.

Example Suppose a particle is described by the wave function

$\displaystyle \psi_{E}\left(r,\theta,\phi\right)=Ae^{-r/a_{0}} \ \ \ \ \ (6)$

where ${A}$ and ${a_{0}}$ are constants. What can we deduce about the system?

First, since ${\psi_{E}}$ is independent of ${\theta}$ and ${\phi}$ we see from 2 that

$\displaystyle L^{2}\psi_{E}=0 \ \ \ \ \ (7)$

so the eigenvalue is ${l=0}$ and the state has no angular momentum. From 3 we therefore have

$\displaystyle -\frac{\hbar^{2}}{2\mu}\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+V\psi=E\psi \ \ \ \ \ (8)$

Working out the derivatives, we have

 $\displaystyle \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)$ $\displaystyle =$ $\displaystyle -\frac{A}{r^{2}}\frac{d}{dr}\left(\frac{r^{2}}{a_{0}}e^{-r/a_{0}}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{-r/a_{0}}\left(-\frac{2}{ra_{0}}+\frac{1}{a_{0}^{2}}\right) \ \ \ \ \ (10)$

Plugging this back into 8 and cancelling terms gives

$\displaystyle -\frac{2}{ra_{0}}+\frac{1}{a_{0}^{2}}=\frac{2\mu}{\hbar^{2}}\left(V-E\right) \ \ \ \ \ (11)$

If ${V\left(r\right)\rightarrow0}$ as ${r\rightarrow\infty}$ we have, in this limit

$\displaystyle E=-\frac{\hbar^{2}}{2\mu a_{0}^{2}} \ \ \ \ \ (12)$

The energy is constant at all values of ${r}$ so we can now find ${V}$ from 11

 $\displaystyle -\frac{2}{ra_{0}}+\frac{1}{a_{0}^{2}}$ $\displaystyle =$ $\displaystyle \frac{2\mu}{\hbar^{2}}\left(V\left(r\right)+\frac{\hbar^{2}}{2\mu a_{0}^{2}}\right)\ \ \ \ \ (13)$ $\displaystyle V\left(r\right)$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{\mu a_{0}r} \ \ \ \ \ (14)$

# Linear combinations of spherical harmonics; probabilities

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.13.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

If we can express a 3-d quantum state in terms of the spherical harmonics, we can calculate directly the probabilities of ${L_{z}}$ having one of its eigenvalues. That is, if we can write a state ${\psi}$ as

$\displaystyle \psi\left(r,\theta,\phi\right)=f\left(r\right)\sum_{m}C_{l}^{m}Y_{l}^{m} \ \ \ \ \ (1)$

for some constant coefficients ${C_{l}^{m}}$ and ${f}$ is some function of ${r}$ alone, then

$\displaystyle P\left(l_{z}=m\hbar\right)=\frac{\left|C_{l}^{m}\right|^{2}}{\sum_{n}\left|C_{l}^{n}\right|^{2}} \ \ \ \ \ (2)$

As an example, suppose we have

$\displaystyle \psi=N\left(x+y+2z\right)e^{-\alpha r} \ \ \ \ \ (3)$

where ${N}$ is a normalization constant. We start by expressing ${x}$, ${y}$ and ${z}$ in terms of ${Y_{1}^{m}}$. We have

 $\displaystyle Y_{1}^{\pm1}$ $\displaystyle =$ $\displaystyle \mp\sqrt{\frac{3}{8\pi}}\sin\theta e^{\pm i\phi}\ \ \ \ \ (4)$ $\displaystyle Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{4\pi}}\cos\theta \ \ \ \ \ (5)$

Using standard spherical-to-rectangular conversions

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\sin\theta\cos\phi\ \ \ \ \ (6)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta\sin\phi\ \ \ \ \ (7)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\cos\theta \ \ \ \ \ (8)$

Therefore

 $\displaystyle \cos\phi$ $\displaystyle =$ $\displaystyle \frac{x}{r\sin\theta}\ \ \ \ \ (9)$ $\displaystyle \sin\phi$ $\displaystyle =$ $\displaystyle \frac{y}{r\sin\theta}\ \ \ \ \ (10)$ $\displaystyle \cos\theta$ $\displaystyle =$ $\displaystyle \frac{z}{r} \ \ \ \ \ (11)$

Plugging these into 4 we have

 $\displaystyle Y_{1}^{\pm1}$ $\displaystyle =$ $\displaystyle \mp\sqrt{\frac{3}{8\pi}}\sin\theta\left(\cos\phi\pm i\sin\phi\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mp\sqrt{\frac{3}{8\pi}}\frac{x\pm iy}{r}\ \ \ \ \ (13)$ $\displaystyle Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{4\pi}}\frac{z}{r}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2}\sqrt{\frac{3}{8\pi}}\frac{z}{r} \ \ \ \ \ (15)$

Inverting these, we have

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{-1}-Y_{1}^{1}\right)\ \ \ \ \ (16)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle -\frac{1}{2i}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{-1}+Y_{1}^{1}\right)\ \ \ \ \ (17)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\sqrt{\frac{8\pi}{3}}rY_{1}^{0} \ \ \ \ \ (18)$

Thus 3 becomes

$\displaystyle \psi=\sqrt{\frac{8\pi}{3}}Nre^{-\alpha r}\left[Y_{1}^{1}\left(-\frac{1}{2}-\frac{1}{2i}\right)+Y_{1}^{-1}\left(\frac{1}{2}-\frac{1}{2i}\right)+Y_{1}^{0}\sqrt{2}\right] \ \ \ \ \ (19)$

Comparing with 1 we find

 $\displaystyle C_{1}^{1}$ $\displaystyle =$ $\displaystyle -\frac{1}{2}-\frac{1}{2i}\ \ \ \ \ (20)$ $\displaystyle C_{1}^{-1}$ $\displaystyle =$ $\displaystyle \frac{1}{2}-\frac{1}{2i}\ \ \ \ \ (21)$ $\displaystyle C_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{2} \ \ \ \ \ (22)$

We have

 $\displaystyle \sum_{n}\left|C_{1}^{n}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{2}+\frac{1}{2}+2=3\ \ \ \ \ (23)$ $\displaystyle P\left(l_{z}=0\right)$ $\displaystyle =$ $\displaystyle \frac{\left|C_{l}^{0}\right|^{2}}{\sum_{n}\left|C_{l}^{n}\right|^{2}}=\frac{2}{3}\ \ \ \ \ (24)$ $\displaystyle P\left(l_{z}=\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|C_{l}^{1}\right|^{2}}{\sum_{n}\left|C_{l}^{n}\right|^{2}}=\frac{1}{6}\ \ \ \ \ (25)$ $\displaystyle P\left(l_{z}=-\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|C_{l}^{-1}\right|^{2}}{\sum_{n}\left|C_{l}^{n}\right|^{2}}=\frac{1}{6} \ \ \ \ \ (26)$

# Angular momentum and parity

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.12.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

The parity operator in 3-d reflects every point directly through the origin, so that a position vector ${\mathbf{r}\rightarrow-\mathbf{r}}$. In rectangular coordinates this means replacing each coordinate by its negative. In spherical coordinates, the angular coordinates change according to

 $\displaystyle \theta$ $\displaystyle \rightarrow$ $\displaystyle \pi-\theta\ \ \ \ \ (1)$ $\displaystyle \phi$ $\displaystyle \rightarrow$ $\displaystyle \pi+\phi \ \ \ \ \ (2)$

If this isn’t obvious, picture reflecting a vector ${\mathbf{r}}$ through the origin. If the original vector makes an angle ${\theta}$ with the ${z}$ (vertical) axis, then the reflected vector makes an angle ${\theta}$ with the ${-z}$ axis, which is equivalent to an angle of ${\pi-\theta}$ with the ${+z}$ axis. The azimuthal angle ${\phi}$ just gets rotated by ${\pi}$ to lie on the other side of the ${z}$ axis.

Using this, we can see that the parity operator ${\Pi}$ commutes with both ${L^{2}}$ and ${L_{z}}$, as follows. Since neither of these operators involves the radial coordinate, we can consider their effect on a function ${f\left(\theta,\phi\right)}$. Under parity, we have

$\displaystyle \Pi f\left(\theta,\phi\right)\rightarrow f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (3)$

Thus the derivatives transform under parity according to

 $\displaystyle \frac{\partial f\left(\theta,\phi\right)}{\partial\theta}$ $\displaystyle \rightarrow$ $\displaystyle -\frac{\partial f\left(\pi-\theta,\pi+\phi\right)}{\partial\theta}\ \ \ \ \ (4)$ $\displaystyle \frac{\partial f\left(\theta,\phi\right)}{\partial\phi}$ $\displaystyle \rightarrow$ $\displaystyle \frac{\partial f\left(\pi-\theta,\pi+\phi\right)}{\partial\phi} \ \ \ \ \ (5)$
 $\displaystyle L^{2}$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]\ \ \ \ \ (6)$ $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (7)$

Thus the combined operation gives

 $\displaystyle L^{2}\Pi f\left(\theta,\phi\right)$ $\displaystyle \rightarrow$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\left(-\frac{\partial}{\partial\theta}\right)\left(\sin\theta\left(-\frac{\partial}{\partial\theta}\right)\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L^{2}f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (11)$

If we apply ${\Pi}$ to ${L^{2}}$, we have

 $\displaystyle \Pi\left[L^{2}f\left(\theta,\phi\right)\right]$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\left(\pi-\theta\right)}\left(-\frac{\partial}{\partial\theta}\right)\left(\sin\left(\pi-\theta\right)\left(-\frac{\partial}{\partial\theta}\right)\right)+\frac{1}{\sin^{2}\left(\pi-\theta\right)}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L^{2}f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (14)$

Thus

$\displaystyle \left[\Pi,L^{2}\right]=0 \ \ \ \ \ (15)$

where in the first line we used ${\sin\left(\pi-\theta\right)=\sin\theta}$.

Since ${L_{z}}$ involves only a derivative with respect to ${\phi}$ which doesn’t change under parity, we have

$\displaystyle \left[\Pi,L_{z}\right]=0 \ \ \ \ \ (16)$

Since ${\Pi}$ commutes with both ${L^{2}}$ and ${L_{z}}$ it is possible to find a set of functions that are simultaneous eigenfunctions of all three operators. These functions turn out to be the same spherical harmonics that we’ve been using all along. We can show this by starting with the top spherical harmonic

$\displaystyle Y_{l}^{l}=\left(-1\right)^{l}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\phi}\sin^{l}\theta \ \ \ \ \ (17)$

where we’ve included the ${\left(-1\right)^{l}}$ to be consistent with Shankar’s equation 12.5.32. Under parity, this transforms as

 $\displaystyle \Pi Y_{l}^{l}$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\left(\pi+\phi\right)}\sin^{l}\left(\pi-\theta\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}e^{il\pi}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\phi}\sin^{l}\theta\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}Y_{l}^{l} \ \ \ \ \ (20)$

where we used ${e^{il\pi}=\left(-1\right)^{l}}$ in the second line. Thus ${Y_{l}^{l}}$ is an eigenfunction of ${\Pi}$ with eigenvalue ${\left(-1\right)^{l}}$.

To show that the other spherical harmonics are also eigenfunctions, we can use the lowering operator ${L_{-}}$. In spherical coordinates, we have

$\displaystyle L_{-}Y_{l}^{m}=\hbar\sqrt{(\ell+m)(\ell-m+1)}Y_{l}^{m-1} \ \ \ \ \ (21)$

The operator can be expressed as

$\displaystyle L_{-}=-\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right] \ \ \ \ \ (22)$

Under parity, we can transform 22 using ${\sin\left(\pi-\theta\right)=\sin\theta}$ and ${\cos\left(\pi-\theta\right)=-\cos\theta}$, so that ${\cot\left(\pi-\theta\right)=-\cot\theta}$. We therefore have

 $\displaystyle \Pi L_{-}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\left(\pi+\phi\right)}\left[-\frac{\partial}{\partial\theta}+i\cot\theta\frac{\partial}{\partial\phi}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L_{-} \ \ \ \ \ (25)$

Thus ${L_{-}}$ is unchanged by parity, which means that from 21, ${Y_{l}^{m-1}}$ has the same parity as ${Y_{l}^{m}}$. Starting with ${Y_{l}^{l}}$ and using the lowering operator successively to reduce the superscript index, we have therefore

$\displaystyle \Pi Y_{l}^{m}=\left(-1\right)^{l}Y_{l}^{m} \ \ \ \ \ (26)$

Thus all spherical harmonics are also eigenfunctions of parity.

# Spherical harmonics from power series – examples for m=0

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.10.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

The total angular momentum operator ${L^{2}}$ can be written in spherical coordinates as

$\displaystyle L^{2}=-\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right] \ \ \ \ \ (1)$

Since ${\left[L^{2},L_{z}\right]=0}$, we can find a basis consisting of simultaneous eigenfunctions of ${L^{2}}$ and ${L_{z}}$. Suppose we call these states ${\left|\alpha\beta\right\rangle }$, where ${\alpha}$ is the eigenvalue of ${L^{2}}$ and ${\beta}$ is the eigenvalue of ${L_{z}}$. In spherical coordinates, we know that

$\displaystyle L_{z}=-i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (2)$

and that its eigenvalues are ${m\hbar}$ for integer values of ${m}$. Thus we can separate the ${\theta}$ and ${\phi}$ dependence in the eigenstates and write

$\displaystyle \psi_{\alpha m}\left(\theta,\phi\right)=P_{\alpha}^{m}\left(\theta\right)e^{im\phi} \ \ \ \ \ (3)$

We therefore have the eigenvalue equation

 $\displaystyle L^{2}\left|\alpha m\right\rangle$ $\displaystyle =$ $\displaystyle \alpha\left|\alpha m\right\rangle \ \ \ \ \ (4)$ $\displaystyle L^{2}\psi_{\alpha m}\left(\theta,\phi\right)$ $\displaystyle =$ $\displaystyle \alpha\psi_{\alpha m}\left(\theta,\phi\right) \ \ \ \ \ (5)$

Combining 3 with 1, we have

 $\displaystyle \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi_{\alpha m}}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}\psi_{\alpha m}}{\partial\phi^{2}}$ $\displaystyle =$ $\displaystyle \alpha\psi_{\alpha m}\ \ \ \ \ (6)$ $\displaystyle \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial P_{\alpha}^{m}}{\partial\theta}\right)-\frac{m^{2}}{\sin^{2}\theta}P_{\alpha}^{m}+\frac{\alpha}{\hbar^{2}}P_{\alpha}^{m}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

We’d like to show that solutions of this equation require that (1)

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \hbar^{2}\ell\left(\ell+1\right)\ \ \ \ \ (8)$ $\displaystyle \left|m\right|$ $\displaystyle \le$ $\displaystyle \ell \ \ \ \ \ (9)$

for ${\ell=0,1,2,\ldots}$. In the problem given in Shankar, we tackle the less demanding case of ${m=0}$ and demonstrate only the result for ${\alpha}$. We begin by transforming 7 using the variable substitution:

$\displaystyle u\equiv\cos\theta \ \ \ \ \ (10)$

This gives us

$\displaystyle du=-\sin\theta\;d\theta \ \ \ \ \ (11)$

so that 7 becomes

 $\displaystyle \frac{-\sin\theta}{\sin\theta}\frac{\partial}{\partial u}\left(-\sin^{2}\theta\frac{\partial P_{\alpha}^{0}}{\partial u}\right)+\frac{\alpha}{\hbar^{2}}P_{\alpha}^{0}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (12)$ $\displaystyle \frac{\partial}{\partial u}\left(\left(1-u^{2}\right)\frac{\partial P_{\alpha}^{0}}{\partial u}\right)+\frac{\alpha}{\hbar^{2}}P_{\alpha}^{0}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \left(1-u^{2}\right)\frac{\partial^{2}P_{\alpha}^{0}}{\partial u^{2}}-2u\frac{\partial P_{\alpha}^{0}}{\partial u}+\frac{\alpha}{\hbar^{2}}P_{\alpha}^{0}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

We can use a power series to solve this by defining

 $\displaystyle P_{\alpha}^{0}\left(u\right)$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}C_{n}u^{n}\ \ \ \ \ (15)$ $\displaystyle \frac{\partial P_{\alpha}^{0}}{\partial u}$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}C_{n}nu^{n-1}\ \ \ \ \ (16)$ $\displaystyle \frac{\partial^{2}P_{\alpha}^{0}}{\partial u^{2}}$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}C_{n}n\left(n-1\right)u^{n-2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}C_{n+2}\left(n+2\right)\left(n+1\right)u^{n} \ \ \ \ \ (18)$

Plugging these into 14 and collecting terms, we get

$\displaystyle P_{\alpha}^{0}\left(u\right)=\sum_{n=0}^{\infty}\left[C_{n+2}\left(n+2\right)\left(n+1\right)+C_{n}\left(-n\left(n-1\right)-2n+\frac{\alpha}{\hbar^{2}}\right)\right]u^{n}=0 \ \ \ \ \ (19)$

If a power series equals zero, the coefficient of each power of ${u}$ must be zero (power series theorem from math), so we get the recurrence relation

 $\displaystyle C_{n+2}$ $\displaystyle =$ $\displaystyle \frac{n\left(n-1\right)+2n-\frac{\alpha}{\hbar^{2}}}{\left(n+2\right)\left(n+1\right)}C_{n}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{n^{2}+n-\frac{\alpha}{\hbar^{2}}}{n^{2}+3n+2}C_{n} \ \ \ \ \ (21)$

For large ${n}$ we have

$\displaystyle C_{n+2}\rightarrow\frac{n^{2}}{n^{2}}C_{n}=C_{n} \ \ \ \ \ (22)$

Since ${u=\cos\theta}$, ${u\in\left[-1,1\right]}$ and the series must converge for all these values. Although the power series ${\sum_{n=0}^{\infty}u^{n}}$ converges if ${\left|u\right|<1}$ (that’s the standard geometric series), it clearly diverges if ${u=1}$. Thus we require the series to terminate, which imposes a condition on ${\alpha}$:

$\displaystyle \alpha=\ell\left(\ell+1\right)\hbar^{2} \ \ \ \ \ (23)$

for some integer value ${\ell=0,1,2,\ldots}$. Since choosing a value for ${\ell}$ can be done only once in any given series, and the recursion relation relates every second coefficient, this implies that either all even coefficients or all odd coefficients must be zero. Thus ${P_{\alpha}^{0}\left(u\right)}$ is either a sum of even powers (making it an even function) or of odd powers (making it an odd function) only.

The first few values of ${P_{\alpha}^{0}\left(u\right)}$ are found by choosing values for ${C_{0}}$ and ${C_{1}}$ and then generating all higher coefficients using 21. If we take

 $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (24)$ $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (25)$

then if we choose ${\ell=0}$ we get

$\displaystyle P_{0}^{0}=1 \ \ \ \ \ (26)$

Taking

 $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (27)$ $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (28)$

and ${\ell=1}$ gives

$\displaystyle P_{1}^{0}=u=\cos\theta \ \ \ \ \ (29)$

Reverting to an even series and taking ${\ell=2}$ we have from 21

 $\displaystyle C_{2}$ $\displaystyle =$ $\displaystyle -\frac{\alpha}{2\hbar^{2}}C_{0}=-\frac{\ell\left(\ell+1\right)}{2}\left(1\right)=-3\ \ \ \ \ (30)$ $\displaystyle P_{2}^{0}$ $\displaystyle =$ $\displaystyle 1-3u^{2}=1-3\cos^{2}\theta \ \ \ \ \ (31)$

These values for ${P_{\ell}^{0}}$ agree with the spherical harmonics ${Y_{\ell}^{0}}$ apart from the constant scaling factors in each case. See Shankar’s equation 12.5.39 for comparison.

# Total angular momentum is Hermitian

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.9.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

The total angular momentum operator ${L^{2}}$ can be written in spherical coordinates as

$\displaystyle L^{2}=-\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right] \ \ \ \ \ (1)$

As ${L^{2}}$ is an observable, it should be Hermitian. We can verify this by showing that

$\displaystyle \left\langle \psi_{2}\left|L^{2}\right|\psi_{1}\right\rangle =\left\langle \psi_{1}\left|L^{2}\right|\psi_{2}\right\rangle ^* \ \ \ \ \ (2)$

In spherical coordinates, this becomes

$\displaystyle \int\psi_{2}^*\left(L^{2}\psi_{1}\right)d\Omega=\left[\int\psi_{1}^*\left(L^{2}\psi_{2}\right)d\Omega\right]^* \ \ \ \ \ (3)$

The element of solid angle ${d\Omega=\sin\theta\;d\theta\;d\phi}$, so the full integral is

$\displaystyle \int\psi_{2}^*\left(L^{2}\psi_{1}\right)d\Omega=\int_{0}^{2\pi}\int_{0}^{\pi}\psi_{2}^*\left(L^{2}\psi_{1}\right)\sin\theta\;d\theta\;d\phi \ \ \ \ \ (4)$

We can verify 3 by showing that it is true for each of the two terms in 1 separately. As usual for these sorts of integrals, we need to use integration by parts. To simplify things, we’ll consider ${-L^{2}/\hbar^{2}}$ so we can deal only with the terms in the brackets in 1. We’ll also use the shorthand notation

 $\displaystyle s$ $\displaystyle \equiv$ $\displaystyle \sin\theta\ \ \ \ \ (5)$ $\displaystyle c$ $\displaystyle \equiv$ $\displaystyle \cos\theta \ \ \ \ \ (6)$

Also, a prime indicates a derivative with respect to ${\theta}$: ${\psi_{1}^{\prime}\equiv\frac{\partial\psi_{1}}{\partial\theta}}$, etc.

For the first term, we have, considering only the integration over ${\theta}$:

 $\displaystyle \int_{0}^{\pi}\psi_{2}^*\frac{1}{s}\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{1}}{\partial\theta}\right)s\;d\theta$ $\displaystyle =$ $\displaystyle \int_{0}^{\pi}\left[\psi_{2}^*c\psi_{1}^{\prime}+\psi_{2}^*s\psi_{1}^{\prime\prime}\right]d\theta\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.\psi_{2}^*c\psi_{1}\right|_{0}^{\pi}+\left.\psi_{2}^*s\psi_{1}^{\prime}\right|_{0}^{\pi}-\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}c\psi_{1}-\psi_{2}^*s\psi_{1}\right]d\theta-\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}s\psi_{1}^{\prime}+\psi_{2}^*c\psi_{1}^{\prime}\right]d\theta \ \ \ \ \ (10)$

The second term in 8 is zero since ${\sin0=\sin\pi=0}$, but we can’t ignore the first term, which is not, in general, zero. Thus we are left with

 $\displaystyle \int_{0}^{\pi}\psi_{2}^*\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{1}}{\partial\theta}\right)\;d\theta$ $\displaystyle =$ $\displaystyle \left.\psi_{2}^*c\psi_{1}\right|_{0}^{\pi}-\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}c\psi_{1}-\psi_{2}^*s\psi_{1}\right]d\theta-\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}s\psi_{1}^{\prime}+\psi_{2}^*c\psi_{1}^{\prime}\right]d\theta \ \ \ \ \ (13)$

We can now integrate the last line by parts again to get rid of the derivatives of ${\psi_{1}}$:

 $\displaystyle -\int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}s\psi_{1}^{\prime}+\psi_{2}^*c\psi_{1}^{\prime}\right]d\theta$ $\displaystyle =$ $\displaystyle -\left.\left(\psi_{2}^*\right)^{\prime}s\psi_{1}\right|_{0}^{\pi}-\left.\psi_{2}^*c\psi_{1}\right|_{0}^{\pi}+\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime\prime}s+\left(\psi_{2}^*\right)^{\prime}c\psi_{1}\right]d\theta+\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime}c-\psi_{2}^*s\psi_{1}\right]d\theta\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left.\psi_{2}^*c\psi_{1}\right|_{0}^{\pi}+\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime\prime}s+\left(\psi_{2}^*\right)^{\prime}c\psi_{1}\right]d\theta+\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime}c-\psi_{2}^*s\psi_{1}\right]d\theta \ \ \ \ \ (19)$

Inserting this back into 11 and cancelling terms, we have

$\displaystyle \int_{0}^{\pi}\psi_{2}^*\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{1}}{\partial\theta}\right)\;d\theta=\int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime\prime}s+\left(\psi_{2}^*\right)^{\prime}c\psi_{1}\right]d\theta \ \ \ \ \ (20)$

Comparing this with 7, we see that

$\displaystyle \int_{0}^{\pi}\psi_{2}^*\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{1}}{\partial\theta}\right)\;d\theta=\left[\int_{0}^{\pi}\psi_{1}^*\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{2}}{\partial\theta}\right)\;d\theta\right]^* \ \ \ \ \ (21)$

Thus the first term in 1 is Hermitian. (As this first term involves no derivatives with respect to ${\phi}$, the integration over ${\phi}$ is automatically Hermitian.)

For the second term in 1, we need to consider only the integral over ${\phi}$, so we have

$\displaystyle \int_{0}^{2\pi}\psi_{2}^*\frac{1}{\sin^{2}\theta}\frac{\partial^{2}\psi_{1}}{\partial\phi^{2}}\sin\theta\;d\phi=\frac{1}{s}\int_{0}^{2\pi}\psi_{2}^*\frac{\partial^{2}\psi_{1}}{\partial\phi^{2}}\;d\phi \ \ \ \ \ (22)$

(As we’re integrating over ${\phi}$, terms in ${\theta}$ act as constants and can be taken outside the integral.) The first integration by parts gives (where a prime now indicates a derivative with respect to ${\phi}$):

 $\displaystyle \int_{0}^{2\pi}\psi_{2}^*\psi_{1}^{\prime\prime}\;d\phi$ $\displaystyle =$ $\displaystyle \left.\psi_{2}^*\psi_{1}^{\prime}\right|_{0}^{2\pi}-\int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime}\psi_{1}^{\prime}\;d\phi\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime}\psi_{1}^{\prime}\;d\phi \ \ \ \ \ (24)$

This time, we’re able to set the integrated term to zero, since ${\phi=0}$ and ${\phi=2\pi}$ refer to the same angle. A second integration by parts gives

 $\displaystyle -\int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime}\psi_{1}^{\prime}\;d\phi$ $\displaystyle =$ $\displaystyle -\left.\left(\psi_{2}^*\right)^{\prime}\psi_{1}\right|_{0}^{2\pi}+\int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime\prime}\psi_{1}\;d\phi\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime\prime}\psi_{1}\;d\phi\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\int_{0}^{2\pi}\psi_{1}^*\psi_{2}^{\prime\prime}\;d\phi\right]^* \ \ \ \ \ (27)$

Thus both terms in 1 are Hermitian, so the complete operator ${L^{2}}$ is also Hermitian.

# Angular momentum – raising and lowering operators from rectangular coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.8.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

To calculate the eigenfunctions of angular momentum, we will need expressions for the raising and lowering operators ${L_{\pm}}$ in spherical coordinates. We’ve seen one way of getting these by working with the gradient in spherical coordinates from the start, but it is also possible to convert the rectangular forms of ${L_{\pm}}$ to spherical coordinates by using the chain rule from calculus. This method is similar to one we used earlier in 2-d. To set the scene, we need the conversion formulas between rectangular and spherical coordinates:

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\sin\theta\cos\phi\ \ \ \ \ (1)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta\sin\phi\ \ \ \ \ (2)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\cos\theta\ \ \ \ \ (3)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle \sqrt{x^{2}+y^{2}+z^{2}}\ \ \ \ \ (4)$ $\displaystyle \theta$ $\displaystyle =$ $\displaystyle \arctan\frac{\sqrt{x^{2}+y^{2}}}{z}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \arctan\frac{q}{z}\ \ \ \ \ (6)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\frac{y}{x} \ \ \ \ \ (7)$

To simplify the notation, we’ve defined

$\displaystyle q\equiv\sqrt{x^{2}+y^{2}}=r\sin\theta \ \ \ \ \ (8)$

We’ll also use shorthand notation for sines and cosines so that

 $\displaystyle s_{\theta}$ $\displaystyle \equiv$ $\displaystyle \sin\theta\ \ \ \ \ (9)$ $\displaystyle c_{\theta}$ $\displaystyle \equiv$ $\displaystyle \cos\theta \ \ \ \ \ (10)$

and similarly for ${\phi}$. We’ll also use the notation ${\partial_{r}}$ to mean the partial derivative with respect to ${r}$, with a similar notation for other derivatives.

The required derivatives are

 $\displaystyle \partial_{x}$ $\displaystyle =$ $\displaystyle \partial_{x}r\cdot\partial_{r}+\partial_{x}\theta\cdot\partial_{\theta}+\partial_{x}\phi\cdot\partial_{\phi}\ \ \ \ \ (11)$ $\displaystyle \partial_{y}$ $\displaystyle =$ $\displaystyle \partial_{y}r\cdot\partial_{r}+\partial_{y}\theta\cdot\partial_{\theta}+\partial_{y}\phi\cdot\partial_{\phi}\ \ \ \ \ (12)$ $\displaystyle \partial_{z}$ $\displaystyle =$ $\displaystyle \partial_{z}r\cdot\partial_{r}+\partial_{z}\theta\cdot\partial_{\theta} \ \ \ \ \ (13)$

The required derivatives are

 $\displaystyle \partial_{x}r$ $\displaystyle =$ $\displaystyle \frac{x}{r}\ \ \ \ \ (14)$ $\displaystyle \partial_{y}r$ $\displaystyle =$ $\displaystyle \frac{y}{r}\ \ \ \ \ (15)$ $\displaystyle \partial_{z}r$ $\displaystyle =$ $\displaystyle \frac{z}{r}\ \ \ \ \ (16)$ $\displaystyle \partial_{x}\theta$ $\displaystyle =$ $\displaystyle \frac{x/q}{z\left(1+\frac{q^{2}}{z^{2}}\right)}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{xz}{qr^{2}}\ \ \ \ \ (18)$ $\displaystyle \partial_{y}\theta$ $\displaystyle =$ $\displaystyle \frac{yz}{qr^{2}}\ \ \ \ \ (19)$ $\displaystyle \partial_{z}\theta$ $\displaystyle =$ $\displaystyle -\frac{q}{r^{2}}\ \ \ \ \ (20)$ $\displaystyle \partial_{x}\phi$ $\displaystyle =$ $\displaystyle \frac{-y/x^{2}}{1+y^{2}/x^{2}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{y}{q^{2}}\ \ \ \ \ (22)$ $\displaystyle \partial_{y}\phi$ $\displaystyle =$ $\displaystyle \frac{x}{q^{2}}\ \ \ \ \ (23)$ $\displaystyle \partial_{z}\phi$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (24)$

Plugging all these into 11 to 13 we have

 $\displaystyle \partial_{x}$ $\displaystyle =$ $\displaystyle \frac{x}{r}\partial_{r}+\frac{xz}{qr^{2}}\partial_{\theta}-\frac{y}{q^{2}}\partial_{\phi}\ \ \ \ \ (25)$ $\displaystyle \partial_{y}$ $\displaystyle =$ $\displaystyle \frac{y}{r}\partial_{r}+\frac{yz}{qr^{2}}\partial_{\theta}+\frac{x}{q^{2}}\partial_{\phi}\ \ \ \ \ (26)$ $\displaystyle \partial_{z}$ $\displaystyle =$ $\displaystyle \frac{z}{r}\partial_{r}-\frac{q}{r^{2}}\partial_{\theta} \ \ \ \ \ (27)$

We can now calculate the components ${L_{x}}$ and ${L_{y}}$:

 $\displaystyle L_{x}$ $\displaystyle =$ $\displaystyle -i\hbar\left(y\partial_{z}-z\partial_{y}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left[\frac{yz}{r}\partial_{r}-\frac{yq}{r^{2}}\partial_{\theta}-\frac{yz}{r}\partial_{r}-\frac{yz^{2}}{qr^{2}}\partial_{\theta}-\frac{xz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(\frac{yq}{r^{2}}+\frac{yz^{2}}{qr^{2}}\right)\partial_{\theta}+\frac{xz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(s_{\theta}^{2}s_{\phi}+\frac{s_{\theta}s_{\phi}c_{\theta}^{2}}{s_{\theta}}\right)\partial_{\theta}+\frac{s_{\theta}c_{\theta}c_{\phi}}{s_{\theta}^{2}}\partial_{\phi}\right]\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(\sin\phi\frac{\partial}{\partial\theta}+\cos\phi\cot\theta\frac{\partial}{\partial\phi}\right)\ \ \ \ \ (32)$ $\displaystyle L_{y}$ $\displaystyle =$ $\displaystyle -i\hbar\left(z\partial_{x}-x\partial_{z}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left[\frac{xz}{r}\partial_{r}+\frac{xz^{2}}{qr^{2}}\partial_{\theta}-\frac{xz}{r}\partial_{r}+\frac{xq}{r^{2}}\partial_{\theta}-\frac{yz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(-\frac{xz^{2}}{qr^{2}}-\frac{xq}{r^{2}}\right)\partial_{\theta}+\frac{yz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(-\frac{s_{\theta}c_{\phi}c_{\theta}^{2}}{s_{\theta}}-s_{\theta}^{2}c_{\phi}\right)\partial_{\theta}+\frac{s_{\theta}c_{\theta}s_{\phi}}{s_{\theta}^{2}}\partial_{\phi}\right]\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(-\cos\phi\frac{\partial}{\partial\theta}+\sin\phi\cot\theta\frac{\partial}{\partial\phi}\right) \ \ \ \ \ (37)$

From this we get the raising and lowering operators

 $\displaystyle L_{\pm}$ $\displaystyle =$ $\displaystyle L_{x}\pm iL_{y}\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(\sin\phi\frac{\partial}{\partial\theta}+\cos\phi\cot\theta\frac{\partial}{\partial\phi}\right)\mp\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle$ $\displaystyle \hbar\left(-\cos\phi\frac{\partial}{\partial\theta}+\sin\phi\cot\theta\frac{\partial}{\partial\phi}\right)\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar e^{\pm i\phi}\frac{\partial}{\partial\theta}\pm i\hbar e^{\pm i\phi}\cot\theta\frac{\partial}{\partial\phi}\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar e^{\pm i\phi}\left(\frac{\partial}{\partial\theta}\pm i\cot\theta\frac{\partial}{\partial\phi}\right) \ \ \ \ \ (42)$

[Admittedly, it’s probably easier and more elegant to use spherical coordinates from the start, but it’s instructive to see how it’s done starting with rectangular coordinates.]

# Rotations in 3-d: Euler angles

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.7.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

Any 3-d rotation can be expressed in terms of the Euler angles. These angles specify a sequence of three successive rotations about the rectangular axes. There are various definitions of Euler angles involving different sets of rotations, but the set used by Shankar in this problem consists of (1) a rotation by ${\gamma}$ about the ${z}$ axis, followed by (2) a rotation by ${\beta}$ about the ${y}$ axis and concluding with (3) a rotation by ${\alpha}$ about the ${z}$ axis. The proof that any rotation can be expressed this way would take us too far afield at this point, so we’ll just accept this for now.

We can see how these work in quantum mechanics by considering the special case of ${j=1}$, for which we derived the formula for a finite rotation here. A state ${\left|\psi\right\rangle }$ is transformed by a rotation ${\boldsymbol{\theta}}$ according to

 $\displaystyle \left|\psi^{\prime}\right\rangle$ $\displaystyle =$ $\displaystyle D^{\left(1\right)}\left[R\right]\left|\psi\right\rangle \ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[I^{\left(1\right)}+\frac{\left(\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\theta-1\right)-\frac{i\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}}{\hbar}\sin\theta\right]\left|\psi\right\rangle \ \ \ \ \ (2)$

For our purposes below, we’ll need the three components of ${\mathbf{J}^{\left(1\right)}}$, which can be copied from Shankar’s equations 12.5.23 and 12.5.24:

 $\displaystyle J_{x}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle J_{y}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle \left(J_{y}^{\left(1\right)}\right)^{2}$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2}\left[\begin{array}{ccc} 1 & 0 & -1\\ 0 & 2 & 0\\ -1 & 0 & 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle J_{z}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \hbar\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle \left(J_{z}^{\left(1\right)}\right)^{2}$ $\displaystyle =$ $\displaystyle \hbar^{2}\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (7)$

Evaluating 2 for the three Euler rotations, we have

 $\displaystyle D_{\gamma}$ $\displaystyle =$ $\displaystyle I^{\left(1\right)}+\frac{\left(J_{z}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\gamma-1\right)-\frac{iJ_{z}^{\left(1\right)}}{\hbar}\sin\gamma\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \cos\gamma-i\sin\gamma & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & \cos\gamma+i\sin\gamma \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} e^{-i\gamma} & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & e^{i\gamma} \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle D_{\beta}$ $\displaystyle =$ $\displaystyle I^{\left(1\right)}+\frac{\left(J_{y}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\beta-1\right)-\frac{iJ_{y}^{\left(1\right)}}{\hbar}\sin\beta\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{ccc} 1+\cos\beta & -\sqrt{2}\sin\beta & 1-\cos\beta\\ \sqrt{2}\sin\beta & 2\cos\beta & -\sqrt{2}\sin\beta\\ 1-\cos\beta & \sqrt{2}\sin\beta & 1+\cos\beta \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle D_{\alpha}$ $\displaystyle =$ $\displaystyle I^{\left(1\right)}+\frac{\left(J_{z}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\alpha-1\right)-\frac{iJ_{z}^{\left(1\right)}}{\hbar}\sin\alpha\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \cos\alpha-i\sin\alpha & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & \cos\alpha+i\sin\alpha \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} e^{-i\alpha} & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & e^{i\alpha} \end{array}\right] \ \ \ \ \ (15)$

The complete rotation is the product of the three matrices:

 $\displaystyle D_{total}$ $\displaystyle =$ $\displaystyle D_{\alpha}D_{\beta}D_{\gamma}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle D_{\alpha}\frac{1}{2}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{-i\gamma} & -\sqrt{2}\sin\beta & \left(1-\cos\beta\right)e^{i\gamma}\\ \sqrt{2}\sin\beta e^{-i\gamma} & 2\cos\beta & -\sqrt{2}\sin\beta e^{i\gamma}\\ \left(1-\cos\beta\right)e^{-i\gamma} & \sqrt{2}\sin\beta & \left(1+\cos\beta\right)e^{i\gamma} \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{-i\gamma}e^{-i\alpha} & -\sqrt{2}\sin\beta e^{-i\alpha} & \left(1-\cos\beta\right)e^{i\gamma}e^{-i\alpha}\\ \sqrt{2}\sin\beta e^{-i\gamma} & 2\cos\beta & -\sqrt{2}\sin\beta e^{i\gamma}\\ \left(1-\cos\beta\right)e^{-i\gamma}e^{i\alpha} & \sqrt{2}\sin\beta e^{i\alpha} & \left(1+\cos\beta\right)e^{i\gamma}e^{i\alpha} \end{array}\right] \ \ \ \ \ (18)$

In the ${\left|jm\right\rangle }$ basis, the state ${\left|11\right\rangle }$ is represented by

$\displaystyle \left|11\right\rangle =\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right] \ \ \ \ \ (19)$

Applying the rotation 18 to this state, we get

 $\displaystyle \left|11^{\prime}\right\rangle$ $\displaystyle =$ $\displaystyle D_{total}\left|11\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{-i\gamma}e^{-i\alpha} & -\sqrt{2}\sin\beta e^{-i\alpha} & \left(1-\cos\beta\right)e^{i\gamma}e^{-i\alpha}\\ \sqrt{2}\sin\beta e^{-i\gamma} & 2\cos\beta & -\sqrt{2}\sin\beta e^{i\gamma}\\ \left(1-\cos\beta\right)e^{-i\gamma}e^{i\alpha} & \sqrt{2}\sin\beta e^{i\alpha} & \left(1+\cos\beta\right)e^{i\gamma}e^{i\alpha} \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\gamma}e^{-i\alpha}\\ \sqrt{2}\sin\beta e^{-i\gamma}\\ \left(1-\cos\beta\right)e^{-i\gamma}e^{i\alpha} \end{array}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{e^{-i\gamma}}{2}\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\alpha}\\ \sqrt{2}\sin\beta\\ \left(1-\cos\beta\right)e^{i\alpha} \end{array}\right] \ \ \ \ \ (23)$

We can work out the average values of the components of ${\mathbf{J}}$ in the rotated state in the same way as in the previous problem, by using 3, 4 and 6. Note that ${\gamma}$ disappears from the matrix elements as it enters only in an overall phase factor. We get (I used Maple to do the tedious matrix multiplications):

 $\displaystyle \left\langle J_{x}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 11^{\prime}\left|J_{x}\right|11^{\prime}\right\rangle \ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4\sqrt{2}}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{i\alpha} & \sqrt{2}\sin\beta & \left(1-\cos\beta\right)e^{-i\alpha}\end{array}\right]\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\alpha}\\ \sqrt{2}\sin\beta\\ \left(1-\cos\beta\right)e^{i\alpha} \end{array}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\sin\beta\cos\alpha\ \ \ \ \ (26)$ $\displaystyle \left\langle J_{y}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 11^{\prime}\left|J_{y}\right|11^{\prime}\right\rangle \ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{4\sqrt{2}}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{i\alpha} & \sqrt{2}\sin\beta & \left(1-\cos\beta\right)e^{-i\alpha}\end{array}\right]\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\alpha}\\ \sqrt{2}\sin\beta\\ \left(1-\cos\beta\right)e^{i\alpha} \end{array}\right]\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\sin\beta\sin\alpha\ \ \ \ \ (29)$ $\displaystyle \left\langle J_{z}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 11^{\prime}\left|J_{z}\right|11^{\prime}\right\rangle \ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{i\alpha} & \sqrt{2}\sin\beta & \left(1-\cos\beta\right)e^{-i\alpha}\end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\alpha}\\ \sqrt{2}\sin\beta\\ \left(1-\cos\beta\right)e^{i\alpha} \end{array}\right]\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\cos\beta \ \ \ \ \ (32)$

Going back to 23, we can confirm our earlier result that it is impossible to rotate the state ${\left|11\right\rangle }$ into just ${\left|10\right\rangle }$. To do so, the state in 23 would have to be a multiple of ${\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]}$ which could happen only if both ${1+\cos\beta}$ and ${1-\cos\beta}$ were zero, which is impossible.

However, if we take ${\beta=\pi}$, then the rotated state 23 becomes

$\displaystyle \left|11^{\prime}\right\rangle =e^{i\left(\alpha-\gamma\right)}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=e^{i\left(\alpha-\gamma\right)}\left|1,-1\right\rangle \ \ \ \ \ (33)$

so apart from a phase factor, it is possible to rotate one eigenstate of ${J_{z}}$ into another.

The only values of ${\beta}$ that produce zero elements in 23 are ${\beta=0}$ and ${\beta=\pi}$, (the values of ${\alpha}$ and ${\gamma}$ produce only phase factors), so for any other value of ${\beta}$, all three elements of 23 are non-zero. Thus a general rotation from any starting state can always be made to produce a rotated state containing all three eigenstates of ${J_{z}}$: ${\left|11\right\rangle }$, ${\left|10\right\rangle }$ and ${\left|1,-1\right\rangle }$.

# Rotations in 3-d: classical and quantum rotations compared

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.6.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

This is an example of how risky it can be to attempt to derive quantum behaviour by using logic based on classical mechanics. In classical mechanics, if we have a system with some total angular momentum with magnitude ${\left|\mathbf{J}\right|}$ and rotate this sytem through any angle, the magnitude of the angular momentum remains the same (although, of course, its direction changes). Based on this fact, we might think that if we start with a quantum state such as ${\left|jm\right\rangle }$ (where ${j}$ is the total angular momentum number and ${m}$ is the number for ${J_{z}}$), we should be able to obtain the other states with the same total angular momentum number ${j}$ by rotating this state through various angles about the appropriate rotation axis. To see that this won’t work, suppose we consider a state with ${j=1}$ and ${m=1}$, that is ${\left|jm\right\rangle =\left|11\right\rangle }$. Classically, such a system has its angular momentum aligned along the ${z}$ axis, so we might think that if we rotate this system by ${\frac{\pi}{2}}$ about, say, the ${x}$ axis, we should get a state with ${m=0}$, since the angular momentum is now aligned along the ${-y}$ axis.

To see if this works, we can use the formula for a finite rotation for a ${j=1}$ state. Since ${j}$ remains constant, a rotation of a state ${\left|jm\right\rangle }$ is given by

$\displaystyle D^{\left(1\right)}\left[R\right]\left|jm\right\rangle \ \ \ \ \ (1)$

where

$\displaystyle D^{\left(1\right)}\left[R\right]=I^{\left(1\right)}+\frac{\left(\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\theta-1\right)-\frac{i\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}}{\hbar}\sin\theta \ \ \ \ \ (2)$

For a rotation by an angle ${\theta}$ about the ${x}$ axis, this formula reduces to

$\displaystyle D^{\left(1\right)}\left[R\left(\theta\hat{\mathbf{x}}\right)\right]=I^{\left(1\right)}+\frac{\left(J_{x}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\theta-1\right)-\frac{iJ_{x}^{\left(1\right)}}{\hbar}\sin\theta \ \ \ \ \ (3)$

We can copy the matrix ${J_{x}^{\left(1\right)}}$ from Shankar’s equation 12.5.23:

$\displaystyle J_{x}^{\left(1\right)}=\frac{\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right] \ \ \ \ \ (4)$

We therefore have

$\displaystyle \left(J_{x}^{\left(1\right)}\right)^{2}=\frac{\hbar^{2}}{2}\left[\begin{array}{ccc} 1 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 1 \end{array}\right] \ \ \ \ \ (5)$

Plugging these into 3 we have

$\displaystyle D^{\left(1\right)}\left[R\left(\theta\hat{\mathbf{x}}\right)\right]=\frac{1}{2}\left[\begin{array}{ccc} 1+\cos\theta & -\sqrt{2}i\sin\theta & \cos\theta-1\\ -\sqrt{2}i\sin\theta & 2\cos\theta & -\sqrt{2}i\sin\theta\\ \cos\theta-1 & -\sqrt{2}i\sin\theta & 1+\cos\theta \end{array}\right] \ \ \ \ \ (6)$

In the ${\left|jm\right\rangle }$ basis, the state ${\left|11\right\rangle }$ is represented by

$\displaystyle \left|11\right\rangle =\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right] \ \ \ \ \ (7)$

Thus a rotation about the ${x}$ axis rotates this state into:

 $\displaystyle \left|\psi\right\rangle$ $\displaystyle =$ $\displaystyle D^{\left(1\right)}\left[R\left(\theta\hat{\mathbf{x}}\right)\right]\left|11\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{ccc} 1+\cos\theta & -\sqrt{2}i\sin\theta & \cos\theta-1\\ -\sqrt{2}i\sin\theta & 2\cos\theta & -\sqrt{2}i\sin\theta\\ \cos\theta-1 & -\sqrt{2}i\sin\theta & 1+\cos\theta \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right] \ \ \ \ \ (10)$

If rotation by some angle ${\theta}$ could change ${\left|11\right\rangle }$ into the state ${\left|10\right\rangle }$, this result would need to be a multiple of ${\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]}$, so we’d need to find ${\theta}$ such that both of the following equations are true:

 $\displaystyle 1+\cos\theta$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle \cos\theta-1$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

This is impossible, so no rotation about the ${x}$ axis can change ${\left|11\right\rangle }$ into the state ${\left|10\right\rangle }$.

However, there is still a correspondence between classical and quantum mechanics if we compare the average values of the components of ${\mathbf{J}}$. That is, we want to find ${\left\langle \mathbf{J}\right\rangle }$ for the state 10. To do this, we need the other two matrix components ${J_{y}^{\left(1\right)}}$ and ${J_{z}^{\left(1\right)}}$. We can get ${J_{y}^{\left(1\right)}}$ from Shankar’s equation 12.5.24:

$\displaystyle J_{y}^{\left(1\right)}=\frac{i\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right] \ \ \ \ \ (13)$

${J_{z}^{\left(1\right)}}$ is just the diagonal matrix:

$\displaystyle J_{z}^{\left(1\right)}=\hbar\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (14)$

We can now calculate the averages for the state 10:

 $\displaystyle \left\langle J_{x}^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|J_{x}^{\left(1\right)}\right|\psi\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4\sqrt{2}}\left[\begin{array}{ccc} 1+\cos\theta & \sqrt{2}i\sin\theta & \cos\theta-1\end{array}\right]\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} i\sin\theta & \sqrt{2}\cos\theta & i\sin\theta\end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (18)$ $\displaystyle \left\langle J_{y}^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|J_{y}^{\left(1\right)}\right|\psi\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{4\sqrt{2}}\left[\begin{array}{ccc} 1+\cos\theta & \sqrt{2}i\sin\theta & \cos\theta-1\end{array}\right]\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} -\sin\theta & -i\sqrt{2} & \sin\theta\end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar\sin\theta\ \ \ \ \ (22)$ $\displaystyle \left\langle J_{z}^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|J_{z}^{\left(1\right)}\right|\psi\right\rangle \ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} 1+\cos\theta & \sqrt{2}i\sin\theta & \cos\theta-1\end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} 1+\cos\theta & 0 & 1-\cos\theta\end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\cos\theta \ \ \ \ \ (26)$

Thus for the average, we have

$\displaystyle \left\langle \mathbf{J}\right\rangle =-\hbar\sin\theta\hat{\mathbf{y}}+\hbar\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (27)$

In this case, a rotation by ${\theta=\frac{\pi}{2}}$ does indeed rotate ${\left\langle \mathbf{J}\right\rangle }$ so that it points along the ${-y}$ axis, just as it would in classical mechanics.