# Spherical Bessel functions – behaviour for small arguments

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.7.

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The general solution for a free particle in spherical coordinates involves the radial function, which turns out to be

$\displaystyle \frac{R_{l+1}}{\rho^{l+1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{l+1}\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (1)$

where ${l}$ is the total angular momentum quantum number and

 $\displaystyle k^{2}$ $\displaystyle \equiv$ $\displaystyle \frac{2\mu E}{\hbar^{2}}\ \ \ \ \ (2)$ $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle kr \ \ \ \ \ (3)$

We can rewrite this as

$\displaystyle R_{l}=\left(-\rho\right)^{l}\left(\frac{1}{\rho}\frac{d}{d\rho}\right)^{l}R_{0} \ \ \ \ \ (4)$

We saw earlier that the solutions for ${l=0}$ are, with ${U_{l}=\rho R_{l}}$

 $\displaystyle U_{0}^{A}\left(\rho\right)$ $\displaystyle =$ $\displaystyle \sin\rho\ \ \ \ \ (5)$ $\displaystyle U_{0}^{B}\left(\rho\right)$ $\displaystyle =$ $\displaystyle -\cos\rho \ \ \ \ \ (6)$

Thus the two solutions for ${l=0}$ are

 $\displaystyle R_{0}^{A}$ $\displaystyle =$ $\displaystyle \frac{\sin\rho}{\rho}\ \ \ \ \ (7)$ $\displaystyle R_{0}^{B}$ $\displaystyle =$ $\displaystyle -\frac{\cos\rho}{\rho} \ \ \ \ \ (8)$

From these starting points, we can generate all the solutions for higher values of ${l}$ using 4. These functions are

 $\displaystyle j_{l}\left(\rho\right)$ $\displaystyle =$ $\displaystyle \left(-\rho\right)^{l}\left(\frac{1}{\rho}\frac{d}{d\rho}\right)^{l}\frac{\sin\rho}{\rho}\ \ \ \ \ (9)$ $\displaystyle n_{l}\left(\rho\right)$ $\displaystyle =$ $\displaystyle -\left(-\rho\right)^{l}\left(\frac{1}{\rho}\frac{d}{d\rho}\right)^{l}\frac{\cos\rho}{\rho} \ \ \ \ \ (10)$

and are known as spherical Bessel functions ${j_{l}}$ and spherical Neumann functions ${n_{l}}$.

The asymptotic behaviour is given by

 $\displaystyle j_{l}$ $\displaystyle \underset{\rho\rightarrow\infty}{\longrightarrow}$ $\displaystyle \frac{1}{\rho}\sin\left(\rho-\frac{l\pi}{2}\right)\ \ \ \ \ (11)$ $\displaystyle n_{l}$ $\displaystyle \underset{\rho\rightarrow\infty}{\longrightarrow}$ $\displaystyle -\frac{1}{\rho}\cos\left(\rho-\frac{l\pi}{2}\right) \ \ \ \ \ (12)$

For ${\rho\rightarrow0}$, we have

 $\displaystyle j_{l}$ $\displaystyle \underset{\rho\rightarrow0}{\longrightarrow}$ $\displaystyle \frac{\rho^{l}}{\left(2l+1\right)!!}\ \ \ \ \ (13)$ $\displaystyle n_{l}$ $\displaystyle \underset{\rho\rightarrow0}{\longrightarrow}$ $\displaystyle -\frac{\left(2l-1\right)!!}{\rho^{l+1}} \ \ \ \ \ (14)$

We can verify the latter equation for ${j_{l}}$ for a couple of cases with small ${l}$. From 9, we can generate a couple of ${j_{l}}$s:

 $\displaystyle j_{0}$ $\displaystyle =$ $\displaystyle \frac{\sin\rho}{\rho}\ \ \ \ \ (15)$ $\displaystyle j_{1}$ $\displaystyle =$ $\displaystyle -\rho\frac{1}{\rho}\frac{d}{dr}\left(\frac{\sin\rho}{\rho}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho\frac{1}{\rho}\left(\frac{\cos\rho}{\rho}-\frac{\sin\rho}{\rho^{2}}\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sin\rho}{\rho^{2}}-\frac{\cos\rho}{\rho}\ \ \ \ \ (18)$ $\displaystyle j_{2}$ $\displaystyle =$ $\displaystyle \left(-\rho\right)^{2}\frac{1}{\rho}\frac{d}{d\rho}\left[\frac{1}{\rho}\frac{d}{dr}\left(\frac{\sin\rho}{\rho}\right)\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(-\rho\right)^{2}\frac{1}{\rho}\frac{d}{d\rho}\left[\frac{1}{\rho}\left(\frac{\cos\rho}{\rho}-\frac{\sin\rho}{\rho^{2}}\right)\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{3}{\rho^{3}}-\frac{1}{\rho}\right)\sin\rho-\frac{3\cos\rho}{\rho^{2}} \ \ \ \ \ (21)$

We can get the limits for ${\rho\rightarrow0}$ by expanding the sine and cosine. That is, we use the limiting forms

 $\displaystyle \sin\rho$ $\displaystyle \rightarrow$ $\displaystyle \rho-\frac{\rho^{3}}{3!}+\ldots\ \ \ \ \ (22)$ $\displaystyle \cos\rho$ $\displaystyle \rightarrow$ $\displaystyle 1-\frac{1}{2}\rho^{2}+\ldots \ \ \ \ \ (23)$

We need to retain enough terms for ${j_{l}}$ so that we get all the terms up to the first power of ${\rho}$ that doesn’t cancel out when we do the algebra. We get

 $\displaystyle j_{0}$ $\displaystyle \rightarrow$ $\displaystyle 1=\frac{\rho^{0}}{1!!}\ \ \ \ \ (24)$ $\displaystyle j_{1}$ $\displaystyle \rightarrow$ $\displaystyle \frac{1}{\rho}-\frac{\rho}{6}-\frac{1}{\rho}\left(1-\frac{1}{2}\rho^{2}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho}{3}=\frac{\rho^{1}}{3!!}\ \ \ \ \ (26)$ $\displaystyle j_{2}$ $\displaystyle \rightarrow$ $\displaystyle \left(\frac{3}{\rho^{3}}-\frac{1}{\rho}\right)\left(\rho-\frac{\rho^{3}}{6}+\frac{\rho^{5}}{120}\right)-\frac{3}{\rho^{2}}\left(1-\frac{1}{2}\rho^{2}+\frac{1}{24}\rho^{4}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle \rightarrow$ $\displaystyle \left(\frac{1}{6}+\frac{1}{40}-\frac{1}{8}\right)\rho^{2}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{20+3-15}{120}\rho^{2}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho^{2}}{15}=\frac{\rho^{2}}{5!!} \ \ \ \ \ (30)$

# Free particle in spherical coordinates – finding the solutions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.6.

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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

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

For a free particle, ${V=0}$ and ${E>0}$, so we have

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (3)$

Defining

 $\displaystyle k^{2}$ $\displaystyle \equiv$ $\displaystyle \frac{2\mu E}{\hbar^{2}}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle kr \ \ \ \ \ (5)$

we convert the equation to

$\displaystyle \left(-\frac{d^{2}}{d\rho^{2}}+\frac{l\left(l+1\right)}{\rho^{2}}\right)U_{l}=U_{l} \ \ \ \ \ (6)$

This equation can be solved by a method similar to that for the harmonic oscillator and its raising and lowering operators. The entire solution is fairly involved, so we’ll start out here by showing how the new raising and lowering operators are defined.

We define

$\displaystyle d_{l}\equiv\frac{d}{d\rho}+\frac{l+1}{\rho} \ \ \ \ \ (7)$

$\displaystyle d_{l}^{\dagger}=-\frac{d}{d\rho}+\frac{l+1}{\rho} \ \ \ \ \ (8)$

To see where the minus sign comes from on the RHS, we need to recall that the momentum operator is defined in one dimension as

$\displaystyle P=-i\hbar\frac{\partial}{\partial x} \ \ \ \ \ (9)$

Since ${P}$ is an observable, it is hermitian, so that ${P^{\dagger}=P}$. Under the hermitian operation ${i\rightarrow-i}$, so we must also have ${\frac{\partial}{\partial x}\rightarrow-\frac{\partial}{\partial x}}$. Thus the first derivative with respect to a position variable is anti-hermitian. If this doesn’t convince you, you can also work out the integral:

$\displaystyle \int_{0}^{\infty}\psi_{2}^*\frac{d}{d\rho}\psi_{1}d\rho=\left.\psi_{2}^*\psi_{1}\right|_{0}^{\infty}-\int_{0}^{\infty}\psi_{1}\frac{d}{d\rho}\psi_{2}^*d\rho \ \ \ \ \ (10)$

Under the usual assumption that ${\psi\rightarrow0}$ at the limits, the integrated term is zero and we have

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

In bracket notation, this is

$\displaystyle \left\langle \psi_{2}\left|\frac{d}{d\rho}\psi_{1}\right.\right\rangle =-\left\langle \frac{d}{d\rho}\psi_{2}\left|\psi_{1}\right.\right\rangle \ \ \ \ \ (13)$

which shows that ${\frac{d}{d\rho}}$ is an anti-hermitian operator.

Returning to 7 and 8, we have

 $\displaystyle d_{l}d_{l}^{\dagger}U_{l}$ $\displaystyle =$ $\displaystyle \left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)U_{l}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(-U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}-\frac{l+1}{\rho^{2}}U_{l}+\frac{l+1}{\rho}U_{l}^{\prime}-\frac{l+1}{\rho}U_{l}^{\prime}+\frac{\left(l+1\right)^{2}}{\rho^{2}}U_{l}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{l\left(l+1\right)}{\rho^{2}}U_{l} \ \ \ \ \ (17)$

Comparing with 6 we see that

$\displaystyle d_{l}d_{l}^{\dagger}U_{l}=U_{l} \ \ \ \ \ (18)$

We can also show that

 $\displaystyle d_{l}^{\dagger}d_{l}U_{l}$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)U_{l}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{l+1}{\rho^{2}}U_{l}-\frac{l+1}{\rho}U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}^{\prime}+\frac{\left(l+1\right)^{2}}{\rho^{2}}U_{l}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{\left(l+1\right)^{2}+l+1}{\rho^{2}}U_{l}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{\left(l+1\right)\left(l+2\right)}{\rho^{2}}U_{l}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle d_{l+1}d_{l+1}^{\dagger}U_{l} \ \ \ \ \ (24)$

Starting from 18 we multiply on the left by ${d_{l}^{\dagger}}$ to get

$\displaystyle d_{l}^{\dagger}d_{l}\left(d_{l}^{\dagger}U_{l}\right)=d_{l}^{\dagger}U_{l} \ \ \ \ \ (25)$

Comparing this with 24 we see that

$\displaystyle d_{l}^{\dagger}U_{l}=c_{l}U_{l+1} \ \ \ \ \ (26)$

where ${c_{l}}$ is a constant.

Thus ${d_{l}^{\dagger}}$ is a raising operator, in that it raises the angular momentum number ${l}$ by 1 when it acts on ${U_{l}}$. By convention, ${c_{l}=1}$ (any adjustments to the constant can be made when normalizing).

We can start the process by looking at 6 with ${l=0}$ which is

$\displaystyle \frac{d^{2}}{d\rho^{2}}U_{l}=-U_{l} \ \ \ \ \ (27)$

This has the two solutions

 $\displaystyle U_{0}^{A}\left(\rho\right)$ $\displaystyle =$ $\displaystyle \sin\rho\ \ \ \ \ (28)$ $\displaystyle U_{0}^{B}\left(\rho\right)$ $\displaystyle =$ $\displaystyle -\cos\rho \ \ \ \ \ (29)$

The minus sign in front of ${\cos\rho}$ is just conventional. Since we require ${U_{0}\left(0\right)=0}$, ${U_{0}^{B}}$ is unacceptable if the region we’re considering include ${\rho=0}$, so we have

$\displaystyle U_{0}\left(\rho\right)=\sin\rho \ \ \ \ \ (30)$

For the general case that excludes ${\rho=0}$, we must include the cosine term as well.

From here, we can generate solutions for higher values of ${l}$ by applying 26. Actually, the radial function that appears in the wave function is given by 2, so it is ${R_{l}}$ that we really want. That is, we want

$\displaystyle R_{l}=\frac{U_{l}}{r}=k\frac{U_{l}}{\rho} \ \ \ \ \ (31)$

As with the constant ${c_{l}}$ in 26, we can absorb ${k}$ into normalization to be done later, so we can generate functions

$\displaystyle R_{l}=\frac{U_{l}}{\rho} \ \ \ \ \ (32)$

Applying 26 we have

 $\displaystyle \rho R_{l+1}$ $\displaystyle =$ $\displaystyle d_{l}^{\dagger}\left(\rho R_{l}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(\rho R_{l}\right)\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{l}-\rho R_{l}^{\prime}+\left(l+1\right)R_{l}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho R_{l}^{\prime}+lR_{l}\ \ \ \ \ (36)$ $\displaystyle R_{l+1}$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l}{\rho}\right)R_{l}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho^{l}\frac{d}{d\rho}\left(\frac{R_{l}}{\rho^{l}}\right) \ \ \ \ \ (38)$

We can convert this into a general formula by writing

$\displaystyle \frac{R_{l+1}}{\rho^{l+1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{l}}{\rho^{l}} \ \ \ \ \ (39)$

Starting at ${l=0}$, we have

$\displaystyle \frac{R_{1}}{\rho^{1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (40)$

For the next step, we have

 $\displaystyle \frac{R_{2}}{\rho^{2}}$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{1}}{\rho^{1}}\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{0}}{\rho^{0}}\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{2}\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (43)$

Thus in general

$\displaystyle \frac{R_{l+1}}{\rho^{l+1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{l+1}\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (44)$

Note that

$\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{l+1}\ne\left(-\frac{1}{\rho}\right)^{l+1}\frac{d^{l+1}}{d\rho^{l+1}} \ \ \ \ \ (45)$

since the factor of ${\frac{1}{\rho}}$ has to be included when taking the derivative.

We’ll explore the nature of these solutions in the next post.

# 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.

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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

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

# Equation display may be misaligned

Today I switched the plugin used to display math equations in Latex to the “Beautiful maths” feature of Jetpack. Briefly, this caused some of the equations to appear out of alignment. In particular, in multiple-line equations, the equals sign may not be centered with respect to the rest of the equation.

If you see this, do a complete reload of the page (Ctrl + F5 on most desktop browsers). On mobiles and tablets without either Ctrl or F5 keys, you will probably have to clear the cache (which you can do on Android devices by selecting History from the menu of the Chrome browser; sorry but I don’t know if this works on other devices or device browsers).

# Spherical harmonics: rotation about the x axis

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.14.

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

Here’s another example of using spherical harmonics to study the behaviour of wave functions in 3-d. Under a rotation by ${\theta_{x}}$ about the ${x}$ axis, the coordinates transform using the rotation matrix

$\displaystyle R\left(\theta_{x}\hat{\mathbf{x}}\right)=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & \cos\theta_{x} & -\sin\theta_{x}\\ 0 & \sin\theta_{x} & \cos\theta_{x} \end{array}\right] \ \ \ \ \ (1)$

This results in the coordinate transformations

 $\displaystyle x$ $\displaystyle \rightarrow$ $\displaystyle x\ \ \ \ \ (2)$ $\displaystyle y$ $\displaystyle \rightarrow$ $\displaystyle y\cos\theta_{x}-z\sin\theta_{x}\ \ \ \ \ (3)$ $\displaystyle z$ $\displaystyle \rightarrow$ $\displaystyle z\cos\theta_{x}+y\sin\theta_{x} \ \ \ \ \ (4)$

Using similar techniques to those for translations, it is found that the wave function ${\psi\left(x,y,z\right)}$ transforms into the wave function at the position obtained by rotating by ${-\theta_{x}}$ (that is, by rotating by ${\theta_{x}}$ in the opposite direction):

$\displaystyle \psi\left(x,y,z\right)\rightarrow\psi_{R}=\psi\left(x,y\cos\theta_{x}+z\sin\theta_{x},z\cos\theta_{x}-y\sin\theta_{x}\right) \ \ \ \ \ (5)$

Suppose we have a wave function given by

$\displaystyle \psi=Aze^{-r^{2}/a^{2}} \ \ \ \ \ (6)$

for some constants ${a}$ and ${A}$. Under this rotation, using 5 it transforms to

$\displaystyle \psi_{R}=A\left(z\cos\theta_{x}-y\sin\theta_{x}\right)e^{-r^{2}/a^{2}} \ \ \ \ \ (7)$

[Note that ${r^{2}}$ remains invariant under rotations about the origin, since the distance of a point from the origin is not affected by a rotation. You can verify this directly if you like by working out ${r^{2}=x^{2}+y^{2}+z^{2}}$ after the rotation.]

Equation 7 differs from the equation given in Shankar, which is

$\displaystyle \psi_{R}=A\left(z\cos\theta_{x}+y\sin\theta_{x}\right)e^{-r^{2}/a^{2}} \ \ \ \ \ (8)$

Curiously, in the errata for Shankar’s book (2006 edition) 7 is listed as the incorrect version, which is ‘corrected’ to 8. In my copy of the book (which doesn’t have a date on the title page), 8 is printed, but I don’t think this is right. In any case, we’ll proceed with the problem.

First, we write 6 in terms of spherical harmonics, using

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{-1}-Y_{1}^{1}\right)\ \ \ \ \ (9)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle -\frac{1}{2i}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{-1}+Y_{1}^{1}\right)\ \ \ \ \ (10)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle \sqrt{\frac{4\pi}{3}}rY_{1}^{0} \ \ \ \ \ (11)$

We have

$\displaystyle \psi=A\sqrt{\frac{4\pi}{3}}rY_{1}^{0}e^{-r^{2}/a^{2}} \ \ \ \ \ (12)$

With the three spherical harmonics ${Y_{1}^{1}}$, ${Y_{1}^{0}}$ and ${Y_{1}^{-1}}$ as the basis, we can write this in vector notation as

$\displaystyle \psi=A\sqrt{\frac{4\pi}{3}}re^{-r^{2}/a^{2}}\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right] \ \ \ \ \ (13)$

A rotation in 3-d for ${\ell=1}$ is given by

$\displaystyle D^{\left(1\right)}\left[R\right]=I^{\left(1\right)}+\frac{\left(\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\theta-1\right)-\frac{i\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}}{\hbar}\sin\theta \ \ \ \ \ (14)$

For ${\hat{\boldsymbol{\theta}}=\theta_{x}\hat{\mathbf{x}}}$, this works out to

$\displaystyle D^{\left(1\right)}\left[R\left(\theta_{x}\hat{\mathbf{x}}\right)\right]=\frac{1}{2}\left[\begin{array}{ccc} 1+\cos\theta_{x} & -\sqrt{2}i\sin\theta_{x} & \cos\theta_{x}-1\\ -\sqrt{2}i\sin\theta_{x} & 2\cos\theta_{x} & -\sqrt{2}i\sin\theta_{x}\\ \cos\theta_{x}-1 & -\sqrt{2}i\sin\theta_{x} & 1+\cos\theta_{x} \end{array}\right] \ \ \ \ \ (15)$

We can use this to transform 13 to get

 $\displaystyle \psi_{R}$ $\displaystyle =$ $\displaystyle D^{\left(1\right)}\left[R\left(\theta_{x}\hat{\mathbf{x}}\right)\right]\psi\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\sqrt{\frac{4\pi}{3}}re^{-r^{2}/a^{2}}\frac{1}{2}\left[\begin{array}{ccc} 1+\cos\theta_{x} & -\sqrt{2}i\sin\theta_{x} & \cos\theta_{x}-1\\ -\sqrt{2}i\sin\theta_{x} & 2\cos\theta_{x} & -\sqrt{2}i\sin\theta_{x}\\ \cos\theta_{x}-1 & -\sqrt{2}i\sin\theta_{x} & 1+\cos\theta_{x} \end{array}\right]\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\sqrt{\frac{\pi}{3}}re^{-r^{2}/a^{2}}\left[\begin{array}{c} -\sqrt{2}i\sin\theta_{x}\\ 2\cos\theta_{x}\\ -\sqrt{2}i\sin\theta_{x} \end{array}\right]\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{-r^{2}/a^{2}}\left\{ \sqrt{\frac{4\pi}{3}}r\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]\cos\theta_{x}-\sqrt{\frac{2\pi}{3}}ri\sin\theta_{x}\left(\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]+\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]\right)\right\} \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{-r^{2}/a^{2}}\left\{ \sqrt{\frac{4\pi}{3}}rY_{1}^{0}\cos\theta_{x}+\frac{1}{2i}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{1}+Y_{1}^{-1}\right)\sin\theta_{x}\right\} \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{-r^{2}/a^{2}}\left(z\cos\theta_{x}-y\sin\theta_{x}\right) \ \ \ \ \ (21)$

where we used 10 to get the last line. This result agrees with 7 and not with the equation 8 given in Shankar, so (provided I got the signs right) it looks like Shankar’s equation is wrong.

# 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.

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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 using the lowering operator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.11.

[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 raising and lowering operators for angular momentum are

$\displaystyle L_{\pm}\equiv L_{x}\pm iL_{y} \ \ \ \ \ (1)$

On a state ${\left|\ell m\right\rangle }$ in the basis of eigenstates of ${L^{2}}$ and ${L_{z}}$, they have the effect:

$\displaystyle L_{\pm}\left|\ell m\right\rangle =\hbar\sqrt{(\ell\mp m)(\ell\pm m+1)}\left|\ell,m\pm1\right\rangle \ \ \ \ \ (2)$

This means that, if we can find the top state ${\left|\ell\ell\right\rangle }$, we can find the state for all lower values of ${m}$ by applying ${L_{-}}$ successively. To illustrate the process we’ll derive the 3 states for ${\ell=1}$. The top state ${\left|11\right\rangle }$ can be obtained by following the derivation given in Shankar from his equation 12.5.28 onwards. In spherical coordinates, the raising and lowering operators have the form

$\displaystyle L_{\pm}=\pm\hbar e^{\pm i\phi}\left[\frac{\partial}{\partial\theta}\pm i\cot\theta\frac{\partial}{\partial\phi}\right] \ \ \ \ \ (3)$

Applying ${L_{+}}$ to the top state ${\left|11\right\rangle }$ must give zero, so if ${\psi_{1}^{1}}$ is the representation of this state in spherical coordinates, we must solve the differential equation

$\displaystyle \left[\frac{\partial}{\partial\theta}+i\cot\theta\frac{\partial}{\partial\phi}\right]\psi_{1}^{1}=0 \ \ \ \ \ (4)$

Since ${\psi_{1}^{1}}$ is also an eigenfunction of ${L_{z}}$ with eigenvalue ${\hbar}$, we know that

$\displaystyle \psi_{1}^{1}=U_{1}^{1}\left(r,\theta\right)e^{i\phi} \ \ \ \ \ (5)$

Thus 4 becomes

$\displaystyle \left(\frac{\partial}{\partial\theta}-\cot\theta\right)U_{1}^{1}=0 \ \ \ \ \ (6)$

This can be solved by writing it in the form

 $\displaystyle \frac{dU_{1}^{1}}{U_{1}^{1}}$ $\displaystyle =$ $\displaystyle \frac{d\left(\sin\theta\right)}{\sin\theta}\ \ \ \ \ (7)$ $\displaystyle \ln U_{1}^{1}$ $\displaystyle =$ $\displaystyle \ln\left(\sin\theta\right)+\ln R\left(r\right)+\ln A \ \ \ \ \ (8)$

where ${R}$ is some unspecified function of ${r}$, and ${A}$ is a constant. We therefore have

 $\displaystyle U_{1}^{1}\left(r,\theta\right)$ $\displaystyle =$ $\displaystyle R\left(r\right)\left(A\sin\theta\right) \ \ \ \ \ (9)$

If we ignore ${R}$ for now, we can normalize over the angular coordinates by requiring

$\displaystyle \int\left|A\sin\theta\right|^{2}d\Omega=1 \ \ \ \ \ (10)$

The element ${d\Omega}$ of solid angle is

$\displaystyle d\Omega=\sin\theta\;d\phi\;d\theta \ \ \ \ \ (11)$

so we have

 $\displaystyle \left|A\right|^{2}\int_{0}^{\pi}\int_{0}^{2\pi}\sin^{3}\theta\;d\phi\;d\theta$ $\displaystyle =$ $\displaystyle 2\pi\left|A\right|^{2}\int_{0}^{\pi}\sin\theta\left(1-\cos^{2}\theta\right)d\theta\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8\pi}{3}\left|A\right|^{2}\ \ \ \ \ (13)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{8\pi}} \ \ \ \ \ (14)$

Thus the spherical harmonic ${Y_{1}^{1}}$ is (using Shankar’s normalization convention of multiplying by ${\left(-1\right)^{\ell}}$):

$\displaystyle Y_{1}^{1}=-\sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi} \ \ \ \ \ (15)$

We can now get ${Y_{1}^{0}}$ by applying ${L_{-}}$ to ${Y_{1}^{1}}$. From 2 we have

 $\displaystyle L_{-}Y_{1}^{1}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\left(1+1\right)\left(1-1+1\right)}Y_{1}^{0}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2}\hbar Y_{1}^{0} \ \ \ \ \ (17)$

From 3 we have

 $\displaystyle L_{-}Y_{1}^{1}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]Y_{1}^{1}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left(-\sqrt{\frac{3}{8\pi}}\right)\left[\cos\theta-i\cot\theta\left(i\sin\theta\right)\right]e^{i\phi}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\hbar\sqrt{\frac{3}{8\pi}}\cos\theta \ \ \ \ \ (20)$

Comparing the last two results gives

 $\displaystyle \sqrt{2}\hbar Y_{1}^{0}$ $\displaystyle =$ $\displaystyle 2\sqrt{\frac{3}{8\pi}}\cos\theta\ \ \ \ \ (21)$ $\displaystyle Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{4\pi}}\cos\theta \ \ \ \ \ (22)$

Repeating the process, we get

 $\displaystyle L_{-}Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\left(1+0\right)\left(1-0+1\right)}Y_{1}^{-1}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2}\hbar Y_{1}^{-1} \ \ \ \ \ (24)$

Also

 $\displaystyle L_{-}Y_{1}^{0}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]Y_{1}^{0}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\sqrt{\frac{3}{4\pi}}\left(-\sin\theta-0\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\frac{3}{4\pi}}\sin\theta e^{-i\phi} \ \ \ \ \ (27)$

Thus

 $\displaystyle \sqrt{2}\hbar Y_{1}^{-1}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\frac{3}{4\pi}}\sin\theta e^{-i\phi}\ \ \ \ \ (28)$ $\displaystyle Y_{1}^{-1}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{8\pi}}\sin\theta e^{-i\phi} \ \ \ \ \ (29)$

Comparing these results with Shankar’s equation 12.5.39 we see that they match. [This exercise is similar to one we did earlier, where we used the raising operator to generate spherical harmonics with higher values of ${m}$.]