Angular Momentum Problems

These solved problems practice angular-momentum algebra as a working tool: ladder operators, orbital eigenfunctions, central-potential labels, coupled bases, and spin–orbit shifts.

Use Angular Momentum Identity Index as the formula companion. For focused lookup, use Spherical Harmonics Quick Reference and Clebsch–Gordan Quick Reference. The canonical derivations live in Ladder Operators, Central Potentials and Rotational Symmetry, and Spin–Orbit Coupling.

Conventions

The angular momentum algebra is

$$[J_i,J_j] = i\hbar\sum_k\epsilon_{ijk}J_k, \qquad \epsilon_{xyz}=+1.$$

The standard basis satisfies

$$J^2\lvert j,m\rangle = \hbar^2j(j+1)\lvert j,m\rangle, \qquad J_z\lvert j,m\rangle = \hbar m\lvert j,m\rangle.$$

With

$$J_\pm=J_x\pm iJ_y,$$

the ladder action is

$$J_\pm\lvert j,m\rangle = \hbar \sqrt{j(j+1)-m(m\pm1)} \lvert j,m\pm1\rangle.$$

For orbital angular momentum, write $j\to\ell$ and $J_i\to L_i$. Spherical harmonics obey

$$L^2Y_\ell^m = \hbar^2\ell(\ell+1)Y_\ell^m, \qquad L_zY_\ell^m = \hbar mY_\ell^m.$$

For adding two angular momenta,

$$\mathbf J=\mathbf J_1+\mathbf J_2,$$

and the allowed total labels are

$$J= \lvert j_1-j_2\rvert, \lvert j_1-j_2\rvert+1, \ldots, j_1+j_2.$$

Skill Map

Problem group	Skills	Preparation
Ladder operators	norms, endpoints, matrix elements	Ladder Operators
Orbital eigenfunctions	$L_z$, parity, spherical harmonics	Spherical Harmonics
Central potentials	good quantum numbers, degeneracy, symmetry breaking	Central Potentials and Rotational Symmetry
Addition	dimension checks, coupled and uncoupled bases	Clebsch–Gordan Coefficients
Spin–orbit coupling	scalar products, $J^2$ identity, level shifts	Spin–Orbit Coupling

Problems

1. Ladder Coefficient from a Norm

Starting from

$$J_-J_+ = J^2-J_z^2-\hbar J_z,$$

derive the coefficient in

$$J_+\lvert j,m\rangle = C_{jm}^{(+)}\lvert j,m+1\rangle$$

for normalized states. Explain why the coefficient vanishes at the top of the multiplet.

Solution

Take the norm of the raised state:

$$\left\lVert J_+\lvert j,m\rangle \right\rVert^2 = \langle j,m|J_-J_+|j,m\rangle.$$

Use the identity in the question and the eigenvalue equations:

$$\begin{aligned} \langle j,m|J_-J_+|j,m\rangle &= \hbar^2j(j+1) - \hbar^2m^2 - \hbar^2m \\ &= \hbar^2 \left[ j(j+1)-m(m+1) \right]. \end{aligned}$$

Since the states are normalized,

$$\left|C_{jm}^{(+)}\right|^2 = \hbar^2 \left[ j(j+1)-m(m+1) \right].$$

With the standard Condon–Shortley phase convention the coefficient is chosen real and nonnegative:

$$C_{jm}^{(+)} = \hbar \sqrt{j(j+1)-m(m+1)}.$$

For $m=j$, the expression under the square root is

$$j(j+1)-j(j+1)=0,$$

so

$$J_+\lvert j,j\rangle=0.$$

The top state is annihilated because there is no allowed state with $m=j+1$ inside the same irreducible multiplet.

2. Transverse Angular Momentum in a Fixed State

For a normalized angular momentum state $\lvert j,m\rangle$, compute

$$\langle J_x^2+J_y^2\rangle.$$

Then show that

$$\langle J_x\rangle=\langle J_y\rangle=0.$$

Solution

Use

$$J^2=J_x^2+J_y^2+J_z^2.$$

Therefore

$$\begin{aligned} \langle J_x^2+J_y^2\rangle &= \langle J^2-J_z^2\rangle \\ &= \hbar^2j(j+1)-\hbar^2m^2 \\ &= \hbar^2 \left[ j(j+1)-m^2 \right]. \end{aligned}$$

For the first moments, write

$$J_x=\frac12(J_++J_-), \qquad J_y=\frac{1}{2i}(J_+-J_-).$$

The ladder operators map $\lvert j,m\rangle$ to states with $m\pm1$, which are orthogonal to $\lvert j,m\rangle$. Hence

$$\langle j,m|J_\pm|j,m\rangle=0,$$

and therefore

$$\langle J_x\rangle=\langle J_y\rangle=0.$$

The result does not mean the transverse components have no fluctuations. Their squared sum is nonzero unless the state is the trivial $j=0$ representation.

3. Orbital Eigenvalue and Parity Check

The normalized spherical harmonic $Y_2^1$ is proportional to

$$\sin\theta\cos\theta\,e^{i\phi}.$$

Use

$$L_z=-i\hbar\frac{\partial}{\partial\phi}$$

to identify its $m$ label. Then determine its parity.

Solution

Acting with $L_z$ gives

$$\begin{aligned} L_zY_2^1 &= -i\hbar \frac{\partial}{\partial\phi} \left[ \text{constant}\cdot \sin\theta\cos\theta\,e^{i\phi} \right] \\ &= -i\hbar \left[ i\, \text{constant}\cdot \sin\theta\cos\theta\,e^{i\phi} \right] \\ &= \hbar Y_2^1. \end{aligned}$$

Thus $m=1$. The $\ell$ label is already given by the subscript $\ell=2$, so

$$L^2Y_2^1 = \hbar^2\,2(2+1)Y_2^1 = 6\hbar^2Y_2^1.$$

The parity of a scalar spherical harmonic is

$$(-1)^\ell.$$

For $\ell=2$, the parity is even:

$$Y_2^1(\pi-\theta,\phi+\pi) = Y_2^1(\theta,\phi).$$

The sign convention in the overall normalization does not affect the eigenvalue or parity conclusions.

4. Central-Potential Labels and Degeneracy

A spinless particle in a generic central potential has a bound-state sector labeled by

$$\lvert n_r,\ell,m\rangle.$$

For fixed $n_r$ and $\ell=3$, how many states are related by rotational symmetry? Which operators form a natural commuting set? If a spin-$1/2$ degree of freedom is added but the Hamiltonian remains spin independent, what is the degeneracy before spin–orbit coupling?

Solution

For fixed $\ell$, the magnetic label is

$$m=-\ell,-\ell+1,\ldots,\ell.$$

For $\ell=3$, this gives

$$m=-3,-2,-1,0,1,2,3,$$

so there are

$$2\ell+1=7$$

states in the orbital multiplet.

For a spinless central Hamiltonian, a natural commuting set is

$$H,\quad L^2,\quad L_z.$$

The choice of $L_z$ is a basis convention inside the rotational multiplet; no physical $z$ direction is selected by a pure central potential.

If a spin-$1/2$ degree of freedom is added without spin-dependent terms, the Hilbert space is the orbital multiplet tensored with a two-dimensional spin space. The degeneracy becomes

$$(2\ell+1)(2s+1) = 7\cdot2 = 14.$$

This degeneracy is not all protected in the same way. The $m$ degeneracy follows from spatial rotations, while the spin degeneracy follows from the absence of spin-dependent interactions.

5. Dimension Check for Addition

Add angular momenta $j_1=2$ and $j_2=3/2$. List the allowed total $J$ values and verify the dimension count.

Solution

The allowed total angular momenta are

$$J= \left|2-\frac32\right|, \left|2-\frac32\right|+1, \ldots, 2+\frac32.$$

Thus

$$J=\frac12,\frac32,\frac52,\frac72.$$

The uncoupled product space has dimension

$$(2j_1+1)(2j_2+1) = (5)(4) = 20.$$

The coupled decomposition has dimension

$$\sum_J(2J+1) = \left(2\cdot\frac12+1\right) + \left(2\cdot\frac32+1\right) + \left(2\cdot\frac52+1\right) + \left(2\cdot\frac72+1\right).$$

Therefore

$$\sum_J(2J+1) = 2+4+6+8 = 20.$$

The dimension check confirms that the coupled basis is a reorganization of the same Hilbert space, not a new Hilbert space.

6. A Coupled State from Lowering

Couple $j_1=1$ and $j_2=1/2$. Starting from

$$\left|\frac32,\frac32\right\rangle = \lvert 1,1\rangle \left|\frac12,\frac12\right\rangle,$$

use the total lowering operator $J_-=J_{1-}+J_{2-}$ to find the total-$J=3/2$, $M=1/2$ state. Then find the orthogonal total-$J=1/2$, $M=1/2$ state in the same subspace.

Solution

On the left side of a coupled-state equation, the labels are total $J,M$ labels. In product states, the two kets carry the separate $j_1,m_1$ and $j_2,m_2$ labels.

On the coupled state,

$$J_-\left|\frac32,\frac32\right\rangle = \hbar \sqrt{ \frac32\left(\frac32+1\right) - \frac32\left(\frac32-1\right) } \left|\frac32,\frac12\right\rangle.$$

The square root is

$$\sqrt{3},$$

so

$$J_-\left|\frac32,\frac32\right\rangle = \hbar\sqrt3 \left|\frac32,\frac12\right\rangle.$$

Now act on the product state:

$$\begin{aligned} J_-\, \lvert1,1\rangle \left|\frac12,\frac12\right\rangle &= (J_{1-}+J_{2-}) \lvert1,1\rangle \left|\frac12,\frac12\right\rangle \\ &= \hbar\sqrt2 \lvert1,0\rangle \left|\frac12,\frac12\right\rangle + \hbar \lvert1,1\rangle \left|\frac12,-\frac12\right\rangle. \end{aligned}$$

Equating the two expressions and dividing by $\hbar\sqrt3$ gives

$$\left|\frac32,\frac12\right\rangle = \sqrt{\frac23} \lvert1,0\rangle \left|\frac12,\frac12\right\rangle + \frac{1}{\sqrt3} \lvert1,1\rangle \left|\frac12,-\frac12\right\rangle.$$

The $M=1/2$ uncoupled subspace is spanned by

$$\lvert1,0\rangle \left|\frac12,\frac12\right\rangle, \qquad \lvert1,1\rangle \left|\frac12,-\frac12\right\rangle.$$

The normalized state orthogonal to $\left|\frac32,\frac12\right\rangle$ is, up to an overall phase,

$$\left|\frac12,\frac12\right\rangle = \frac{1}{\sqrt3} \lvert1,0\rangle \left|\frac12,\frac12\right\rangle - \sqrt{\frac23} \lvert1,1\rangle \left|\frac12,-\frac12\right\rangle.$$

The overall sign is conventional, but once a Condon–Shortley convention is chosen it must be used consistently with the tables.

7. Spin–Orbit Shifts for a p Multiplet

Consider the effective Hamiltonian

$$H_{\mathrm{SO}} = A\,\frac{\mathbf L\cdot\mathbf S}{\hbar^2}$$

on a fixed orbital multiplet with $\ell=1$ and spin $s=1/2$. Find the possible energy shifts and their degeneracies.

Solution

The total angular momenta are

$$j=\ell+\frac12=\frac32, \qquad j=\ell-\frac12=\frac12.$$

Use

$$\mathbf L\cdot\mathbf S = \frac12 \left( J^2-L^2-S^2 \right).$$

On a coupled state,

$$\frac{\mathbf L\cdot\mathbf S}{\hbar^2} = \frac12 \left[ j(j+1)-\ell(\ell+1)-s(s+1) \right].$$

For $\ell=1$, $s=1/2$, and $j=3/2$,

$$\begin{aligned} \frac{\mathbf L\cdot\mathbf S}{\hbar^2} &= \frac12 \left[ \frac32\frac52 -2 -\frac12\frac32 \right] \\ &= \frac12. \end{aligned}$$

For $j=1/2$,

$$\begin{aligned} \frac{\mathbf L\cdot\mathbf S}{\hbar^2} &= \frac12 \left[ \frac12\frac32 -2 -\frac12\frac32 \right] \\ &= -1. \end{aligned}$$

Thus the shifts are

$$\Delta E_{j=3/2}=\frac{A}{2}, \qquad \Delta E_{j=1/2}=-A.$$

Their degeneracies are

$$2j+1=4 \quad \text{and} \quad 2j+1=2.$$

The original product space has dimension

$$(2\ell+1)(2s+1)=3\cdot2=6,$$

which is exactly $4+2$.

8. Choosing the Natural Basis

Let two angular momenta $\mathbf J_1$ and $\mathbf J_2$ have fixed quantum numbers $j_1$ and $j_2$. Consider

$$H=E_0+A\,\frac{\mathbf J_1\cdot\mathbf J_2}{\hbar^2}.$$

Which basis diagonalizes the interaction? Give the energy shift for a sector of total angular momentum $J$.

Solution

The interaction is a scalar product of the two angular momenta. It is not generally diagonal in the uncoupled basis

$$\lvert j_1,m_1\rangle \lvert j_2,m_2\rangle,$$

because terms such as $J_{1+}J_{2-}$ and $J_{1-}J_{2+}$ can mix product states with the same total $M=m_1+m_2$.

The coupled basis

$$\lvert j_1,j_2;J,M\rangle$$

is the natural basis. With

$$\mathbf J=\mathbf J_1+\mathbf J_2,$$

one has

$$\mathbf J_1\cdot\mathbf J_2 = \frac12 \left( J^2-J_1^2-J_2^2 \right).$$

Therefore

$$\frac{\mathbf J_1\cdot\mathbf J_2}{\hbar^2} = \frac12 \left[ J(J+1)-j_1(j_1+1)-j_2(j_2+1) \right]$$

on the sector with total angular momentum $J$. The energy shift is

$$\Delta E_J = \frac{A}{2} \left[ J(J+1)-j_1(j_1+1)-j_2(j_2+1) \right].$$

Rotational symmetry leaves the $M$ states inside a fixed $J$ multiplet degenerate unless another term breaks the symmetry.

Common Mistakes

• Treating $m$ as an angular momentum rather than the dimensionless label whose eigenvalue is $\hbar m$.
• Forgetting that $J_\pm$ changes $m$ but not $j$.
• Using $L_z=-i\hbar\partial_\phi$ with the wrong sign convention for $e^{im\phi}$.
• Counting only allowed $J$ values instead of their multiplicities $2J+1$.
• Assuming every central-potential degeneracy is caused by ordinary rotational symmetry.
• Using the uncoupled basis after a scalar coupling has made the coupled basis diagonal.
• Reading a spin–orbit energy ordering from $j$ alone without the sign and radial coefficient of the Hamiltonian.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• D. M. Brink and G. R. Satchler, Angular Momentum, 3rd ed., Oxford University Press, 1993.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.