Eigenvalues of J² and Jz

The angular momentum algebra forces the simultaneous eigenvalues of $J^2$ and $J_z$ to have the form

$$J^2|j,m\rangle = \hbar^2j(j+1)|j,m\rangle,$$

and

$$J_z|j,m\rangle = \hbar m|j,m\rangle,$$

with

$$j=0,\frac12,1,\frac32,\ldots, \qquad m=-j,-j+1,\ldots,j.$$

The derivation uses only the commutation relations, ladder operators, and positivity of norms. Orbital angular momentum then adds an extra condition: ordinary single-valued spatial wavefunctions allow only integer orbital labels $\ell=0,1,2,\ldots$. Spin and total angular momentum can have half-integer $j$.

Setup

The angular momentum components obey

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

Define

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

and

$$J_\pm=J_x\pm iJ_y.$$

Since

$$[J^2,J_z]=0,$$

we choose simultaneous eigenstates. Before knowing the allowed eigenvalues, write them as

$$J^2|\lambda,m\rangle = \hbar^2\lambda|\lambda,m\rangle,$$

and

$$J_z|\lambda,m\rangle = \hbar m|\lambda,m\rangle.$$

Here $\lambda$ and $m$ are dimensionless labels. The goal is to show that $\lambda=j(j+1)$ and that the allowed $m$ values form a finite ladder from $-j$ to $j$.

Ladder Operators Preserve J²

The ladder commutators are

$$[J_z,J_\pm]=\pm\hbar J_\pm,$$

and

$$[J^2,J_\pm]=0.$$

If $|\lambda,m\rangle$ is a simultaneous eigenstate, then

$$J_z(J_\pm|\lambda,m\rangle) = \hbar(m\pm1)(J_\pm|\lambda,m\rangle),$$

while

$$J^2(J_\pm|\lambda,m\rangle) = \hbar^2\lambda(J_\pm|\lambda,m\rangle).$$

Thus $J_+$ and $J_-$ move the $m$ label by one unit while keeping the same $J^2$ eigenvalue.

Positivity Bounds the Ladder

For a normalized simultaneous eigenstate,

$$\langle J^2-J_z^2\rangle = \langle J_x^2+J_y^2\rangle \geq0.$$

Using the eigenvalues gives

$$\hbar^2(\lambda-m^2)\geq0,$$

so

$$m^2\leq\lambda.$$

The $m$ ladder is bounded. Since ladder operations change $m$ in integer steps, repeated raising and lowering must stop.

Let $m_+$ be the highest $m$ value in a fixed ladder and $m_-$ the lowest. Then

$$J_+|\lambda,m_+\rangle=0, \qquad J_-|\lambda,m_-\rangle=0.$$

Norms Fix the Endpoints

Use the operator identities

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

and

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

Then

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

and

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

At the top,

$$\|J_+|\lambda,m_+\rangle\|^2=0,$$

so

$$\lambda=m_+(m_+ +1).$$

At the bottom,

$$\|J_-|\lambda,m_-\rangle\|^2=0,$$

so

$$\lambda=m_-(m_- -1).$$

Equating the two endpoint expressions gives

$$m_+(m_+ +1) = m_-(m_- -1).$$

Rearrange:

$$(m_+ +m_-)(m_+-m_-+1)=0.$$

The second factor is positive because $m_+\geq m_-$. Therefore

$$m_-=-m_+.$$

Define

$$j=m_+.$$

Then

$$m_-=-j,$$

and

$$\lambda=j(j+1).$$

This gives the standard eigenvalue

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

Allowed m Values

The ladder runs in unit steps from $-j$ to $j$:

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

Since the number of steps from $-j$ to $j$ is $2j$, this number must be a nonnegative integer. Therefore

$$2j=0,1,2,\ldots,$$

or

$$j=0,\frac12,1,\frac32,\ldots.$$

For fixed $j$, the number of allowed $m$ values is

$$2j+1.$$

This is the dimension of the irreducible angular momentum multiplet.

Normalized Ladder Action

After relabeling $\lambda=j(j+1)$, the norm formulas give the standard action

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

up to phase conventions for the basis states. With the usual Condon-Shortley convention, the square-root coefficient is chosen real and nonnegative.

The coefficient vanishes at the endpoints:

$$J_+|j,j\rangle=0, \qquad J_-|j,-j\rangle=0.$$

Examples

$j$	allowed $m$ values	multiplet dimension
$0$	$0$	$1$
$1/2$	$-1/2,1/2$	$2$
$1$	$-1,0,1$	$3$
$3/2$	$-3/2,-1/2,1/2,3/2$	$4$
$2$	$-2,-1,0,1,2$	$5$

For $j=1/2$, the ladder has only two states. For $j=1$, it has three states. The formula $2j+1$ is not a mnemonic; it is the length of the finite ladder forced by the algebra.

Orbital Versus Spin

The algebraic derivation permits both integer and half-integer $j$. Which values occur depends on the physical representation of rotations.

For orbital angular momentum of an ordinary scalar wavefunction, rotations act on spatial coordinates. In spherical coordinates,

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

An $L_z$ eigenfunction has angular dependence

$$e^{im\phi}.$$

Single-valuedness under

$$\phi\mapsto\phi+2\pi$$

requires

$$e^{i2\pi m}=1,$$

so

$$m\in\mathbb Z.$$

Since $m$ runs from $-\ell$ to $\ell$, orbital angular momentum has

$$\ell=0,1,2,\ldots.$$

Spin is different. Spin states transform under representations of $SU(2)$, the double cover of $SO(3)$. Half-integer spinors can change sign under a $2\pi$ rotation while representing the same physical ray. This is why spin-$1/2$ is allowed even though scalar orbital wavefunctions have integer $\ell$.

Physical Interpretation

The magnitude associated with $J^2$ is not $\hbar j$ but

$$\sqrt{j(j+1)}\,\hbar.$$

The $z$-component is

$$\hbar m.$$

For fixed $j$, the different $m$ states are different projections of the same total angular momentum multiplet along the chosen quantization axis. Choosing $z$ is a convention; another axis could be chosen, but one cannot generally assign sharp values to $J_x$, $J_y$, and $J_z$ simultaneously.

Common Mistakes

• Writing the $J^2$ eigenvalue as $\hbar^2j^2$ instead of $\hbar^2j(j+1)$.
• Forgetting that $m$ changes by integer steps even when $j$ is half-integer.
• Treating $j$ and $m$ as operator eigenvalues rather than dimensionless labels.
• Assuming the algebra alone makes orbital angular momentum integer-valued; the orbital integer condition comes from the spatial representation.
• Applying $J_+$ or $J_-$ outside the allowed ladder endpoints.
• Confusing the number of states, $2j+1$, with the largest $m$ value, $j$.

Cross-Links

• Angular Momentum Algebra
• Ladder Operators
• Orbital Angular Momentum
• Spherical Harmonics
• Central Potentials and Rotational Symmetry
• Spin-1/2 Hilbert Space
• Spin as Intrinsic Angular Momentum
• SO(3) and SU(2) Preview
• SU(2)
• SU(2) vs SO(3)
• Angular Momentum Formula Card

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• 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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Starting from $J_-J_+=J^2-J_z^2-\hbar J_z$, derive the norm of $J_+|j,m\rangle$.

Solution

For normalized $|j,m\rangle$,

$$\|J_+|j,m\rangle\|^2 = \langle j,m|J_-J_+|j,m\rangle.$$

Substitute the identity:

$$\|J_+|j,m\rangle\|^2 = \langle j,m| J^2-J_z^2-\hbar J_z |j,m\rangle.$$

Using

$$J^2|j,m\rangle = \hbar^2j(j+1)|j,m\rangle,$$

and

$$J_z|j,m\rangle = \hbar m|j,m\rangle,$$

gives

$$\|J_+|j,m\rangle\|^2 = \hbar^2\left[ j(j+1)-m(m+1) \right].$$

2. List the allowed $m$ values for $j=5/2$ and count the states.

Solution

The values are

$$m=-\frac52,-\frac32,-\frac12,\frac12,\frac32,\frac52.$$

There are

$$2j+1=2\cdot\frac52+1=6$$

states.

3. Explain why the algebra permits $j=1/2$ but orbital angular momentum of a scalar wavefunction does not have $\ell=1/2$.

Solution

The algebra only requires $2j$ to be a nonnegative integer, so half-integer representations are allowed. Orbital angular momentum acts on spatial wavefunctions. Since $L_z=-i\hbar\,\partial/\partial\phi$, an $L_z$ eigenfunction contains $e^{im\phi}$. Single-valuedness under $\phi\mapsto\phi+2\pi$ requires $e^{i2\pi m}=1$, so $m$ must be integer. Therefore orbital labels $\ell$ are integers. Half-integer spin instead comes from spinor representations of $SU(2)$.

4. Show that $J_+|j,j\rangle=0$ and $J_-|j,-j\rangle=0$ using the normalized ladder coefficient.

Solution

For the top state,

$$J_+|j,j\rangle = \hbar \sqrt{j(j+1)-j(j+1)} \,|j,j+1\rangle = 0.$$

For the bottom state,

$$J_-|j,-j\rangle = \hbar \sqrt{j(j+1)-(-j)(-j-1)} \,|j,-j-1\rangle = 0.$$

The formal outside-ladder kets do not represent physical states in the multiplet because the coefficient already vanishes.