Ladder-Operator Action

Purpose

Angular-momentum ladder operators move between projection states inside one fixed-$j$ multiplet. Define

$$J_+ = J_x+iJ_y, \qquad J_- = J_x-iJ_y.$$

Their normalized action is

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

This card collects direct actions, repeated powers, matrix elements, and state-construction formulas. The norm derivation belongs at Ladder Operators, and the surrounding commutator and rotation structure is summarized on the Angular Momentum Algebra card.

Convention and labels

The dimensionful angular-momentum algebra is

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

The basis states satisfy

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

with

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

The standard phase convention chooses every nonzero square-root coefficient below to be real and positive. Rephasing basis states changes individual matrix entries but not the algebra or measurable probabilities.

At a glance

Task	Formula
Raise one step	$J_+\lvert j,m\rangle=\hbar\sqrt{(j-m)(j+m+1)}\lvert j,m+1\rangle$
Lower one step	$J_-\lvert j,m\rangle=\hbar\sqrt{(j+m)(j-m+1)}\lvert j,m-1\rangle$
Top endpoint	$J_+\lvert j,j\rangle=0$
Bottom endpoint	$J_-\lvert j,-j\rangle=0$
Recover Cartesian components	$J_x=(J_++J_-)/2$, $J_y=(J_+-J_-)/(2i)$
Preserve multiplet	$[J^2,J_\pm]=0$
Change projection	$[J_z,J_\pm]=\pm\hbar J_\pm$

Single-step action

The raising and lowering coefficients can be written in either form:

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

or

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

and

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

or

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

The factored forms make endpoint zeros and positivity transparent. The action changes $m$ by one but does not change $j$:

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

At the endpoints,

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

Any requested state with $m$ outside $[-j,j]$ is absent, not a new member of the representation.

Product identities and norms

The useful operator products are

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

Therefore

$$\langle j,m\rvert J_-J_+\lvert j,m\rangle = \hbar^2(j-m)(j+m+1)$$

and

$$\langle j,m\rvert J_+J_-\lvert j,m\rangle = \hbar^2(j+m)(j-m+1).$$

These are squared norms of the raised and lowered states. They must be nonnegative, which is one route to the finite range of $m$.

The symmetric combination gives

$$\frac12 \left( J_+J_-+J_-J_+ \right) = J_x^2+J_y^2 = J^2-J_z^2.$$

The antisymmetric combination gives

$$[J_+,J_-] = 2\hbar J_z.$$

Repeated raising and lowering

For a nonnegative integer $r$ with $r\le j-m$,

$$\begin{aligned} (J_+)^r\lvert j,m\rangle &= \hbar^r \left[ \frac{(j-m)!(j+m+r)!} {(j-m-r)!(j+m)!} \right]^{1/2} \\ &\qquad\times \lvert j,m+r\rangle. \end{aligned}$$

For $r\le j+m$,

$$\begin{aligned} (J_-)^r\lvert j,m\rangle &= \hbar^r \left[ \frac{(j+m)!(j-m+r)!} {(j+m-r)!(j-m)!} \right]^{1/2} \\ &\qquad\times \lvert j,m-r\rangle. \end{aligned}$$

All factorial arguments are nonnegative integers because $2j$ and $j\pm m$ are integers. If $r$ exceeds the stated bound, the result is zero after the ladder reaches an endpoint. Do not interpret the factorial formula outside its range by analytically continuing the factorials.

These expressions are useful in high-order matrix elements and in constructing an entire irreducible representation from one extremal state.

Constructing normalized multiplets

Starting from a normalized highest-weight state,

$$\begin{aligned} \lvert j,m\rangle &= \left[ \frac{(j+m)!} {(2j)!(j-m)!} \right]^{1/2} \\ &\qquad\times \left( \frac{J_-}{\hbar} \right)^{j-m} \lvert j,j\rangle. \end{aligned}$$

Starting from the bottom state,

$$\begin{aligned} \lvert j,m\rangle &= \left[ \frac{(j-m)!} {(2j)!(j+m)!} \right]^{1/2} \\ &\qquad\times \left( \frac{J_+}{\hbar} \right)^{j+m} \lvert j,-j\rangle. \end{aligned}$$

These formulas fix the relative phases to match the positive ladder coefficients. A different phase convention must be propagated consistently into spherical harmonics and Clebsch–Gordan coefficients.

Matrix elements

In a fixed-$j$ basis,

$$\langle j,m'\rvert J_z\lvert j,m\rangle = \hbar m\,\delta_{m'm},$$ $$\begin{aligned} \langle j,m'\rvert J_+\lvert j,m\rangle &= \hbar \sqrt{(j-m)(j+m+1)} \\ &\qquad\times \delta_{m',m+1}, \end{aligned}$$

and

$$\begin{aligned} \langle j,m'\rvert J_-\lvert j,m\rangle &= \hbar \sqrt{(j+m)(j-m+1)} \\ &\qquad\times \delta_{m',m-1}. \end{aligned}$$

The Cartesian matrix elements follow from

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

Thus $J_x$ and $J_y$ connect only $m'=m\pm1$, whereas $J_z$ is diagonal in this basis. These are matrix sparsity statements, not the full selection rules for every vector or tensor operator.

Matrix assembly

Order the basis by descending projection:

$$\left( \lvert j,j\rangle, \lvert j,j-1\rangle, \ldots, \lvert j,-j\rangle \right).$$

Then:

1. place $\hbar m$ on the diagonal of $J_z$;
2. place the raising coefficient in row $m+1$, column $m$ of $J_+$;
3. set $J_-=(J_+)^\dagger$;
4. form $J_x$ and $J_y$ from the linear combinations above;
5. verify $J^2=\hbar^2j(j+1)I$.

For $j=3/2$, this gives

$$J_+ = \hbar \begin{pmatrix} 0&\sqrt3&0&0\\ 0&0&2&0\\ 0&0&0&\sqrt3\\ 0&0&0&0 \end{pmatrix}.$$

The complete low-spin fixed matrices are collected at Spin Matrices. Always check its basis order before comparing entries.

Orbital differential realization

For ordinary orbital angular momentum acting on angular wavefunctions,

$$L_\pm = \hbar e^{\pm i\phi} \left( \pm\frac{\partial}{\partial\theta} + i\cot\theta \frac{\partial}{\partial\phi} \right).$$

With the standard spherical-harmonic phases,

$$L_\pm Y_\ell^m = \hbar \sqrt{ \ell(\ell+1)-m(m\pm1) } Y_\ell^{m\pm1}.$$

Here $j$ is replaced by the integer orbital label $\ell$. The differential formula depends on the coordinate and phase conventions; the abstract ladder action is representation-independent.

Use Spherical Harmonics for normalization, phases, and angular wavefunctions.

Spin-one-half specialization

For $j=1/2$,

$$S_+ = \hbar \begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad S_- = \hbar \begin{pmatrix}0&0\\1&0\end{pmatrix}.$$

If

$$\sigma_\pm = \frac{\sigma_x\pm i\sigma_y}{2},$$

then

$$S_\pm=\hbar\sigma_\pm.$$

The dimensionless $\sigma_\pm$ and dimensionful $S_\pm$ are not interchangeable. The corresponding basis conventions are on the Spin-Half Matrices card.

Angular-momentum ladders are not oscillator ladders

Feature	Angular momentum $J_\pm$	Oscillator $a,a^\dagger$
Preserved label	$j$	no fixed finite multiplet label
Changed label	$m\to m\pm1$	$n\to n\mp1$ or $n\to n+1$
Commutator	$[J_+,J_-]=2\hbar J_z$	$[a,a^\dagger]=1$
Ladder length	finite, $2j+1$ states	unbounded above in the ideal oscillator
Coefficient	depends on both $j$ and $m$	$\sqrt n$ or $\sqrt{n+1}$
Units	angular momentum	dimensionless in the standard convention

The notation “raising” refers to a chosen eigenvalue label, not to a universal operator algebra. See Harmonic Oscillator Ladder Operators for the oscillator case.

Calculation workflow

1. Identify the representation label $j$ and verify that $m$ is allowed.
2. Decide whether the operator is $J_+$ or $J_-$ before choosing the sign in the coefficient.
3. Prefer the factored coefficient to expose endpoint zeros.
4. For several steps, use the repeated-action formula or multiply successive coefficients while checking the endpoint.
5. For matrices, state the basis order and place row and column labels before inserting numbers.
6. Check $J_-=(J_+)^\dagger$ and $J^2=\hbar^2j(j+1)I$.
7. Keep basis phases consistent when the result will be combined with Clebsch–Gordan coefficients or spherical harmonics.

Common mistakes

• Assuming $J_\pm$ changes $j$ rather than $m$.
• Losing the factor of $\hbar$ in a dimensionful convention.
• Pairing the upper sign in $m\pm1$ with the wrong coefficient.
• Continuing beyond $m=\pm j$ instead of obtaining zero.
• Applying a repeated-action factorial formula outside its allowed range.
• Reversing row and column placement when constructing $J_+$.
• Using matrices written in descending $m$ order with a basis ordered in the opposite direction.
• Confusing angular-momentum ladders with harmonic-oscillator creation and annihilation operators.
• Mixing $S_\pm$ with dimensionless $\sigma_\pm$.
• Changing basis-state phases without updating coupling coefficients.

Canonical links

• Eigenvalues of J² and Jz derives the finite $m$ range.
• Angular Momentum Algebra owns the rotation-generator interpretation.
• Ladder Operators as Lie-Algebra Tools generalizes highest-weight methods.
• Addition of Angular Momentum uses ladder operators to construct coupled states.
• Angular-Momentum Problems supplies worked applications.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• 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.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.

Exercises

1. Evaluate $(J_+)^4\lvert2,-2\rangle$.

Solution

Use $j=2$, $m=-2$, and $r=4$ in the repeated-raising formula:

$$\begin{aligned} (J_+)^4\lvert2,-2\rangle &= \hbar^4 \left[ \frac{4!\,4!}{0!\,0!} \right]^{1/2} \lvert2,2\rangle \\ &= 24\hbar^4\lvert2,2\rangle. \end{aligned}$$

One more raising operation gives zero.

2. Compute the norm of $J_+\lvert j,m\rangle$ and identify when it vanishes.

Solution

Because $J_+^\dagger=J_-$,

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

Using $J_-J_+=J^2-J_z^2-\hbar J_z$ gives

$$\left\lVert J_+\lvert j,m\rangle\right\rVert^2 = \hbar^2(j-m)(j+m+1).$$

Within the allowed range, this vanishes at $m=j$, the highest-weight state.

3. Construct $J_x$ for $j=3/2$ in the descending-$m$ basis.

Solution

From the displayed $J_+$ matrix, $J_-=J_+^\dagger$. Therefore

$$J_x = \frac{\hbar}{2} \begin{pmatrix} 0&\sqrt3&0&0\\ \sqrt3&0&2&0\\ 0&2&0&\sqrt3\\ 0&0&\sqrt3&0 \end{pmatrix}.$$

It is Hermitian and connects only states whose $m$ values differ by one.

4. Apply $L_-$ to $Y_2^1$ and then apply $L_+$ to the result.

Solution

The first action is

$$L_-Y_2^1 = \hbar \sqrt{2(3)-1(0)} Y_2^0 = \hbar\sqrt6,Y_2^0.$$

Then

$$L_+Y_2^0 = \hbar \sqrt{2(3)-0(1)} Y_2^1 = \hbar\sqrt6,Y_2^1.$$

Combining the factors,

$$L_+L_-Y_2^1 = 6\hbar^2Y_2^1.$$

This agrees with $L_+L_-=L^2-L_z^2+\hbar L_z$ evaluated at $\ell=2$, $m=1$.