---
title: The Angular-Momentum Algebra and Ladder Operators
module: Angular Momentum
moduleNumber: 6
lessonNumber: 2
order: 602
summary: >
  The eigenvalues of angular momentum follow from the commutation relations alone,
  with no reference to coordinates or wavefunctions. Raising and lowering operators
  built from the components generate finite multiplets, force the total quantum
  number to be a non-negative integer or half-integer, and fix the matrix elements
  of every component. The half-integer values excluded by orbital motion appear
  here, and they are what spin realizes.
topics: [Angular Momentum]
sources:
  - book: Sakurai & Napolitano
    ref: "§3.5 Eigenvalues and Eigenstates of Angular Momentum"
  - book: Griffiths & Schroeter
    ref: "§4.3.1 Ladder Operators"
  - book: Shankar
    ref: "Ch. 12 §12.5 — The Eigenvalue Problem of L²"
draft: false
---

The [spherical-harmonic](/quantum-mechanics/angular-momentum/orbital-angular-momentum-and-spherical-harmonics)
treatment solved a differential equation and read off the spectrum. That route
ties angular momentum to a spatial wavefunction and delivers only integer $\ell$.
A purely algebraic method uses nothing but the commutation relations
$[J_i, J_j] = i\hbar\varepsilon_{ijk}J_k$, taken as the definition of _any_
angular momentum. It reproduces the orbital spectrum, fixes every matrix element,
and additionally admits half-integer quantum numbers. The half-integers are not a
mathematical curiosity; they are the values that
[spin](/quantum-mechanics/spin/spin-half-pauli-matrices-and-stern-gerlach)
takes, and no differential equation on the sphere can reach them.

Write $\vec J$ for a generic angular momentum, meaning any triple of Hermitian
operators obeying the algebra. Orbital $\vec L$, spin $\vec S$, and the total
$\vec J = \vec L + \vec S$ are all instances.

## The algebra and its Casimir operator

The defining relations and the one combination that commutes with the whole set:

$$
[J_x, J_y] = i\hbar\,J_z,
\qquad
[J_y, J_z] = i\hbar\,J_x,
\qquad
[J_z, J_x] = i\hbar\,J_y,
$$

and, as derived for the orbital case and unchanged here,

$$
J^2 = J_x^2 + J_y^2 + J_z^2,
\qquad
[J^2, J_i] = 0 \;\;\text{for every } i.
$$

$J^2$ is the **Casimir operator** of the algebra: it commutes with every element,
so it takes a fixed value on any irreducible multiplet. Since $J^2$ and $J_z$
commute and are both Hermitian, they possess a common orthonormal eigenbasis.
Label the joint eigenstates by their eigenvalues, writing them provisionally as

$$
J^2\,\ket{a, b} = \hbar^2\,a\,\ket{a, b},
\qquad
J_z\,\ket{a, b} = \hbar\,b\,\ket{a, b},
$$

with $a, b$ dimensionless. The task is to determine the allowed pairs $(a, b)$
from the algebra.

> **Definition (Standard angular-momentum basis).** The simultaneous
> orthonormal eigenstates $\ket{a, b}$ of the commuting pair $\{J^2, J_z\}$, one
> label for the magnitude and one for the $z$-projection. Every representation of
> the angular-momentum algebra decomposes into these states.

The two labels play different roles. $J^2$ is constant across an entire
multiplet, so it sorts the state space into blocks; $J_z$ resolves the states
_within_ a block. A complete set of commuting observables for a rotational
problem therefore reads $\{J^2, J_z\}$ plus whatever radial or internal labels
the full Hamiltonian needs. Because $J^2$ commutes with all three components, and
in a rotationally invariant problem with $H$ as well, the $2j+1$ states of a
multiplet are degenerate in energy: rotating a stationary state about any axis
produces another stationary state of the same energy. That degeneracy is the
dynamical fingerprint of the symmetry, and lifting it requires an interaction
that singles out a direction, such as an external field.

$$
% caption: Two labels organize the state space: the total-square eigenvalue
% selects a row (one multiplet) and the projection eigenvalue selects a column
% (one state within it).
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=0.95]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-3.3,0) -- (3.3,0) node[anchor=west, black] {$m$};
\draw[->, black] (0,-0.4) -- (0,3.9) node[anchor=south, black] {$J^{2}$};
\foreach \j in {0,1,2} {
  \node[anchor=east, font=\scriptsize] at (-3.5, \j*1.3) {$j = \j$};
  \foreach \m in {-2,-1,0,1,2} {
    \ifnum\m>\j \else \ifnum\m<-\j \else
      \fill[acc] (\m, \j*1.3) circle (2.8pt);
    \fi\fi
  }
}
\foreach \m/\lab in {-2/{-2}, -1/{-1}, 0/{0}, 1/{+1}, 2/{+2}}
  \node[anchor=north, font=\scriptsize, black!70] at (\m, -0.15) {\lab};
\end{tikzpicture}
$$

## Ladder operators

Define the non-Hermitian combinations

$$
J_\pm = J_x \pm i\,J_y,
\qquad
J_\pm^\dagger = J_\mp,
$$

the analogs of the harmonic-oscillator
[ladder operators](/quantum-mechanics/oscillator-and-symmetry/ladder-operators-and-the-number-states).
Their commutators with $J_z$ and $J^2$ follow directly from the algebra:

$$
[J_z, J_\pm]
  = [J_z, J_x] \pm i\,[J_z, J_y]
  = i\hbar J_y \pm i(-i\hbar J_x)
  = \pm\hbar\,(J_x \pm i J_y)
  = \pm\hbar\,J_\pm,
$$
$$
[J^2, J_\pm] = 0,
\qquad
[J_+, J_-] = 2\hbar\,J_z .
$$

The relation $[J_z, J_\pm] = \pm\hbar\,J_\pm$ is the entire mechanism. Apply $J_z$
to the state $J_\pm\ket{a,b}$ and commute it through:

$$
J_z\,(J_\pm\ket{a,b})
  = (J_\pm J_z \pm \hbar J_\pm)\ket{a,b}
  = \hbar(b \pm 1)\,(J_\pm\ket{a,b}).
$$

So $J_\pm\ket{a,b}$ is again a $J_z$ eigenstate, with eigenvalue raised or lowered
by one unit of $\hbar$. Because $[J^2, J_\pm] = 0$, the value of $a$ is untouched:
the ladder moves within a fixed-magnitude multiplet.

> **Theorem (Ladder action).** $J_+$ raises the $J_z$ eigenvalue by $\hbar$ and
> $J_-$ lowers it by $\hbar$, while both preserve the $J^2$ eigenvalue:
> $$
> J_\pm\,\ket{a, b} = c_\pm(a,b)\,\hbar\,\ket{a, b \pm 1}
> $$
> for scalars $c_\pm$ fixed below. Repeated application walks a single multiplet
> up and down in unit steps.

$$
% caption: Within one multiplet of fixed magnitude, the raising and lowering
% operators step between adjacent projection states in units of hbar.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% rungs
\foreach \y/\lab in {0/{b = -j}, 1.1/{-j+1}, 2.2/{...}, 3.3/{+j-1}, 4.4/{b = +j}} {
  \draw[black, thick] (-1.4,\y) -- (1.4,\y);
  \node[anchor=west, font=\scriptsize] at (1.7,\y) {\lab};
}
% raising arrows (left side, up)
\foreach \y in {0,1.1,2.2,3.3} {
  \draw[->, acc, very thick] (-0.9,\y+0.12) -- (-0.9,\y+0.98);
}
% lowering arrows (right side, down)
\foreach \y in {1.1,2.2,3.3,4.4} {
  \draw[->, black, very thick, dashed] (0.9,\y-0.12) -- (0.9,\y-0.98);
}
\node[acc, anchor=south, font=\scriptsize] at (-0.9,4.55) {raising};
\node[black, anchor=north, font=\scriptsize] at (0.9,-0.55) {lowering};
% top and bottom annotations
\node[anchor=east, font=\scriptsize, black!70] at (-1.6,4.4) {top};
\node[anchor=east, font=\scriptsize, black!70] at (-1.6,0) {bottom};
\end{tikzpicture}
$$

## Termination and the eigenvalue spectrum

The ladder cannot run forever. The projection is bounded by the magnitude, because
$J_x^2 + J_y^2 = J^2 - J_z^2$ is a sum of squares of Hermitian operators and hence
has non-negative expectation:

$$
\bra{a,b}\,(J^2 - J_z^2)\,\ket{a,b} = \hbar^2(a - b^2) \ge 0
\quad\Longrightarrow\quad
b^2 \le a .
$$

For fixed $a$ the allowed $b$ lie in a bounded interval, so the ladder must
terminate at both ends. Let $b_{\max}$ be the largest and $b_{\min}$ the smallest.
Termination means the step off each end produces the zero vector:

$$
J_+\,\ket{a, b_{\max}} = 0,
\qquad
J_-\,\ket{a, b_{\min}} = 0 .
$$

To turn these into equations for $a$, use the operator identities obtained by
multiplying out $J_\mp J_\pm$:

$$
J_\mp\,J_\pm = J_x^2 + J_y^2 \pm i[J_x, J_y] = J^2 - J_z^2 \mp \hbar J_z,
$$

so

$$
J^2 = J_\mp J_\pm + J_z^2 \pm \hbar J_z .
$$

Apply the top-rung equation $J_+\ket{a,b_{\max}} = 0$ to
$J^2 = J_- J_+ + J_z^2 + \hbar J_z$:

$$
\hbar^2 a = 0 + \hbar^2 b_{\max}^2 + \hbar^2 b_{\max}
\quad\Longrightarrow\quad
a = b_{\max}(b_{\max} + 1).
$$

Apply the bottom-rung equation $J_-\ket{a,b_{\min}} = 0$ to
$J^2 = J_+ J_- + J_z^2 - \hbar J_z$:

$$
\hbar^2 a = 0 + \hbar^2 b_{\min}^2 - \hbar^2 b_{\min}
\quad\Longrightarrow\quad
a = b_{\min}(b_{\min} - 1).
$$

Equating the two expressions for $a$ gives
$b_{\max}(b_{\max}+1) = b_{\min}(b_{\min}-1)$, whose only solution with
$b_{\min} \le b_{\max}$ is

$$
b_{\min} = -\,b_{\max}.
$$

Finally, $J_+$ raises $b$ in integer steps from $b_{\min}$ to $b_{\max}$, so
$b_{\max} - b_{\min} = 2\,b_{\max}$ must be a non-negative integer. Writing
$j \equiv b_{\max}$:

$$
2j \in \{0, 1, 2, \dots\}
\quad\Longrightarrow\quad
j \in \left\{0, \tfrac{1}{2}, 1, \tfrac{3}{2}, 2, \dots\right\}.
$$

> **Theorem (Angular-momentum spectrum).** The commutation relations alone force
> $$
> J^2\,\ket{j, m} = \hbar^2\,j(j+1)\,\ket{j, m},
> \qquad
> J_z\,\ket{j, m} = \hbar\,m\,\ket{j, m},
> $$
> with $j \in \{0, \tfrac12, 1, \tfrac32, \dots\}$ and, for each $j$, the
> projection $m$ running over the $2j+1$ values
> $m = -j, -j+1, \dots, j-1, j$. Integer $j$ recovers the orbital $\ell$;
> half-integer $j$ is new and is realized by spin.

The relabelling $a = j(j+1)$, $b = m$ matches the orbital result exactly for
integer $j$, and each multiplet has dimension $2j+1$.

$$
% caption: The number of projection states is 2j+1: a single state for j = 0,
% a doublet for j = one-half, a triplet for j = 1, a quartet for j = three-halves.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% four columns, one per j value
\foreach \x/\jlab/\n in {0/{j = 0}/1, 2.4/{j = 1/2}/2, 4.8/{j = 1}/3, 7.2/{j = 3/2}/4} {
  \node[anchor=south, font=\scriptsize] at (\x, 3.4) {\jlab};
}
% j=0: one rung at center
\draw[very thick] (-0.5,1.5) -- (0.5,1.5);
% j=1/2: two rungs
\draw[very thick] (1.9,1.05) -- (2.9,1.05);
\draw[very thick] (1.9,1.95) -- (2.9,1.95);
% j=1: three rungs
\draw[very thick] (4.3,0.6) -- (5.3,0.6);
\draw[very thick] (4.3,1.5) -- (5.3,1.5);
\draw[very thick] (4.3,2.4) -- (5.3,2.4);
% j=3/2: four rungs
\draw[very thick] (6.7,0.15) -- (7.7,0.15);
\draw[very thick] (6.7,1.05) -- (7.7,1.05);
\draw[very thick] (6.7,1.95) -- (7.7,1.95);
\draw[very thick] (6.7,2.85) -- (7.7,2.85);
% count labels
\foreach \x/\c in {0/1, 2.4/2, 4.8/3, 7.2/4}
  \node[anchor=north, font=\scriptsize, black!70] at (\x, -0.2) {$2j+1 = \c$};
\end{tikzpicture}
$$

## Matrix elements and representations

The scalars $c_\pm$ in the ladder action come from normalization. The squared
norm of $J_+\ket{j,m}$ is

$$
\norm{J_+\ket{j,m}}^2
  = \bra{j,m}J_- J_+\ket{j,m}
  = \bra{j,m}\,(J^2 - J_z^2 - \hbar J_z)\,\ket{j,m}
  = \hbar^2\big[j(j+1) - m(m+1)\big].
$$

Choosing real, positive phases gives the standard result:

$$
J_\pm\,\ket{j, m} = \hbar\sqrt{j(j+1) - m(m \pm 1)}\;\ket{j, m \pm 1}.
$$

The radical vanishes exactly at $m = +j$ for $J_+$ and $m = -j$ for $J_-$,
confirming that the ladder terminates and no state escapes the multiplet. From
$J_x = \tfrac12(J_+ + J_-)$ and $J_y = \tfrac{1}{2i}(J_+ - J_-)$, the matrix
elements of every component follow:

$$
\bra{j, m'}J_x\ket{j, m}
  = \frac{\hbar}{2}\Big[
    \sqrt{j(j+1) - m(m+1)}\,\delta_{m', m+1}
    + \sqrt{j(j+1) - m(m-1)}\,\delta_{m', m-1}
  \Big],
$$

with $J_y$ carrying the same entries times $\mp i$. In the $\ket{j,m}$ basis
$J_z$ is diagonal, $J_\pm$ are single off-diagonal bands, and $J_x, J_y$ are the
symmetric and antisymmetric combinations of those bands. Each fixed $j$ gives a
$(2j+1)$-dimensional irreducible representation of the algebra.

$$
% caption: In the fixed-j basis the projection operator is diagonal while the
% raising and lowering operators occupy the neighbouring off-diagonal bands.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% draw a 4x4 grid
\foreach \i in {0,1,2,3,4} {
  \draw[black] (0,\i) -- (4,\i);
  \draw[black] (\i,0) -- (\i,4);
}
% diagonal (Jz) top-left to bottom-right; grid rows go top=high index
\foreach \i in {0,1,2,3} {
  \fill[acc!30] (\i,3-\i) rectangle (\i+1,4-\i);
}
% super-diagonal (raising) and sub-diagonal (lowering) bands
\foreach \i in {0,1,2} {
  \fill[black] (\i+1,3-\i) rectangle (\i+2,4-\i);   % one above diagonal
  \fill[black!22] (\i,2-\i) rectangle (\i+1,3-\i);      % one below diagonal
}
\node[anchor=south, font=\scriptsize] at (2,4.1) {matrix in the (j, m) basis};
\node[acc, anchor=west, font=\scriptsize] at (4.3,3.5) {diagonal: $J_{z}$};
\node[anchor=west, font=\scriptsize, black] at (4.3,2.7) {above diagonal: raising};
\node[anchor=west, font=\scriptsize, black] at (4.3,1.9) {below diagonal: lowering};
\end{tikzpicture}
$$

### The uncertainty relation among components

The non-commutativity has a quantitative face. The
[generalized uncertainty relation](/quantum-mechanics/formalism/commutators-and-the-generalized-uncertainty-principle)
applied to $J_x, J_y$ with $[J_x, J_y] = i\hbar J_z$ reads

$$
\sigma_{J_x}\,\sigma_{J_y} \ge \frac{\hbar}{2}\,\big|\langle J_z\rangle\big|.
$$

Evaluate both sides in a state $\ket{j,m}$. The projection has mean
$\langle J_z\rangle = m\hbar$, while symmetry about the $z$-axis forces
$\langle J_x\rangle = \langle J_y\rangle = 0$ and
$\langle J_x^2\rangle = \langle J_y^2\rangle
= \tfrac12\langle J^2 - J_z^2\rangle
= \tfrac{\hbar^2}{2}\big[j(j+1) - m^2\big]$. The variances are equal, so

$$
\sigma_{J_x}\,\sigma_{J_y} = \frac{\hbar^2}{2}\big[j(j+1) - m^2\big]
\;\ge\; \frac{\hbar^2}{2}\,|m|,
$$

the inequality reducing to $j(j+1) \ge m^2 + |m| = |m|(|m|+1)$, which holds for all
$|m| \le j$ and is saturated at $m = \pm j$. The **stretched states** $m = \pm j$
are therefore the minimum-uncertainty states of the transverse components, the
closest an angular-momentum eigenstate comes to pointing along the axis. This is
the algebraic content of the vector-model cones: even the top rung keeps a
residual transverse spread $\sigma_{J_x} = \sigma_{J_y} = \hbar\sqrt{j/2}$.

### The spin-half and spin-one representations

For $j = \tfrac12$ the basis is $\{\ket{\tfrac12, \tfrac12}, \ket{\tfrac12, -\tfrac12}\}$
and the matrices are $\tfrac{\hbar}{2}$ times the
[Pauli matrices](/quantum-mechanics/spin/spin-half-pauli-matrices-and-stern-gerlach):

$$
J_x = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\quad
J_y = \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix},
\quad
J_z = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.
$$

For $j = 1$ the basis is $\{\ket{1,1}, \ket{1,0}, \ket{1,-1}\}$. The radical gives
$\sqrt{1\cdot 2 - 1\cdot 0} = \sqrt 2$ for the $m = 0 \leftrightarrow \pm 1$ steps,
so

$$
J_x = \frac{\hbar}{\sqrt 2}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix},
\quad
J_y = \frac{\hbar}{\sqrt 2}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix},
\quad
J_z = \hbar\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}.
$$

> **Worked example.** Verify $[J_x, J_y] = i\hbar J_z$ for $j = 1$. Multiplying
> the two matrices gives
> $$
> J_x J_y = \frac{\hbar^2}{2}\begin{pmatrix} i & 0 & -i \\ 0 & 0 & 0 \\ i & 0 & -i \end{pmatrix},
> \qquad
> J_y J_x = \frac{\hbar^2}{2}\begin{pmatrix} -i & 0 & -i \\ 0 & 0 & 0 \\ i & 0 & i \end{pmatrix},
> $$
> and subtracting leaves
> $$
> J_x J_y - J_y J_x
>   = \hbar^2\begin{pmatrix} i & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -i \end{pmatrix}
>   = i\hbar \cdot \hbar\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}
>   = i\hbar\,J_z,
> $$
> the algebra reproduced by the explicit $3 \times 3$ matrices. The half-integer
> and integer representations satisfy the same commutators; only the dimension
> differs.

### Why half-integers survive here but not for orbital motion

Both integer and half-integer $j$ solve the algebra, yet the
[spherical harmonics](/quantum-mechanics/angular-momentum/orbital-angular-momentum-and-spherical-harmonics)
carry only integer $\ell$. The algebra is silent about single-valuedness; it knows
only the commutators. When $\vec J = \vec L$ is built from $\vec r \times \vec p$,
the extra requirement that the wavefunction be a single-valued function of the
azimuth $\varphi$ removes the half-integer rungs, because those states pick up a
factor $-1$ under a $2\pi$ rotation. Spin carries no coordinate wavefunction and
faces no such constraint, so nature uses the half-integer representations for it.
This is the algebraic reason spin cannot be reduced to orbital motion of anything.

The two-index bookkeeping $\ket{j,m}$ and the ladder rules developed here are the
tools the [addition of angular momenta](/quantum-mechanics/angular-momentum/addition-of-angular-momenta-and-clebsch-gordan)
uses to combine two such multiplets into the coupled basis, where the same
lowering operator constructs the total-spin states of composite systems.[^sakurai-eig][^gs-ladder]

[^sakurai-eig]: Sakurai & Napolitano, _Modern Quantum Mechanics_, 3rd ed. (Cambridge, 2021), §3.5 — the algebraic determination of the $J^2$ and $J_z$ spectrum from the ladder operators, including the termination argument and the $\sqrt{j(j+1) - m(m\pm 1)}$ matrix elements. Publisher: https://doi.org/10.1017/9781108587280
[^gs-ladder]: Griffiths & Schroeter, _Introduction to Quantum Mechanics_, 3rd ed. (Cambridge, 2018), §4.3.1 — ladder operators $L_\pm = L_x \pm iL_y$, the spectrum $\ell(\ell+1)\hbar^2$ and $m\hbar$, and the appearance of half-integer solutions in the abstract algebra. Publisher: https://doi.org/10.1017/9781316995433