2.2.1 Angular Momentum Algebra

2.2.1 Angular Momentum Algebra#

Prompts

  • What are the fundamental commutation relations for angular momentum, and how do they encode the geometry of 3D rotations?

  • Define the Casimir operator \(\hat{J}^2\) and explain why it commutes with all components of \(\hat{\boldsymbol{J}}\). Why does this allow simultaneous eigenstates \(\vert j, m\rangle\)?

  • How do the ladder operators \(\hat{J}_\pm\) connect eigenstates with different \(m\)? Derive their normalization coefficients.

  • Show that angular momentum quantization (\(j = 0, \frac{1}{2}, 1, \ldots\)) follows purely from the algebra, without solving a differential equation.

Lecture Notes#

Overview#

In Chapter 1, we studied a single qubit — the spin-1/2 system with two eigenstates. This raises a natural question: what about higher spin? What constrains the possible values of angular momentum, and how do they arise?

The answer is remarkable: angular momentum quantization follows entirely from commutation relations, without solving any differential equation. All forms of angular momentum — orbital (\(\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}\)), spin (\(\boldsymbol{S}\)), and total (\(\boldsymbol{J} = \boldsymbol{L} + \boldsymbol{S}\)) — satisfy the same algebra. This algebra is the Lie algebra of SU(2), the group of rotations we met in §1.3.3. The physical content is simple: rotations in 3D do not commute, and angular momentum operators are their generators.

Angular Momentum Algebra#

Defining Relation

The three components of any angular momentum operator \(\hat{\boldsymbol{J}}=(\hat J_x,\hat J_y,\hat J_z)\) satisfy

(38)#\[ [\hat J_i,\hat J_j]=\mathrm{i}\hbar\,\epsilon_{ijk}\hat J_k \]

Explicitly,

\[ [\hat J_x,\hat J_y]=\mathrm{i}\hbar\hat J_z, \quad [\hat J_y,\hat J_z]=\mathrm{i}\hbar\hat J_x, \quad [\hat J_z,\hat J_x]=\mathrm{i}\hbar\hat J_y. \]

For orbital angular momentum these follow from \([\hat r_i,\hat p_j]=\mathrm{i}\hbar\delta_{ij}\); for spin they are postulated as the rotation-generator algebra.

Derivation: Orbital Angular Momentum Commutators

Using \(\hat L_i=\epsilon_{imn}\hat r_m\hat p_n\) and the canonical commutators \([\hat r_a,\hat p_b]=\mathrm{i}\hbar\delta_{ab}\), \([\hat r_a,\hat r_b]=[\hat p_a,\hat p_b]=0\):

\[\begin{split} \begin{split} [\hat L_i,\hat L_j] &=\epsilon_{imn}\epsilon_{jpq}[\hat r_m\hat p_n,\hat r_p\hat p_q]\\ &=\epsilon_{imn}\epsilon_{jpq}\Bigl( \hat r_m[\hat p_n,\hat r_p]\hat p_q +\hat r_p\hat r_m[\hat p_n,\hat p_q] +[\hat r_m,\hat r_p]\hat p_q\hat p_n +\hat r_p[\hat r_m,\hat p_q]\hat p_n \Bigr)\\ &=\epsilon_{imn}\epsilon_{jpq}\Bigl( -\mathrm{i}\hbar\,\delta_{np}\,\hat r_m\hat p_q +\mathrm{i}\hbar\,\delta_{mq}\,\hat r_p\hat p_n \Bigr)\\ &=-\mathrm{i}\hbar\,\epsilon_{imn}\epsilon_{jnq}\hat r_m\hat p_q +\mathrm{i}\hbar\,\epsilon_{inq}\epsilon_{jpm}\hat r_p\hat p_n. \end{split} \end{split}\]

Using \(\epsilon_{abc}\epsilon_{ade}=\delta_{bd}\delta_{ce}-\delta_{be}\delta_{cd}\) gives

\[ [\hat L_i,\hat L_j]=\mathrm{i}\hbar\,(\hat r_i\hat p_j-\hat r_j\hat p_i) =\mathrm{i}\hbar\,\epsilon_{ijk}\hat L_k. \]

\(\checkmark\)

Casimir Operator#

Casimir Operator

The total angular momentum squared

(39)#\[ \hat J^2=\hat J_x^2+\hat J_y^2+\hat J_z^2 \]

commutes with every component:

(40)#\[ [\hat J^2,\hat J_i]=0\qquad(i=x,y,z). \]

Derivation: \([\hat J^2,\hat J_z]=0\)

\[ [\hat J^2,\hat J_z]=[\hat J_x^2,\hat J_z]+[\hat J_y^2,\hat J_z]. \]

Using \([\hat A^2,\hat B]=\hat A[\hat A,\hat B]+[\hat A,\hat B]\hat A\):

\[ [\hat J_x^2,\hat J_z]= -\mathrm{i}\hbar(\hat J_x\hat J_y+\hat J_y\hat J_x), \]
\[ [\hat J_y^2,\hat J_z]= +\mathrm{i}\hbar(\hat J_y\hat J_x+\hat J_x\hat J_y), \]

which cancel exactly.

Ladder Operators#

Ladder Operators

Define

(41)#\[ \hat J_+=\hat J_x+\mathrm{i}\hat J_y, \qquad \hat J_-=\hat J_x-\mathrm{i}\hat J_y. \]

They satisfy

\[ [\hat J_z,\hat J_\pm]=\pm\hbar\hat J_\pm, \qquad [\hat J_+,\hat J_-]=2\hbar\hat J_z, \qquad [\hat J^2,\hat J_\pm]=0. \]

Derivation: Ladder Operator Commutation Relations

From \(\hat J_\pm=\hat J_x\pm\mathrm{i}\hat J_y\) and \([\hat J_i,\hat J_j]=\mathrm{i}\hbar\epsilon_{ijk}\hat J_k\):

\[\begin{split} \begin{split} [\hat J_z,\hat J_+] &=[\hat J_z,\hat J_x]+\mathrm{i}[\hat J_z,\hat J_y] =\mathrm{i}\hbar\hat J_y+\hbar\hat J_x =\hbar\hat J_+,\\ [\hat J_z,\hat J_-] &=[\hat J_z,\hat J_x]-\mathrm{i}[\hat J_z,\hat J_y] =\mathrm{i}\hbar\hat J_y-\hbar\hat J_x =-\hbar\hat J_-. \end{split} \end{split}\]
\[\begin{split} \begin{split} [\hat J_+,\hat J_-] &=[\hat J_x+\mathrm{i}\hat J_y,\hat J_x-\mathrm{i}\hat J_y] =-\mathrm{i}[\hat J_x,\hat J_y]+\mathrm{i}[\hat J_y,\hat J_x]\\ &=-2\mathrm{i}[\hat J_x,\hat J_y] =-2\mathrm{i}(\mathrm{i}\hbar\hat J_z)=2\hbar\hat J_z. \end{split} \end{split}\]

And by linearity,

\[ [\hat J^2,\hat J_\pm]=[\hat J^2,\hat J_x\pm\mathrm{i}\hat J_y] =[\hat J^2,\hat J_x]\pm\mathrm{i}[\hat J^2,\hat J_y]=0. \]

These commutation relations imply two useful operator identities:

(42)#\[\begin{split} \begin{split} \hat J_-\hat J_+ &= \hat J^2-\hat J_z^2-\hbar\hat J_z,\\ \hat J_+\hat J_- &= \hat J^2-\hat J_z^2+\hbar\hat J_z. \end{split} \end{split}\]

Derivation: Ladder Product Identities

From

\[ \hat J_x=\frac{1}{2}(\hat J_++\hat J_-), \qquad \hat J_y=\frac{1}{2\mathrm{i}}(\hat J_+-\hat J_-), \]

we get

\[ \hat J_x^2+\hat J_y^2=\frac{1}{2}(\hat J_+\hat J_-+\hat J_-\hat J_+)=\hat J^2-\hat J_z^2. \]

Using \([\hat J_+,\hat J_-]=2\hbar\hat J_z\),

\[ \hat J_+\hat J_-=\hat J_-\hat J_+ + 2\hbar\hat J_z. \]

Solving these two equations gives (42).

Angular Momentum Representation Theory#

Because \([\hat J^2,\hat J_z]=0\), we can choose simultaneous eigenstates and build irreducible multiplets algebraically.

Common Eigenstates of \(\hat J^2\) and \(\hat J_z\)#

Simultaneous Eigenstates

Let \(\vert j,m\rangle\) satisfy

(43)#\[\begin{split} \begin{split} \hat J^2\vert j,m\rangle &= \hbar^2 j(j+1)\vert j,m\rangle,\\ \hat J_z\vert j,m\rangle &= \hbar m\vert j,m\rangle. \end{split} \end{split}\]

At this stage, \(j,m\in\mathbb{R}\) are labels; quantization is derived below.

Ladder Operator Action#

Ladder Action on \(\vert j,m\rangle\)

(44)#\[\begin{split} \begin{split} \hat J_+\vert j,m\rangle &= \hbar\sqrt{j(j+1)-m(m+1)}\,\vert j,m+1\rangle,\\ \hat J_-\vert j,m\rangle &= \hbar\sqrt{j(j+1)-m(m-1)}\,\vert j,m-1\rangle. \end{split} \end{split}\]

So \(\hat J_+\) raises \(m\) by one and \(\hat J_-\) lowers \(m\) by one.

Derivation: Ladder-Action Coefficients

From \([\hat J_z,\hat J_\pm]=\pm\hbar\hat J_\pm\) and \([\hat J^2,\hat J_\pm]=0\), the states \(\hat J_\pm\vert j,m\rangle\) lie in the same \(j\) multiplet with \(m\to m\pm1\).

For normalization, use (42):

\[ \|\hat J_+\vert j,m\rangle\|^2 =\langle j,m\vert\hat J_-\hat J_+\vert j,m\rangle =\hbar^2\bigl[j(j+1)-m(m+1)\bigr], \]
\[ \|\hat J_-\vert j,m\rangle\|^2 =\langle j,m\vert\hat J_+\hat J_-\vert j,m\rangle =\hbar^2\bigl[j(j+1)-m(m-1)\bigr]. \]

Taking square roots yields (44).

Angular Momentum Quantization#

Quantization Rules

(45)#\[ j\in\left\{0,\tfrac12,1,\tfrac32,2,\ldots\right\}, \qquad m\in\{-j,-j+1,\ldots,j\}. \]

For each \(j\), the multiplet dimension is \(2j+1\).

Proof: Quantum bootstrap argument

The quantum bootstrap is an algebraic method: constrain quantum numbers by positivity of norms. For any operator \(\hat A\) and state \(\vert\psi\rangle\),

\[ \|\hat A\vert\psi\rangle\|^2\ge 0. \]

Applied to angular momentum, this principle fixes the allowed values of \(j\) and \(m\).

From (43), take simultaneous eigenstates \(\vert j,m\rangle\) of \(\hat J^2\) and \(\hat J_z\).

(1) Positivity constraints. Apply positivity to \(\hat A=\hat J_+\) and \(\hat A=\hat J_-\) on \(\vert j,m\rangle\):

\[ \|\hat J_+\vert j,m\rangle\|^2\ge0, \qquad \|\hat J_-\vert j,m\rangle\|^2\ge0. \]

Using (44), these become

(46)#\[\begin{split} \begin{split} \text{(raising)}\;& j(j+1)-m(m+1)\ge 0,\\ \text{(lowering)}\;& j(j+1)-m(m-1)\ge 0. \end{split} \end{split}\]

So \(m\) is bounded: \(-j\le m\le j\).

(2) Ladder unit-step structure. From \([\hat J_z,\hat J_\pm]=\pm\hbar\hat J_\pm\), each ladder action changes \(m\) by exactly one unit. Therefore all \(m\) values in one multiplet differ by integers.

(3) Finite-dimensional irreducibility. In a finite-dimensional irrep, the ladder must terminate at top and bottom states:

\[ \hat J_+\vert j,m_{\max}\rangle=0, \qquad \hat J_-\vert j,m_{\min}\rangle=0. \]

By (44), vanishing coefficients imply

\[ m_{\max}=j, \qquad m_{\min}=-j. \]

Hence the ladder runs

\[ m=-j,-j+1,\ldots,j, \]

with \(N=2j+1\) states.

Since \(N\in\mathbb N\), we must have \(2j\in\mathbb N_0\). Therefore

\[ j\in\left\{0,\tfrac12,1,\tfrac32,\ldots\right\}, \qquad m\in\{-j,-j+1,\ldots,j\}. \]

So quantization follows from algebra + positivity + finite-dimensional irreducibility.

Integer vs. half-integer: Orbital angular momentum (\(\hat{\boldsymbol{L}}=\hat{\boldsymbol{r}}\times\hat{\boldsymbol{p}}\)) gives only integer \(\ell\) from single-valued wavefunctions, while spin admits half-integer \(j\).

Matrix Representations#

For fixed \(j\), the operators act on a \((2j+1)\)-dimensional space.

Spin-1/2: Pauli Matrices

For \(j=1/2\):

\[\begin{split} \hat J_z=\frac{\hbar}{2} \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \quad \hat J_x=\frac{\hbar}{2} \begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad \hat J_y=\frac{\hbar}{2} \begin{pmatrix}0&-\mathrm{i}\\\mathrm{i}&0\end{pmatrix}. \end{split}\]

Spin-1: 3\(\times\)3 Matrices

For \(j=1\):

\[\begin{split} \hat J_z=\hbar \begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}, \quad \hat J_x=\frac{\hbar}{\sqrt2} \begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}, \quad \hat J_y=\frac{\hbar}{\sqrt2} \begin{pmatrix}0&-\mathrm{i}&0\\\mathrm{i}&0&-\mathrm{i}\\0&\mathrm{i}&0\end{pmatrix}. \end{split}\]

Summary#

  • All angular momenta satisfy the same commutation relations \([\hat{J}_i, \hat{J}_j] = \mathrm{i}\hbar\epsilon_{ijk}\hat{J}_k\) — the Lie algebra \(\mathfrak{su}(2)\).

  • The Casimir operator \(\hat{J}^2\) commutes with all components, enabling simultaneous eigenstates \(\vert j, m\rangle\).

  • Quantization (\(j = 0, \frac{1}{2}, 1, \ldots\); \(m = -j, \ldots, j\)) follows from algebra alone.

  • Ladder operators \(\hat{J}_\pm\) shift \(m\) by \(\pm 1\) with known normalization.

  • Representations: spin-1/2 recovers Pauli matrices; spin-1 gives \(3\times 3\) matrices; general \(j\) gives \((2j+1)\)-dimensional irreducible representations.

Homework#

1. Verify the commutation relations \([\hat{J}_x, \hat{J}_y] = \mathrm{i}\hbar \hat{J}_z\) (and cyclic) by explicit computation for the spin-1/2 representation \(\hat{J}_i = \frac{\hbar}{2}\hat{\sigma}^i\).

2. Show that \([\hat{J}^2, \hat{J}_z] = 0\) by expanding \(\hat{J}^2 = \hat{J}_x^2 + \hat{J}_y^2 + \hat{J}_z^2\) and using the fundamental commutation relations. Why is this result essential for labeling states by \(j\) and \(m\) simultaneously?

3. Using the definitions \(\hat{J}_\pm = \hat{J}_x \pm \mathrm{i}\hat{J}_y\), derive \([\hat{J}_+, \hat{J}_-] = 2\hbar\hat{J}_z\) and \([\hat{J}_z, \hat{J}_\pm] = \pm\hbar\hat{J}_\pm\).

4. Use \(\hat{J}_+\vert j, m\rangle = \hbar\sqrt{(j-m)(j+m+1)}\vert j, m+1\rangle\) to explain why \(\hat{J}_+\vert 1, 1\rangle = 0\). Compute \(\hat{J}_-\vert 1, 0\rangle\) and verify using the spin-1 matrix representation.

5. For a state \(\vert j, m\rangle\), compute \(\langle \hat{J}_x\rangle\) and \(\langle \hat{J}_y\rangle\) using \(\hat{J}_x = \frac{1}{2}(\hat{J}_+ + \hat{J}_-)\). Explain physically why both vanish.

6. Show that the identity \(\hat{J}_-\hat{J}_+ = \hat{J}^2 - \hat{J}_z^2 - \hbar\hat{J}_z\) holds by expanding the left side. Use it to derive \(\|\hat{J}_+\vert j, m\rangle\|^2 = \hbar^2(j-m)(j+m+1)\).

7. Explain why orbital angular momentum only has integer \(\ell\) (hint: single-valuedness of \(Y_\ell^m(\theta, \phi)\) on the sphere) while spin can be half-integer. How does this connect to the spin-statistics theorem?