Lesson 2.7 · 2. Quantum Mechanics

Quantum Angular Momentum

Quantum angular momentum explains atomic structure, particle spin, and much of spectroscopy. Unlike a classical vector that can point in any direction, its components cannot all have definite values at once. The geometry of rotations becomes an algebra of operators.

Orbital angular momentum and spin

Orbital angular momentum $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ comes from spatial motion. Spin $\mathbf{S}$ is intrinsic and does not describe a tiny rotating sphere. Both obey the same rotation algebra.

The Rotation Algebra

For any angular momentum $\mathbf{J}$, the components satisfy:

[J_i,J_j]=i\hbar\,\varepsilon_{ijk}J_k

Because $[J_x,J_y]\neq0$, exact values cannot be assigned to all three components simultaneously. However, $J^2=J_x^2+J_y^2+J_z^2$ commutes with every component. We therefore use common eigenstates of $J^2$ and $J_z$:

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

The number $j$ is $0,\frac12,1,\frac32,\ldots$, and $m$ takes the $2j+1$ values from $-j$ to $j$.

Ladder Operators

The operators $J_\pm=J_x\pm iJ_y$ change $m$ without changing $j$:

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

The ladder must end at $m=\pm j$, or the norm would become negative. This algebraic requirement produces angular-momentum quantization.

Spin One Half

For an electron, $s=1/2$. In the $S_z$ basis, the operators are represented by Pauli matrices:

S_i=\frac{\hbar}{2}\sigma_i,\quad \sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad \sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}

A pure spin-one-half state corresponds to a direction on the Bloch sphere:

|\psi\rangle=\cos\frac{\theta}{2}|\uparrow\rangle+e^{i\phi}\sin\frac{\theta}{2}|\downarrow\rangle

Why the angle is halved

Spin-one-half states form a representation of $SU(2)$, the double cover of the rotation group $SO(3)$. A $2\pi$ rotation multiplies the ket by $-1$, and a $4\pi$ rotation restores it exactly. The global sign is invisible for an isolated state but becomes observable through interference.

Adding Angular Momenta

For two systems, $\mathbf{J}=\mathbf{J}_1+\mathbf{J}_2$. The allowed total values are:

j=|j_1-j_2|,\ |j_1-j_2|+1,\ldots,j_1+j_2

Two spin-one-half systems give a triplet with $j=1$ and a singlet with $j=0$:

|0,0\rangle=\frac{|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle}{\sqrt2}

The singlet is invariant under common rotations and is the entangled state used in Bell tests.

Orbital Angular Momentum

For a spatial wavefunction, the common eigenfunctions of $L^2$ and $L_z$ are the spherical harmonics $Y_\ell^m(\theta,\phi)$. Here $\ell$ is an integer because the orbital wavefunction must return to the same value after a full rotation. A spherical harmonic has parity $(-1)^\ell$.

Exercises

List all $m$ values for $j=2$.
Show that the singlet has eigenvalue zero under $J_z=S_{1z}+S_{2z}$.
Which total angular momenta result from combining $j_1=1$ and $j_2=1/2$?

Key Takeaways

Angular-momentum commutators encode the quantum geometry of rotations.
The states $|j,m\rangle$ diagonalize $J^2$ and one component, usually $J_z$.
Spin is intrinsic and is not classical mechanical rotation.
Adding angular momenta creates superpositions described by Clebsch-Gordan coefficients.
The two-spin singlet connects rotational symmetry and entanglement.

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