05 Jun 2026

Angular Momentum and Rotations

Angular momentum as the generator of rotations, rotation operators, and transformation of quantum states.

msc semester-i quantum-mechanics rotations angular-momentum

Angular momentum is the generator of rotations in quantum mechanics. This means that a rotation of a state is produced by applying an exponential operator built from the angular momentum operator.

Rotation about an axis

For a rotation by angle $\theta$ about a unit vector $\hat{\mathbf n}$, the rotation operator is

\[\boxed{ U(R)=\exp\left(-\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf J\right). }\]

For rotations about the coordinate axes,

\[U_z(\phi)=\exp\left(-\frac{i}{\hbar}\phi J_z\right),\] \[U_y(\theta)=\exp\left(-\frac{i}{\hbar}\theta J_y\right),\] \[U_x(\alpha)=\exp\left(-\frac{i}{\hbar}\alpha J_x\right).\]

The negative sign appears because the operator acts on the quantum state.

Infinitesimal rotation

For a very small angle $\delta\theta$ about the $z$-axis,

\[U_z(\delta\theta) \simeq 1-\frac{i}{\hbar}\delta\theta J_z.\]

Thus $J_z$ is the generator of rotations about the $z$-axis. Similarly, $J_x$ and $J_y$ generate rotations about the $x$ and $y$ axes.

Transformation of operators

If a state transforms as

\[\lvert \psi'\rangle=U\lvert\psi\rangle,\]

then an operator transforms as

\[A'=UAU^\dagger.\]

For a vector operator $\mathbf V$, the angular momentum generators satisfy

\[[J_i,V_j]=i\hbar\epsilon_{ijk}V_k.\]

This relation means that $\mathbf V$ transforms as a vector under rotations.

Rotation of angular momentum eigenstates

Since

\[J_z\lvert j,m\rangle=\hbar m\lvert j,m\rangle,\]

a rotation about the $z$-axis gives

\[U_z(\phi)\lvert j,m\rangle = e^{-im\phi}\lvert j,m\rangle.\]

Thus an eigenstate of $J_z$ only gains a phase under rotation about the $z$-axis.

For rotations about other axes, states with different $m$ values are mixed.

Spin-one-half rotation

For spin one-half,

\[S_i=\frac{\hbar}{2}\sigma_i.\]

A rotation about the $z$-axis is

\[U_z(\phi)=\exp\left(-\frac{i}{2}\phi\sigma_z\right).\]

Using $\sigma_z^2=I$,

\[U_z(\phi)= \begin{pmatrix} e^{-i\phi/2}&0\\ 0&e^{i\phi/2} \end{pmatrix}.\]

For $\phi=2\pi$,

\[U_z(2\pi)=-I.\]

Therefore a spin-one-half state changes sign under a $2\pi$ rotation and returns exactly to itself only after a $4\pi$ rotation. This is one of the most striking features of spinors.

Rotational invariance and conservation

If the Hamiltonian is invariant under rotations, then

\[[H,\mathbf J]=0.\]

This implies conservation of angular momentum. For a spherically symmetric potential,

\[V=V(r),\]

the Hamiltonian commutes with $L^2$ and $L_z$:

\[[H,L^2]=0,\qquad [H,L_z]=0.\]

Hence angular momentum quantum numbers can be used to label the energy eigenstates.

Main points

• Angular momentum generates rotations.
• A finite rotation is represented by an exponential operator.
• Rotations about $z$ multiply $\lvert j,m\rangle$ by $e^{-im\phi}$.
• Spin-one-half states acquire a minus sign under a $2\pi$ rotation.
• Rotational symmetry gives conservation of angular momentum.

Practice questions

1. Derive the infinitesimal rotation operator about the $z$-axis.
2. Show that $U_z(\phi)\lvert j,m\rangle=e^{-im\phi}\lvert j,m\rangle$.
3. Find the spin-one-half rotation matrix about the $z$-axis.
4. Explain why spinors require a $4\pi$ rotation to return to the same state.
5. Why does spherical symmetry imply conservation of angular momentum?