05 Jun 2026
Angular Momentum and Rotations
Angular momentum as the generator of rotations, rotation operators, and transformation of quantum states.
In quantum mechanics, a spatial symmetry is represented by an operator acting on states. For rotations, the corresponding generator is angular momentum. This means that a finite rotation of a state is produced by an exponential operator built from the angular momentum operator.
The connection is the quantum version of a familiar classical idea: angular momentum is the quantity associated with rotational motion and rotational symmetry.
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.
Finite rotations are obtained by applying many infinitesimal rotations successively. This is why the exponential form appears.
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.
Worked spinor rotation
For a general spin-one-half state in the $S_z$ basis,
\[\chi= \begin{pmatrix} a\\ b \end{pmatrix},\]a rotation by $\phi$ about the $z$-axis gives
\[\chi'= U_z(\phi)\chi = \begin{pmatrix} e^{-i\phi/2}&0\\ 0&e^{i\phi/2} \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix}.\]Therefore
\[\boxed{ \chi'= \begin{pmatrix} ae^{-i\phi/2}\\ be^{i\phi/2} \end{pmatrix}. }\]For $\phi=2\pi$,
\[\chi'=-\chi.\]This sign change has no effect on a single probability, but it is physically important in interference.
Infinitesimal operator change
For a small rotation about $z$,
\[U\simeq 1-\frac{i}{\hbar}\delta\phi J_z.\]The transformed operator is
\[A'=UAU^\dagger.\]Keeping only first-order terms,
\[\delta A=A'-A =-\frac{i}{\hbar}\delta\phi[J_z,A].\]This formula is often the shortest way to connect commutators with rotations.
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
- Derive the infinitesimal rotation operator about the $z$-axis.
- Show that $U_z(\phi)\lvert j,m\rangle=e^{-im\phi}\lvert j,m\rangle$.
- Find the spin-one-half rotation matrix about the $z$-axis.
- Explain why spinors require a $4\pi$ rotation to return to the same state.
- Why does spherical symmetry imply conservation of angular momentum?
Discussion