Rotations and Orbital Angular Momentum

Rotating a quantum state connects three descriptions: a spatial transformation, a unitary operator, and angular-momentum generators. Their algebra determines the allowed $j,m$ multiplets. For ordinary scalar wavefunctions, the orbital realization gives integer $\ell$, spherical harmonics, and the angular structure of central-potential states. Symmetry determines that structure, but does not by itself solve the radial dynamics or fix every degeneracy.

Helpful background. Spatial symmetries fixes the active rotation, while generators and commutators explain its infinitesimal form.

Enter this chapter

Question	Read next
How do vectors and wavefunctions rotate?	Rotations in three dimensions and SO(3) versus SU(2)
What algebra do the generators obey?	Angular-momentum operators and their algebra
Why are the multiplets finite?	Ladder operators and J², Jz eigenvalues
How do the operators act on spatial wavefunctions?	Orbital angular momentum and position-space representation
How does the angular basis arise?	Spherical coordinates and spherical harmonics
What does symmetry determine in a spectrum?	Central potentials, hydrogen angular structure, and the rigid rotor
How does orbital motion couple to a field?	Orbital magnetic moments

The generator depends on what rotates

With active rotations,

$$U(\hat{\mathbf n},\theta)=e^{-i\theta\hat{\mathbf n}\cdot\mathbf J/\hbar}.$$

For a scalar spatial wavefunction, $(U(\mathcal R)\psi)(\mathbf r)=\psi(\mathcal R^{-1}\mathbf r)$ and $\mathbf J=\mathbf L=\mathbf R\times\mathbf P$. For an internal spin, the generator is $\mathbf S$. If both rotate, it is $\mathbf L+\mathbf S$. These realizations share an algebra without describing the same physical motion.

Orbital angular momentum depends on the chosen origin. Spin does not arise from a position-space cross product. The spin chapter develops that distinct realization after the common algebra below.

The algebra gives compatible labels

The defining relations are

$$[J_i,J_j] = i\hbar \sum_k\epsilon_{ijk}J_k.$$

Equivalently,

$$\begin{aligned} [J_x,J_y]&=i\hbar J_z, \\ [J_y,J_z]&=i\hbar J_x, \\ [J_z,J_x]&=i\hbar J_y. \end{aligned}$$

The components cannot generally be sharp simultaneously. The rotationally invariant quadratic operator

$$J^2=J_x^2+J_y^2+J_z^2$$

commutes with every component:

$$[J^2,J_i]=0.$$

Thus one chooses $J^2$ and one component, conventionally $J_z$, as a compatible set. Their simultaneous eigenstates satisfy

$$\begin{aligned} J^2|j,m\rangle &= \hbar^2j(j+1)|j,m\rangle, \\ J_z|j,m\rangle &= \hbar m|j,m\rangle. \end{aligned}$$

The labels $j,m$ are dimensionless. The factors of $\hbar$ belong to the operator eigenvalues.

Ladders determine the multiplets

Define

$$J_\pm=J_x\pm iJ_y.$$

They obey

$$[J_z,J_\pm]=\pm\hbar J_\pm, \qquad [J^2,J_\pm]=0.$$

Therefore $J_\pm$ changes $m$ by one while preserving $j$. With the standard phase convention,

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

Positivity of norms and termination of the ladder imply

$$j=0,\frac12,1,\frac32,\ldots,$$

and, for fixed $j$,

$$m=-j,-j+1,\ldots,j.$$

The multiplet dimension is $2j+1$. The algebra allows both integer and half-integer $j$. Additional global and wavefunction conditions determine which occur in a given realization. Ordinary scalar orbital wavefunctions have integer $\ell$, whereas spinor states can carry half-integer $j$.

The orbital realization lives on the sphere

For ordinary scalar wavefunctions on three-dimensional space,

$$\mathbf L=-i\hbar\mathbf r\times\boldsymbol\nabla, \qquad L_z=-i\hbar\partial_\phi, \qquad L^2=-\hbar^2\Delta_{S^2}.$$

The angular eigenfunctions are $Y_\ell^m(\theta,\phi)$, orthonormal under $d\Omega=\sin\theta\,d\theta\,d\phi$. Regularity and single-valuedness give integer $\ell\geq0$ and $m=-\ell,\ldots,\ell$. Spinors are different objects and can carry half-integer representations. The spherical-harmonic article states the normalization and phase convention explicitly.

An orbital picture usually displays $|Y_\ell^m|^2$ or a real linear combination. These are different from the complex wavefunction itself: density discards phase, and real combinations can change nodal patterns. For instance, $Y_1^1\propto\sin\theta\,e^{i\phi}$ has no azimuthal nodal plane, whereas a real $p_x$ angular function does. Keep the chosen basis in mind when interpreting a picture.

Rotation symmetry fixes magnetic degeneracy

For a spinless particle with a rotation-invariant domain and boundary conditions,

$$H=\frac{\mathbf P^2}{2m}+V(r),$$

rotational invariance gives

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

Stationary states can be chosen as

$$\psi_{\alpha\ell m}(r,\theta,\phi) = R_{\alpha\ell}(r) Y_\ell^m(\theta,\phi).$$

The energy is independent of $m$ in a fully rotationally invariant spinless problem. For each fixed $\ell$, symmetry therefore gives a $(2\ell+1)$-dimensional magnetic multiplet. A generic central potential can still have energy dependence on $\ell$; rotational symmetry alone does not force degeneracy between different $\ell$ values.

The radial potential controls bound states, scattering, and accidental degeneracies. Symmetry controls the angular organization. Keeping those roles separate prevents overclaiming what rotation invariance alone determines.

Hydrogen has an additional degeneracy

The ideal spinless Coulomb eigenfunctions separate as

$$\psi_{n\ell m} = R_{n\ell}(r)Y_\ell^m(\theta,\phi),$$

with

$$\begin{gathered} n=1,2,\ldots, \\ \ell=0,1,\ldots,n-1, \\ m=-\ell,\ldots,\ell. \end{gathered}$$

Angular momentum algebra determines the $m$ values once $\ell$ is fixed. The radial Coulomb equation and normalizability determine which $\ell$ occur in a given principal shell. The ideal energy’s independence of $\ell$ is an additional Coulomb degeneracy, associated with hidden symmetry, not a generic consequence of $SO(3)$ alone.

The spatial number of states in shell $n$ is

$$\sum_{\ell=0}^{n-1}(2\ell+1)=n^2.$$

Including spin-$1/2$ gives $2n^2$ basis states in the ideal shell. Spin-dependent interactions can split their energies; they do not remove states from that counting.

A field reveals which symmetry remains

For an electron in a weak uniform field $\mathbf B=B\hat{\mathbf z}$, the leading orbital Zeeman term is

$$H_Z=\frac{\mu_B B}{\hbar}L_z, \qquad \mu_B=\frac{e\hbar}{2m_e},\quad e>0.$$

The plus sign follows from the electron’s negative charge: its orbital magnetic moment is opposite to $\mathbf L$. The shift is $\mu_B Bm$, so the field splits the magnetic multiplet and preserves only axial rotational symmetry. This linear-in-field model omits the diamagnetic term and spin effects. Isotropic changes to a rotor Hamiltonian, by contrast, can change its $J$-dependent energies while preserving the magnetic degeneracy. The perturbation’s symmetry, not merely its presence, decides which degeneracy survives.

Exercises

1. Compute $J_+|3/2,1/2\rangle$ with the standard normalization.

Solution

The coefficient is

$$\hbar \sqrt{ j(j+1)-m(m+1) }.$$

For $j=3/2$ and $m=1/2$,

$$\begin{aligned} j(j+1)&=\frac{15}{4}, \\ m(m+1)&=\frac{3}{4}. \end{aligned}$$

Thus

$$J_+|3/2,1/2\rangle = \hbar\sqrt3 |3/2,3/2\rangle.$$

2. Explain why an orbital factor $e^{im\phi}$ requires integer $m$ for an ordinary scalar wavefunction, and why this does not exclude half-integer spin.

Solution

Single-valuedness of the scalar spatial wavefunction requires

$$e^{im(\phi+2\pi)}=e^{im\phi}.$$

Therefore

$$e^{i2\pi m}=1,$$

which gives integer $m$. This condition applies to the orbital position-space factor. A spinor transforms in an internal $SU(2)$ representation and can change sign under a $2\pi$ rotation while representing the same isolated ray. Its half-integer labels are not scalar orbital Fourier indices.

3. For a generic spinless central potential, how many states are symmetry-degenerate for fixed radial label and fixed $\ell=2$? Does rotational symmetry require degeneracy with $\ell=1$?

Solution

For fixed $\ell$, the allowed values are

$$m=-\ell,-\ell+1,\ldots,\ell.$$

At $\ell=2$, there are $2\ell+1=5$ values:

$$m=-2,-1,0,1,2.$$

Rotational invariance makes the energy independent of $m$ within this multiplet. It does not generally make different $\ell$ sectors degenerate. Degeneracy between $\ell=2$ and $\ell=1$ would require additional dynamics or hidden symmetry, as in special idealized potentials.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• A. Messiah, Quantum Mechanics, Dover, 1999.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.