Orbital Angular Momentum

Orbital angular momentum is the angular momentum associated with rotations of spatial wavefunctions. The corresponding spatial rotation group is $SO(3)$. For one particle in three dimensions, it is defined by

$$\mathbf L=\mathbf R\times\mathbf P.$$

It is distinct from spin: orbital angular momentum acts on position dependence, while spin acts on internal degrees of freedom.

Classical Analogy and Quantum Definition

Classically, angular momentum is $\mathbf r\times\mathbf p$. Quantum mechanically, position and momentum become operators:

$$L_i=\sum_{j,k}\epsilon_{ijk}R_jP_k.$$

In Cartesian components,

$$L_x=YP_z-ZP_y, \qquad L_y=ZP_x-XP_z, \qquad L_z=XP_y-YP_x.$$

Each product uses different Cartesian components, which commute: $[R_j,P_k]=0$ for $j\ne k$. Thus no ordering ambiguity is introduced by the cross product. The formulas act first on a suitable common domain, such as smooth rapidly decreasing wavefunctions. With $P_j=-i\hbar\partial_j$, they become differential operators.

Position-Space Representation

In position space,

$$L_x = -i\hbar \left( y\frac{\partial}{\partial z} -z\frac{\partial}{\partial y} \right),$$ $$L_y = -i\hbar \left( z\frac{\partial}{\partial x} -x\frac{\partial}{\partial z} \right),$$

and

$$L_z = -i\hbar \left( x\frac{\partial}{\partial y} -y\frac{\partial}{\partial x} \right).$$

These formulas show explicitly that orbital angular momentum differentiates the angular dependence of a wavefunction.

The derivation and the angular form of $L^2$ are developed in Position-Space Representation.

Commutators

Orbital angular momentum satisfies the same angular momentum algebra:

$$[L_i,L_j] = i\hbar\sum_k\epsilon_{ijk}L_k.$$

It also rotates position and momentum as vectors:

$$[L_i,R_j] = i\hbar\sum_k\epsilon_{ijk}R_k,$$

and

$$[L_i,P_j] = i\hbar\sum_k\epsilon_{ijk}P_k.$$

Generator of Orbital Rotations

A spatial rotation by angle $\theta$ about $\hat{\mathbf n}$ acts on the orbital wavefunction through

$$U_L(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf L \right).$$

For a rotation about the $z$ axis,

$$U_L(\hat z,\theta) = e^{-i\theta L_z/\hbar}.$$

In spherical coordinates, $L_z$ becomes

$$L_z=-i\hbar\frac{\partial}{\partial\phi}.$$

This is why eigenfunctions of $L_z$ contain factors $e^{im\phi}$.

The one-dimensional Particle on a Ring: First Encounter uses this operator in its simplest angular setting.

Distinction from Spin

Orbital angular momentum is built from spatial operators. Spin is not. A spin-$1/2$ state can have angular momentum even when there is no spatial orbit. Conversely, a scalar wavefunction can carry orbital angular momentum through its angular dependence.

For a particle with spin, total angular momentum is

$$\mathbf J=\mathbf L+\mathbf S.$$

The same algebra applies to $\mathbf J$, but the Hilbert space includes both position and spin degrees of freedom. The one-particle coupled basis is developed in Addition of Orbital and Spin Angular Momentum.

Central Potentials

For a scalar particle with no magnetic vector potential, in a central potential with rotation-invariant domain,

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

rotational invariance gives

$$[H,L_i]=0.$$

This is why central-potential eigenstates can be labeled by $\ell$ and $m$, as explained in Central Potentials and Rotational Symmetry. The corresponding radial-angular separation is derived in Angular and Radial Separation.

Canonical and mechanical angular momentum

Here $\mathbf P$ is the canonical generator of translations. In a magnetic background the mechanical momentum is $\boldsymbol\Pi=\mathbf P-q\mathbf A$, and $\mathbf R\times\boldsymbol\Pi$ describes mechanical orbital angular momentum. It need not equal the generator $\mathbf L=\mathbf R\times\mathbf P$, nor be conserved. Rotations of the full electromagnetic problem may require a compensating gauge transformation. The relation to magnetic moments is discussed in Magnetic Moments from Orbital Motion.

Exercises

1. Show that $L_z=-i\hbar\,\partial/\partial\phi$ in the $xy$ plane using $x=r\cos\phi$ and $y=r\sin\phi$.

Solution

Start from

$$L_z=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right).$$

The angular derivative is

$$\frac{\partial}{\partial\phi} = -y\frac{\partial}{\partial x} +x\frac{\partial}{\partial y}.$$

Therefore $L_z=-i\hbar\,\partial/\partial\phi$.

2. Why can orbital angular momentum have only integer $\ell$ values for ordinary single-valued wavefunctions on the sphere?

Solution

The $L_z$ dependence is $e^{im\phi}$. Single-valuedness under $\phi\mapsto\phi+2\pi$ requires $e^{i2\pi m}=1$, so $m$ is an integer. The ladder structure then gives integer $\ell$ for orbital angular momentum.

3. A normalizable wavefunction has the form $\psi(\mathbf r)=f(r)(x+iy)$ with a smooth decaying radial function. Find its $L_z$ eigenvalue. Compare with the real combination $f(r)x$.

Solution

The radial function is unchanged by $L_z$. Acting on the remaining factor gives

$$L_z(x+iy)=-i\hbar(ix-y)=\hbar(x+iy).$$

Thus this state has $m=+1$. But $x=[(x+iy)+(x-iy)]/2$ combines $m=+1$ and $m=-1$, so $f(r)x$ is not a definite-$L_z$ state. A real orbital can have angular-momentum fluctuations without a classical orbit.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter 3.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.