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Orbital Angular Momentum

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

L=R×P.\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.

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

Li=∑j,kϵijkRjPk.L_i=\sum_{j,k}\epsilon_{ijk}R_jP_k.

In Cartesian components,

Lx=YPz−ZPy,Ly=ZPx−XPz,Lz=XPy−YPx.L_x=YP_z-ZP_y, \qquad L_y=ZP_x-XP_z, \qquad L_z=XP_y-YP_x.

With Pj=−iℏ∂jP_j=-i\hbar\partial_j, these become differential operators on wavefunctions.

In position space,

Lx=−iℏ(y∂∂z−z∂∂y),L_x = -i\hbar \left( y\frac{\partial}{\partial z} -z\frac{\partial}{\partial y} \right), Ly=−iℏ(z∂∂x−x∂∂z),L_y = -i\hbar \left( z\frac{\partial}{\partial x} -x\frac{\partial}{\partial z} \right),

and

Lz=−iℏ(x∂∂y−y∂∂x).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 L2L^2 are developed in Position-Space Representation.

Orbital angular momentum satisfies the same angular momentum algebra:

[Li,Lj]=iℏ∑kϵijkLk.[L_i,L_j] = i\hbar\sum_k\epsilon_{ijk}L_k.

It also rotates position and momentum as vectors:

[Li,Rj]=iℏ∑kϵijkRk,[L_i,R_j] = i\hbar\sum_k\epsilon_{ijk}R_k,

and

[Li,Pj]=iℏ∑kϵijkPk.[L_i,P_j] = i\hbar\sum_k\epsilon_{ijk}P_k.

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

UL(n^,θ)=exp⁡(−iℏθ n^⋅L).U_L(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf L \right).

For a rotation about the zz axis,

UL(z^,θ)=e−iθLz/ℏ.U_L(\hat z,\theta) = e^{-i\theta L_z/\hbar}.

In spherical coordinates, LzL_z becomes

Lz=−iℏ∂∂ϕ.L_z=-i\hbar\frac{\partial}{\partial\phi}.

This is why eigenfunctions of LzL_z contain factors eimϕe^{im\phi}.

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

Orbital angular momentum is built from spatial operators. Spin is not. A spin-1/21/2 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

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

The same algebra applies to J\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.

For a central potential,

H=P22m+V(r),H=\frac{\mathbf P^2}{2m}+V(r),

rotational invariance gives

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

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

  • Treating orbital angular momentum as a literal classical trajectory of the particle.
  • Confusing orbital angular momentum L\mathbf L with total angular momentum J\mathbf J.
  • Applying orbital formulas directly to spin states.
  • Forgetting that boundary conditions and domains matter for differential operators.
  • Assuming every angular momentum quantum number is orbital; spin allows half-integer labels.
  • 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.
  • A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
  1. Show that Lz=−iℏ ∂/∂ϕL_z=-i\hbar\,\partial/\partial\phi in the xyxy plane using x=rcos⁡ϕx=r\cos\phi and y=rsin⁡ϕy=r\sin\phi.
Solution

Start from

Lz=−iℏ(x∂∂y−y∂∂x).L_z=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right).

The angular derivative is

∂∂ϕ=−y∂∂x+x∂∂y.\frac{\partial}{\partial\phi} = -y\frac{\partial}{\partial x} +x\frac{\partial}{\partial y}.

Therefore Lz=−iℏ ∂/∂ϕL_z=-i\hbar\,\partial/\partial\phi.

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

The LzL_z dependence is eimϕe^{im\phi}. Single-valuedness under ϕ↦ϕ+2π\phi\mapsto\phi+2\pi requires ei2πm=1e^{i2\pi m}=1, so mm is an integer. The ladder structure then gives integer ℓ\ell for orbital angular momentum.