Position-Space Representation

In position space, orbital angular momentum becomes a set of first-order differential operators. This is the concrete bridge between the abstract generator $\mathbf L=\mathbf R\times\mathbf P$ and wave mechanics: orbital angular momentum differentiates how a wavefunction changes under rotations of its spatial argument.

The compact vector formula is

$$\mathbf L = -i\hbar\,\mathbf r\times\nabla.$$

This page derives the Cartesian component formulas, explains the special role of $L_z=-i\hbar\,\partial/\partial\phi$, and records the angular form of $L^2$ used by spherical harmonics. The group-theoretic role of $\mathbf L$ belongs to Orbital Angular Momentum; the coordinate conventions for $r,\theta,\phi$ are fixed in Spherical Coordinates.

Coordinate Representation

For a spinless particle in $\mathbb R^3$, the position representation sends a state to a wavefunction

$$\psi(\mathbf r)=\langle\mathbf r|\psi\rangle.$$

The position and momentum operators act as

$$(X_i\psi)(\mathbf r)=x_i\psi(\mathbf r),$$

and

$$(P_i\psi)(\mathbf r) = -i\hbar\frac{\partial\psi}{\partial x_i}(\mathbf r),$$

on a suitable domain. Therefore the orbital angular momentum operator

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

acts as a differential operator.

Cartesian Components

Using

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

one obtains

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

and

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

These formulas are the position-space form of the same generator relation

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

Each component differentiates along the infinitesimal rotation field around the corresponding axis.

The inverse argument in an active rotation, $(U(\mathcal R)\psi)(\mathbf r)=\psi(\mathcal R^{-1}\mathbf r)$, fixes the minus sign. The expansion is derived in Rotations in Three Dimensions.

Azimuthal Derivative

In the $xy$ plane,

$$x=\rho\cos\phi, \qquad y=\rho\sin\phi.$$

The angular derivative at fixed $\rho$ is

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

Therefore

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

This is the simplest way to see why $L_z$ eigenfunctions have azimuthal dependence

$$e^{im\phi}.$$

Acting on such a factor gives

$$L_z e^{im\phi} = \hbar m\,e^{im\phi}.$$

For ordinary single-valued scalar wavefunctions, $e^{im(\phi+2\pi)}=e^{im\phi}$ requires $m\in\mathbb Z$. The full angular momentum ladder then gives integer orbital labels $\ell$.

Angular Form of L Squared

The operator

$$L^2=L_x^2+L_y^2+L_z^2$$

contains only angular derivatives. In spherical coordinates,

$$L^2 = -\hbar^2 \left[ \frac{1}{\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \frac{\partial}{\partial\theta} \right) + \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial\phi^2} \right].$$

Equivalently,

$$L^2=-\hbar^2\Delta_{S^2},$$

where $\Delta_{S^2}$ is the Laplacian on the unit sphere. This is why the eigenfunctions of $L^2$ are functions on the sphere, namely the spherical harmonics.

The radial coordinate does not appear in this angular operator. Radial dynamics enters through the Hamiltonian, not through the angular generator itself.

Radial Functions and Central Potentials

Any function that depends only on

$$r=\sqrt{x^2+y^2+z^2}$$

is unchanged by rotations. Therefore

$$L_i f(r)=0$$

for each component $L_i$. More generally, for a central potential $V(r)$,

$$[L_i,V(r)]=0.$$

This is the position-space reason central potentials are rotationally invariant. When the kinetic energy is also rotationally invariant, one obtains

$$[H,L_i]=0$$

for the central Hamiltonian. The separation into radial and angular pieces is developed in Angular and Radial Separation and summarized from the symmetry side in Central Potentials and Rotational Symmetry.

Domains and Boundary Conditions

Differential operators are not fully specified by formulas alone. Their domains matter.

On the line or in $\mathbb R^3$, square-integrability and differentiability conditions are part of the operator definition. On angular coordinates, periodicity and regularity matter. For example, the operator

$$-i\hbar\frac{\partial}{\partial\phi}$$

is self-adjoint on the circle domain of square-integrable functions with a square-integrable weak first derivative and matching endpoint values, $f(2\pi)=f(0)$. If the boundary condition is changed, the spectrum can change. This is why the particle on a ring is the cleanest first model of angular quantization.

For ordinary orbital angular momentum on the sphere, regularity at the poles and single-valuedness around $\phi$ select the usual spherical harmonics.

Common Mistakes

• Dropping the minus sign in $\mathbf L=-i\hbar\,\mathbf r\times\nabla$.
• Forgetting that $L_z=-i\hbar\,\partial/\partial\phi$ uses the azimuthal angle, not the polar angle.
• Treating the Cartesian differential formulas as if they apply to spin states.
• Ignoring domains and boundary conditions for angular derivatives.
• Forgetting the $\sin\theta$ factors in $L^2$ on the sphere.
• A purely radial scalar wavefunction really does have zero orbital angular momentum. An attached spin state can still carry intrinsic angular momentum; the Cartesian derivatives above do not act on that spin factor.

Exercises

1. Derive $L_z=-i\hbar\,\partial/\partial\phi$ from the Cartesian formula.

Solution

Start from

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

In polar coordinates in the $xy$ plane,

$$x=\rho\cos\phi, \qquad y=\rho\sin\phi.$$

At fixed $\rho$,

$$\frac{\partial x}{\partial\phi}=-\rho\sin\phi=-y, \qquad \frac{\partial y}{\partial\phi}=\rho\cos\phi=x.$$

Thus

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

Therefore

$$x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} = \frac{\partial}{\partial\phi},$$

so

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

2. Show that $L_z e^{im\phi}=\hbar m e^{im\phi}$.

Solution

Using $L_z=-i\hbar\,\partial/\partial\phi$,

$$L_z e^{im\phi} = -i\hbar \frac{\partial}{\partial\phi} e^{im\phi} = -i\hbar (im)e^{im\phi} = \hbar m e^{im\phi}.$$

3. Verify directly that $L_z f(r)=0$ for a radial function $f(r)$.

Solution

Use the chain rule:

$$\frac{\partial f}{\partial x} = f'(r)\frac{x}{r}, \qquad \frac{\partial f}{\partial y} = f'(r)\frac{y}{r}.$$

Then

$$\left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right)f(r) = x f'(r)\frac{y}{r} - y f'(r)\frac{x}{r} =0.$$

Therefore $L_z f(r)=0$. The same geometric argument applies to $L_x$ and $L_y$: radial functions are unchanged by rotations.

4. Apply $L^2$ to $Y(\theta,\phi)=\cos\theta$. Identify its orbital label.

Solution

There is no azimuthal derivative. The remaining term gives

$$L^2\cos\theta =-\frac{\hbar^2}{\sin\theta}\frac{d}{d\theta} (-\sin^2\theta) =2\hbar^2\cos\theta.$$

Thus $\ell(\ell+1)=2$, or $\ell=1$, and $m=0$. Multiplication by $\sqrt{3/(4\pi)}$ normalizes this function to $Y_1^0$.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• 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.