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.

Infinitesimal Rotation Check

For a small active rotation about the $z$ axis,

$$\mathcal R_z(-\delta\theta)\mathbf r = \bigl( x+\delta\theta\,y,\, y-\delta\theta\,x,\, z \bigr) + O(\delta\theta^2),$$

because the wavefunction action uses the inverse argument. Thus

$$\begin{aligned} (U_z(\delta\theta)\psi)(x,y,z) &= \psi(x+\delta\theta\,y,\, y-\delta\theta\,x,\, z) \\ &= \psi - \delta\theta \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right)\psi + O(\delta\theta^2). \end{aligned}$$

Comparing with

$$U_z(\delta\theta) = I-\frac{i}{\hbar}\delta\theta L_z + O(\delta\theta^2)$$

gives the same $L_z$ formula. This is a useful sign check: the inverse argument in the wavefunction transformation is what produces the standard generator convention.

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 a periodic domain appropriate to wavefunctions on a circle. 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.
• Thinking $L_i f(r)=0$ means all angular momentum is zero; it only says a purely radial scalar function has no angular dependence.

Cross-Links

• Coordinate Representation
• Position and Momentum Representations
• Angular Momentum Operators
• Orbital Angular Momentum
• Angular Momentum Algebra
• Eigenvalues of J² and Jz
• Spherical Coordinates
• Wave-Mechanics Spherical Coordinates
• Spherical Harmonics
• Central Potentials and Rotational Symmetry
• Angular and Radial Separation
• Particle on a Ring: First Encounter
• Spherical Harmonics Math Reference

References

• 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. Messiah, Quantum Mechanics, Dover, 1999.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.

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. Use the angular form of $L^2$ to show that a constant function on the sphere has zero orbital angular momentum.

Solution

If $Y(\theta,\phi)=C$ is constant, then

$$\frac{\partial Y}{\partial\theta}=0, \qquad \frac{\partial Y}{\partial\phi}=0.$$

Every derivative in

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

therefore vanishes, so

$$L^2Y=0.$$

This is the $\ell=0$ spherical harmonic sector.