Spherical Coordinates

Spherical coordinates are the coordinate system adapted to rotations about a point. For angular momentum, their main role is not merely geometric convenience: they separate distance from direction, turn $L_z$ into an azimuthal derivative, and turn $L^2$ into the Laplacian on the unit sphere.

This page is the symmetry-side bridge. General coordinate conventions, volume elements, gradients, and the full Laplacian are treated in Spherical Coordinates in the Wave Mechanics volume. Here the focus is the angular-momentum structure that leads to spherical harmonics and central-potential quantum numbers.

Coordinate Convention

The convention is the standard physics convention:

x=r\sin\theta\cos\phi,

y=r\sin\theta\sin\phi,

and

z=r\cos\theta.

The ranges are

r\ge0, \qquad 0\le\theta\le\pi, \qquad 0\le\phi<2\pi.

Here $\theta$ is the polar angle measured from the positive $z$ axis, and $\phi$ is the azimuthal angle in the $xy$ plane. The solid-angle measure is

d\Omega=\sin\theta\,d\theta\,d\phi.

For angular wavefunctions, the natural inner product is therefore

\langle Y_1,Y_2\rangle_{S^2} = \int_0^{2\pi}\int_0^\pi Y_1(\theta,\phi)^* Y_2(\theta,\phi) \sin\theta\,d\theta\,d\phi.

The factor $\sin\theta$ belongs to the geometry of the sphere. It is not a normalization convention that can be dropped.

Lz as Azimuthal Generator

The $z$ component of orbital angular momentum is

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

In spherical coordinates this becomes

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

Thus azimuthal phase factors are $L_z$ eigenfunctions:

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

For ordinary scalar wavefunctions, single-valuedness under $\phi\mapsto\phi+2\pi$ requires

e^{i2\pi m}=1,

so $m$ is an integer. This is an orbital statement about scalar wavefunctions on space; it is not the rule for intrinsic spinors.

L Squared on the Sphere

The total orbital angular-momentum magnitude is

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

In spherical coordinates, it contains no radial derivatives:

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].

The bracketed operator is the Laplacian on the unit sphere:

\Delta_{S^2} = \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}.

Therefore

L^2=-\hbar^2\Delta_{S^2}.

This formula is the essential reason spherical harmonics appear in every central-potential problem, not only in the hydrogen atom.

Laplacian Split

The three-dimensional Laplacian separates into radial and angular pieces:

\nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r} \right) + \frac{1}{r^2}\Delta_{S^2}.

Using $L^2=-\hbar^2\Delta_{S^2}$ , this can be written as

\nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r} \right) - \frac{L^2}{\hbar^2r^2}.

For a spinless particle with central potential $V(r)$ ,

H = -\frac{\hbar^2}{2M}\nabla^2 + V(r),

so the Hamiltonian becomes

H = -\frac{\hbar^2}{2M} \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r} \right) + \frac{L^2}{2Mr^2} + V(r).

The term $L^2/(2Mr^2)$ is the angular kinetic energy. After separation it becomes the centrifugal contribution in the radial equation.

Separation of Variables

For a central potential, try

\psi(r,\theta,\phi)=R(r)Y(\theta,\phi).

Since $V(r)$ depends only on $r$ , the angular equation is an eigenvalue problem on the sphere:

L^2Y=\hbar^2\lambda Y.

Because $L_z$ commutes with $L^2$ , one also chooses

L_zY=\hbar mY.

Regularity on the sphere and single-valuedness in $\phi$ select

\lambda=\ell(\ell+1), \qquad \ell=0,1,2,\ldots,

and

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

The angular eigenfunctions are the spherical harmonics:

Y(\theta,\phi)=Y_\ell^m(\theta,\phi).

Thus central-potential wavefunctions take the angular-radial form

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

where $\alpha$ denotes the remaining radial or spectral labels.

What the Coordinates Explain

Spherical coordinates make three facts visible:

the radial coordinate $r$ is invariant under rotations;
$L_z$ differentiates the azimuthal angle $\phi$ ;
$L^2$ differentiates only the angular variables $\theta$ and $\phi$ .

This is why rotational symmetry lets one solve the angular problem once and reuse it for many radial potentials. The angular basis is universal; the radial equation depends on the specific $V(r)$ .

For the rigid rotor or a particle constrained to a sphere, the radial coordinate is fixed. Then the angular kinetic energy is the whole kinetic-energy operator, and the spectrum is controlled directly by $L^2$ .

Common Mistakes

Calling $\theta$ the azimuthal angle; in the convention here, $\theta$ is polar and $\phi$ is azimuthal.
Normalizing angular wavefunctions with $d\theta\,d\phi$ instead of $\sin\theta\,d\theta\,d\phi$ .
Forgetting the $1/\sin^2\theta$ factor in the angular Laplacian.
Treating the coordinate singularities at $\theta=0,\pi$ as physical singularities.
Assuming the $m$ degeneracy solves the radial problem; it only labels orientation within a fixed $\ell$ multiplet.
Confusing ordinary scalar single-valuedness in $\phi$ with spinor behavior under physical $2\pi$ rotations.

Cross-Links

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.
D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2012.

Exercises

Starting from $L^2=-\hbar^2\Delta_{S^2}$ , rewrite the three-dimensional Laplacian in terms of $L^2$ .

Solution

The spherical-coordinate Laplacian is

\nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r} \right) + \frac{1}{r^2}\Delta_{S^2}.

Since

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

substitution gives

\nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r} \right) - \frac{L^2}{\hbar^2r^2}.

Show why $m$ must be an integer for ordinary scalar orbital wavefunctions.

Solution

The $L_z$ eigenvalue equation gives angular dependence

Y(\theta,\phi)\propto e^{im\phi}.

For an ordinary scalar wavefunction, the same point is reached when $\phi\mapsto\phi+2\pi$ , so

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

This requires

e^{i2\pi m}=1,

so $m\in\mathbb Z$ .

Explain why $m$ does not appear in the radial equation for a central potential.

Solution

The radial equation depends on the angular eigenvalue of $L^2$ :

L^2Y_\ell^m=\hbar^2\ell(\ell+1)Y_\ell^m.

For fixed $\ell$ , every $m=-\ell,\ldots,\ell$ has the same $L^2$ eigenvalue. A central potential contains no preferred axis, so it cannot distinguish the different orientations inside the same $\ell$ multiplet. The radial equation therefore depends on $\ell$ , but not on $m$ .

Why is the singular factor $1/\sin^2\theta$ in $L^2$ not by itself a physical singularity at the poles?

Solution

At $\theta=0$ and $\theta=\pi$ , the azimuthal coordinate $\phi$ is undefined because all azimuths label the same direction. The singular coefficients reflect this coordinate degeneracy. Physical scalar wavefunctions must be regular on the sphere, and the allowed spherical harmonics satisfy the corresponding regularity conditions. The coordinate expression is singular, but the underlying sphere is smooth at the poles.