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Spherical Coordinates

Spherical coordinates are the natural coordinate system for central potentials, angular wavefunctions, rigid rotors, and partial-wave expansions. They are useful because they separate distance from direction, but they are also a common source of errors: the volume element, unit vectors, and Laplacian all change.

This page fixes the coordinate convention used in this volume:

r≥0,0≤θ≤π,0≤ϕ<2π.r\ge0, \qquad 0\le\theta\le\pi, \qquad 0\le\phi\lt 2\pi.

Here θ\theta is the polar angle measured down from the positive zz axis, and ϕ\phi is the azimuthal angle in the xyxy plane.

Spherical-coordinate convention with radius, polar angle, and azimuth

The convention used here is the standard physics convention: rr is distance from the origin, θ\theta is the polar angle from the positive zz axis, and ϕ\phi is the azimuthal angle of the projection into the xyxy plane.

The Cartesian coordinates are

x=rsin⁡θcos⁡ϕ,x=r\sin\theta\cos\phi, y=rsin⁡θsin⁡ϕ,y=r\sin\theta\sin\phi,

and

z=rcos⁡θ.z=r\cos\theta.

Conversely,

r=x2+y2+z2,r=\sqrt{x^2+y^2+z^2}, cos⁡θ=zr,\cos\theta=\frac{z}{r},

and

ϕ=atan2⁡(y,x).\phi=\operatorname{atan2}(y,x).

The function atan2⁡\operatorname{atan2} chooses the correct quadrant for the azimuth. At r=0r=0, the angles are undefined because every direction labels the same physical point.

The spherical unit vectors are

e^r=(sin⁡θcos⁡ϕ, sin⁡θsin⁡ϕ, cos⁡θ),\hat{\mathbf e}_r = (\sin\theta\cos\phi,\, \sin\theta\sin\phi,\, \cos\theta), e^θ=(cos⁡θcos⁡ϕ, cos⁡θsin⁡ϕ, −sin⁡θ),\hat{\mathbf e}_\theta = (\cos\theta\cos\phi,\, \cos\theta\sin\phi,\, -\sin\theta),

and

e^ϕ=(−sin⁡ϕ, cos⁡ϕ, 0).\hat{\mathbf e}_\phi = (-\sin\phi,\, \cos\phi,\, 0).

The vector e^r\hat{\mathbf e}_r points outward. The vector e^θ\hat{\mathbf e}_\theta points in the direction of increasing polar angle, and e^ϕ\hat{\mathbf e}_\phi points in the direction of increasing azimuth.

Unlike Cartesian unit vectors, these unit vectors depend on position. Differentiating a vector field in spherical coordinates therefore requires differentiating both components and basis vectors.

The volume element is

d3r=r2sin⁡θ dr dθ dϕ.d^3r = r^2\sin\theta\,dr\,d\theta\,d\phi.

The solid-angle element is

dΩ=sin⁡θ dθ dϕ.d\Omega = \sin\theta\,d\theta\,d\phi.

Thus

d3r=r2 dr dΩ.d^3r=r^2\,dr\,d\Omega.

A normalized wavefunction obeys

∫0∞∫0π∫02π∣ψ(r,θ,ϕ)∣2r2sin⁡θ dϕ dθ dr=1.\int_0^\infty \int_0^\pi \int_0^{2\pi} \lvert\psi(r,\theta,\phi)\rvert^2 r^2\sin\theta\,d\phi\,d\theta\,dr =1.

This is the same normalization as ∫R3∣ψ(r)∣2 d3r=1\int_{\mathbb R^3}\lvert\psi(\mathbf r)\rvert^2\,d^3r=1, written in spherical coordinates. The factor r2sin⁡θr^2\sin\theta is not optional; it is the Jacobian of the coordinate transformation.

For a spherically symmetric probability density, the probability of finding the particle between rr and r+drr+dr is

dP=4πr2∣ψ(r)∣2 dr.dP = 4\pi r^2\lvert\psi(r)\rvert^2\,dr.

For a separated central-potential wavefunction

ψ(r,θ,ϕ)=R(r)Y(θ,ϕ),\psi(r,\theta,\phi) = R(r)Y(\theta,\phi),

with

∫S2∣Y(θ,ϕ)∣2 dΩ=1,\int_{S^2}\lvert Y(\theta,\phi)\rvert^2\,d\Omega=1,

the radial normalization is

∫0∞∣R(r)∣2r2 dr=1.\int_0^\infty \lvert R(r)\rvert^2r^2\,dr =1.

This is why the radial probability density is ∣R(r)∣2r2\lvert R(r)\rvert^2r^2, not just ∣R(r)∣2\lvert R(r)\rvert^2. If u(r)=rR(r)u(r)=rR(r), then the same condition becomes

∫0∞∣u(r)∣2 dr=1.\int_0^\infty \lvert u(r)\rvert^2\,dr =1.

For a scalar function f(r,θ,ϕ)f(r,\theta,\phi),

∇f=e^r∂f∂r+e^θ1r∂f∂θ+e^ϕ1rsin⁡θ∂f∂ϕ.\nabla f = \hat{\mathbf e}_r\frac{\partial f}{\partial r} + \hat{\mathbf e}_\theta \frac{1}{r} \frac{\partial f}{\partial\theta} + \hat{\mathbf e}_\phi \frac{1}{r\sin\theta} \frac{\partial f}{\partial\phi}.

The factors 1/r1/r and 1/(rsin⁡θ)1/(r\sin\theta) appear because a change in angle corresponds to an arc length, not a Cartesian distance. Near the poles, sin⁡θ=0\sin\theta=0, so the coordinate expression is singular even though the underlying space is not.

For scalar wavefunctions, the spherical-coordinate Laplacian is

∇2f=1r2∂∂r(r2∂f∂r)+1r2sin⁡θ∂∂θ(sin⁡θ∂f∂θ)+1r2sin⁡2θ∂2f∂ϕ2.\nabla^2 f = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2\frac{\partial f}{\partial r} \right) + \frac{1}{r^2\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta\frac{\partial f}{\partial\theta} \right) + \frac{1}{r^2\sin^2\theta} \frac{\partial^2 f}{\partial\phi^2}.

It is often useful to separate the radial and angular parts:

∇2=1r2∂∂r(r2∂∂r)+1r2ΔS2,\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},

where the unit-sphere angular Laplacian is

ΔS2=1sin⁡θ∂∂θ(sin⁡θ∂∂θ)+1sin⁡2θ∂2∂ϕ2.\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}.

In quantum mechanics,

L^2=−ℏ2ΔS2,\hat L^2=-\hbar^2\Delta_{S^2},

so the Laplacian may also be written as

∇2=1r2∂∂r(r2∂∂r)−L^2ℏ2r2.\nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r} \right) - \frac{\hat L^2}{\hbar^2r^2}.

This identity is the coordinate bridge to Angular and Radial Separation, angular momentum, and the radial Schrödinger equation.

For a central potential,

V(r)=V(r),V(\mathbf r)=V(r),

try a product form

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

The angular equation is the eigenvalue problem on the unit sphere:

ΔS2Yℓm=−ℓ(ℓ+1)Yℓm.\Delta_{S^2}Y_\ell^m = -\ell(\ell+1)Y_\ell^m.

The angular functions are spherical harmonics,

Yℓm(θ,ϕ),Y_\ell^m(\theta,\phi),

with

ℓ=0,1,2,…,m=−ℓ,…,ℓ.\ell=0,1,2,\ldots, \qquad m=-\ell,\ldots,\ell.

The radial function then obeys a one-dimensional equation on the half-line, with the angular eigenvalue producing the centrifugal term. The separation step is derived in Angular and Radial Separation, and the deeper radial boundary conventions belong to Radial Schrödinger Equation.

Spherical coordinates are singular at the origin and at the polar axis:

  • at r=0r=0, every direction labels the same point;
  • at θ=0\theta=0 and θ=π\theta=\pi, the azimuth ϕ\phi is undefined;
  • the factors 1/sin⁡θ1/\sin\theta in derivative formulas reflect a coordinate singularity, not a physical singularity.

Physical scalar wavefunctions must be single-valued and regular where the space itself is regular. In particular,

ψ(r,θ,ϕ+2π)=ψ(r,θ,ϕ)\psi(r,\theta,\phi+2\pi) = \psi(r,\theta,\phi)

for ordinary scalar wavefunctions. Spinor sign changes under rotations are a different issue and belong to the spin-volume discussion of SU(2)SU(2).

  • Swapping the physics convention for θ\theta and ϕ\phi without saying so.
  • Forgetting the Jacobian r2sin⁡θr^2\sin\theta in normalization and expectation values.
  • Replacing the spherical Laplacian by only ∂2/∂r2\partial^2/\partial r^2 for radial functions.
  • Treating the poles as physical singularities rather than coordinate singularities.
  • Forgetting that spherical unit vectors depend on angle.
  • Using arctan⁡(y/x)\arctan(y/x) instead of atan2⁡(y,x)\operatorname{atan2}(y,x) and losing the quadrant of ϕ\phi.
  • D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
  • G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2012.
  • M. L. Boas, Mathematical Methods in the Physical Sciences, 3rd ed., Wiley, 2006.
  1. Compute the volume of a ball of radius aa using the spherical volume element.
Solution

Use

d3r=r2sin⁡θ dr dθ dϕ.d^3r=r^2\sin\theta\,dr\,d\theta\,d\phi.

Then

V=∫0ar2 dr∫0πsin⁡θ dθ∫02πdϕ.V = \int_0^a r^2\,dr \int_0^\pi\sin\theta\,d\theta \int_0^{2\pi}d\phi.

The three factors are

a33,2,2π.\frac{a^3}{3}, \qquad 2, \qquad 2\pi.

Thus

V=4πa33.V=\frac{4\pi a^3}{3}.
  1. For a radial function f(r)f(r), simplify the spherical Laplacian.
Solution

If ff depends only on rr, all angular derivatives vanish. Therefore

∇2f(r)=1r2ddr(r2dfdr).\nabla^2 f(r) = \frac{1}{r^2} \frac{d}{dr} \left( r^2\frac{df}{dr} \right).

Equivalently,

∇2f(r)=d2fdr2+2rdfdr.\nabla^2 f(r) = \frac{d^2f}{dr^2} + \frac{2}{r}\frac{df}{dr}.

The second term is the part often missed by treating rr as if it were a Cartesian coordinate.

  1. Show that the unit-sphere area is 4π4\pi.
Solution

The area element on the unit sphere is

dΩ=sin⁡θ dθ dϕ.d\Omega=\sin\theta\,d\theta\,d\phi.

Thus

∫S2dΩ=∫0πsin⁡θ dθ∫02πdϕ=2⋅2π=4π.\int_{S^2}d\Omega = \int_0^\pi\sin\theta\,d\theta \int_0^{2\pi}d\phi = 2\cdot2\pi = 4\pi.