Separation of Variables in Spherical Coordinates

A hydrogen atom's electron feels one force only: the Coulomb pull toward a proton sitting at a point. That force depends purely on distance, not direction. This single geometric fact is why the three-dimensional The Hydrogen Atom Schrödinger Equation can be pulled apart into three one-dimensional problems. Anyone who has solved a textbook ODE can handle each one. The trick is not calculus wizardry. It is a change of coordinates so well-matched to the symmetry of the problem that the equation practically separates itself. This matters because the Schrödinger equation in Cartesian coordinates (x,y,z)(x,y,z) for the hydrogen atom is a genuinely coupled partial differential equation. The potential energy term −e2/4πε0x2+y2+z2-e^{2}/4\pi\varepsilon_{0}\sqrt{x^{2}+y^{2}+z^{2}} tangles all three variables together in a square root. No amount of staring at it produces a clean split. Switch to spherical coordinates (r,θ,ϕ)(r,\theta,\phi), however, and the potential becomes a function of rr alone. Suddenly the equation has a structure begging to be factored. One piece only cares about how far the electron is from the nucleus. The other piece only cares about which direction it points. This is the same separation-of-variables logic used for the particle in a box and the vibrating string. Here it is elevated to three dimensions and curvilinear coordinates. The payoff is enormous. Once separated, the angular problem turns out to be identical to the angular momentum problem solved for any central-force system. Its solutions are the spherical harmonics Yℓm(θ,ϕ)Y_{\ell}^{m}(\theta,\phi). These are valid for every atom, every diatomic rotor, every central potential you will ever meet. The radial problem, in turn, becomes an ordinary differential equation in rr alone. Its solutions (once solved) generate the familiar 1s,2s,2p,3d,…1s, 2s, 2p, 3d,\dots radial profiles. Crucially, they also yield the Bohr energy levels En=−13.6 eV/n2E_{n}=-13.6\,\text{eV}/n^{2}. These levels fall out with no extra assumptions. This article does one job: show exactly how the full three-variable equation splits into a radial equation and an angular equation. It also identifies the separation constants that link them. These are the same ℓ(ℓ+1)\ell(\ell+1) and mm that reappear as quantum numbers throughout atomic and molecular structure. The angular equation's solutions (The Angular Wavefunctions and Spherical Harmonics) and the radial equation's solutions (The Radial Equation and Radial Wavefunctions) are each substantial stories of their own. Here the goal is to derive, cleanly and completely, the machinery that hands those problems to you.

Key takeaways
  • The hydrogen Schrödinger equation in Cartesian coordinates does not separate; switching to spherical coordinates (r,θ,ϕ)(r,\theta,\phi) makes the potential depend on rr alone, enabling separation.
  • The wavefunction is factored as ψ(r,θ,ϕ)=R(r) Y(θ,ϕ)\psi(r,\theta,\phi) = R(r)\,Y(\theta,\phi), or further as R(r)Θ(θ)Φ(ϕ)R(r)\Theta(\theta)\Phi(\phi).
  • Substituting into the PDE and dividing by ψ\psi splits the equation into an rr-only piece and a (θ,ϕ)(\theta,\phi)-only piece, each equal to the same constant.
  • That constant is forced to equal ℓ(ℓ+1)\ell(\ell+1) by the requirement that angular solutions stay finite and single-valued on the sphere.
  • A second separation splits the angular equation itself into a ϕ\phi-only equation (giving quantum number mm) and a θ\theta-only equation (the associated Legendre equation).
  • Key idea: because the Coulomb potential depends only on rr, the full 3-D Schrödinger equation factors into one radial ODE and one universal angular ODE shared by every central-force problem.

Where this lives today

The angular equation produced by this separation — with its universal spherical-harmonic solutions — reappears essentially unchanged in every problem with spherical or near-spherical symmetry across chemistry and physics. Molecular rotational spectroscopy of diatomic and linear molecules uses precisely the same Θ(θ)Φ(ϕ)\Theta(\theta)\Phi(\phi) angular equation (the rigid rotor is mathematically identical to the hydrogen angular problem with V(r)V(r) removed entirely), so the microwave and far-infrared rotational transitions used to determine bond lengths in gas-phase molecules trace directly back to this separation. In multi-electron atomic structure theory, the central-field approximation (used in Hartree–Fock and density functional theory for atoms) treats each electron as moving in an effective spherically averaged potential from the nucleus and all other electrons; this approximation is only tractable because the resulting single-particle equations separate exactly the way shown here, producing atomic orbitals labeled by the same n,ℓ,mn,\ell,m quantum numbers used for hydrogen. Nuclear physics uses the identical angular machinery for nucleons in a spherical shell-model potential, explaining nuclear magic numbers through the same ℓ(ℓ+1)\ell(\ell+1) centrifugal structure. Modern computational chemistry codes that build Gaussian-type or Slater-type atomic basis sets explicitly encode spherical harmonics as the angular factor of every basis function, a direct legacy of this separation. Even scanning-tunneling-microscopy image simulations of atomic orbitals on surfaces, and NMR/EPR calculations involving angular momentum operators, ultimately rest on the same universal angular equation first isolated by separating the Laplacian in spherical coordinates.

Ask the tutor: Can you show step-by-step how the radial equation, once solved for the Coulomb potential, produces the quantized energies En=−13.6 eV/n2E_n = -13.6\,\text{eV}/n^2?

The physical picture

Picture the electron not as a point orbiting on a track but as a smeared-out wave ψ(r,θ,ϕ)\psi(r,\theta,\phi) filling three-dimensional space around the proton. The full time-independent Schrödinger equation is

−ℏ22μ∇2ψ+V(r)ψ=Eψ-\frac{\hbar^{2}}{2\mu}\nabla^{2}\psi + V(r)\psi = E\psi

with V(r)=−e24πε0rV(r) = -\dfrac{e^{2}}{4\pi\varepsilon_{0}r} and μ\mu the reduced mass of the electron–proton pair. Written out in Cartesian coordinates, ∇2=∂2/∂x2+∂2/∂y2+∂2/∂z2\nabla^{2} = \partial^{2}/\partial x^{2} + \partial^{2}/\partial y^{2} + \partial^{2}/\partial z^{2}, and r=x2+y2+z2r = \sqrt{x^{2}+y^{2}+z^{2}} appears inside a square root buried in VV. Every attempt to write ψ=X(x)Y(y)Z(z)\psi = X(x)Y(y)Z(z) fails, because the potential refuses to split along xx, yy, zz individually. The fix is to stop fighting the coordinate system and instead choose one that mirrors the physics. In spherical coordinates, a point is located by its distance from the origin r∈[0,∞)r\in[0,\infty), its polar angle θ∈[0,π]\theta\in[0,\pi] measured from the zz-axis, and its azimuthal angle ϕ∈[0,2π)\phi\in[0,2\pi) measured around the zz-axis. The enormous advantage: V(r)=−e2/4πε0rV(r) = -e^{2}/4\pi\varepsilon_{0}r depends on only one of these three variables. All the angular dependence has been swept entirely out of the potential and now lives purely in the kinetic-energy (Laplacian) operator. That operator, in spherical coordinates, reads

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

Every angular term carries a prefactor of 1/r21/r^{2} — this is the structural signature that makes separation possible, because multiplying through by r2r^{2} cleanly isolates a radial derivative term from a purely angular operator (recognizable as −L^2/ℏ2-\hat L^{2}/\hbar^{2}, the squared angular momentum operator in disguise).

Figure
Spherical coordinates: r, θ, φ and how the Laplacian's angular part factors out
z y x electron at (r, θ, φ) θ (polar) φ (azimuthal) nucleus V(r) depends on r only: V(r) = −e²/4πε₀r ∇² splits as: radial part (∂/∂r terms) × 1 + angular part (∂/∂θ, ∂/∂φ) × 1/r² ⇒ ψ = R(r) Y(θ,φ) splits Schrödinger eqn into one radial ODE + one angular ODE, linked by constant ℓ(ℓ+1)

The strategy, then, is the standard separation-of-variables playbook applied twice. First split off the radial dependence: guess ψ(r,θ,ϕ)=R(r)Y(θ,ϕ)\psi(r,\theta,\phi) = R(r)Y(\theta,\phi). Substituting into the Schrödinger equation and dividing everything by R(r)Y(θ,ϕ)R(r)Y(\theta,\phi) produces one side that depends only on rr and one side that depends only on (θ,ϕ)(\theta,\phi); since rr and (θ,ϕ)(\theta,\phi) are independent variables, both sides must equal the same constant. Second, split Y(θ,ϕ)=Θ(θ)Φ(ϕ)Y(\theta,\phi) = \Theta(\theta)\Phi(\phi) and repeat the argument — the ϕ\phi-only piece and the θ\theta-only piece must again share a common constant. Three coupled variables become three uncoupled ordinary differential equations, cascading from the general PDE down to single-variable problems you can solve with power series or known special functions. The constants that appear at each stage are not arbitrary — physical requirements (finiteness, periodicity, single-valuedness of the wavefunction) force them to be quantized. The first separation constant becomes ℓ(ℓ+1)\ell(\ell+1) with ℓ=0,1,2,…\ell = 0,1,2,\dots (the orbital angular momentum quantum number); the second becomes m2m^{2} with m=−ℓ,…,+ℓm = -\ell,\dots,+\ell (the magnetic quantum number). These are not put in by hand — they emerge as the only way the separated equations can have physically acceptable solutions.

Figure
The cascade of separations: PDE → radial ODE + angular PDE → radial ODE + polar ODE + azimuthal ODE
Schrödinger PDE in (r, θ, φ) 3 coupled variables ψ = R(r) Y(θ,φ) ⇒ divide by ψ radial eqn = angular eqn = ℓ(ℓ+1) Radial ODE in R(r) alone contains E and ℓ(ℓ+1) Angular PDE: Y = Θ(θ)Φ(φ) splits again ⇒ constant m² Φ(φ) ODE → m Θ(θ) ODE → ℓ,m

The core ideas

The starting point is the full time-independent Schrödinger equation for a single particle of reduced mass μ\mu moving in a central potential V(r)V(r):

−ℏ22μ[1r2∂∂r ⁣(r2∂ψ∂r)+1r2sin⁡θ∂∂θ ⁣(sin⁡θ∂ψ∂θ)+1r2sin⁡2θ∂2ψ∂ϕ2]+V(r)ψ=Eψ-\frac{\hbar^{2}}{2\mu}\left[\frac{1}{r^{2}}\frac{\partial}{\partial r}\!\left(r^{2}\frac{\partial \psi}{\partial r}\right) + \frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial \theta}\!\left(\sin\theta\frac{\partial \psi}{\partial \theta}\right) + \frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}\psi}{\partial \phi^{2}}\right] + V(r)\psi = E\psi

where:

  • ℏ\hbar is the reduced Planck constant, μ\mu the reduced mass of the two-body system,
  • r,θ,ϕr,\theta,\phi are spherical coordinates centered on the nucleus,
  • V(r)V(r) is any potential depending on distance alone (Coulomb for hydrogen, but the separation works for any central potential),
  • EE is the total energy eigenvalue,
  • ψ=ψ(r,θ,ϕ)\psi = \psi(r,\theta,\phi) is the full three-dimensional wavefunction.

Reading: this equation says the electron's total energy is the sum of a radial kinetic contribution, an angular kinetic contribution (built entirely from θ\theta- and ϕ\phi-derivatives), and the potential energy — and because VV carries no angular dependence, the angular kinetic terms are the only place angles ever appear. Assume a product solution:

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

where:

  • R(r)R(r) carries all the radial dependence,
  • Y(θ,ϕ)Y(\theta,\phi) carries all the angular dependence, later itself factored as Θ(θ)Φ(ϕ)\Theta(\theta)\Phi(\phi).

Reading: this is an ansatz — a proposed form of solution, justified after the fact by the equation splitting cleanly and by the completeness of the resulting basis of product functions. Substituting and multiplying through by −2μr2/(ℏ2RY)-2\mu r^{2}/(\hbar^{2}RY) produces the separated form:

1Rddr ⁣(r2dRdr)−2μr2ℏ2[V(r)−E]=1Y[−1sin⁡θ∂∂θ ⁣(sin⁡θ∂Y∂θ)−1sin⁡2θ∂2Y∂ϕ2]\frac{1}{R}\frac{d}{dr}\!\left(r^{2}\frac{dR}{dr}\right) - \frac{2\mu r^{2}}{\hbar^{2}}\big[V(r)-E\big] = \frac{1}{Y}\left[\frac{-1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right) - \frac{1}{\sin^{2}\theta}\frac{\partial^{2}Y}{\partial\phi^{2}}\right]

where:

  • the left side depends only on rr,
  • the right side depends only on θ\theta and ϕ\phi,
  • both sides must therefore equal the same constant, conventionally written ℓ(ℓ+1)\ell(\ell+1).

Reading: an expression that depends purely on rr can equal an expression that depends purely on angles, for all values of r,θ,ϕr,\theta,\phi, only if both are equal to one fixed number — this is the entire logical engine of separation of variables. Setting each side equal to ℓ(ℓ+1)\ell(\ell+1) produces two decoupled equations. The radial equation:

1r2ddr ⁣(r2dRdr)−ℓ(ℓ+1)r2R+2μℏ2[E−V(r)]R=0\frac{1}{r^{2}}\frac{d}{dr}\!\left(r^{2}\frac{dR}{dr}\right) - \frac{\ell(\ell+1)}{r^{2}}R + \frac{2\mu}{\hbar^{2}}\big[E-V(r)\big]R = 0

where:

  • ℓ=0,1,2,…\ell = 0,1,2,\dots is the orbital angular momentum quantum number, forced to be a non-negative integer by the angular boundary conditions,
  • ℓ(ℓ+1)ℏ2/r2\ell(\ell+1)\hbar^{2}/r^{2} acts exactly like an extra repulsive "centrifugal" potential term,
  • this single ODE, solved for V(r)=−e2/4πε0rV(r) = -e^{2}/4\pi\varepsilon_{0}r, yields the radial wavefunctions and the Bohr energies.

And the angular equation:

1sin⁡θ∂∂θ ⁣(sin⁡θ∂Y∂θ)+1sin⁡2θ∂2Y∂ϕ2+ℓ(ℓ+1)Y=0\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right) + \frac{1}{\sin^{2}\theta}\frac{\partial^{2}Y}{\partial\phi^{2}} + \ell(\ell+1)Y = 0

where:

  • this equation contains no reference to V(r)V(r) or EE at all — it is universal for every central-force problem, from the hydrogen atom to a rigid rotor to a particle on a sphere,
  • its solutions, the spherical harmonics Yℓm(θ,ϕ)Y_{\ell}^{m}(\theta,\phi), are shared verbatim across all these systems.

Reading: the entire chemical identity of the potential (Coulomb, harmonic, square well, anything central) survives only in the radial equation; the angular equation is pure geometry, blind to what force is actually acting.

Derivation ASplitting the full PDE into radial and angular equations

Goal. Starting from the spherical-coordinate Schrödinger equation, show rigorously that assuming ψ=R(r)Y(θ,ϕ)\psi = R(r)Y(\theta,\phi) forces both a radial ODE and an angular PDE, linked by one separation constant, with no approximation involved.

Step 1 — Write the Laplacian in spherical coordinates and substitute the product ansatz. Begin from

−ℏ22μ[1r2∂∂r ⁣(r2∂ψ∂r)+1r2sin⁡θ∂∂θ ⁣(sin⁡θ∂ψ∂θ)+1r2sin⁡2θ∂2ψ∂ϕ2]+V(r)ψ=Eψ.-\frac{\hbar^{2}}{2\mu}\left[\frac{1}{r^{2}}\frac{\partial}{\partial r}\!\left(r^{2}\frac{\partial\psi}{\partial r}\right) + \frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right) + \frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}\psi}{\partial\phi^{2}}\right] + V(r)\psi = E\psi.

Since Y(θ,ϕ)Y(\theta,\phi) does not depend on rr, and R(r)R(r) does not depend on θ,ϕ\theta,\phi, each derivative acts on only one factor:

∂2ψ∂r2∣radial term=Yddr ⁣(r2dRdr)/r2 (schematically),∂ψ∂θ=R ∂Y∂θ,∂2ψ∂ϕ2=R ∂2Y∂ϕ2.\frac{\partial^{2}\psi}{\partial r^{2}}\bigg|_{\text{radial term}} = Y\frac{d}{dr}\!\left(r^{2}\frac{dR}{dr}\right)\Big/r^{2}\ \text{(schematically)},\quad \frac{\partial\psi}{\partial\theta} = R\,\frac{\partial Y}{\partial\theta},\quad \frac{\partial^{2}\psi}{\partial\phi^{2}} = R\,\frac{\partial^{2}Y}{\partial\phi^{2}}.

Substituting ψ=RY\psi = RY into the full equation gives

−ℏ22μ[Yr2ddr ⁣(r2dRdr)+Rr2sin⁡θ∂∂θ ⁣(sin⁡θ∂Y∂θ)+Rr2sin⁡2θ∂2Y∂ϕ2]+V(r)RY=ERY.-\frac{\hbar^{2}}{2\mu}\left[\frac{Y}{r^{2}}\frac{d}{dr}\!\left(r^{2}\frac{dR}{dr}\right) + \frac{R}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right) + \frac{R}{r^{2}\sin^{2}\theta}\frac{\partial^{2}Y}{\partial\phi^{2}}\right] + V(r)RY = ERY.

Step 2 — Divide through by ψ=RY\psi = RY and multiply by −2μr2/ℏ2-2\mu r^{2}/\hbar^{2}. Dividing every term by RYRY turns products into ratios:

1r2Rddr ⁣(r2dRdr)+1r2Ysin⁡θ∂∂θ ⁣(sin⁡θ∂Y∂θ)+1r2Ysin⁡2θ∂2Y∂ϕ2=2μℏ2[V(r)−E].\frac{1}{r^{2}R}\frac{d}{dr}\!\left(r^{2}\frac{dR}{dr}\right) + \frac{1}{r^{2}Y\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right) + \frac{1}{r^{2}Y\sin^{2}\theta}\frac{\partial^{2}Y}{\partial\phi^{2}} = \frac{2\mu}{\hbar^{2}}\big[V(r)-E\big].

Multiply both sides by r2r^{2} and rearrange so all radial terms sit on one side:

1Rddr ⁣(r2dRdr)−2μr2ℏ2[V(r)−E]=−1Ysin⁡θ∂∂θ ⁣(sin⁡θ∂Y∂θ)−1Ysin⁡2θ∂2Y∂ϕ2.\frac{1}{R}\frac{d}{dr}\!\left(r^{2}\frac{dR}{dr}\right) - \frac{2\mu r^{2}}{\hbar^{2}}\big[V(r)-E\big] = -\frac{1}{Y\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right) - \frac{1}{Y\sin^{2}\theta}\frac{\partial^{2}Y}{\partial\phi^{2}}.

The left side is a function of rr alone; the right side is a function of θ,ϕ\theta,\phi alone. This is the crucial structural moment: the r2r^{2} multiplication has completely detached the angular terms from any residual 1/r21/r^{2} factor, leaving a true function of rr equated to a true function of angles.

Step 3 — Argue both sides must be a constant. Fix θ\theta and ϕ\phi at arbitrary values and vary rr: the right side, having no rr-dependence, stays fixed while the left side is free to change with rr — for the equality to hold for every rr, the left side cannot actually vary; it must be a constant, call it KK. By the identical argument run in reverse (fix rr, vary θ,ϕ\theta,\phi), the right side is the same constant KK. Physical convention (matched to the eventual normalizability of angular solutions) writes K=ℓ(ℓ+1)K = \ell(\ell+1), with ℓ\ell to be determined as a non-negative integer once the angular equation is solved.

Step 4 — State the two resulting equations. The radial equation is

1r2ddr ⁣(r2dRdr)+[2μℏ2(E−V(r))−ℓ(ℓ+1)r2]R=0,\frac{1}{r^{2}}\frac{d}{dr}\!\left(r^{2}\frac{dR}{dr}\right) + \left[\frac{2\mu}{\hbar^{2}}\big(E-V(r)\big) - \frac{\ell(\ell+1)}{r^{2}}\right]R = 0,

and the angular equation is

−1sin⁡θ∂∂θ ⁣(sin⁡θ∂Y∂θ)−1sin⁡2θ∂2Y∂ϕ2=ℓ(ℓ+1)Y.-\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right) - \frac{1}{\sin^{2}\theta}\frac{\partial^{2}Y}{\partial\phi^{2}} = \ell(\ell+1)Y.

Both are now exact consequences of the original PDE plus the product ansatz — no approximation has entered anywhere in this derivation.

Go deeper · Derivation B · 1 minSecond separation: splitting the angular equation into θ- and φ-only ODEs
Derivation BSecond separation: splitting the angular equation into θ- and φ-only ODEs

Goal. Show that the angular equation for Y(θ,ϕ)Y(\theta,\phi) further separates into a one-variable equation in ϕ\phi (giving quantum number mm) and a one-variable equation in θ\theta (the associated Legendre equation), by the same logic applied one level down.

Step 1 — Propose the second product ansatz. Write Y(θ,ϕ)=Θ(θ)Φ(ϕ)Y(\theta,\phi) = \Theta(\theta)\Phi(\phi) and substitute into

1sin⁡θ∂∂θ ⁣(sin⁡θ∂Y∂θ)+1sin⁡2θ∂2Y∂ϕ2+ℓ(ℓ+1)Y=0.\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right) + \frac{1}{\sin^{2}\theta}\frac{\partial^{2}Y}{\partial\phi^{2}} + \ell(\ell+1)Y = 0.

Since Φ\Phi does not depend on θ\theta and Θ\Theta does not depend on ϕ\phi, this becomes

Φsin⁡θddθ ⁣(sin⁡θdΘdθ)+Θsin⁡2θd2Φdϕ2+ℓ(ℓ+1)ΘΦ=0.\frac{\Phi}{\sin\theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \frac{\Theta}{\sin^{2}\theta}\frac{d^{2}\Phi}{d\phi^{2}} + \ell(\ell+1)\Theta\Phi = 0.

Step 2 — Divide by ΘΦ\Theta\Phi and multiply by sin⁡2θ\sin^{2}\theta. Dividing through by ΘΦ\Theta\Phi gives

sin⁡θΘddθ ⁣(sin⁡θdΘdθ)+ℓ(ℓ+1)sin⁡2θ+1Φd2Φdϕ2=0.\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \ell(\ell+1)\sin^{2}\theta + \frac{1}{\Phi}\frac{d^{2}\Phi}{d\phi^{2}} = 0.

Isolate the ϕ\phi-only piece:

1Φd2Φdϕ2=−[sin⁡θΘddθ ⁣(sin⁡θdΘdθ)+ℓ(ℓ+1)sin⁡2θ].\frac{1}{\Phi}\frac{d^{2}\Phi}{d\phi^{2}} = -\left[\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \ell(\ell+1)\sin^{2}\theta\right].

The left side depends only on ϕ\phi; the right side depends only on θ\theta. By the same argument as before, both sides equal a single constant, conventionally written −m2-m^{2}.

Step 3 — Solve the resulting φ equation and quantize m. The azimuthal equation is

d2Φdϕ2=−m2Φ⟹Φ(ϕ)=eimϕ.\frac{d^{2}\Phi}{d\phi^{2}} = -m^{2}\Phi \quad\Longrightarrow\quad \Phi(\phi) = e^{im\phi}.

Physical single-valuedness demands Φ(ϕ+2π)=Φ(ϕ)\Phi(\phi + 2\pi) = \Phi(\phi), i.e. eim(ϕ+2π)=eimϕe^{im(\phi+2\pi)} = e^{im\phi}, which holds only if e2πim=1e^{2\pi i m}=1. This forces mm to be an integer: m=0,±1,±2,…m = 0,\pm1,\pm2,\dots. This is the origin of the quantization of mm — it comes directly from requiring the wavefunction to return to the same value after a full turn around the zz-axis, not from any energy argument.

Step 4 — Write the remaining θ equation. Substituting −m2-m^{2} back gives the polar equation

1sin⁡θddθ ⁣(sin⁡θdΘdθ)+[ℓ(ℓ+1)−m2sin⁡2θ]Θ=0.\frac{1}{\sin\theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \left[\ell(\ell+1) - \frac{m^{2}}{\sin^{2}\theta}\right]\Theta = 0.

This is precisely the associated Legendre equation, whose polynomial solutions Pℓm(cos⁡θ)P_{\ell}^{m}(\cos\theta) exist and remain finite over the physical range θ∈[0,π]\theta\in[0,\pi] only when ℓ\ell is a non-negative integer and ∣m∣≤ℓ|m|\le \ell. This is exactly where the constraint ℓ(ℓ+1)\ell(\ell+1) (rather than some arbitrary constant) is retroactively justified — it is the unique family of constants for which Θ(θ)\Theta(\theta) stays finite at the poles θ=0,π\theta = 0,\pi.

Worked examples

Example 1 — Foundational: verifying the separation constant algebraically for ℓ=2\ell = 2. Take the angular equation with the separation constant already fixed at ℓ(ℓ+1)\ell(\ell+1) for ℓ=2\ell=2, so ℓ(ℓ+1)=2×3=6\ell(\ell+1) = 2\times3 = 6. Confirm directly that a known angular solution satisfies it. Take m=0m=0 and the associated Legendre polynomial P20(cos⁡θ)=12(3cos⁡2θ−1)P_{2}^{0}(\cos\theta) = \tfrac{1}{2}(3\cos^{2}\theta - 1), a valid Θ(θ)\Theta(\theta) for ℓ=2,m=0\ell=2, m=0. Let x=cos⁡θx = \cos\theta, so Θ=12(3x2−1)\Theta = \tfrac12(3x^{2}-1), and recall the polar equation in this substitution (the Legendre equation) reads

ddx ⁣[(1−x2)dΘdx]+ℓ(ℓ+1)Θ=0(m=0).\frac{d}{dx}\!\left[(1-x^{2})\frac{d\Theta}{dx}\right] + \ell(\ell+1)\Theta = 0 \quad (m=0).

Compute dΘdx=3x\dfrac{d\Theta}{dx} = 3x, so (1−x2)dΘdx=3x−3x3(1-x^{2})\dfrac{d\Theta}{dx} = 3x - 3x^{3}. Differentiate again:

ddx(3x−3x3)=3−9x2.\frac{d}{dx}(3x-3x^{3}) = 3 - 9x^{2}.

Now add ℓ(ℓ+1)Θ=6×12(3x2−1)=9x2−3\ell(\ell+1)\Theta = 6\times\tfrac12(3x^{2}-1) = 9x^{2}-3:

(3−9x2)+(9x2−3)=0.(3-9x^{2}) + (9x^{2}-3) = 0.

The equation is satisfied identically for all xx, confirming that ℓ=2\ell=2 and Θ=P20(cos⁡θ)\Theta = P_2^0(\cos\theta) are mutually consistent, exactly as the separation argument predicts. As a limiting check: setting ℓ=0\ell=0 instead gives Θ=\Theta = const (since P00=1P_0^0=1), and indeed ddx[(1−x2)⋅0]+0⋅1=0\frac{d}{dx}[(1-x^2)\cdot 0] + 0\cdot 1 = 0 trivially — the ss-orbital angular part is simply a constant, as it must be since ss orbitals are spherically symmetric.

Example 2 — Applied: reading off the centrifugal term for a 3d3d electron in hydrogen. For a hydrogen 3d3d orbital, n=3n=3, ℓ=2\ell=2. The radial equation derived above contains the term

ℓ(ℓ+1)ℏ22μr2\frac{\ell(\ell+1)\hbar^{2}}{2\mu r^{2}}

acting as an effective repulsive potential added to V(r)V(r) (this grouping is standard once the radial equation is rearranged into an effective one-dimensional Schrödinger form for u(r)=rR(r)u(r)=rR(r)). Evaluate this centrifugal term numerically at r=3a0r = 3a_{0} (roughly the most probable radius region for n=3n=3 states), where a0=5.29×10−11 ma_0 = 5.29\times10^{-11}\,\text{m} is the Bohr radius. With ℓ=2\ell=2: ℓ(ℓ+1)=6\ell(\ell+1) = 6.

ℓ(ℓ+1)ℏ22μr2=6×(1.055×10−34 J ⋅ s)22×(9.109×10−31 kg)×(3×5.29×10−11 m)2\frac{\ell(\ell+1)\hbar^{2}}{2\mu r^{2}} = \frac{6\times(1.055\times10^{-34}\,\text{J}\,\cdot\,\text{s})^{2}}{2\times(9.109\times10^{-31}\,\text{kg})\times(3\times5.29\times10^{-11}\,\text{m})^{2}}

(using μ≈me\mu \approx m_e to three figures, since the reduced-mass correction is under 0.06%). Compute the numerator:

6×(1.055×10−34)2=6×1.113×10−68=6.680×10−68 J2s2.6\times(1.055\times10^{-34})^{2} = 6\times1.113\times10^{-68} = 6.680\times10^{-68}\ \text{J}^2\text{s}^2.

Compute r2r^2: r=1.587×10−10 mr = 1.587\times10^{-10}\,\text{m}, so r2=2.519×10−20 m2r^2 = 2.519\times10^{-20}\,\text{m}^2. Denominator: 2×9.109×10−31×2.519×10−20=4.590×10−50 kg ⋅ m22\times9.109\times10^{-31}\times2.519\times10^{-20} = 4.590\times10^{-50}\,\text{kg}\,\cdot\,\text{m}^2.

6.680×10−684.590×10−50=1.455×10−18 J=9.08 eV.\frac{6.680\times10^{-68}}{4.590\times10^{-50}} = 1.455\times10^{-18}\,\text{J} = 9.08\,\text{eV}.

This centrifugal contribution (about 9.1 eV at this radius) is comparable in magnitude to the total hydrogen ground-state energy scale (13.6 eV), illustrating why dd orbitals (ℓ=2\ell=2) are pushed to much larger characteristic radii than ss orbitals of the same nn — the angular momentum barrier physically excludes the electron from small rr. As a limiting check, for an ss orbital (ℓ=0\ell=0) this term vanishes identically at every rr, consistent with ss-electron densities having no centrifugal exclusion from the nucleus (and hence the famous nonzero probability of an ss electron being found at r=0r=0).

Example 3 — Competition-level: showing the separation constant cannot be a non-half-integer-related irrational number by boundary-condition analysis. Suppose (for contradiction, as a rigor check) that the separation constant in the polar equation were not ℓ(ℓ+1)\ell(\ell+1) for integer ℓ\ell, but instead an arbitrary constant λ\lambda, and ask what value of λ\lambda is forced by requiring Θ(θ)\Theta(\theta) finite at both poles θ=0,π\theta=0,\pi for the m=0m=0 case. The Legendre equation with general λ\lambda is

ddx ⁣[(1−x2)dΘdx]+λΘ=0,x=cos⁡θ∈[−1,1].\frac{d}{dx}\!\left[(1-x^{2})\frac{d\Theta}{dx}\right] + \lambda\Theta = 0,\qquad x=\cos\theta\in[-1,1].

This is a Sturm–Liouville problem with singular endpoints at x=±1x=\pm1 (since the coefficient (1−x2)(1-x^2) vanishes there). Attempt a power-series (Frobenius) solution about x=0x=0:

Θ(x)=∑k=0∞akxk.\Theta(x) = \sum_{k=0}^{\infty} a_k x^k.

Substituting into the ODE and collecting powers of xkx^{k} produces the recursion

ak+2=k(k+1)−λ(k+1)(k+2) ak.a_{k+2} = \frac{k(k+1)-\lambda}{(k+1)(k+2)}\,a_k.

For generic (non-quantized) λ\lambda, this recursion never terminates — the series continues to all orders and, by a standard asymptotic ratio test as k→∞k\to\infty (the ratio ak+2/ak→1a_{k+2}/a_k \to 1), the series diverges at x=±1x=\pm1, i.e. exactly at the poles θ=0,π\theta = 0,\pi. A divergent Θ\Theta at the poles is physically unacceptable (the wavefunction must remain finite everywhere on the sphere, including the poles, since these are perfectly ordinary points in three-dimensional space — nothing physically singular happens there). The only way to avoid divergence is for the series to terminate after a finite number of terms, turning it into a polynomial. Setting the numerator to zero at k=ℓk=\ell (some non-negative integer) gives

ℓ(ℓ+1)−λ=0⟹λ=ℓ(ℓ+1),ℓ=0,1,2,3,…\ell(\ell+1) - \lambda = 0 \quad\Longrightarrow\quad \lambda = \ell(\ell+1), \qquad \ell = 0,1,2,3,\dots

So the demand for a finite, physically sensible wavefunction at the poles — not an ad hoc assumption — is exactly what forces the separation constant to take the discrete values 0,2,6,12,20,…0, 2, 6, 12, 20,\dots (i.e. ℓ(ℓ+1)\ell(\ell+1) for integer ℓ\ell) rather than any continuous or irrational value. This is the rigorous origin of orbital angular momentum quantization purely from the separation-of-variables machinery, before any operator algebra with L^2\hat L^2 is even introduced.

Limiting cases, scope & common traps

SituationWhat happens to the separationInterpretation
ℓ=0\ell = 0Centrifugal term vanishes; Θ,Φ\Theta,\Phi are both constantsSpherically symmetric ss orbitals; Y00=1/4πY_0^0 = 1/\sqrt{4\pi}
V(r)V(r) replaced by any other central potential (harmonic, square well)Angular equation is identical; only the radial equation changesSpherical harmonics are universal to all central-force problems
m=0m=0 vs. m≠0m\ne0Φ(ϕ)\Phi(\phi) constant vs. oscillatory eimϕe^{im\phi}Governs whether ψ\psi has nodes in azimuthal angle (relevant to px,pyp_x,p_y vs. pzp_z-type combinations)
Two-electron atomsSeparation into radial × angular parts fails for the full wavefunctionElectron–electron repulsion 1/r121/r_{12} depends on both electrons' coordinates jointly; only single-particle central problems separate this way
r→∞r\to\inftyRadial equation reduces to R′′≈−(2μE/ℏ2)RR'' \approx -(2\mu E/\hbar^2)RDetermines whether E<0E<0 (bound, exponentially decaying RR) or E>0E>0 (continuum, oscillatory RR)
Free particle (V=0V=0) in spherical coordinatesSame angular equation; radial equation becomes spherical Bessel equationConfirms angular separation depends only on the shape of the Laplacian, not the specific potential

The chemist behind it

The separation-of-variables method for the Laplacian in spherical coordinates predates quantum mechanics by more than a century, developed for problems in gravitation, electrostatics, and heat conduction. Its formal quantum-mechanical application to the hydrogen atom was carried out by Erwin Schrödinger himself in his 1926 papers, drawing on classical mathematical physics techniques already well established by Adrien-Marie Legendre (whose polynomials solve the polar equation) and Pierre-Simon Laplace (whose equation in spherical coordinates is the direct ancestor of the angular equation here).

Origins · 1 minErwin Schrödinger and how this was discovered
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Erwin Schrödinger
1887–1961 · Vienna and Zurich, 1920s wave mechanics

Schrödinger developed wave mechanics in a concentrated burst of work during 1925–1926, formulating what is now his eponymous equation and, in his second 1926 paper, explicitly solving the hydrogen atom by separating the Laplacian in spherical coordinates exactly as shown here. He had trained thoroughly in classical continuum mechanics and the mathematics of vibrating systems, and he approached the electron not as a particle on rails but as a de Broglie-style matter wave obeying a wave equation. The hydrogen solution — with its radial and angular quantum numbers, correct energy levels, and correct degeneracies — was the single most persuasive piece of evidence in 1926 that wave mechanics reproduced (and extended) the empirical successes of the older Bohr–Sommerfeld quantum theory, while resolving conceptual problems that theory could not address, such as why ℓ=0\ell=0 states carry zero angular momentum despite Bohr's picture of circular orbits.

History
- 1782 — Legendre introduces the polynomials that bear his name while studying gravitational potentials of spheroids.
- 1880s — Spherical harmonics become standard tools in classical potential theory (electrostatics, heat conduction) well before quantum mechanics.
- 1913 — Bohr's model gives hydrogen energy levels En=−13.6 eV/n2E_n = -13.6\,\text{eV}/n^2 without any separation-of-variables treatment or angular quantum numbers.
- 1926 — Schrödinger publishes wave mechanics and solves the hydrogen atom by separating variables in spherical coordinates, reproducing Bohr's energies and adding ℓ\ell, mm quantum numbers naturally.
- 1927 onward — The method becomes the standard textbook route to atomic orbitals, later generalized to molecular central-field and multi-electron approximate treatments.
Why does the potential energy have to depend only on r for this separation to work?

The core algebraic step in the derivation multiplies through by r2r^2 and rearranges so that every angular derivative ends up on one side of the equation and every radial term (including V(r)V(r)) ends up on the other. If VV depended on θ\theta or ϕ\phi as well as rr, it would contaminate the angular side of the equation, and the left side would no longer be a pure function of rr alone — the argument that "a function of rr can equal a function of angles only if both are constant" would break down entirely. Physically, this is why the technique generalizes to any central (spherically symmetric) potential — Coulomb, harmonic oscillator, square well — but fails immediately for non-central potentials such as a molecule in an external electric field along a fixed lab-frame direction, or two interacting electrons whose repulsion depends on the angle between their position vectors.

Is the separation constant ℓ(ℓ+1) related to angular momentum, and if so how?

Yes, directly. The angular part of the Laplacian, once multiplied by −ℏ2-\hbar^2, is exactly the operator representation of L^2\hat L^2, the square of the total orbital angular momentum operator. The angular equation derived here is therefore equivalent to the eigenvalue equation L^2Y=ℓ(ℓ+1)ℏ2Y\hat L^2 Y = \ell(\ell+1)\hbar^2 Y, meaning that solving this angular equation is simultaneously solving for the eigenstates of squared angular momentum. This is why the separation constant carries physical meaning beyond convenience: ℓ(ℓ+1)ℏ2\ell(\ell+1)\hbar^2 is literally the squared magnitude of the electron's orbital angular momentum in that state, and mℏm\hbar is the projection of angular momentum along the chosen zz-axis.

Does this separation work for multi-electron atoms like helium or carbon?

Not directly for the exact many-electron wavefunction, because the electron–electron repulsion term e2/4πε0r12e^2/4\pi\varepsilon_0 r_{12} depends on the distance between two electrons, which mixes the coordinates of both particles in a way that cannot be written as a function of one electron's rr alone. However, the central-field approximation used throughout computational chemistry sidesteps this by replacing the true repulsion with an averaged, spherically symmetric effective potential felt by each electron individually; under that approximation, each electron's equation separates exactly as shown here, which is precisely why multi-electron atomic orbitals are still labeled with quantum numbers n,ℓ,mn,\ell,m borrowed from the hydrogen solution.

What would happen if we tried to separate variables using ψ = f(x)g(y)h(z) instead of spherical coordinates?

The attempt fails at the very first algebraic step. Substituting ψ=f(x)g(y)h(z)\psi=f(x)g(y)h(z) into the Cartesian Schrödinger equation and dividing by ψ\psi leaves the potential term V(x,y,z)=−e2/4πε0x2+y2+z2V(x,y,z) = -e^2/4\pi\varepsilon_0\sqrt{x^2+y^2+z^2} sitting in the equation as an inseparable function of all three variables simultaneously — it cannot be pushed entirely onto one side because the square root ties xx, yy, and zz together irreducibly. There is no rearrangement that isolates a function of xx alone equated to a function of (y,z)(y,z) alone. This is precisely why the choice of coordinate system is not cosmetic: it must be matched to the symmetry of the potential for separation of variables to be possible at all.

Why is the angular equation described as "universal" across different central potentials?

Because the angular equation, once separated, contains no trace of V(r)V(r), EE, or the reduced mass at all — every one of those physical specifics is swept entirely into the radial equation during the first separation step. The angular equation is purely a statement about the geometry of derivatives on a sphere, so its solutions (the spherical harmonics YℓmY_\ell^m) are identical whether the underlying radial problem is a hydrogen atom, a three-dimensional harmonic oscillator, an infinite spherical well, or a rigid diatomic rotor with V(r)=0V(r)=0. Only the radial equation — and therefore the allowed energies and radial node structure — changes from system to system.

How many quantum numbers come out of this separation, and where does the third one (n) come from?

The separation itself directly produces two quantum numbers: ℓ\ell (from requiring finite Θ(θ)\Theta(\theta) at the poles) and mm (from requiring single-valued Φ(ϕ)\Phi(\phi) under 2π2\pi rotation). The third quantum number, the principal quantum number nn, does not come from the angular separation at all — it emerges later, from actually solving the radial equation for the specific Coulomb potential, where the requirement that R(r)→0R(r)\to 0 as r→∞r\to\infty (normalizability of a bound state) forces the energy EE to take discrete values indexed by an integer n>ℓn > \ell. That step belongs to solving the radial equation itself, not to the separation process described here.

Does this separation-of-variables method apply outside chemistry, in other areas of physics?

Yes — this is one of the most recycled derivations in all of mathematical physics. The identical spherical-coordinate Laplacian separation is used to solve Laplace's equation for electrostatic potentials around charged spheres, to find the normal modes of a vibrating spherical drumhead or planetary body, to compute multipole expansions in electrodynamics, and to solve the angular part of the Schrödinger equation for nucleons in nuclear shell-model potentials. Whenever a physical system has a central point of symmetry and the governing PDE is linear, this same two-stage separation — first splitting off rr, then splitting θ\theta from ϕ\phi — is the standard route to a solution, with spherical harmonics as the universal angular output.

Before this

Where this leads