Separation of the Central-Potential Schrodinger Equation

Derivation

Separation of the Central-Potential Schrodinger Equation

D-217 Home PU-301 Threads energy · symmetry · waves Depends on Spherical Harmonics as Angular Momentum Eigenstates

Statement

For a single particle of mass \(m\) in a central potential \(V(r)\) that depends only on the radial distance \(r=|\vec r|\), the time-independent Schrodinger equation \(\hat H\psi=E\psi\) separates in spherical coordinates as \(\psi(r,\theta,\phi)=R(r)\,Y_\ell^m(\theta,\phi)\). The angular factor is a spherical harmonic (an eigenfunction of \(\hat L^2\) with eigenvalue \(\hbar^2\ell(\ell+1)\)), and the radial factor obeys a one-dimensional equation. With the substitution \(u(r)=rR(r)\) this becomes exactly a 1D Schrodinger equation on the half-line \(r\ge 0\) with the effective potential \(V_{\text{eff}}(r)=V(r)+\dfrac{\hbar^2\ell(\ell+1)}{2mr^2}\).

Why it matters

Every exactly solvable bound-state problem of atomic and nuclear physics — hydrogen, the isotropic 3D oscillator, the deuteron, the spherical well — is reached through this reduction. It converts a partial differential equation in three variables into an ordinary differential equation in one, and it does so before the potential is even specified, so the entire angular structure of the spectrum is fixed by rotational symmetry alone.

The reduction also makes the physics visible: the centrifugal term \(\hbar^2\ell(\ell+1)/2mr^2\) is a repulsive barrier of purely quantum-kinematic origin that keeps states of nonzero angular momentum away from the origin. Reading the radial equation as a 1D problem lets one import all of the intuition of one-dimensional quantum mechanics — nodes, turning points, tunnelling — directly into three dimensions.

Assumptions

The potential is central, \(V=V(r)\).If \(V\) depends on \(\theta\) or \(\phi\), \(\hat H\) does not commute with \(\hat L^2\) and \(\hat L_z\); the angular and radial motions no longer separate and \(Y_\ell^m\) is not an eigenfunction of \(\hat H\).

The wavefunction is separable, \(\psi=R(r)Y_\ell^m(\theta,\phi)\).General \(\psi\) is a superposition over \(\ell,m\); separation gives the basis states. Dropping it means solving the full 3D PDE with no reduction in dimension.

The Hamiltonian is rotationally invariant, so \([\hat H,\hat L^2]=[\hat H,\hat L_z]=0\).This is what guarantees simultaneous eigenstates of \(\hat H,\hat L^2,\hat L_z\) exist. Without it the labels \(\ell,m\) are not good quantum numbers and the degeneracy structure collapses.

\(R(r)\) is regular enough that \(u=rR\) satisfies \(u(0)=0\).Needed for \(\psi\) to be normalizable and for the kinetic operator to be self-adjoint on the half-line; if relaxed one admits spurious \(1/r\) solutions that are not square-integrable at the origin.

Derivation

\[-\frac{\hbar^2}{2m}\nabla^2\psi + V(r)\,\psi = E\,\psi\]

The time-independent Schrodinger equation with the central Hamiltonian \(\hat H=-\frac{\hbar^2}{2m}\nabla^2+V(r)\). A

\[\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{r^2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right]\]

The Laplacian written in spherical coordinates \((r,\theta,\phi)\); a pure coordinate identity. A

\[\hat L^2=-\hbar^2\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right]\ \Longrightarrow\ \nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right)-\frac{\hat L^2}{\hbar^2 r^2}\]

The bracketed angular operator is exactly \(-\hat L^2/\hbar^2\), the orbital-angular-momentum operator expressed in coordinates. This isolates all \(\theta,\phi\) dependence inside \(\hat L^2\). B

\[\psi(r,\theta,\phi)=R(r)\,Y_\ell^m(\theta,\phi),\qquad \hat L^2\,Y_\ell^m=\hbar^2\ell(\ell+1)\,Y_\ell^m\]

Separation ansatz. The spherical harmonics are the eigenfunctions of \(\hat L^2\) (assumed prior result), with \(\ell=0,1,2,\dots\) The angular equation is already solved. B

\[-\frac{\hbar^2}{2m}\,\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right)Y_\ell^m+\frac{\hbar^2\ell(\ell+1)}{2mr^2}R\,Y_\ell^m+V(r)R\,Y_\ell^m=E\,R\,Y_\ell^m\]

Substitute the ansatz into step 1 using step 3, and let \(\hat L^2\) act on \(Y_\ell^m\) via its eigenvalue. Every term now carries the common factor \(Y_\ell^m\). B

\[-\frac{\hbar^2}{2m}\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right)+\left[V(r)+\frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]R=E\,R\]

Divide through by \(Y_\ell^m(\theta,\phi)\) (nonzero almost everywhere). The result is an ODE in \(r\) alone — the radial equation — with the potential augmented by the centrifugal term. A

\[u(r)\equiv rR(r)\quad\Longrightarrow\quad \frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right)=\frac{1}{r}\frac{d^2u}{dr^2}\]

Reduced radial function. The identity follows from \(\frac{d}{dr}(r^2R')=r^2R''+2rR'\) and \(u''=(rR)''=rR''+2R'\), so \(r^2R''+2rR'=r\,u''\). This is the algebraic heart of the reduction. C

\[-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2}+\left[V(r)+\frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]u=E\,u,\qquad u(0)=0\]

Multiply the step-6 equation by \(r\) and apply step 7. The kinetic term becomes a plain second derivative: a 1D Schrodinger equation on \(r\ge0\). Regularity of \(R\) at the origin forces \(u(0)=0\). A

Result

\[-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2}+V_{\text{eff}}(r)\,u=E\,u,\qquad V_{\text{eff}}(r)=V(r)+\frac{\hbar^2\ell(\ell+1)}{2mr^2},\qquad u=rR,\ \ u(0)=0\]

Reading. The full 3D problem for a central potential collapses to a single one-dimensional Schrodinger equation for \(u(r)\) on the half-line, moving in the effective potential \(V_{\text{eff}}\). The only trace of three-dimensionality is (i) the boundary condition \(u(0)=0\), and (ii) the repulsive centrifugal term \(\hbar^2\ell(\ell+1)/2mr^2\), which is added to the true potential and grows with angular momentum. The energy \(E\) and the quantum numbers \(\ell,m\) are inherited unchanged; \(m\) does not appear in the radial equation, giving the \((2\ell+1)\)-fold degeneracy.

Units check. Each term must be an energy \(\times u\). \(\hbar^2/(2m)\) has units \(\mathrm{J^2\,s^2/kg}=\mathrm{J\,m^2}\) (since \(\mathrm{J=kg\,m^2\,s^{-2}}\)); times \(d^2u/dr^2\) (units of \(u\cdot\mathrm{m^{-2}}\)) gives \(\mathrm{J}\cdot u\). The centrifugal term \(\hbar^2\ell(\ell+1)/(2mr^2)\) has units \(\mathrm{J\,m^2/m^2}=\mathrm{J}\), matching \(V(r)\) and \(E\). Consistent.

Limiting cases

\(\ell=0\) (s states): the centrifugal term vanishes and the equation is literally the 1D Schrodinger equation for \(u\) in \(V(r)\) on the half-line with \(u(0)=0\) — identical to a 1D problem with an infinite wall at \(r=0\).
\(V(r)\to 0\) (free particle): \(V_{\text{eff}}\) is purely centrifugal; solutions are spherical Bessel functions \(j_\ell(kr)\) with \(E=\hbar^2k^2/2m\), regular at the origin.
Large \(\ell\): the centrifugal barrier \(\propto\ell(\ell+1)\) dominates near the origin, pushing the wavefunction outward — the classical limit of high angular momentum orbiting at large radius.
\(r\to\infty\) with \(E<0\): if \(V\to0\), the equation reduces to \(u''=\kappa^2u\) with \(\kappa=\sqrt{-2mE}/\hbar\), giving bound-state decay \(u\sim e^{-\kappa r}\).

Breaks when

Non-central potentials. If \(V\) depends on angle (e.g. a molecule in an external field, spin–orbit coupling \(\propto\vec L\cdot\vec S\), or the tensor force between nucleons), \([\hat H,\hat L^2]\ne0\), \(\ell\) is no longer a good quantum number, and different \(\ell\) channels mix — the clean separation fails.
Singular potentials at the origin. If \(V(r)\) diverges faster than the centrifugal term (worse than \(-1/r^2\) with a large enough coefficient), the boundary condition \(u(0)=0\) is insufficient to fix the solution: the operator loses a unique self-adjoint extension and the spectrum becomes ambiguous ("fall to the centre").
Relativistic or spin-dependent regimes. For the Dirac equation the separation uses spinor spherical harmonics with quantum number \(\kappa\) instead of \(\ell\); the simple scalar radial equation above is only the nonrelativistic limit.
Coupled multiparticle systems. For two or more interacting particles the coordinate must first be reduced (reduced mass \(\mu\), centre-of-mass separation); if the interaction is not expressible through a single relative \(|\vec r_1-\vec r_2|\), no central reduction exists.

Failure modes

Forgetting the \(u=rR\) Jacobian. Writing the kinetic term as \(-\frac{\hbar^2}{2m}R''\) instead of using \(\frac{1}{r^2}(r^2R')'\) drops the \(2R'/r\) term and gives the wrong equation; the substitution \(u=rR\) is what removes it cleanly.
Using \(\ell^2\) instead of \(\ell(\ell+1)\). The \(\hat L^2\) eigenvalue is \(\hbar^2\ell(\ell+1)\), not \(\hbar^2\ell^2\); the difference shifts every centrifugal barrier and every energy level.
Normalizing \(u\) as if it were \(R\). The normalization is \(\int_0^\infty |u|^2\,dr=1\) (the \(r^2\) from the volume element is absorbed into \(u=rR\)); students who keep an extra \(r^2\) double-count it.
Imposing \(R(0)=0\) instead of \(u(0)=0\). Regular s-state radial functions are finite and nonzero at the origin (\(R(0)\ne0\)); it is \(u=rR\) that must vanish there.
Putting \(m\) (magnetic quantum number) into the radial equation. The radial equation contains only \(\ell\); expecting \(m\) to appear leads to spurious extra structure and misses the \((2\ell+1)\) degeneracy.

Discussion

The reduction is a direct consequence of rotational symmetry via Noether's theorem: because \(\hat H\) is invariant under rotations, the three components of \(\vec L\) are conserved, \(\hat L^2\) and \(\hat L_z\) commute with \(\hat H\), and one may diagonalize all three simultaneously. The angular eigenvalue problem is universal — the same spherical harmonics appear for hydrogen, for the oscillator, for any \(V(r)\) — so all of the potential-specific physics is quarantined into the single radial equation. This is why the angular momentum barrier is "kinematic": it comes from the geometry of the Laplacian, not from any force.

The centrifugal term has a transparent classical analogue. In classical mechanics a particle of angular momentum \(L\) moving in \(V(r)\) obeys energy conservation \(E=\tfrac12 m\dot r^2 + V(r)+\tfrac{L^2}{2mr^2}\); the last term is the centrifugal potential. Quantum mechanics reproduces it exactly with \(L^2\to\hbar^2\ell(\ell+1)\). The correspondence is so tight that turning points, allowed regions and WKB quantization of the radial motion all carry over with \(V\to V_{\text{eff}}\).

The condition \(u(0)=0\) is more than a convenience: it is what makes the radial kinetic operator essentially self-adjoint for \(\ell\ge1\) and for well-behaved \(V\). For \(\ell=0\) the origin is a genuine boundary and \(u(0)=0\) plays the role of the hard wall that a 1D problem on the half-line requires. When \(V\) is attractive and singular as \(-\alpha/r^2\), the two independent solutions near the origin behave as \(r^s\) with \(s\) complex once \(\alpha\) exceeds a critical value; the boundary condition no longer selects a unique physical solution, and the system requires an additional short-distance input (a self-adjoint extension, or a regularizing cutoff) to define its spectrum — the quantum-mechanical "fall to the centre."

Common misconceptions. The effective potential is a bookkeeping device, not a new force: no physical push acts on the particle from \(\hbar^2\ell(\ell+1)/2mr^2\); it is the angular kinetic energy re-expressed as a radial potential. Likewise, separation does not mean the particle "chooses" a definite \(\ell\) — a general state is a superposition of separated pieces; separation supplies the complete basis in which any central-potential state is expanded.

Worked examples

Example 1 — Magnitude of the centrifugal barrier for a \(p\)-electron.

\[V_{\text{cf}}(r)=\frac{\hbar^2\ell(\ell+1)}{2m_e r^2},\qquad \ell=1\ \Rightarrow\ \ell(\ell+1)=2\]

Take the centrifugal piece of \(V_{\text{eff}}\) for an electron in a \(p\) state; symbols first. A

\[r=a_0=5.29\times10^{-11}\,\mathrm{m},\quad \hbar=1.055\times10^{-34}\,\mathrm{J\,s},\quad m_e=9.11\times10^{-31}\,\mathrm{kg}\]

Evaluate the barrier at the Bohr radius, the natural atomic length scale. A

\[V_{\text{cf}}=\frac{(1.055\times10^{-34})^2\,(2)}{2(9.11\times10^{-31})(5.29\times10^{-11})^2}=\frac{2.23\times10^{-68}}{5.10\times10^{-51}}\]

Insert numbers: numerator \(\hbar^2\cdot2\), denominator \(2m_e a_0^2=1.822\times10^{-30}\times2.80\times10^{-21}\). A

\[V_{\text{cf}}\approx 4.37\times10^{-18}\,\mathrm{J}\approx 27.3\,\mathrm{eV}\]

Reading. The centrifugal barrier for an \(\ell=1\) electron at the Bohr radius is of order one hartree (27.2 eV) — comparable to the Coulomb binding itself, which is exactly why angular momentum controls the shape and energy ordering of atomic orbitals. Units check: \(\mathrm{(J\,s)^2/(kg\,m^2)=J}\).

Example 2 — Ground state of an \(\ell=0\) electron in an infinite spherical well of radius \(a=0.100\,\mathrm{nm}\).

\[-\frac{\hbar^2}{2m_e}\frac{d^2u}{dr^2}=E\,u,\qquad u(0)=u(a)=0\quad(\ell=0,\ V=0\ \text{inside})\]

For \(\ell=0\) the centrifugal term vanishes; inside the well \(V=0\), so \(V_{\text{eff}}=0\) and the radial equation is a 1D box for \(u\). B

\[u_n(r)=\sin\!\left(\frac{n\pi r}{a}\right),\qquad E_n=\frac{\hbar^2\pi^2 n^2}{2m_e a^2},\quad n=1,2,\dots\]

The boundary conditions select \(\sin(k r)\) with \(k=n\pi/a\); this is the standard particle-in-a-box quantization, now for the reduced function \(u\). B

\[E_1=\frac{(1.055\times10^{-34})^2\,\pi^2}{2(9.11\times10^{-31})(1.00\times10^{-10})^2}=\frac{1.099\times10^{-67}}{1.822\times10^{-50}}\]

Ground state \(n=1\); insert \(\hbar,m_e,a\). A

\[E_1\approx 6.03\times10^{-18}\,\mathrm{J}\approx 37.7\,\mathrm{eV},\qquad R_1(r)=\frac{u_1(r)}{r}=\frac{1}{r}\sin\!\left(\frac{\pi r}{a}\right)\]

Reading. The \(\ell=0\) confined electron has ground-state energy \(\sim38\) eV; the physical radial function \(R_1=u_1/r=\mathrm{sinc}(\pi r/a)\) is finite and nonzero at the origin (\(R_1(0)=\pi/a\)), even though \(u_1(0)=0\). This is the concrete illustration that \(u(0)=0\), not \(R(0)=0\), is the correct regularity condition. Units check: \(\mathrm{J^2 s^2/(kg\,m^2)=J}\).

Problems

Write the effective potential for a hydrogen atom (\(V(r)=-e^2/4\pi\varepsilon_0 r\)) with \(\ell=2\), and find the radius \(r_{\min}\) at which \(V_{\text{eff}}\) is minimum. Use \(m=m_e\).
Solution
\(V_{\text{eff}}(r)=-\dfrac{e^2}{4\pi\varepsilon_0 r}+\dfrac{\hbar^2\ell(\ell+1)}{2m_e r^2}\) with \(\ell(\ell+1)=6\). Set \(dV_{\text{eff}}/dr=0\): \(\dfrac{e^2}{4\pi\varepsilon_0 r^2}-\dfrac{\hbar^2\ell(\ell+1)}{m_e r^3}=0\Rightarrow r_{\min}=\dfrac{4\pi\varepsilon_0\hbar^2\ell(\ell+1)}{m_e e^2}=\ell(\ell+1)\,a_0\). With \(\ell(\ell+1)=6\): \(r_{\min}=6a_0=6\times5.29\times10^{-11}=3.17\times10^{-10}\,\mathrm{m}\). (The bound state itself sits near \(n^2a_0\); here \(n\ge3\) for \(\ell=2\), consistent in scale.)
An \(\ell=1\) electron sits at \(r=2a_0\). By what numerical factor is its centrifugal barrier smaller than the value found in Worked Example 1 (at \(r=a_0\))?
Solution
\(V_{\text{cf}}\propto 1/r^2\), so at \(r=2a_0\) the barrier is \((1/2)^2=1/4\) of its value at \(a_0\). Thus \(V_{\text{cf}}(2a_0)=27.3/4\approx 6.8\,\mathrm{eV}\). Factor \(=4\) smaller.
For the free particle (\(V=0\)), verify by direct substitution that \(R(r)=j_0(kr)=\dfrac{\sin kr}{kr}\) solves the \(\ell=0\) radial equation, and identify \(E\).
Solution
For \(\ell=0\), \(u=rR=\dfrac{\sin kr}{k}\). Then \(u''=-k\sin kr=-k^2u\), so \(-\dfrac{\hbar^2}{2m}u''=\dfrac{\hbar^2k^2}{2m}u=Eu\) with \(E=\dfrac{\hbar^2k^2}{2m}\). Also \(u(0)=0\) since \(\sin(0)=0\). Hence \(R=j_0(kr)\) is a valid regular solution with \(E=\hbar^2k^2/2m\).
Show that the normalization \(\int_0^\infty |R(r)|^2 r^2\,dr=1\) is equivalent to \(\int_0^\infty |u(r)|^2\,dr=1\).
Solution
Since \(u=rR\), \(|u|^2=r^2|R|^2\). Therefore \(\int_0^\infty|u|^2\,dr=\int_0^\infty r^2|R|^2\,dr\). The \(r^2\) of the volume element \(d^3r=r^2\,dr\,d\Omega\) (with \(\int|Y_\ell^m|^2 d\Omega=1\)) is exactly absorbed into \(u\). Hence the two normalizations coincide.
The isotropic 3D harmonic oscillator has \(V(r)=\tfrac12 m\omega^2 r^2\). Write its radial equation for general \(\ell\), and, for \(\ell=0\), verify that \(u(r)=r\,e^{-m\omega r^2/2\hbar}\) is a solution and find its energy.
Solution
Radial equation: \(-\dfrac{\hbar^2}{2m}u''+\left[\tfrac12 m\omega^2 r^2+\dfrac{\hbar^2\ell(\ell+1)}{2mr^2}\right]u=Eu\). For \(\ell=0\), try \(u=r\,e^{-\alpha r^2/2}\) with \(\alpha=m\omega/\hbar\). Then \(R=u/r=e^{-\alpha r^2/2}\), the Gaussian ground state. Compute: \(u'=(1-\alpha r^2)e^{-\alpha r^2/2}\) and \(u''=(-3\alpha r+\alpha^2 r^3)e^{-\alpha r^2/2}=(\alpha^2 r^2-3\alpha)\,u\). So \(-\dfrac{\hbar^2}{2m}u''=-\dfrac{\hbar^2}{2m}(\alpha^2 r^2-3\alpha)\,u\). With \(\alpha=m\omega/\hbar\): \(\dfrac{\hbar^2\alpha^2}{2m}=\tfrac12 m\omega^2\), cancelling the potential term, leaving \(\dfrac{\hbar^2}{2m}\cdot3\alpha=\dfrac{3}{2}\hbar\omega\). Hence \(E=\tfrac32\hbar\omega\), the correct ground-state energy of the 3D oscillator.