Separation of the Central-Potential Schrodinger Equation

Derivation

Separation of the Central-Potential Schrodinger Equation

D-296 Home PU-306 Threads energy · matter · waves · symmetry Depends on Separation of Variables and Stationary States, angular-momentum-eigenstates, spherical-harmonics

Statement

For a single particle of mass \(m\) moving in a spherically symmetric potential \(V(r)\), the stationary Schrödinger equation \(\hat{H}\psi = E\psi\) separates in spherical coordinates: every bound eigenstate factorises as \(\psi(r,\theta,\phi) = R(r)\,Y_\ell^{m_\ell}(\theta,\phi)\), where \(Y_\ell^{m_\ell}\) is a spherical harmonic labelled by \(\ell = 0,1,2,\dots\) and \(m_\ell = -\ell,\dots,+\ell\), and the radial factor obeys the one-dimensional equation \(-\frac{\hbar^2}{2m}\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) + \left[V(r) + \frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]R = E\,R\), in which the term \(\frac{\hbar^2\ell(\ell+1)}{2mr^2}\) is the centrifugal (angular-momentum) barrier added to the true potential to form an effective potential \(V_{\text{eff}}(r)\).

Why it matters

This separation is the structural backbone of atomic, nuclear and molecular quantum mechanics. It reduces a three-dimensional partial differential equation to a family of ordinary differential equations indexed by \(\ell\), so that the hydrogen atom, the isotropic oscillator, the spherical well and the deuteron are all solved by the same machinery. The angular part is universal — the spherical harmonics are fixed once and for all by rotational symmetry — while the radial equation carries all the information about the specific force law.

It also makes the physical role of angular momentum transparent: orbital motion contributes a repulsive \(1/r^2\) term that pushes the wavefunction away from the origin, competing with an attractive \(V(r)\). The balance sets the size and energy scale of every atom, and explains why \(s\)-states (\(\ell=0\)) can be finite at the nucleus while higher \(\ell\) cannot.

Assumptions

The potential is central, \(V=V(r)\), depending only on \(|\vec r|\).If \(V\) depends on angle, \(\hat H\) no longer commutes with \(\hat{\vec L}^2\) and the angular and radial parts couple; \(\ell\) and \(m_\ell\) cease to be good quantum numbers and no single product \(R\,Y_\ell^{m_\ell}\) solves the equation.

The particle is non-relativistic with a well-defined mass \(m\).Drop it and one needs the Dirac equation; spin-orbit coupling mixes the orbital and spin angular momenta, and the clean \((\ell,m_\ell)\) labelling is replaced by total-\(j\) states.

The wavefunction is single-valued, square-integrable and (for bound states) regular at the origin.Without single-valuedness \(m_\ell\) need not be an integer; without regularity at \(r=0\) the irregular radial solution \(\sim r^{-(\ell+1)}\) survives and produces non-normalisable or unphysical states.

The Laplacian is taken in flat three-dimensional space and the reduced mass has been used for two-body problems.In curved space or non-inertial frames extra terms appear; ignoring the reduction to relative coordinates gives the wrong energy scale for atoms and molecules.

Derivation

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

Time-independent Schrödinger equation for a particle in a potential; assumed prior result. A

\[ \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) - \frac{\hat{\vec L}^2}{\hbar^2 r^2} \]

The Laplacian in spherical coordinates splits into a radial piece and an angular piece; the angular differential operator is exactly \(\hat{\vec L}^2/(\hbar^2 r^2)\), where \(\hat{\vec L}^2\) is the squared orbital angular momentum. B

\[ \left[-\frac{\hbar^2}{2m}\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{\hat{\vec L}^2}{2m r^2} + V(r)\right]\psi = E\,\psi \]

Substitute the spherical Laplacian into step 1 and multiply the angular term by \(-\hbar^2/2m\); the \(\hbar^2\) cancels, leaving \(\hat{\vec L}^2/(2mr^2)\). A

\[ \hat{\vec L}^2\,Y_\ell^{m_\ell}(\theta,\phi) = \hbar^2\,\ell(\ell+1)\,Y_\ell^{m_\ell}(\theta,\phi) \]

The spherical harmonics are the simultaneous eigenfunctions of \(\hat{\vec L}^2\) and \(\hat L_z\); assumed prior result from angular-momentum eigenstates. This is the key fact that lets the angular dependence be removed algebraically. B

\[ \psi(r,\theta,\phi) = R(r)\,Y_\ell^{m_\ell}(\theta,\phi) \]

Separation ansatz. It is legal because \([\hat H,\hat{\vec L}^2]=0\) and \([\hat H,\hat L_z]=0\) for a central potential, so \(\hat H\), \(\hat{\vec L}^2\) and \(\hat L_z\) share a complete common eigenbasis; the angular factor may be taken to be a single \(Y_\ell^{m_\ell}\). C

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

Insert the ansatz (step 5) into step 3 and act with \(\hat{\vec L}^2\) on \(Y_\ell^{m_\ell}\) using step 4. The radial derivatives pass through \(Y_\ell^{m_\ell}\), which is a constant with respect to \(r\). A

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

Divide both sides by the common non-zero factor \(Y_\ell^{m_\ell}(\theta,\phi)\); no angular dependence remains, confirming the separation is consistent. A

\[ u(r)\equiv r\,R(r)\;\Rightarrow\; -\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[V(r) + \frac{\hbar^2\ell(\ell+1)}{2m r^2}\right]u = E\,u \]

Substitute \(R=u/r\); the identity \(\frac{1}{r^2}\frac{d}{dr}(r^2 R') = \frac{1}{r}u''\) collapses the first term into a plain second derivative, casting the radial problem into the form of a one-dimensional Schrödinger equation on \(0\le r<\infty\) with boundary condition \(u(0)=0\). C

Result

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

Reading. A three-dimensional central-force problem is exactly equivalent to a one-dimensional problem for the function \(u=rR\) on the half-line \(r\ge 0\), moving in the true potential \(V(r)\) plus a purely repulsive centrifugal barrier proportional to \(\ell(\ell+1)\). The angular structure is entirely carried by the spherical harmonic \(Y_\ell^{m_\ell}\) and does not affect the energy, which depends only on \(V\) and \(\ell\), not on \(m_\ell\) — the origin of the \((2\ell+1)\)-fold degeneracy of each level.

Units check. Each term must be an energy (joules). \(\frac{\hbar^2}{2m}\frac{d^2}{dr^2}\) has dimensions \(\frac{(\text{J·s})^2}{\text{kg}}\cdot\text{m}^{-2} = \frac{\text{J}^2\text{s}^2}{\text{kg·m}^2}\); since \(\text{J}=\text{kg·m}^2\text{s}^{-2}\), this is \(\text{J}\cdot\frac{\text{kg·m}^2\text{s}^{-2}\cdot\text{s}^2}{\text{kg·m}^2}=\text{J}\). The barrier \(\frac{\hbar^2\ell(\ell+1)}{2mr^2}\) has \(\frac{(\text{J·s})^2}{\text{kg·m}^2}=\text{J}\) with \(\ell(\ell+1)\) dimensionless. All terms are energies. Consistent.

Limiting cases

\(\ell=0\) (\(s\)-states): the centrifugal barrier vanishes, \(V_{\text{eff}}=V\), and \(u''=\frac{2m}{\hbar^2}(V-E)u\) is a bare 1D problem; \(R=u/r\) can be finite at \(r=0\).
Large \(\ell\): the barrier \(\propto \ell(\ell+1)/r^2\) dominates at small \(r\), forcing the wavefunction outward and raising the energy; classically this is the conservation of angular momentum keeping the particle away from the centre.
\(r\to\infty\) with \(V\to 0\): the equation reduces to \(u''=-\frac{2mE}{\hbar^2}u\); bound states (\(E<0\)) decay as \(e^{-\kappa r}\) with \(\kappa=\sqrt{-2mE}/\hbar\), scattering states (\(E>0\)) oscillate.
Free particle (\(V=0\) everywhere): the radial solutions are the spherical Bessel functions \(j_\ell(kr)\) with \(k=\sqrt{2mE}/\hbar\); the separation still holds, giving the partial-wave basis used in scattering theory.

Breaks when

Non-central potentials. An external electric or magnetic field, or a molecular (non-spherical) core, breaks rotational symmetry: \([\hat H,\hat{\vec L}^2]\neq 0\), the single-\(Y_\ell^{m_\ell}\) ansatz fails, and states become superpositions of different \(\ell\) (Stark mixing, crystal-field splitting).
Relativistic or spin-orbit regimes. When \(v/c\) is not small, spin-orbit coupling \(\propto \hat{\vec L}\cdot\hat{\vec S}\) mixes orbital and spin angular momenta; only the total \(\hat{\vec J}\) is conserved, and one must separate the Dirac equation with quantum numbers \((j,\ell)\) rather than \((\ell,m_\ell)\).
Momentum-dependent or non-local potentials. If \(V\) depends on \(\hat{\vec p}\) or is non-local (e.g. exchange terms in Hartree–Fock, velocity-dependent nuclear forces), the simple \(1/r^2\) centrifugal term is modified and the radial equation is no longer a plain 1D Schrödinger equation.
Strongly singular potentials at the origin. If \(V(r)\) diverges faster than \(-1/r^2\) as \(r\to 0\), the centrifugal barrier can no longer regulate the behaviour, the regularity boundary condition \(u(0)=0\) is insufficient, and the Hamiltonian loses self-adjointness (fall-to-the-centre).

Failure modes

Writing \(\ell^2\) instead of \(\ell(\ell+1)\) in the barrier — the eigenvalue of \(\hat{\vec L}^2\) is \(\hbar^2\ell(\ell+1)\), not \(\hbar^2\ell^2\).
Forgetting the Jacobian in the radial derivative, writing \(-\frac{\hbar^2}{2m}R''\) instead of \(-\frac{\hbar^2}{2m}\frac{1}{r^2}(r^2R')'\); the two differ by first-derivative terms and give wrong energies.
Normalising \(R\) as if the volume element were \(dr\) rather than \(r^2\,dr\); the correct condition is \(\int_0^\infty |R|^2 r^2\,dr=1\), equivalently \(\int_0^\infty |u|^2\,dr=1\).
Imposing \(R(0)=0\) for every \(\ell\). The physical condition is \(u(0)=0\); for \(\ell=0\), \(R\) itself may be finite and non-zero at the origin.
Claiming the energy depends on \(m_\ell\). Because \(V_{\text{eff}}\) contains only \(\ell\), all \(2\ell+1\) values of \(m_\ell\) share one energy in a central field.
Dropping the irregular solution by hand without justification; its exclusion is a physical (normalisability/regularity) requirement, not an algebraic one.

Discussion

The separation is a direct consequence of symmetry, and the appearance of \(\ell(\ell+1)\) is Noether's theorem in quantum dress: rotational invariance of \(\hat H\) means the three generators of rotations, the components of \(\hat{\vec L}\), commute with \(\hat H\); \(\hat{\vec L}^2\) is the associated Casimir invariant, and its quantised eigenvalue \(\hbar^2\ell(\ell+1)\) is exactly what surfaces as the centrifugal term. The energy's independence of \(m_\ell\) is the statement that no direction is special. Any lifting of that degeneracy — Zeeman splitting in a magnetic field, for instance — is a direct diagnostic of broken rotational symmetry.

Physically, the substitution \(u=rR\) is more than algebra: it maps the radial problem onto genuine one-dimensional intuition, complete with an effective potential whose minimum sets the equilibrium radius and whose curvature sets vibrational spacings in molecules. The centrifugal barrier is the quantum echo of the classical statement that a particle with nonzero angular momentum cannot reach the force centre; here it manifests as the boundary condition \(R\sim r^\ell\) near the origin, which suppresses the probability density more strongly for higher \(\ell\).

The universality of the angular factor is what makes the periodic table computable. Whether the radial force is the \(-1/r\) Coulomb attraction, the harmonic \(+r^2\) confinement of a quantum dot, or a short-range nuclear well, the shells are labelled by the same \(\ell\), and the same selection rules \(\Delta\ell=\pm 1\) govern dipole transitions because they follow from the angular integrals of the \(Y_\ell^{m_\ell}\) alone, independent of \(V(r)\).

At the level of functional analysis, the reduction is a decomposition of the Hilbert space \(L^2(\mathbb{R}^3)\) into an orthogonal sum of sectors \(L^2(0,\infty;\,r^2dr)\otimes\{Y_\ell^{m_\ell}\}\), on each of which \(\hat H\) acts as an ordinary Sturm–Liouville operator. The regularity condition \(u(0)=0\) is precisely the choice of self-adjoint extension that makes \(\hat H\) Hermitian on the half-line; for potentials less singular than \(1/r^2\) this extension is unique (limit-point case at the origin), which is the rigorous reason the naive boundary condition suffices for atoms but fails for a \(-\alpha/r^2\) potential with \(\alpha\) above a critical strength.

Common misconceptions. The centrifugal term is not part of the "real" potential — it is a bookkeeping device that packages the angular kinetic energy into a radial form; it exists even for a free particle. And separability is a property of the symmetry, not of solvability: the equation always separates for any central \(V\), even when the resulting radial equation has no closed-form solution.

Worked examples

Height of the centrifugal barrier for a \(2p\) electron. Estimate the maximum of the centrifugal energy \(\frac{\hbar^2\ell(\ell+1)}{2m_e r^2}\) at the Bohr-scale radius \(r=a_0=5.29\times10^{-11}\,\text{m}\), for \(\ell=1\). A

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

\[ E_{\text{cf}}=\frac{(1.055\times10^{-34})^2\,(2)}{2\,(9.11\times10^{-31})\,(5.29\times10^{-11})^2} \]

\[ =\frac{(1.113\times10^{-68})(2)}{2(9.11\times10^{-31})(2.80\times10^{-21})}=\frac{2.226\times10^{-68}}{5.10\times10^{-51}}\approx 4.4\times10^{-18}\,\text{J} \]

\[ E_{\text{cf}}\approx 4.4\times10^{-18}\,\text{J}\approx 27\,\text{eV} \]

Reading. At the Bohr radius the centrifugal barrier for a \(p\)-electron is tens of electron-volts — comparable to the Coulomb binding — confirming that angular momentum is dynamically significant at atomic scales, not a small correction.

Ground-state energy of an \(\ell=0\) infinite spherical well. A particle of mass \(m\) is confined to \(r<a\) with \(V=0\) inside and \(V=\infty\) at \(r=a\). For \(\ell=0\) the radial equation for \(u=rR\) is a 1D infinite well; find the lowest energy. B

\[ -\frac{\hbar^2}{2m}\frac{d^2u}{dr^2}=E\,u,\qquad u(0)=0,\;u(a)=0 \]

\[ u(r)=A\sin(kr),\quad k=\frac{\sqrt{2mE}}{\hbar};\qquad u(a)=0\Rightarrow ka=n\pi,\;n=1,2,\dots \]

\[ E_n=\frac{\hbar^2 k^2}{2m}=\frac{n^2\pi^2\hbar^2}{2m a^2},\qquad E_1=\frac{\pi^2\hbar^2}{2m a^2} \]

\[ E_1=\frac{\pi^2\hbar^2}{2m a^2} \]

Reading. The \(u(0)=0\) boundary condition — automatic from \(u=rR\) — makes the \(s\)-state of the spherical well identical to the odd (sine) modes of a 1D box of width \(a\); the cosine modes are excluded precisely because they would give \(R=u/r\) singular at the origin. Numerically, for an electron in \(a=1.0\times10^{-10}\,\text{m}\): \(E_1=\frac{\pi^2(1.055\times10^{-34})^2}{2(9.11\times10^{-31})(10^{-20})}\approx 6.0\times10^{-18}\,\text{J}\approx 38\,\text{eV}\).

Problems

(A) Write down \(V_{\text{eff}}(r)\) for the hydrogen atom, \(V(r)=-\frac{e^2}{4\pi\varepsilon_0 r}\), and state its behaviour as \(r\to0\) for \(\ell=0\) versus \(\ell\ge1\).
Solution
\(V_{\text{eff}}(r)=-\dfrac{e^2}{4\pi\varepsilon_0 r}+\dfrac{\hbar^2\ell(\ell+1)}{2m_e r^2}\). For \(\ell=0\) there is no barrier and \(V_{\text{eff}}=-\dfrac{e^2}{4\pi\varepsilon_0 r}\to-\infty\) as \(r\to0\). For \(\ell\ge1\) the \(+1/r^2\) barrier dominates the \(-1/r\) attraction at small \(r\), so \(V_{\text{eff}}\to+\infty\); there is a potential well with a minimum at finite \(r\). Setting \(dV_{\text{eff}}/dr=0\) gives \(r_{\min}=\dfrac{4\pi\varepsilon_0\hbar^2\ell(\ell+1)}{m_e e^2}=\ell(\ell+1)a_0\).
(A) A particle has orbital quantum number \(\ell=3\). Compute the eigenvalue of \(\hat{\vec L}^2\) and the numerical factor \(\ell(\ell+1)\) in the centrifugal term. How many degenerate \(m_\ell\) values share the same radial energy?
Solution
\(\ell(\ell+1)=3\cdot4=12\), so \(\hat{\vec L}^2\) eigenvalue \(=12\hbar^2=12(1.055\times10^{-34})^2=1.34\times10^{-67}\,\text{J}^2\text{s}^2\). The magnitude \(|\vec L|=\sqrt{12}\,\hbar=3.65\times10^{-34}\,\text{J·s}\). Degeneracy \(=2\ell+1=7\) (\(m_\ell=-3,\dots,+3\)), all with identical energy because \(V_{\text{eff}}\) depends only on \(\ell\).
(B) Show that near the origin the regular radial solution behaves as \(R(r)\sim r^\ell\). (Hint: keep only the dominant \(1/r^2\) terms in the radial equation.)
Solution
Near \(r=0\) the centrifugal and kinetic terms dominate any finite \(V\) and \(E\). Using \(u=rR\): \(-\frac{\hbar^2}{2m}u''+\frac{\hbar^2\ell(\ell+1)}{2mr^2}u\approx0\Rightarrow u''=\frac{\ell(\ell+1)}{r^2}u\). Try \(u\sim r^s\): \(s(s-1)=\ell(\ell+1)\Rightarrow s=\ell+1\) or \(s=-\ell\). The regular choice is \(s=\ell+1\), giving \(u\sim r^{\ell+1}\) and hence \(R=u/r\sim r^\ell\). The rejected root \(s=-\ell\) gives \(R\sim r^{-(\ell+1)}\), non-normalisable/singular, and is discarded.
(B) For a 3D isotropic harmonic oscillator, \(V(r)=\tfrac12 m\omega^2 r^2\), write the effective potential and find the radius \(r_0\) minimising \(V_{\text{eff}}\) for \(\ell=2\).
Solution
\(V_{\text{eff}}=\tfrac12 m\omega^2 r^2+\dfrac{\hbar^2\ell(\ell+1)}{2mr^2}\). Minimise: \(\dfrac{dV_{\text{eff}}}{dr}=m\omega^2 r-\dfrac{\hbar^2\ell(\ell+1)}{mr^3}=0\Rightarrow r^4=\dfrac{\hbar^2\ell(\ell+1)}{m^2\omega^2}\), so \(r_0=\left(\dfrac{\hbar^2\ell(\ell+1)}{m^2\omega^2}\right)^{1/4}\). For \(\ell=2\), \(\ell(\ell+1)=6\): \(r_0=\left(\dfrac{6\hbar^2}{m^2\omega^2}\right)^{1/4}=6^{1/4}\sqrt{\dfrac{\hbar}{m\omega}}\approx1.57\sqrt{\hbar/m\omega}\).
(C) A deuteron model treats the neutron–proton system with reduced mass \(\mu=m_N/2\) (\(m_N=1.67\times10^{-27}\,\text{kg}\)) in an \(\ell=0\) attractive square well of depth \(V_0\) and radius \(a=2.1\times10^{-15}\,\text{m}\). Using the \(u\)-equation, derive the transcendental condition fixing the bound-state energy and evaluate the minimum depth \(V_0\) that binds one \(\ell=0\) state (\(E\to0^-\)).
Solution
For \(\ell=0\), \(u=rR\) obeys \(-\frac{\hbar^2}{2\mu}u''+V u=Eu\) with \(u(0)=0\). Inside (\(r<a\), \(V=-V_0\)): \(u=A\sin(k_1 r)\), \(k_1=\sqrt{2\mu(V_0-|E|)}/\hbar\). Outside (\(r>a\), \(V=0\), \(E=-|E|\)): \(u=Be^{-\kappa r}\), \(\kappa=\sqrt{2\mu|E|}/\hbar\). Matching \(u\) and \(u'\) at \(a\) gives \(k_1\cot(k_1 a)=-\kappa\). The threshold for the first bound state is \(E\to0^-\Rightarrow\kappa\to0\Rightarrow\cot(k_1 a)=0\Rightarrow k_1 a=\dfrac{\pi}{2}\), i.e. \(\sqrt{2\mu V_0}\,a/\hbar=\pi/2\). Thus \(V_0^{\min}=\dfrac{\pi^2\hbar^2}{8\mu a^2}\). Numerically, \(\mu=8.35\times10^{-28}\,\text{kg}\): \(V_0^{\min}=\dfrac{\pi^2(1.055\times10^{-34})^2}{8(8.35\times10^{-28})(2.1\times10^{-15})^2}=\dfrac{1.098\times10^{-67}}{2.94\times10^{-56}}\approx3.7\times10^{-12}\,\text{J}\approx23\,\text{MeV}\), the correct order of magnitude for the nuclear well depth.