Central Potentials and Rotational Symmetry

A central potential is a scalar potential that depends only on distance from one fixed origin,

$$V(\mathbf r)=V(r), \qquad r=\lvert\mathbf r\rvert.$$

For a spinless particle with Hamiltonian

$$H=\frac{\mathbf P^2}{2m}+V(r),$$

this is not merely a convenient coordinate choice. It is a symmetry statement: rotations leave the Hamiltonian invariant. The consequence is that energy eigenstates can be organized by orbital angular momentum,

$$L^2\psi=\hbar^2\ell(\ell+1)\psi, \qquad L_z\psi=\hbar m\psi,$$

with

$$\ell=0,1,2,\ldots, \qquad m=-\ell,-\ell+1,\ldots,\ell.$$

This page explains the symmetry engine behind central-potential problems. The explicit radial equation, boundary conditions, and model-specific spectra live in Central Potentials and the Radial Schrödinger Equation.

Scope

The symmetry content of a central potential is:

• rotations are generated by orbital angular momentum $\mathbf L$;
• the Hamiltonian commutes with every component $L_i$;
• $H$, $L^2$, and one chosen component such as $L_z$ may be diagonalized together;
• the angular dependence is described by spherical harmonics;
• the magnetic quantum number $m$ labels a rotational multiplet, not a different radial problem.

The dynamical content is separate. The functional form of $V(r)$ determines whether the spectrum has bound states, continuum states, special accidental degeneracies, or no bound states at all.

Rotational Invariance

Let

$$\mathbf L=\mathbf R\times\mathbf P$$

be orbital angular momentum. Its components rotate position and momentum as vectors:

$$[L_i,R_j] = i\hbar\sum_k\epsilon_{ijk}R_k,$$

and

$$[L_i,P_j] = i\hbar\sum_k\epsilon_{ijk}P_k.$$

Therefore scalar combinations built from dot products are rotationally invariant. In particular,

$$[L_i,\mathbf R^2]=0, \qquad [L_i,\mathbf P^2]=0.$$

Since $r=(\mathbf R^2)^{1/2}$, a sufficiently well-defined function of $r$ also commutes with $\mathbf L$:

$$[L_i,V(r)]=0.$$

For the central-potential Hamiltonian,

$$H=\frac{\mathbf P^2}{2m}+V(r),$$

one obtains

$$[H,L_i]=0, \qquad i=x,y,z.$$

Equivalently, if

$$U(\mathcal R) = \exp\left( -\frac{i}{\hbar}\boldsymbol\theta\cdot\mathbf L \right)$$

is the unitary implementing a spatial rotation $\mathcal R$, then

$$U(\mathcal R)HU(\mathcal R)^\dagger=H.$$

The infinitesimal commutator and the finite rotation statement are the same symmetry expressed in two languages.

Good Quantum Numbers

Because every $L_i$ commutes with $H$, the Casimir $L^2$ also commutes with $H$:

$$[H,L^2]=0.$$

Since

$$[L^2,L_z]=0,$$

we may choose stationary states that are simultaneous eigenstates of $H$, $L^2$, and $L_z$:

$$\begin{aligned} H|\alpha,\ell,m\rangle &= E_{\alpha\ell}|\alpha,\ell,m\rangle, \\ L^2|\alpha,\ell,m\rangle &= \hbar^2\ell(\ell+1)|\alpha,\ell,m\rangle, \\ L_z|\alpha,\ell,m\rangle &= \hbar m|\alpha,\ell,m\rangle. \end{aligned}$$

The label $\alpha$ denotes whatever additional information is needed to distinguish states with the same angular labels. For bound central potentials it is often a radial quantum number $n_r$; for hydrogen it is often traded for a principal quantum number $n$; for scattering states it may be an energy, wavenumber, and boundary convention.

The set $\{H,L^2,L_z\}$ is therefore a natural commuting set, but it need not be complete in the presence of degeneracies. Extra labels may be required when several independent states have the same $E$, $\ell$, and $m$.

Why Only One Component Is Chosen

Rotational symmetry gives

$$[H,L_x]=[H,L_y]=[H,L_z]=0.$$

It does not make the angular momentum components commute with one another:

$$[L_x,L_y]=i\hbar L_z.$$

Thus an energy eigenstate can be chosen to have definite $L^2$ and definite $L_z$, but not generally definite $L_x$, $L_y$, and $L_z$ simultaneously. The $z$ axis is a convention for labeling a basis inside the multiplet. In a pure central potential there is no physical $z$ direction.

Separation into Angular and Radial Parts

In position representation, the same symmetry appears as angular-radial separation:

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

The spherical harmonic carries the representation-theoretic data:

$$L^2Y_\ell^m = \hbar^2\ell(\ell+1)Y_\ell^m, \qquad L_zY_\ell^m = \hbar mY_\ell^m.$$

The radial factor carries the dynamics of the particular potential. For a generic bound central potential one expects energies of the form

$$E=E_{n_r\ell},$$

not $E=E_n$ independent of $\ell$. The derivation of the separated radial equation is given in Angular and Radial Separation, and the half-line boundary conventions are collected in the Radial Schrödinger Equation.

Magnetic Degeneracy

The commutator

$$[H,L_i]=0$$

implies

$$[H,L_\pm]=0, \qquad L_\pm=L_x\pm iL_y.$$

If $|\alpha,\ell,m\rangle$ is an energy eigenstate, then

$$H(L_\pm|\alpha,\ell,m\rangle) = E_{\alpha\ell}(L_\pm|\alpha,\ell,m\rangle).$$

The ladder operators change $m$ while leaving $\ell$ fixed:

$$L_\pm|\alpha,\ell,m\rangle = \hbar \sqrt{\ell(\ell+1)-m(m\pm1)} |\alpha,\ell,m\pm1\rangle.$$

Therefore all nonzero states in the same fixed-$\ell$ ladder have the same energy. For each $\ell$ this gives

$$2\ell+1$$

states with different $m$ labels. This is the magnetic degeneracy of a central potential. The word “magnetic” refers to the magnetic quantum number $m$, not to the presence of a magnetic field.

This degeneracy is best understood as a representation statement. A rotationally invariant Hamiltonian acts the same way on every orientation inside an irreducible angular-momentum multiplet. The radial dynamics may change from one $\ell$ sector to another, but it cannot distinguish $m=-\ell$ from $m=\ell$ without a preferred direction.

What Symmetry Does Not Imply

Rotational symmetry guarantees less than many first encounters with the hydrogen atom suggest.

Statement	Guaranteed by central symmetry?	Reason
$m$ degeneracy within a fixed $\ell$ sector	yes	rotations relate the $2\ell+1$ orientations
integer orbital $\ell$ for scalar wavefunctions	yes	single-valued functions on the sphere
degeneracy between different $\ell$ values	no	different $\ell$ sectors have different radial equations
hydrogenic $n^2$ bound-state degeneracy	no	it uses the special Coulomb problem
conservation of total spin	not part of this scalar model	spin requires extra Hilbert-space factors

For a generic central potential,

$$E=E_{n_r\ell}$$

is the natural expectation. The ideal Coulomb potential has a stronger degeneracy:

$$E_n = -\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2} \frac{1}{n^2},$$

which is independent of $\ell$ as well as $m$. That stronger pattern comes from hidden structure beyond ordinary rotational invariance and is the standard example in Accidental Symmetry and Degeneracy of the Hydrogen Atom.

Perturbations and Surviving Symmetry

A small noncentral perturbation can break part of the angular-momentum labeling. For example,

$$H = \frac{\mathbf P^2}{2m} +V(r) +\epsilon Z$$

selects the $z$ axis. The perturbation $\epsilon Z$ is invariant under rotations around the $z$ axis but not under arbitrary rotations. Consequently,

$$[H,L_z]=0,$$

while generally

$$[H,L^2]\ne0.$$

The label $m$ may remain useful, but $\ell$ is no longer protected. This is the symmetry reason external fields split and mix central-potential multiplets.

Similarly, a spin-orbit interaction of the form

$$f(r)\,\mathbf L\cdot\mathbf S$$

is rotationally invariant under simultaneous rotations of orbital and spin degrees of freedom, but it need not commute with $\mathbf L$ and $\mathbf S$ separately. In that problem the protected angular momentum is the total

$$\mathbf J=\mathbf L+\mathbf S,$$

not orbital angular momentum alone. The angular-momentum-addition analysis is developed in Spin–Orbit Coupling.

Worked Example: Degeneracy from Ladder Operators

Suppose $H$ is central and has a normalized eigenstate $|\alpha,\ell,m\rangle$ with

$$H|\alpha,\ell,m\rangle = E|\alpha,\ell,m\rangle.$$

Since $[H,L_+]=0$,

$$\begin{aligned} H(L_+|\alpha,\ell,m\rangle) &= L_+H|\alpha,\ell,m\rangle \\ &= E(L_+|\alpha,\ell,m\rangle). \end{aligned}$$

If $m<\ell$, the raised state is nonzero and proportional to $|\alpha,\ell,m+1\rangle$. It has the same energy $E$. Repeating the argument connects all states from $m=-\ell$ to $m=\ell$. At the endpoints the ladder coefficient vanishes, so the argument stops naturally.

This proof uses symmetry only. It says nothing about the actual numerical value of $E$, which requires solving the radial problem.

Physical Interpretation

A central potential has no preferred direction. Different values of $m$ are different orientations of the same angular pattern relative to an arbitrarily chosen axis. Rotating the entire state changes $m$ components inside the same multiplet but does not change the energy.

The label $\ell$ is different. It changes the total amount of orbital angular momentum and hence the centrifugal contribution to radial motion. That is why $\ell$ is a dynamical sector label rather than merely an orientation label.

The central-potential problem is therefore a clean example of the general rule:

$$\text{symmetry} \quad\Longrightarrow\quad \text{commuting generators, multiplets, and degeneracies}.$$

It is also a warning: symmetries constrain spectra, but they do not usually solve the radial dynamics by themselves.

Common Mistakes

• Treating “central” as synonymous with “Coulomb.” The Coulomb potential is one special central potential.
• Assuming rotational symmetry makes the energy independent of $\ell$. It guarantees independence of $m$, not generic independence of $\ell$.
• Thinking that choosing $L_z$ means the system has a physical $z$ axis. The axis is a labeling convention unless an external field or boundary condition selects it.
• Forgetting that $L_x$, $L_y$, and $L_z$ do not commute with each other even though each commutes with a central Hamiltonian.
• Treating spherical harmonics as hydrogen-specific functions rather than universal angular momentum eigenfunctions.
• Applying spin-orbit conclusions to a spinless central Hamiltonian without changing the conserved angular momentum from $\mathbf L$ to $\mathbf J$.

Cross-Links

• Angular Momentum Algebra
• States, Observables, and Hamiltonians
• Degeneracy and Multiplets
• Simultaneous Eigenstates and Good Quantum Numbers
• Eigenvalues of J² and Jz
• Orbital Angular Momentum
• Position-Space Representation
• Spherical Coordinates
• Spherical Harmonics
• Hydrogen Atom Angular Structure
• Rigid Rotor
• Spin–Orbit Coupling
• Commutators and Conservation Laws
• Explicit Symmetry Breaking
• Accidental Symmetry
• Hidden Symmetry
• Complete Sets of Commuting Observables
• Wave-Mechanics Central Potentials
• Angular and Radial Separation
• Radial Schrödinger Equation
• Degeneracy of the Hydrogen Atom

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.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.

Exercises

1. Use $[L_i,R_j]=i\hbar\sum_k\epsilon_{ijk}R_k$ to show that $[L_i,\mathbf R^2]=0$.

Solution

Write

$$\mathbf R^2=\sum_j R_jR_j.$$

Then

$$[L_i,\mathbf R^2] = \sum_j \left( [L_i,R_j]R_j +R_j[L_i,R_j] \right).$$

Substitute the vector commutator:

$$[L_i,\mathbf R^2] = i\hbar \sum_{j,k} \epsilon_{ijk} \left( R_kR_j+R_jR_k \right).$$

The position components commute, so the factor in parentheses is symmetric in $j$ and $k$. Contracting a symmetric tensor with the antisymmetric $\epsilon_{ijk}$ gives zero. Therefore

$$[L_i,\mathbf R^2]=0.$$

2. Prove that all $m$ states in a fixed $\ell$ multiplet have the same energy for a central Hamiltonian.

Solution

For a central Hamiltonian, $[H,L_i]=0$, hence $[H,L_\pm]=0$. If

$$H|\alpha,\ell,m\rangle = E|\alpha,\ell,m\rangle,$$

then

$$H(L_\pm|\alpha,\ell,m\rangle) = L_\pm H|\alpha,\ell,m\rangle = E(L_\pm|\alpha,\ell,m\rangle).$$

Whenever the laddered state is nonzero, it is proportional to $|\alpha,\ell,m\pm1\rangle$. Repeated raising or lowering connects every allowed $m$ value in the fixed $\ell$ multiplet, so all of them share the same energy.

3. Consider $H=\mathbf P^2/(2m)+V(r)+\epsilon Z$. Which angular-momentum labels remain protected?

Solution

The term $\epsilon Z$ is invariant under rotations about the $z$ axis, so $L_z$ remains conserved:

$$[H,L_z]=0.$$

It is not invariant under general rotations. For example, using

$$[L_x,Z]=-i\hbar Y,$$

one finds that $[H,L_x]\ne0$ when $\epsilon\ne0$. Therefore $L^2$ is not generally conserved, and $\ell$ is not protected. The magnetic label $m$ can remain a good label, but the full $2\ell+1$ degeneracy is generally split.

4. Why does rotational symmetry not by itself explain the full hydrogen degeneracy?

Solution

Rotational symmetry explains degeneracy among different $m$ values for fixed radial label and fixed $\ell$. A generic central potential has energies of the form

$$E=E_{n_r\ell}.$$

The ideal Coulomb bound-state energy depends only on

$$n=n_r+\ell+1,$$

so states with different $\ell$ but the same $n$ are also degenerate. That stronger degeneracy is not forced by ordinary rotations; it is special to the Coulomb problem and its additional hidden symmetry.