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Today, on a very special episode of Science After Sunclipse, Mary Sue discovers that for a quantum-mechanical system with a central potential, the eigenfunctions of the Hamiltonian can be separated into radial and angular factors, and the angular dependence can be understood using angular momentum operators.

Way back, we showed that the momentum operator was the generator of spatial translations. Thanks to the time we spent studying 3D rotations, we can deduce an analogous fact about angular momentum. Recall that the commutator of two rotations about axes [tex]i[/tex] and [tex]j[/tex] is

[tex][J_i,J_j] = i\epsilon_{ijk}J_k.[/tex]

As Prof. Freedman would say, we learned in elementary school that in classical mechanics, the angular momentum of a point particle can be written

[tex] \vec{L} = \vec{x} \times \vec{p}.[/tex]

Let’s invent the corresponding expression for a quantum particle, by re-interpreting the variables [tex]\vec{x}[/tex] and [tex]\vec{p}[/tex] as operators which satisfy the canonical commutation relation

[tex][x_i,p_j] = i\hbar \delta_{ij}.[/tex]

To work with the angular momentum, we’ll express the cross product using the Levi-Civita symbol:

[tex]L_k = \epsilon_{ijk} x_i p_j.[/tex]

Now, we’ll show that the components of the angular momentum satisfy the same commutation relation as the generators of spatial rotations (up to a constant we don’t really have to care about). To see what this means, let’s start by finding the commutator of [tex]L_1[/tex] and [tex]L_2[/tex]. By the definition of angular momentum, we get

[tex][L_1,L_2] = [x_2 p_3 - x_3 p_2,x_3 p_1 - x_1 p_3],[/tex]

the right-hand side of which can be simplified to

[tex][x_2 p_3,x_3 p_1] + [x_3 p_2,x_1 p_3].[/tex]

Position and momentum operators commute as long as they do not have the same index, so we can pull some of the operators outside of the brackets,

[tex]x_2 [p_3,x_3] p_1 + x_1 [x_3,p_3] p_2,[/tex]

and then apply the canonical commutator to find

[tex]i\hbar(x_1 p_2 – x_2 p_1).[/tex]

Comparing this with the definition of angular momentum we wrote above, we see that

[tex][L_1,L_2] = i\hbar L_3.[/tex]

Notice that we have the commutator of two operators giving us some constants times the third operator. Swapping the indices on the left-hand side changes the sign of the commutator, and (of course) repeating an index makes the commutator work out to zero. By running through the above argument with different subscripts, we can show that in general, the commutator takes the familiar form

[tex][L_i,L_j] = i\hbar \epsilon_{ijk} L_k.[/tex]

This is just the relation we had before when we studied rotation generators, except for Planck’s constant (and when we work in “natural units,” we set Planck’s constant to 1 anyway).

When we looked at spatial translations, we worked with Taylor expansions and “infinitesimal generators.” We can do the analogous thing here, although the derivation will not be necessary for the work which follows. Take a particle whose state is specified by a scalar wavefunction [tex]\psi(\vec{x})[/tex] and perform a rotation

[tex]\vec{x} \rightarrow \vec{x}^\prime = R\vec{x},[/tex]


[tex]R(\theta,\hat{n}) = R(\vec{\omega})[/tex]

is a rotation of some small angle [tex]\theta[/tex] around the axis defined by [tex]\hat{n}[/tex]. The wave function transforms as

[tex]\psi(\vec{x}) \rightarrow \tilde{\psi}(\vec{x}) [/tex]

such that

[tex]\tilde{\psi}(\vec{x}^\prime) = \psi(\vec{x}).[/tex]

It may not be obvious at first glance, but for a small rotation, the transformed wavefunction can be written as a first-order correction to the original, plus terms higher-order in the angle [tex]\theta[/tex]:

[tex]\tilde{\psi}(\vec{x}) = \psi(\vec{x}) – (\vec{\omega}\times\vec{x})\cdot\grad\psi(\vec{x}) + \mathcal{O}(\theta^2).[/tex]

After some algebraic manipulation, one gets that

[tex]\ket{\tilde{\psi}} = \idmat – \frac{i}{\hbar}\vec{\omega} \cdot \vec{L}}\ket{\psi}[/tex]

where [tex]\vec{L} = \vec{x} \times \vec{p}[/tex].


Our problem of interest has an intrinsic spherical symmetry: the interaction potential does not depend upon direction, only distance. However, we’ve been working in Cartesian coordinates. Let’s shift ourselves into a coordinate system which reflects the symmetry of the problem, so we can get at its essence more easily. Mary Sue!


How can we turn the Cartesian coordinates [tex]x_i[/tex] into spherical coordinates?

“Um. . . Like this?”

[tex]x_1 = r\sin\theta\cos\phi,[/tex]

[tex]x_2 = r\sin\theta\sin\phi,[/tex]

[tex]x_3 = r\cos\theta.[/tex]

Excellent, although if we were mathematicians, our [tex]\theta[/tex] and [tex]\phi[/tex] would be defined the other way round. Now, Mary Sue, I’d like you to show for homework —


— for homework, show that with this change of variables, the Cartesian components of the angular momentum can be written in the following way:

[tex]L_1 = i\hbar\left(\sin\phi \partial_\theta + \frac{\cos\phi}{\tan\theta}\partial_\phi\right).[/tex]

[tex]L_2 = i\hbar\left(-\cos\phi\partial_\theta + \frac{\sin\phi}{\tan\theta}\partial_\phi\right).[/tex]

[tex]L_3 = -i\hbar\partial_\phi.[/tex]

In addition, we’ll be working with the square of the magnitude of the angular momentum,

[tex]\vec{L}^2 = L_1^2 + L_2^2 + L_3^2,[/tex]

which in spherical coordinates works out to be

[tex]\vec{L}^2 = -\hbar^2\left(\partial_\theta^2 + \frac{1}{\tan\theta}\partial_\theta + \frac{1}{\sin^2\theta}\partial_\phi^2\right).[/tex]


“Excuse me, Professor Science?”

Yes, Mary Sue?

“This is where we pick one component of the angular momentum, like say [tex]L_3[/tex], and then simultaneously diagonalize [tex]L_3[/tex] and [tex]\vec{L}^2[/tex], right?”

Exactly. And what are the eigenfunctions we find?

“They’re the spherical harmonics — we did this in high school. They’re labeled by two integers, the quantum numbers [tex]l[/tex] and [tex]m[/tex].”

So, drawing on our high-school education, we can write

[tex]\vec{L}^2 Y_l^m = l(l+1) \hbar^2 Y_l^m,[/tex]

[tex]L_3 Y_l^m = m\hbar Y_l^m.[/tex]

Recall, in addition, that [tex]l[/tex] can take the values 0, 1, 2, 3, …, while for any value of [tex]l[/tex], the number [tex]m[/tex] ranges from [tex]-l[/tex] to [tex]+l[/tex].

Spherical harmonics occur in classical electromagnetism as well as in quantum mechanics, since the Laplacian also appears in equations for the electrical potential. The detailed investigation of the spherical harmonics is written up in lots of places, so we’ll move on for now.


So far, we have manipulated our original Schrödinger Equation for a two-particle system into the following form, where [tex]\psi[/tex] is the wavefunction which encapsulates the relative degrees of freedom of the two particles; for the hydrogen atom, this is almost, but not exactly, the “wavefunction of the electron.”

[tex]\left(-\frac{\hbar^2}{2\mu} \grad^2 + V(r)\right)\psi = E\psi.[/tex]

The spherical symmetry of our problem leads us to work in spherical coordinates. In that coordinate system, the Laplacian takes the following form:

[tex]\grad^2 = \frac{1}{r} \partial_r^2 r + \frac{1}{r^2}\left(\partial_\theta^2 + \frac{1}{\tan\theta}\partial_\theta + \frac{1}{\sin^2\theta}\partial_\phi^2\right).[/tex]

Comparing this to the expression for [tex]\vec{L}^2[/tex] whose derivation we sketched above, we can rewrite our Hamiltonian in terms of the “electron’s” angular momentum:

[tex]H = -\frac{\hbar^2}{2\mu} \frac{1}{r}\partial_r^2 r + \frac{1}{2\mu r^2}\vec{L}^2 + V(r).[/tex]

Note that the only term which acts on the angular dependency of the wavefunction is the one containing [tex]\vec{L}^2[/tex]. If we separate variables and write the wavefunction as the product of a radial function and a spherical harmonic,

[tex]\psi(\vec{x}) = R_l(r) Y_l^m(\theta,\phi),[/tex]

then the total function is an eigenfunction of the Hamiltonian

[tex]H\psi = E\psi,[/tex]

but it is also an eigenfunction of [tex]\vec{L}^2[/tex]:

[tex]\vec{L}^2\psi = l(l + 1)\hbar^2 \psi.[/tex]

For a given value of [tex]l[/tex], then, our Schrödinger Equation can be written

[tex]\left[-\frac{\hbar^2}{2\mu} \frac{1}{r}\partial_r^2 r + \frac{l(l + 1)\hbar^2}{2\mu r^2} + V(r)\right]R_l(r) = ER_l(r).[/tex]

We can clean this up a little by scaling by [tex]r[/tex], thusly:

[tex]R_l(r) = \frac{1}{r} u_l(r).[/tex]

With this transformation, the Schrödinger Equation takes the form

[tex]\left[-\frac{\hbar^2}{2\mu} \partial_r^2 + \frac{l(l + 1)\hbar^2}{2\mu r^2} + V(r)\right]u_l(r) = Eu_l(r).[/tex]

This is just what we’d get if we wrote a Schrödinger Equation for a single particle moving in one dimension with the potential

[tex]V_{\rm eff.} (r) = V(r) + \frac{l(l + 1) \hbar^2}{2\mu r^2}.[/tex]

We call this the effective potential. It is just the interaction potential we wrote originally, plus a contribution which depends upon the angular momentum quantum number [tex]l[/tex], a contribution which is known as the angular momentum barrier. This is the quantum analogue of our classical intuition that a particle with large angular momentum spinning around inside a potential well is less likely to be found near the center of the well than is a particle with smaller angular momentum. In other words, the effective potential is where “centrifugal force” sneaks in.

All the work we’ve done so far applies equally well to any two-body system where the interaction potential depends upon the separation distance. We could, for example, use this machinery to attack the 3D harmonic oscillator (two massive particles connected by a massless spring). Instead of doing that, however, we’re going to attack the hydrogenic atoms, where a single electron moves in the Coulomb field of the nucleus. For a nucleus containing [tex]Z[/tex] protons, the interaction potential will be

[tex]V(r) = -\frac{Ze^2}{r}.[/tex]

Therefore, when the electron is in an eigenstate of angular momentum quantum number [tex]l[/tex], its behavior will obey the Schrödinger Equation with the Hamiltonian

[tex]H_l = -\frac{\hbar^2}{2\mu} \partial_r^2 -\frac{Ze^2}{r} + \frac{l(l + 1) \hbar^2}{2\mu r^2}.[/tex]

Next time, we will apply the machinery of SUSY QM and shape invariance to this family of Hamiltonians and deduce both the eigenvalues and the eigenfunctions of the hydrogenic atom.



  1. I’ve seen most of this derivation before, of course (in Pauling & Wilson), but I guess I never really thought about the interpretation of the equation as giving an angular momentum barrier. Nice.

  2. They’re the spherical harmonics — we did this in high school.

    I might be missing some subtle joke here, but at the risk of looking foolish I’ll take this literally and ask: Do you really study spherical harmonics in US high schools?

  3. It’s a joke my group theory professor always made. The most advanced math I got in high school was, if I recall correctly, a double integral.

  4. He’s not the only one to say that sort of thing. I have distinct memories (traumas?) of Serge Lang screaming (yes, screaming) at me that any high-school student knows that when you see a periodic function you should have an irresistable compulsion to hit it with a Fourier transform.