# SUSY QM 5: The Hydrogen Atom

Last time, we found that the problem of the hydrogen atom could be split into a radial part and an angular part. Thanks to spherical symmetry, the angular part could be studied using angular momentum operators and spherical harmonics. We found that the 3D behavior of the electron could be reinterpreted as a 1D wavefunction of a particle in an effective potential which was the two-body interaction potential plus a “barrier” term which depended upon the angular momentum quantum number. Today, we’re going to solve the radial part of the problem and thereby find the eigenstates and eigenenergies of the hydrogen atom.

The technique we’ll employ has a certain charm, because we solved the first part, the angular dependence, using commutator relations, while as we shall see, the radial dependence can be solved with anticommutator relations.

THE FAMILY OF COULOMB HAMILTONIANS

We ended up with a family of Hamiltonians labeled by the angular momentum quantum number:

$H_l^{\rm Coul} = -\frac{\hbar^2}{2\mu} \partial_r^2 -\frac{Ze^2}{r} + \frac{l(l + 1) \hbar^2}{2\mu r^2}.$

For each eigenstate of any $H_l^{\rm Coul}$, the hydrogen atom has a set of states labeled by the quantum number $m$, which takes the values

$m = 0, \pm 1, \pm 2, \ldots, \pm l.$

However, because the original problem was spherically symmetric, the operator $L_3$ does not appear in the overall Hamiltonian (there’s no preferred axis), and so the energy will not depend upon $m$. Geometrical symmetry implies that states of different $m$ but the same $l$ are degenerate; as we proceed, we’ll see an additional example of symmetry implying degeneracy unfold before us.

Carrying around all those constants in $H_l^{\rm Coul}$ would be a pain and not particularly useful, so let’s get rid of them. By appropriately rescaling our variables (craftily choosing “better rulers” for distance and energy) we can rewrite $H_l^{\rm Coul}$ in a cleaner form, preserving the essential features of the Hamiltonians without all that dimensionful baggage:

$H_l^{\rm Coul} = -\partial_x^2 + \frac{l(l+1)}{x^2} – \frac{1}{x}.$

[I originally screwed this bit up; see the comments below.]

Now, we have a family of related Hamiltonians, labeled by a nonnegative integer. Why does this situation sound familiar? When we studied supersymmetry and shape invariance, we saw just such a hierarchy of Hamiltonians appear! So, let’s see if those ideas are applicable to this problem.

We should note one complication right away: when we explored SUSY QM and shape-invariant partner potentials, we worked with an original Hamiltonian whose ground-state energy was zero. Here, however, we’re looking at an electron bound by an atom, so the energies of the eigenstates will be negative. Shifts of the energy scale aren’t going to pose profound difficulties, but we should be on our guard.

LIVING UP TO OUR SUPERPOTENTIAL

Earlier, we took a superpotential and from it derived two partner potentials. Thus, if $W$ was the superpotential, then

$A = \partial_x + W(x)$

and

$A^\dag = -\partial_x + W(x),$

so our partner Hamiltonians would be

$H^{(1)} = A^\dag A = -\partial_x^2 + W^2(x) – \partial_x W(x)$

and

$H^{(2)} = AA^\dag = -\partial_x^2 + W^2(x) + \partial_x W(x).$

To study $H_l^{\rm Coul}$ in this framework, we’ll need a superpotential which, when put through these steps, yields both the $1/x$ Coulomb law and the angular momentum barrier. The superpotential will, therefore, depend upon $l$. We choose the following:

$W_l = -\frac{(l + 1)}{x} + \frac{1}{2(l + 1)}.$

This gives us a pair of operators for each value of $l$:

$A_l = \partial_x – \frac{l + 1}{x} + \frac{1}{2(l + 1)},$

and

$A^\dag_l = -\partial_x – \frac{l + 1}{x} + \frac{1}{2(l + 1)}.$

The first Hamiltonian is given by one ordering,

$H_l^{(1)} = A_l^\dag A_l,$

which works out to be

$H_l^{(1)} = -\partial_x^2 + \frac{l(l+1)}{x^2} – \frac{1}{x} + \frac{1}{4(l+1)^2},$

and when we define a Hamiltonian with the other ordering,

$H_l^{(2)} = A_l A^\dag_l,$

we get as the partner Hamiltonian

$H_l^{(2)} = -\partial_x^2 + \frac{(l+1)(l+2)}{x^2} – \frac{1}{x} + \frac{1}{4(l+1)^2}.$

By comparing these two expressions, we get that

$H_l^{(2)} = H_{l + 1}^{(1)} + \frac{1}{4(l + 1)^2} – \frac{1}{4(l + 2)^2}.$

Here, we see the shape-invariance criterion: $H_l^{(2)}$ is a Hamiltonian which depends upon the parameter $l$, and we can reach $H_l^{(2)}$ by taking its partner $H_l^{(1)}$, tweaking a parameter and adding an offset which does not depend upon $x$. We can see the energy-axis shift mentioned above if we rewrite these operators in terms of the Coulomb Hamiltonian. First, we see that

$H_l^{(1)} = H_l^{\rm Coul} + \frac{1}{4(l + 1)^2},$

so the operators $H_l^{(1)}$ and $H_l^{\rm Coul}$ will have the same eigenstates, although their eigenvalues will be separated by a term dependent upon $l$. We also have

$H_l^{(2)} = H_{l + 1}^{\rm Coul} + \frac{1}{4(l + 1)^2}.$

BUILDING UP THE STATES

We are now in a position to employ the diagrammatic approach we built up earlier. Let’s label the eigenstates of a particular Hamiltonian in the hierarchy by $\nu$, such that $\nu = 0$ for the ground state. Each state in our diagram will then be labeled by $l$ and $\nu$, which we write in Dirac notation as $|\nu, l\rangle$. The isospectrality between partner potentials implies that state $\nu$ of Hamiltonian $H_l^{(1)}$ has the same energy as state $\nu – 1$ of $H_l^{(2)}$. The state annihilated by $A_l$ is defined by

$A_l |0, l\rangle = 0,$

which tells us that

$H_l^{(1)} |0, l\rangle = 0.$

In turn, because of the energy shift, the Coulomb Hamiltonian satisfies

$H_l^{\rm Coul} |0, l\rangle = -\frac{1}{4(l + 1)^2} |0, l\rangle.$

Likewise, for $l + 1$,

$H_{l+1}^{(1)} |0, l + 1\rangle = 0.$

Using the shape-invariance condition yields

$H_l^{(2)} |0, l + 1\rangle = \left[\frac{1}{4(l + 1)^2} – \frac{1}{4(l + 2)^2}\right]|0, l + 1\rangle.$

By the other key property, isospectrality, we know that $H_l^{(1)}$ has a state of the same energy, one notch higher in $\nu$:

$H_l^{(1}} |1, l\rangle = \left[\frac{1}{4(l + 1)^2} – \frac{1}{4(l + 2)^2}\right]|1, l\rangle.$

Again, the physical energy of the hydrogen atom can be found by adding a constant:

$H_l^{\rm Coul} |1, l\rangle = -\frac{1}{4(l + 2)^2} |1, l\rangle.$

Notice how as we moved from $\nu = 0$ to $\nu = 1$, the value in the denominator went from $l + 1$ to $l + 2$? The rest of the states can be built by working “leftwards” in the state diagram:

$|\nu, l\rangle = A^\dag_l A^\dag_{l + 1} \cdots A^\dag_{l + \nu – 1} |0, l + \nu\rangle.$

This just means repeating the process we went through earlier, meaning that the denominator of the eigenvalue will depend on $l + \nu + 1$. Note that the energy of the state labeled by $\nu$ and $l$ depends only upon their sum. If we restore the constants we scaled away earlier, we get that

$E_{\nu l} = -\frac{\mu e^4 Z^2}{2\hbar^2} \frac{1}{(l + \nu + 1)^2},$

so we can call $l + \nu + 1$ by the new name $n$ and write

$E_n = -\frac{\mu e^4 Z^2}{2\hbar^2} \frac{1}{n^2}.$

This is the energy of the $n{\rm th}$ eigenstate of the hydrogenic Hamiltonian. We could work out explicit expressions for states of our choice, but before we plug-and-chug any specific examples, let’s take a breather and look at a picture.

THE SLIGHTLY BIGGER PICTURE

The diagram we drew before now looks like this:

The eigenstate diagram shows more symmetry than we had expected on geometrical considerations alone: the only quantum number necessary for determining the energy is $n$, which corresponds to the energy-level index in Bohr’s model of the atom. If our interaction potential were not shape invariant, this degeneracy would be broken.

This is the way I learned to solve the hydrogenic atom in the misty days of my undergraduacy. The only textbook I know of which takes an approach like this is Ohanian’s Principles of Quantum Mechanics; other than a handful of universities, most schools attack the problem by plowing into Schrödinger’s second-order differential equation and eventually finding a recursion relation for the Laguerre polynomials. Prof. Rajagopal‘s lecture notes call the standard method “much more painful,” and as for why most textbooks follow that route, “Go figure.” I suspect that too many teachers of quantum mechanics have been bitten by the Matrix Zombie and think that mathematics beyond differential equations is just too hard for introductory classes. Rather than making the time investment necessary to use “more advanced” techniques, they solve problems in laborious and rather unilluminating ways.

Unfortunately, MIT’s OpenCourseWare project doesn’t provide the lecture notes we used, or any later editions thereof; the site for 8.05 Quantum Physics II just lists the sections of textbooks which should be read, instead of providing actual juicy PDFs. This post, in particular, was based on the 8.05 material, while my earlier overview of the general superalgebra machinery mostly follows Fred Cooper, Avinash Khare and Uday Sukhatme’s review article, “Supersymmetry and Quantum Mechanics” (1994). As that review explains, Schrödinger himself solved for the hydrogen atom eigenstates with a method rather like this, in 1940; many years later, the supersymmetric context of that “factorization” method was discovered.

From here, we can go in several directions. After perhaps working a few examples, we can head towards the relativistic regime and find SUSY-based solutions to the Dirac Equation. Also, we can look back at classical mechanics and relate these ideas to the Laplace-Runge-Lenz vector, an avenue which will eventually lead us to superalgebras with central charge and BPS bounds. I’m also strongly tempted to look at the application of SUSY to diffusion problems via the Fokker-Planck Equation.

SUSY QM SERIES:

## 3 thoughts on “SUSY QM 5: The Hydrogen Atom”

1. Okay, you chose $x^2=\frac{2\mu}{\hbar^2}r^2$ and that rescales the differential term and the abgular momentum barrier, but why does it handle the potential term, getting rid of the $Ze^2$?

(did I get the TeX right this time?)

2. Argh. This is what I get for writing too fast and not proofreading the whole thing. The scaling which should be done is, writing $a$ for the Bohr radius,

$x = r\left(\frac{2Z}{a}\right),$

to rescale the distance, and

$\tilde{E}_{\nu l} = \frac{\hbar^2}{2\mu e^4 Z^2} E_{\nu l},$

to rescale the energy.

3. Uh-oh, I just got a referral to this page from a Google search, {dirac equation and susy qm}. Looks like I better get my act together and write that next post! In the meantime, an introduction can be found in the review article cited above.