SUSY QM 3: The Two-Body Problem

Our goal in this series is the solution of the hydrogen atom using the methods of supersymmetric quantum mechanics. Last time, we constructed the following picture of the procedure:

If the potential we wish to study satisfies a certain criterion, which we called “shape invariance,” we can construct a hierarchy of Hamiltonians, each missing the lowest-energy eigenstate of the last, and find the complete spectrum of the original Hamiltonian by “working leftward” in the state diagram. We shall see that with the hydrogen atom, each state in the diagram corresponds to a physical eigenstate of the system, but in order to get there, we have to turn the three-dimensional Coulomb potential of the hydrogen atom into the kind of problem we can study with the SUSY QM machinery we’ve built up so far. Two steps will be necessary to do this: first, moving to the center-of-mass reference frame, and second, separating the radial and angular dependencies. In this post, we’ll tackle the first of those two tasks.

While the SUSY part isn’t widely taught, these preliminary steps are more familiar. This brief note is based on Chapter VII of Cohen-Tannoudji, Diu and Laloë.

Let’s work in general terms for the time being. We’ll consider two particles, whose position operators shall be [tex]\vec{x}_1[/tex] and [tex]\vec{x}_2[/tex], and whose momentum operators are [tex]\vec{p}_1[/tex] and [tex]\vec{p}_2[/tex]. Supposing that the interaction between the particles depends only on the displacement between them, we can write a Hamiltonian for the two-body system as follows:

[tex]H = \frac{\vec{p}_1^2}{2m_1} + \frac{\vec{p}_2^2}{2m_2} + V(\vec{x}_1 – \vec{x}_2).[/tex]

We can simplify this problem by dividing up its degrees of freedom between the center-of-mass motion and the relative motion of the particles. Following classical intuition, we introduce new variables,

[tex]\vec{x}_{\rm CM} = \frac{m_1\vec{x}_1 + m_2\vec{x}_2}{m_1 + m_2},[/tex]

[tex]\vec{x} = \vec{x}_1 – \vec{x}_2,[/tex]

[tex]\vec{p}_{\rm CM} = \vec{p}_1 + \vec{p}_2,[/tex]

[tex]\vec{p} = \frac{m_2\vec{p}_1 – m_1\vec{p}_2}{m_1 + m_2}.[/tex]

Knowing the commutators of the original position and momentum operators, it’s not too hard to show that for any components [tex]x_{{\rm CM}j}[/tex] and [tex]p_{{\rm CM}k}[/tex] of the center-of-mass operators, the commutator is just

[tex]\comm{x_{{\rm CM}j}}{p_{{\rm CM}k}} = i\hbar\delta_{jk}.[/tex]

The same holds for the relative degrees of freedom, too:

[tex]\comm{x_j}{p_k} = i\hbar\delta_{jk}.[/tex]

In other words, our new operators are just as good position and momentum operators as the old.

By chugging through a little algebra, we can rewrite our original two-body Hamiltonian using the new operators:

[tex]H = \frac{\vec{p}_{\rm CM}^2}{2M} + \frac{\vec{p}^2}{2\mu} + V(\vec{x}).[/tex]

Here, we’ve written [tex]M[/tex] for the sum of the masses,

[tex]M = m_1 + m_2,[/tex]

and [tex]\mu[/tex] for the reduced mass

[tex]\mu = \left(\frac{1}{m_1} + \frac{1}{m_2}\right)^{-1}.[/tex]

When it’s written this way, it’s easy to see that the Hamiltonian [tex]H[/tex] splits into two parts:

[tex]H = H_{\rm CM} + H_r,[/tex]


[tex]H_{\rm CM} = \frac{\vec{p}_{\rm CM}^2}{2M}[/tex]


[tex]H_r = \frac{\vec{p}^2}{2\mu} + V(\vec{x}).[/tex]

What’s more, these two parts commute with each other,

[tex]\comm{H_{\rm CM}}{H_r} = 0,[/tex]

so we can pick a basis which simultaneously diagonalizes both of them. An eigenstate [tex]\ket{\psi}[/tex] of [tex]H[/tex] satisfies the equation

[tex]H\ket{\psi} = E\ket{\psi},[/tex]

and the same ket is an eigenket of both [tex]H_{\rm CM}[/tex] and [tex]H_r[/tex]:

[tex]H_{\rm CM}\ket{\psi} = E_{\rm CM}\ket{\psi},[/tex]

[tex]H_r\ket{\psi} = E_r\ket{\psi},[/tex]

with [tex]E = E_{\rm CM} + E_r[/tex]. Eigenstates of the total Hamiltonian will be tensor products of eigenstates of the CM and relative parts:

[tex]\ket{\psi} = \ket{\chi}_{\rm CM} \otimes \ket{\omega}_r,[/tex]

let’s say. We can rewrite the CM Schrödinger equation above as a statement about position-space wavefunctions, if we take derivatives with respect to CM coordinates:

[tex]-\frac{\hbar^2}{2M} \partial_{\rm CM}^2 \chi(\vec{x}_{\rm CM}) = E_{\rm CM} \chi(\vec{x}_{\rm CM}).[/tex]

This is just the Schrödinger Equation for a free particle. We know this equation rather well! For example, we can say that the momentum eigenstates in position space are plane waves,

[tex]\chi(\vec{x}_{\rm CM}) = (2\pi\hbar)^{-3/2} \exp\left(\frac{i}{\hbar} \vec{p}_{\rm CM} \cdot \vec{x}_{\rm CM}\right),[/tex]

with energies

[tex]E_{\rm CM} = \frac{\vec{p}_{\rm CM}^2}{2M}.[/tex]

We’ve seen that the system of two coupled particles splits into a center-of-mass part, whose energy is just the translational kinetic energy of a free particle, and a part which embodies the relative motion of the two particles. In classical mechanics, we saw the same thing: for example, in the Earth-Moon system, both bodies are orbiting around their common center of mass (which is actually beneath the Earth’s surface). For the hydrogen atom, our two particles will be a proton and an electron, bound together by the Coulomb interaction between them. Note that the mass of the proton, [tex]m_p[/tex], is much larger than that of the electron, [tex]m_e[/tex], so that the “reduced mass” is very close to the electron mass:

[tex]\mu = \frac{m_e m_p}{m_e + m_p} = \frac{m_e}{1 + m_e/m_p} \approx m_e\left(1 – \frac{m_e}{m_p}\right).[/tex]

The correction term, [tex]m_e/m_p[/tex], is roughly one part in 1800. This means that the CM frame is almost coincident with the rest frame of the proton. We’ll be a bit sloppy, but excusably so, if we refer to the relative degrees of freedom as the “electron” part of the system and the CM as the “nucleus.”

Our next step will explore the realization that the Coulomb problem we wish to study is spherically symmetric. The interaction potential is a function of the distance between the two particles, nothing else:

[tex]V(\vec{x}_1,\vec{x}_2) = V(|\vec{x}_1 – \vec{x}_2|).[/tex]

This symmetry property will allow us to write the Hamiltonian for the relative degrees of freedom (the “electron” part) in terms of the 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]

This will be our starting point for next time.

(I’ve heard that my mathy posts aren’t showing up well in Google Reader, thanks in part to my gloomy color scheme. I’ll try to get this debugged, as part of the theme upgrades and other big changes I’ve got in the works.)