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A couple months ago, I stumbled across an amusing bit of academic woo: “Quantum Mind and Social Science.” The misrepresentations, false dichotomies and nons sequitur of that piece prompted me to wonder what a good litmus test for knowing quantum mechanics might look like. Joshua offered a simple criterion: be able to pick the Schrödinger Equation out of a line-up. At a slightly higher level, I suggested being able to describe in the Heisenberg picture the time evolution of a harmonic oscillator coherent state, and explaining why states of the hydrogen atom with the same n but different angular momentum number l are degenerate. You can’t discuss the relationship between classical and quantum physics without bringing up coherent states eventually, and a good grounding in the basics should include the Schrödinger and Heisenberg pictures. (That’s why I wrote problem 5 in this homework assignment.)

The excited states of the hydrogen atom are our prototype for understanding how the periodic table works, and it’s often the first place one runs into the mathematics of angular momentum. Unfortunately, too many standard treatments of introductory QM say that hydrogen has “accidental degeneracies”: these states have the same energies as those states for no spectacularly interesting reason. But we are trained to associate degeneracies with symmetries — when two sets of eigenstates have the same eigenvalues, we expect some symmetry to be at work. So, is there a symmetry in the hydrogen atom above and beyond the familiar rotational kind, a symmetry which They haven’t been telling us about?

I’d like to explore this topic over a few posts. First, I’ll build up some very general machinery for solving problems, and then I’ll apply those techniques to the hydrogen atom; by that point, we should have a fair amount of knowledge with which we can move in any one of several interesting directions. To begin, let’s familiarize ourselves with the behavior of a superalgebra.

INTRODUCTION

In fundamental quantum mechanics, we learn that an algebra of operators is defined by commutation relations among those operators. For example, the canonical operators of position and momentum have the commutator [tex]\comm{x}{p} = i\hbar[/tex]. A more intricate case is the algebra of angular momentum operators, which we encountered when exploring the rotational symmetries of 3D space. To generalize this concept, we define an anticommutator, which relates operators in the same way as an ordinary commutator, but with the opposite sign:

[tex]\{A,B\} \equiv AB + BA.[/tex]

If operators are related by anticommutators as well as commutators, we say that they are part of a superalgebra. Let’s say we have a quantum system described by a Hamiltonian [tex]\mathcal{H}[/tex] and a set of [tex]N[/tex] self-adjoint operators [tex]Q_i[/tex], each of which commutes with the Hamiltonian. We shall call this system supersymmetric if the following anticommutator is valid for all [tex]i,j=1,2,\ldots,N[/tex]:

[tex]\{Q_i,Q_j\} = \mathcal{H}\delta_{ij}.[/tex]

If this is the case, then we call the operators [tex]Q_i[/tex] the system’s supercharges. [tex]\mathcal{H}[/tex] will be termed the SUSY Hamiltonian, SUSY being a convenient abbreviation for whichever variation of “supersymmetry” is grammatically appropriate. A SUSY algebra is characterized by its number of supercharges, which we typically denote [tex]N[/tex].

Because the [tex]N = 2[/tex] case exemplifies many properties of general SUSY theories, it is worthwhile to work it out in some detail. We require two supercharges, [tex]Q_1 = Q_1^\dag[/tex] and [tex]Q_2 = Q_2^\dag[/tex]. The SUSY algebra we defined a moment ago implies the following relations:

[tex]Q_1 Q_2 = – Q_2 Q_1,\ \mathcal{H} = 2Q_1^2 = 2Q_2^2 = Q_1^2 + Q_2^2.[/tex]

It is sometimes more convenient to work with a “complex” supercharge that is not self-adjoint. (The convention we choose depends upon the given information we have to work with!) If we make linear combinations of our supercharges,

[tex]Q = \frac{1}{\sqrt{2}}(Q_1 + iQ_2),\ Q^\dag = \frac{1}{\sqrt{2}}(Q_1 – iQ_2),[/tex]

then the SUSY algebra implies [tex]\{Q},Q^\dag\} = \mathcal{H}[/tex]. To make this a little more concrete, we can realize a specific incarnation of the superalgebra: let [tex]H_1[/tex] be some Hamiltonian of interest, and suppose that we can factor [tex]H_1[/tex] into the product of an operator and its adjoint:

[tex]H_1 = A^\dag A.[/tex]

Note that this is almost the form of the harmonic oscillator Hamiltonian, except for an energy shift:

[tex]H_{\rm SHO} = \hbar\omega\left(a^\dag a + \frac{1}{2}\right).[/tex]

So, this is not an unfamiliar form for a Hamiltonian. Swapping the order of the factors gives another operator, which you can verify is also Hermitian:

[tex]H_2 = AA^\dag.[/tex]

With [tex]A[/tex] in hand, define the two operators
[tex]Q = \left(\begin{array}{cc} 0 & 0 \\ A & 0 \\ \end{array}\right)[/tex]
and
[tex]Q^\dag = \left(\begin{array}{cc} 0 & A^\dag \\ 0 & 0 \\ \end{array}\right).[/tex]
Matrix arithmetic verifies that

[tex]\{Q,Q^\dag\} = \left(\begin{array}{cc} A^\dag A & 0 \\ 0 & AA^\dag \\ \end{array}\right),[/tex]

so we can say that the anticommutator of our two charges gives a Hamiltonian [tex]\mathcal{H}[/tex] which is block diagonal,

[tex]\mathcal{H} = \left(\begin{array}{cc} H_1 & 0 \\ 0 & H_2 \\ \end{array} \right).[/tex]

[tex]H_1[/tex] and [tex]H_2[/tex] can be considered two Hamiltonians acting on subspaces of the original Hilbert space associated with [tex]\mathcal{H}[/tex].

PARTNER POTENTIALS

What exactly is so special about operators of the forms [tex]A^\dag A[/tex] and [tex]AA^\dag[/tex]? Given a Hamiltonian for some system, [tex]H_1[/tex], if it can be factored into the product of two operators [tex]A^\dag A[/tex], then we can construct another Hamiltonian [tex]H_2 = AA^\dag[/tex] which has almost exactly the same energy eigenvalue spectrum. These “isospectral” Hamiltonians may not describe the same physics, and their respective potentials [tex]V_1(x)[/tex] and [tex]V_2(x)[/tex] may look radically different. As usual, a degeneracy in the energy levels corresponds to a symmetry; in this case, the symmetry is the SUSY between our two Hamiltonians.

First, let’s take a look at the eigenstates of Hamiltonian number 1. These states satisfy the relationship

[tex]H_1 \ket{\psi_n^{(1)}} = A^\dag A \ket{\psi_n^{(1)}} = E_n^{(1)} \ket{\psi_n^{(1)}}.[/tex]

Now, a surprising thing happens: the operator [tex]A[/tex] maps the eigenstates of Hamiltonian 1 into eigenstates of Hamiltonian 2. Look:

[tex]H_2 A\ket{\psi_n^{(1)}} = AA^\dag A \ket{\psi_n^{(1)}},[/tex]

but by the equation just above, this means that

[tex]H_2 A\ket{\psi_n^{(1)}} = E_n^{(1)} A\ket{\psi_n^{(1)}}.[/tex]

The same logic works in the opposite direction, connecting eigenstates of [tex]H_2[/tex] with those of [tex]H_1[/tex]. The eigenstates behind door number 2 satisfy

[tex]H_2 \ket{\psi_n^{(2)}} = AA^\dag \ket{\psi_n^{(2)}} = E_n^{(2)}\ket{\psi_n^{(2)}},[/tex]

so by acting with the operator [tex]A^\dag[/tex],

[tex]H_1 A^\dag\ket{\psi_n^{(2)}} = A^\dag AA^\dag \ket{\psi_n^{(2)}} = E_n^{(2)}A^\dag \ket{\psi_n^{(2)}}.[/tex]

We have shown that [tex]H_1[/tex] and [tex]H_2[/tex] are isospectral. For every eigenstate of one, there lurks an eigenstate of the other with the same energy. One exception is important: if

[tex]A \ket{\psi_n^{(1)}} = 0,[/tex]

that is, if [tex]H_1[/tex] has a zero-energy ground state, the proof does not work, and there is no need for [tex]H_2[/tex] to have a zero-energy ground state. In fact, as we’ll see momentarily, only one of [tex]H_1[/tex] and [tex]H_2[/tex] may have a zero-energy ground state; for consistency, we usually arrange matters so that [tex]H_1[/tex] has the extra eigenstate.

SUPERPOTENTIALS

Most of the time, we find ourselves dealing with Hamiltonians of the form

[tex]H = \frac{p^2}{2m} + V(x),[/tex]

which, knowing that [tex]p = -i\hbar \partial_x[/tex], we can also write as

[tex]H = -\frac{\hbar^2}{2m} \partial_x^2 + V(x).[/tex]

If we want to factor this [tex]H[/tex] into an operator and its adjoint, we should probably start with an operator which is linear in the derivative of [tex]x[/tex], thus:

[tex]A = \frac{\hbar}{\sqrt{2m}} \partial_x + W(x).[/tex]

Here, [tex]W(x)[/tex] is some real function of [tex]x[/tex] which we shall call the superpotential. Taking the adjoint of [tex]A[/tex] flips the sign on the derivative; you should deduce why by observing that the momentum [tex]p[/tex] is observable and therefore self-adjoint.

[tex]A^\dag = -\frac{\hbar}{\sqrt{2m}} \partial_x + W(x).[/tex]

We can connect the superpotential to [tex]H_1[/tex]‘s ordinary potential,

[tex]V_1(x) = W^2(x) – \frac{\hbar}{\sqrt{2m}} \partial_x W(x).[/tex]

This relationship is known as the Riccati Equation. If we reverse the order of our operators, it turns out that [tex]H_2[/tex] is a Hamiltonian with a new potential [tex]V_2(x)[/tex], given by

[tex]V_2(x) = W^2(x) + \frac{\hbar}{\sqrt{2m}} \partial_x W(x).[/tex]

We recognize this as the Riccati Equation with a change of sign. [tex]V_1(x)[/tex] and [tex]V_2(x)[/tex] are known as partner potentials, related through the superpotential [tex]W(x)[/tex].

With one more step, we can relate the superpotential to the ground state wavefunction. Note that the ground state of [tex]H_1[/tex] is annihilated by [tex]A[/tex], satisfying the relation

[tex]A\ket{\psi_0^{(1)}} = 0.[/tex]

Looking back at the form of [tex]A[/tex], we see that this is a first-order differential equation, and we can write its solution as an exponential.

[tex]\psi_0^{(1)}(x) \propto \exp\left[-\frac{\sqrt{2m}}{\hbar} \int_0^x W(y) dy\right].[/tex]

Note that the zero-energy ground state of [tex]H_2[/tex] would be annihilated by [tex]A^\dag[/tex] and would therefore be proportional to

[tex]\exp\left[\frac{\sqrt{2m}}{\hbar} \int_0^x W(y) dy\right].[/tex]

Only one of these two expressions can give a normalizable state: if one behaves nicely, the other will blow up. That’s why only one of the two partner Hamiltonians can have a ground state of zero energy.

Often, these equations are shown in “natural units” where [tex]\hbar = 2m = 1[/tex]. This can always be done by changing the units of [tex]x[/tex], and it makes successive steps in the calculations much cleaner. I think it nice to see the equations with all the original units in place at least once; in following sections, however, when units are not illuminating I will set unnecessary constants to unity.

READING

SUSY QM SERIES

2 Comments

    • Jeremy Henty
    • Posted Sunday, 13 January 2008 at 00:52 am
    • Permalink

    It’s not true in general that only one of H_1, H_2 can have a zero eigenvalue, consider A = (0 0 ; 1 0) , A^dag = (0 1 ; 0 0) , H_1 = (1 0 ; 0 0) , H_2 = (0 0 ; 0 1). (These are 2×2 matrices with the ‘;’ separating the rows.)

    • Jeremy Henty
    • Posted Sunday, 13 January 2008 at 01:01 am
    • Permalink

    Whoops, OK, I see that H_1 and H_2 can’t both have zero eigenvalues when A is given by the superpotential Ansatz you write down later. Didn’t see that coming.


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