Shape-Invariant Potentials in N=2 SUSY QM

[quantumdude]

Hi folks,

A while ago I started a thread on N=2 SUSY QM, but unfortunately (fortunately?) the progress in my research is outpacing the progress in that discussion. So I'm leaving that one on the back burner to skip ahead to shape-invariant potentials. I'm working from Chapter 5 of "Supersymmetric Methods in Quantum and Statistical Physics", by G. Junker (Springer, 1996) as well as the paper An Algebraic Approach to Shape Invariance, by A. B. Balantekin. The link is to the arXiv article, but it was also published in Phys. Rev. A.

Here's the lowdown.

The N=2 SUSY Hamiltonian is given by

$$H=\left( \frac{p^2}{2m} + \Phi ^2(x) \right) \otimes 1 + \frac{\hbar}{\sqrt{2m}}\Phi^{\prime}(x)\otimes\sigma_3$$.

This Hamiltonian can be written in matrix form as follows.

$$H = \left[\begin{array}{cc}H_+ & 0\\0 & H_-\end{array}\right]$$,

where $H_{\pm}=\frac{p^2}{2m}+\Phi^2(x)\pm\frac{\hbar}{\sqrt{2m}}\Phi^{\prime}(x)$ are the so-called partner Hamiltonians. We further define the partner potentials $V_{\pm}$ as follows.

$$V_{\pm}=\Phi^2(x)\pm\frac{\hbar}{\sqrt{2m}}$$.

Now we get to the shape-invariance part.

First, we introduce a set of parameters, denoted by the collective index $a_1$, and a map $F: a_1 \mapsto a_2 = F(a_1)$. This set of parameters can include anything (strength, difuseness, etc) that doesn't depend on $x$.

Definition
The partner potentials $V_{\pm}(a_1,x)$ are called shape-invariant if they are related by

$$V_+(a_1,x)=V_-(a_2,x)+R(a_1)$$,

where $R(a_1)$ is a remainder that does not depend on $x$.

If you're still with me after all that, then here's my first question. How is one supposed to know how to introduce the parameters $a_1$ into the partner potentials? For example, on page 60 of Junker's book he gives the following example.

$$\Phi(a_1,x)=\frac{\hbar}{\sqrt{2m}}a_1\tanh(x)$$

It turns out that the partner potentials corresponding to this SUSY potential are shape-invariant (I will show this if anyone expresses interest). But why put the $a_1$ in that spot and not, say, in the exponent of the tanh function? Of course the obvious answer would be, "because if you do that then you won't get shape-invariant partner potentials". But how would one know that to begin with? In other words, how can one construct shape-invariant potentials?

 

[Greg Bernhardt]

@Tom Mattson did you find any more insight on this topic? Do you want to turn it into an Insight?