5.1. Supersymmetric Quantum Mechanics

For a quantum mechanical system, it is always possible to find a partener potential.

Suppose we have a Hamiltonian \(H_1\) in quantum mechanics, which is defined as

\[H_1 = - \frac{\hbar}{2m}\partial_x^2 + V_1(x).\]

By decomposing it into

\[H_1 = A^\dagger A,\]

we can define the two operators

\[\begin{split}A = & \frac{\hbar}{\sqrt{2m}} \partial_x + W(x),\\ A&\dagger = & -\frac{\hbar}{\sqrt{2m}}\partial_x + W(x).\end{split}\]

The quantity \(W(x)\) is so called superpotential. The partener Hamiltonian of \(H_1\) is

\[H_2 = A A^\dagger,\]

using which one find the corresponding potential is

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

The interesting part is that the superpotential is closely related to ground state wave functions of the orginal system

\[W(x) = -\frac{\hbar}{\sqrt{2m}} \frac{ \psi_0'(x) }{\psi_0(x)}.\]

Proof

HERE

One can also find the relation between the new wave function \(\psi^{(2)}\) and the original one \(\psi^{(1)}\),

\[\begin{split}\psi_n^{(2)} =& \frac{1}{\sqrt{ E^{(1)}_{n+1} }} A \psi_{n+1}^{(1)}, \\ \psi_{n+1}^{(1)} = & \frac{1}{\sqrt{E^{(2)}_n}} A^\dagger \psi_n^{(2)}.\end{split}\]

Meanwhile the energy levels are also related

\[E^{(1)}_{n+1} = E^{(2)}_n.\]

Proof

QED.

5.1.1. References and Notes

1. An Introduction to Supersymmetry in Quantum Mechanical Systems by T. Wellman 2. Cooper, F., Khare, A., & Sukhatme, U. (1995). Supersymmetry and quantum mechanics. Physics Reports, 251(5-6), 267–385. doi:10.1016/0370-1573(94)00080-M.