Supersymmetric Quantum Mechanics ======================================== For a quantum mechanical system, it is always possible to find a partener potential. Suppose we have a Hamiltonian :math:`H_1` in quantum mechanics, which is defined as .. math:: H_1 = - \frac{\hbar}{2m}\partial_x^2 + V_1(x). By decomposing it into .. math:: H_1 = A^\dagger A, we can define the two operators .. math:: A = & \frac{\hbar}{\sqrt{2m}} \partial_x + W(x),\\ A&\dagger = & -\frac{\hbar}{\sqrt{2m}}\partial_x + W(x). The quantity :math:`W(x)` is so called superpotential. The partener Hamiltonian of :math:`H_1` is .. math:: H_2 = A A^\dagger, using which one find the corresponding potential is .. math:: 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 .. math:: W(x) = -\frac{\hbar}{\sqrt{2m}} \frac{ \psi_0'(x) }{\psi_0(x)}. .. admonition:: Proof :class: toggle HERE One can also find the relation between the new wave function :math:`\psi^{(2)}` and the original one :math:`\psi^{(1)}`, .. math:: \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)}. Meanwhile the energy levels are also related .. math:: E^{(1)}_{n+1} = E^{(2)}_n. .. admonition:: Proof :class: toggle QED. 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.