Editing Supersymmetric quantum mechanics (section)

== Shape invariance ==

Suppose <math>W</math> is real for all real <math>x</math>. Then we can simplify the expression for the Hamiltonian to
: <math>H = \frac{(p)^2}{2}+\frac{{W}^2}{2}+\frac{W'}{2}(bb^\dagger-b^\dagger b)</math>

There are certain classes of superpotentials such that both the bosonic and fermionic Hamiltonians have similar forms. Specifically
: <math> V_{+} (x, a_1 ) = V_{-} (x, a_2) + R(a_1)</math>
where the <math>a</math>'s are parameters. For example, the hydrogen atom potential with angular momentum <math>l</math> can be written this way.
: <math> \frac{-e^2}{4\pi \epsilon_0} \frac{1}{r} + \frac{h^2 l (l+1)} {2m} \frac{1}{r^2} - E_0</math>

This corresponds to <math>V_{-}</math> for the superpotential
: <math>W = \frac{\sqrt{2m}}{h} \frac{e^2}{2 4\pi \epsilon_0 (l+1)} - \frac{h(l+1)}{r\sqrt{2m}}</math>
: <math>V_+ = \frac{-e^2}{4\pi \epsilon_0} \frac{1}{r} + \frac{h^2 (l+1) (l+2)} {2m} \frac{1}{r^2} + \frac{e^4 m}{32 \pi^2 h^2 \epsilon_0^2 (l+1)^2}</math>

This is the potential for <math>l+1</math> angular momentum shifted by a constant. After solving the <math>l=0</math> ground state, the supersymmetric operators can be used to construct the rest of the bound state spectrum.

In general, since <math>V_-</math> and <math>V_+</math> are partner potentials, they share the same energy spectrum except the one extra ground energy. We can continue this process of finding partner potentials with the shape invariance condition, giving the following formula for the energy levels in terms of the parameters of the potential
: <math> E_n=\sum\limits_{i=1}^n R(a_i) </math>
where <math>a_i</math> are the parameters for the multiple partnered potentials.