Supersymmetry in Quantum Mechanics (Does it apply?) Suppose we try to apply supersymmetry in quantum mechanics to a particular potential. If you come up with two partner potentials, and two partner Hamiltonians, and then look at the energy of the ground state, is it true that at most one of them can be zero?
If they are both zero, does that mean that supersymmetry is just not preserved by the vacuum states of the partner Hamiltonians? Can we still apply supersymmetry to this system if the energy of both ground states are zero?
 A: Note that you begin with a superpotential $W$ which gives you a generalization of the creation/anihilation  operator as $A^{+} = -\frac{d}{dx} + W(x)$ and $A^{-}= +\frac{d}{dx} + W(x)$. You will get 2 hamiltonians $H_- = A^+ A^-$ and $H_+ = A^- A^+$. So you obtain 2 different potentials $V_{+}$ and  $V_{-}$ for superpartners $V_{-} (x)= W(x)^2 - \frac{dW}{dx}$ and $V_{+} (x)= W(x)^2 + \frac{dW}{dx}$.
Then, suppose that $\psi^-$ is a solution of $A^-\psi^-(x) = 0$ with $\psi^-$ normalizable, you will find that $H_- \psi_- = A^+A^-\psi^- = 0$, so $\psi^-$ is an eigenstate of $H_- $, with eigenvalue 0.
You have $\psi^-(x) \sim e^{-\int^x_{x_0} W(x) dx}$ (from the definition of $A^-$)
With the same reasoning, you will find that a $\psi^+$ solution of $A^+\psi^+(x) = 0$ will give you $\psi^+(x) \sim e^{+\int^x_{x_0} W(x) dx}$ which would be an eigenstate of $H_+ $, with eigenvalue 0.
But you have a problem, you can see that you cannot have, at the same time $\psi^+$ and $\psi^-$ normalizable, because $\psi^+(x) \sim \frac{1}{\large \psi^-(x)}$.
So, one of the functions $\psi^+$ or $\psi^-$ has to vanish.
So, you have, at most, one ground state with zero energy.
[EDIT]
The practical link with supersymmetry is as follows. We define a 2 -dimensional space, with $\psi = (\psi_-, \psi_+)$. One of the $\psi$ states is a bosonic state, the other is a fermionic state.
We define the supersymmetric generators:
$Q^- =\left[ \begin{array}{cccc} 
0 & 0  \\ 
A^- & 0 
\end{array} \right]$
$Q^+ =\left[ \begin{array}{cccc} 
0 & A^+  \\ 
0 & 0 
\end{array} \right]$
The hamiltonian is $H = Q^-Q^+ + Q^+Q^-$
$H =\left[ \begin{array}{cccc} 
A^+A^- & 0  \\ 
0 & A^-A^+ 
\end{array} \right] = \left[ \begin{array}{cccc} 
H_- & 0  \\ 
0 & H_+ 
\end{array} \right]$
It can be easily seen that $(Q^-)^2=(Q^+)^2 = 0$ and $[H,Q^-] = [H,Q^+] = 0$ 
Unbroken supersymmetry corresponds to a ground state $|0>$ such as $<0|H|0> = 0$, and this imply $Q^+|0> = Q^-|0> = 0$
In this case, one of the functions $\psi^+$ or $\psi^-$ is normalizable, and the other vanishes. For instance, suppose $\psi^-$ is normalizable, then it means that $A^- \psi^- =0$
The case of (spontaneously) broken supersymmetry, is when  $<0|H|0> \neq 0$, which implies $Q^+|0> \neq 0$ or $Q^-|0> \neq 0$. In this case, neither of the ground states $\psi^+$ nor $\psi^-$ are normalizable.
