# How to derive the simplest 1D Superpotential Hamiltonian?

In the superpotential wiki article there are definitions of two supersymmetric operators: $$Q_1=\frac{1}{2}\left[(p-iW)b+(p+iW)b^\dagger\right] \\ Q_2=\frac{i}{2}\left[(p-iW)b-(p+iW)b^\dagger\right]$$

Then the following Hamiltonian is defined and presented:

$$H=\{Q_1,Q_1\}=\{Q_2,Q_2\}=\frac{p^2}{2}+\frac{W^2}{2}+\frac{W'}{2}(bb^\dagger-b^\dagger b)$$

Where $$W' = \frac{dW(x)}{dx}$$ and $$\{b,b^\dagger\}=1$$ and $$[b,b^\dagger]=0$$.

1. Why does $$=\{Q_1,Q_1\}=\{Q_2,Q_2\}=\frac{p^2}{2}+\frac{W^2}{2}+\frac{W'}{2}(bb^\dagger-b^\dagger b)$$ hold?

2. What's the motivation/justification to define the superpotential as such? Also why do $$Q_1, Q_2$$ map "bosonic" states into "fermionic" states and vice versa?

3. Lastly, why does it take the form: $$H = \frac{p^2}{2}+\frac{W^2}{2} \pm \frac{W'}{2}$$?

• – Qmechanic Feb 3 at 14:52