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  The simplest 1D Superpotential Hamiltonian derivation

+ 3 like - 0 dislike

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

and then somehow the following Hamiltonian is defined and derived: $$H=\{Q_1,Q_1\}=\{Q_2,Q_2\}=\frac{p^2}{2}+\frac{W^2}{2}+\frac{W'}{2}(bb^\dagger-b^\dagger b)$$

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

Why does $=\{Q_1,Q_1\}=\{Q_2,Q_2\}=\frac{p^2}{2}+\frac{W^2}{2}+\frac{W'}{2}(bb^\dagger-b^\dagger b)$? I can't see how they got this expression.

What's the motivation to define the superpotential is such away? Also why does $Q_1, Q_2$ map "bosonic" states into "fermionic" states and vice versa? Lastly, why does it take the form: $H = \frac{p^2}{2}+\frac{W^2}{2} \pm \frac{W'}{2}$

This post imported from StackExchange Physics at 2017-11-18 23:19 (UTC), posted by SE-user 0x90

asked Nov 17, 2017 in Theoretical Physics by 0x90 (65 points) [ revision history ]
edited Nov 19, 2017 by Dilaton

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