I have seen some books, e.g. by Fred Cooper (Supersymmetry in Quantum Mechanics), define:

$A = \frac{\hbar}{\sqrt{2m}} \frac{d}{dx} + W(x)$, $A^\dagger = \frac{-\hbar}{\sqrt{2m}} \frac{d}{dx} + W(x)$,

$V_1(x) = W^2(x) - \frac{\hbar}{\sqrt{2m}} W'(x)$, $V_2(x) = W^2(x) + \frac{\hbar}{\sqrt{2m}} W'(x)$.

In other places, I have seen it defined:

$A = \frac{\hbar}{\sqrt{2m}} \frac{d}{dx} + W'(x)$, $A^\dagger = \frac{-\hbar}{\sqrt{2m}} \frac{d}{dx} + W'(x)$

$V_1(x) = W'(x)^2 - \frac{\hbar}{\sqrt{2m}} W''(x)$, $V_2(x) = W'(x)^2 + \frac{\hbar}{\sqrt{2m}} W''(x)$.

It seems like these two definitions will give you completely different Hamiltonians and partner potentials. Could someone comment on if these two conventions are equivalent, and if so, how?

  • $\begingroup$ I will recommand you a beautiful little book, Supersymmetric Quantum Mechanics: An Introduction, by Asim Gangopadhyaya , Jeffry V Mallow , Constantin Rasinariu. (World Scientific). This is very well written, and full of amazing things, like shape invariance and its consequences, which allow to obtain very easily the spectrum and wave functions for some potentials (like hydrogen atom). $\endgroup$
    – Trimok
    Jun 4, 2013 at 7:54
  • $\begingroup$ Hi @userØØ7, when you do edits, please only make substantial edits, and please always check the outcome! E.g. when you change in-line eq. to display-style eq., don't forget to put the comma/full stop inside the equation, to prevent them from ending up 'flying' in the left margin. $\endgroup$
    – Qmechanic
    Jun 4, 2013 at 13:35

1 Answer 1


The second convention only differs from the first one by using the symbol $W'$ for what is called $W$ in the first convention. There is a one-to-one correspondence between (reasonable enough) functions and their derivatives so the translation between the two conventions is completely trivial.

  • $\begingroup$ I guess I was curious whether, if given some superpotential $W(x)$, we would produce supersymmetry in one case, but not the other. Do the two different conventions mandate that $W(x)$ be defined differently in the two cases? $\endgroup$
    – Randy
    Jun 4, 2013 at 6:41
  • 1
    $\begingroup$ Dear @Randy, apologies, I don't understand what you don't understand about it now after the answer. The symbol $W$ (or the word "superpotential" that you may associate with it) means different things in the two conventions, do you understand this claim? If it is $x^3$ in one of them, it is $3x^2$ in the other. There is no "the" given superpotential that is the same for both conventions. These are, as you correctly said at the beginning but you ignored it everywhere else, just two conventions to describe the very same thing. As long as $W$ may be integrated/differentiated, one may switch. $\endgroup$ Jun 4, 2013 at 6:49

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