I recently read "An Introduction to Supersymmetry in Quantum Mechanical Systems" by T. Wellman (amongst other sources) in an effort to find out what a superpotential actually is and how it relates to the potentials of particles/fields). Here's the link: http://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CDkQFjAA&url=http%3A%2F%2Fphysics.brown.edu%2Fphysics%2Fundergradpages%2Ftheses%2FSeniorThesis_Wellman.pdf&ei=ulm-UPSCLZOY1AWjwYHYDw&usg=AFQjCNGrg_2jv5NZ7b6k4Fs7er34jgtw3w&sig2=yjQYy1Lf_gZVS-RRefUCsQ

It occurred to me that the Fermionic Hamiltonian and the Bosonic Hamiltonian can be formulated without supersymmetry. On page 13, Wellman expresses these Hamiltonians in terms of the superpotential W in a way that is purely algebraic and doesn't require supersymmetry. In other words we have a bunch of terms that give us the two Hamiltonians; these terms are then simply replaced by W's (see equations 3.2 to 3.9). We only actually get supersymmetry when the separate assertion is made that Q operators exist that transform between our fermionic and bosonic states.

So would it be true to say that non-supersymmetric theories contain superpotentials, W, within their Hamiltonians in the same way that supersymmetric theories do? If this is the case the superpotential is just a useful function that is especially helpfully when we consider supersymmetric theories? I.e. superpotentials exist with or without supersymmetry.


So it seems that I may have found the answer to my own question after consulting Professor Tim Jones of Liverpool University.

It seems that superpotentials mainly serve as a way of writing Lagrangians/Hamiltonians in a more compact way. They're just a form of notation. These Lagrangians/Hamiltonians don't necessarily have to be supersymmetric and we could have our non-supersymmetric theory written in terms of superpotentials (if we wanted to). However, the main application of superpotentials are in supersymmetry, hence the name.


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