Undergrad Is Supersymmetry required by String theory?

[TheHeraclitus]

TL;DR

Supersymmetry is often mentioned together with String theory, but do all String theories require it?

From what I understood Supersymmetry means there are more particles than we currently know about and they are predicted by (some/all, I do not know) versions of String theory.
Is it so important to String theory or can it work without SUSY?

Thank you!

 

[haushofer]

Susy is, as far as we know, required for the particle spectrum to be consistent. Without susy you get e.g. tachyonic particles which are hard to interpret as physical particles. So yes.

 

[Demystifier]

Susy is, as far as we know, required for the particle spectrum to be consistent. Without susy you get e.g. tachyonic particles which are hard to interpret as physical particles. So yes.

Non-perturbative formulation of bosonic string-field theory reinterprets tachyonic state as an unstable state, very much like tachyonic Higgs field before symmetry breaking is reinterpreted as an unstable state. By a process of tachyon condenzation the unstable state (local maximum of the potential) settles down into a stable state (local minimum of the potential) and excitations around the stable state are no longer tachyonic. So I would say no.

But we still need susy in string theory because that's the only known way (as fas as I am aware) to incorporate fermions into string theory, and we need fermions because they are observed in nature.

 

[ohwilleke]

I have seen multiple knowledgeable theoretical physicists state authoritatively that supersymmetry is necessary as a low energy approximation of string theory.

To some extent, however, it boils down to definitions.

If you use a fairly strict definition of string theory or M-theory, then this is probably correct.

If you use a loose definition of string theory that, for example, borrows heavily from its mathematical methods without embracing the entire discipline's canonical form for it, perhaps you could get a stringy theory that doesn't have supersymmetry as a low energy approximation.