Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Heterotic (HO and HE) string is found by tensoring the left movers of the bosonic string theory state and the right movers of the Type II string theory state: $$|\psi_{\operatorname{H}}\rangle=|\psi_{\operatorname{B}}\rangle\otimes|\psi_{\operatorname {II}}\rangle$$

The bosonic state is of course based on the Polyakov Action. Now, what if we had another state (I'll call it the k-heterotic state) which is formed by tensoring the fermionic string theory state and the type II state. $$|\psi_{\operatorname{H}}\rangle=|\psi_{\operatorname{F}}\rangle\otimes|\psi_{\operatorname {II}}\rangle$$ The fermionic string theory state is based on the Dirac Action: $$S_F=S_{RNS}-S_B$$ Why is this theory not consistent? Whether the bosonic state is chosen or the fermionic state, how does that matter? Is it because the Dirac Action is imaginary (since it has an i outside the integral if one expands out the Dirac matrix fermionic field product)? Does it have anything to do with that?

share|cite|improve this question
If you only consider the right-movers, the "fermonic string" and the "type II string" is the same thing! The word "fermionic" doesn't mean that it only has fermions; it means that it also has fermions. Taking only fermions from type II strings would have a wrong central charge i.e. conformal anomaly! Your difference of the two actions is completely incomprehensible to me; perhaps you thought that the fermionic string should only have fermions but that's not the case. And which imaginary action? In the Minkowski space, all actions are always real, they have to be real due to unitarity. – Luboš Motl May 26 '13 at 7:21
I thought there was the Fermionic String Action (Dirac Action) given by:$S = \frac{{iT}}{{{c_0}}}\int_{}^{} {\sqrt { - \det {h_{\alpha \beta }}} {h_{\alpha \beta }}\psi _ - ^\mu \left( {\frac{{\partial \psi _ + ^\nu }}{{\partial \tau }} - \frac{{\partial \psi _ + ^\nu }}{{\partial \sigma }}} \right){g_{\mu \nu }}{\rm{d}}\sigma {\rm{ d}}\tau } $ – centralcharge May 26 '13 at 7:37
The indices $\alpha,\beta$ above aren't even correctly contracted. And again, no, "fermionic string" is an obsolete name for type I/II D=10 string that was used up to the mid 1980s and it contains both bosons and fermions. Moreover, you're using "Dirac action" in a strange way - the Dirac action is primarily the action for any spin-1/2 field, especially in d=4. – Luboš Motl May 27 '13 at 8:35
up vote 0 down vote accepted

I guess Lubos Motl's comment really refers to the terminology used in my post.

If I try to insist on what I meant by "fermionic string", the string formed by $S=S_{RNS}-S_P$, the massless free Dirac Action $S=\iint\limits_{S} i\hbar\gamma^\mu\partial_\mu\psi \mbox{ d}^2\xi$ , then I guess it would simply mean that the theory is inconsistent. The only way I can see that this is so, is that the "fermionic string" again is an inconsistent string theory. I think I get why this is so.

If there are no fields $X^\mu$ in the action, then the string worldsheet can't get embedded into spacetime at all (!). This theory would then not exist.

So the answer boils down to "The (purely) fermionic string is not studied because it's not even a consistent theory, since the string worldsheet wouldn't be embedded into spacetime."

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.