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  Why isn't there a Heterotic string theory which tensors the fermionic state with the Type II state?

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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?

asked May 26, 2013 in Theoretical Physics by dimension10 (1,985 points) [ revision history ]
retagged May 11, 2014 by dimension10
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.

This post imported from StackExchange Physics at 2014-03-09 09:13 (UCT), posted by SE-user Luboš Motl
I thought there was the Fermionic String Action (Dirac Action) given by:$$S = \frac{{iT}}{{{c_0}}}\int_{}^{} {\sqrt { - \det {h_{\alpha \beta }}} {h_{\alpha \beta }}\psi _ - ^\mu \left( {\frac{{\partial \psi _ + ^\nu }}{{\partial \tau }} - \frac{{\partial \psi _ + ^\nu }}{{\partial \sigma }}} \right){g_{\mu \nu }}{\rm{d}}\sigma {\rm{ d}}\tau } $$

This post imported from StackExchange Physics at 2014-03-09 09:13 (UCT), posted by SE-user Dimensio1n0
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.

This post imported from StackExchange Physics at 2014-03-09 09:13 (UCT), posted by SE-user Luboš Motl

1 Answer

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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."

answered Sep 28, 2013 by dimension10 (1,985 points) [ revision history ]
edited Apr 25, 2014 by dimension10

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