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

+ 0 like - 0 dislike
329 views

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
retagged May 11, 2014
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

+ 0 like - 0 dislike

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 (1,985 points)
edited Apr 25, 2014

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.