The question is the following and it is related to the article of Martinec, Shenker and Friedan, "Conformal invariance, supersymmetry and string theory" (and to many others actually, but just to be concrete). I'm not able to recover the OPE between the spin fields $$ S_{\alpha}(z)S_{\beta}(w)\sim(z-w)^{-3/4}\gamma_{\alpha\beta}^{\mu}\psi_{\mu} $$ from the definition, in the bosonized picture, $$ S_{\alpha}(z)=\prod_{I=1}^5e^{i/2(\epsilon_IH_I(z))},\quad H_I(z)H_J(w)\sim \ln(z-w),\quad \epsilon_{I}=\pm 1 $$ The $2^5$ combination of the bosonized spin field gives the 32 component of a spinor in 10 dimensional flat space. I guess it's a standard representation that can be found in almost every string theory textbook. Probably the first thing I would like to know is where the power $-3/4$ comes from, since the index structure should come from a group theoretic argument.


You should use the basic OPE:

\begin{equation} \langle e^{i\lambda_i H_I(z_i) }e^{i\lambda_j H_I(z_j) }\rangle= (z_i-z_j)^{ \lambda_i \cdot \lambda_j} \end{equation}

for every bosonic field $H_I$, depending on the dimensionality of your spin field. In your case $\lambda_i = \frac{1}{2}$ and the leading contribution in 10d for positive chirality spinors should come from a combination like: $\frac{1}{2}(+++++)\frac{1}{2}(+----)$.

Remember that there are subleading terms that are less divergent or regular, for example the one from $\frac{1}{2}(+++++)\frac{1}{2}(+++--)$ or $\frac{1}{2}(+++++)\frac{1}{2}(+++++)$.

Here the notation depict the five $\lambda_i$ and $\lambda_j$.

| cite | improve this answer | |
  • $\begingroup$ Thanks. That is already how I was using to work out the OPE but it exactly this the problem. In principle that OPE can be valid for every polarization so I cannot understand how there is a fix power, namely $(z-w)^{-3/4}$ for every polarization. $\endgroup$ – Andrea89 Nov 16 '17 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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