0
$\begingroup$

We know that in the RNS formalism of the superstring, there is a unique ground state corresponding to the NS sector. This is defined by $\varphi_{r}^{\mu} | 0 \rangle_{NS} = 0$ for all $r > 0$. Here $r$ runs over the half-integers. The states with $r < 0$ then act as raising operators.

In all books it is claimed that, because there is a unique ground state in the NS sector, this state corresponds to a spin $0$ state in spacetime. By acting with raising operators, one then obtains spacetime bosons. I don't understand why the ground state should be spin zero? What is the reason for this?

I read somewhere that this can be checked by applying the Lorentz generators $$ \Sigma^{\mu \nu} = - \frac{i}{2} \sum_{r \in \mathbb{Z} + \nu} [ \varphi_r^{\mu}, \varphi_{-r}^{\nu} ] $$ to the ground state. But I don't really understand this argument. Can someone elaborate please?

Thank you in advance.

$\endgroup$
3
$\begingroup$

One can verify that for $\mu\neq \nu$, $$\Sigma^{\mu\nu}|0\rangle_{NS}=-\frac{i}{2}\sum_{r\in \mathbb{Z+\frac{1}{2}}}[\varphi_r^{\mu}, \varphi_{-r}^{\nu}]|0\rangle_{NS}=-i\sum_{r\in \mathbb{Z+\frac{1}{2}}}\varphi_r^{\mu}\varphi_{-r}^{\nu}|0\rangle_{NS}=i\sum_{r\geq \frac{1}{2}}\varphi_{-r}^{\nu}\varphi_r^{\mu}|0\rangle_{NS}-i\sum_{r\leq \frac{1}{2}}\varphi_r^{\mu}\varphi_{-r}^{\nu}|0\rangle_{NS}=0,$$ where in the second step one uses $\{\varphi_r^{\mu}, \varphi_{-s}^{\nu}\}=\eta^{\mu\nu}\delta_{r,s}$, and in the last step one uses the definition of $|0\rangle_{NS}$.

Since $\Sigma^{\mu\nu}$ is the spacetime Lorentz generator, it indicates that $|0\rangle_{NS}$ transforms trivially under Lorentz transformation and thus a scalar.

$\endgroup$
  • $\begingroup$ +1. I believe here the asker's doubt wasn't the algebra, but why applying this operator gives you the correct result, so only the last line was necessary. $\endgroup$ – Bruce Lee Jun 2 '18 at 19:37

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.