1
$\begingroup$

Im following the GSW book. Specifically equations (5.2.39), (5.2.40) $$H=\frac{1}{2p^-}((p^i)^2+2N)$$

$$N=\sum_{m=1}^\infty(\alpha_{-m}^i\alpha_m^i+mS_{-m}^aS_{m}^a)$$ are not clear for me. I know, off course, that doing $H=p^-$ (and forming p^2 in light-cone coordinates) I can get $M^2=2N$, but ...

... how I can get the fermionic oscillators?

I want to get the mass formula, I know the procedure for the bosonic string but when I tried to do the same

$$p^-\sim \int \frac{\partial}{\partial_\tau X^-}L_{LC} \sim \int \partial_\tau X^-$$

I just got the same formula as for the bosonic string. I know that I'm using the wrong virasoro constraint for the Green-Schwarz string but I cannot find that in the book. I read that the vanishing of the ground state mass directly comes from the cancellation between fermions' and bosons' zero-point energies, that was the reason I started this calculation.

$\endgroup$
1
$\begingroup$

The supersymmetric string has more gauge symmetries than the bosonic string, namely the Kappa symmetry. Gauging this symmetry by fixing half of the fermionic variables to zero

$$ (\gamma_+\theta)=0 $$

and doing a field redefinition

$$ \theta\rightarrow\sqrt{p^+}\theta $$

you will end up with an action of the form:

$$ S=\int d^2z\left(\partial x^i \bar\partial x_i + \theta^\alpha\bar\partial\theta^\alpha\right) $$

where both $i$ and $\alpha$ runs from $1$ to $8$, and are respectively a vectorial index and a chiral spinorial index. The conformal weight of $\theta^\alpha$ is $1/2$ and it has periodic boundary conditions, so it is in the Ramond sector.

In order to see why the fermionic varibles that was, after the gauging fixing, worldsheet scalars (i.e. conformal weight zero), turn into worldsheet spinors (i.e. conformal weight $1/2$) just note that a field redefinition was made, involving multiplication of the old thetas with square roots of $p^+$.

Since the fermionic variables carry conformal weight they will contribute with the $L_0$ generator of the Virasoro algebra. Note that even if a field has conformal weight zero, the derivatives of this field will carry nonzero conformal weight so they will also contribute with $L_0$.

$\endgroup$

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.