1-loop partition function (Ramond-sector) When calculating a part of the trace for the partition function of the fermionic Ramond-sector in light-cone coordinates, I'd like to understand how we get to the result $\left(\frac{\theta \left[1/2;0\right](0,\tau)}{\eta (\tau)} \right)^4$.
My problem is that I get a global $q^{1/4}$ for which I don't know how it should be compensated:
\begin{eqnarray}
Tr_R(q^{L_0}) &= q^{8/24} Tr_R\left[ q^{\frac{1}{2}\sum_{I=2}^9\sum_{r\in \mathbb{Z}}r:\psi_{-r}^I\psi_{r}^I:}\right] = \\\\
q^{8/24} \prod_{r\in \mathbb{N}} (1+q^r)^8 &= \frac{q^{12/24}}{(\eta (\tau ))^4} \prod_{r\in \mathbb{N}} (1-q^r)^4 (1+q^r)^8 =\\\\
&= q^{1/4} \left(\frac{\theta \left[1/2;0\right](0,\tau)}{\eta (\tau)} \right)^4\\\\
&\neq \left(\frac{\theta \left[1/2;0\right](0,\tau)}{\eta (\tau)} \right)^4
\end{eqnarray}
 A: Thank you for your answer. I fortunately already found the mistakes in my computation:

*

*From the second to the third line I simply misinterpreted the identities for the $\theta$-functions.


*I forgot the $\sqrt{2}$ contribution, which would've also lead to the identity you used.
As a remark:

*

*Regarding 1.: My central charges in the computation are chosen correctly, I omitted the argument how I got to $q^{8/24}$, which is the same as the $q^{1/3}$ that you end up with in your calculation, because $1/16-1/48 = 1/24$. The reason for my (1/24) exponent is that I derived it directly from regularizing $L_0$ when implementing normal ordering, which doesn't distinguish the contributions coming from conformality and the Ramond-sector.


*Regarding 2.:
The $1-q^r$ doesn't come from the contributions of the trace, it's just the part that cancels out the $\eta$ term I added on the right side of our second line. You can then apply the product expansions of the $\theta$-functions to see that this is correct.


*Regarding 3.: It is true that I do not take the conformal factor into account explicitly, but it already contributes as stated in "Regarding 1".
A: First, I'm going to work out the calculation to help you be clear about getting the desired result and how similar computations are performed in the literature.
The character of the evolution operator for a free fermion in the Ramond sector on the torus is: $$Tr_{R}[q^{L_{0}-c/24}].$$ Now recall that the central charge of the free fermion CFT is $c=1/2$ and that the conformal weight of a fermion in the Ramond sector is $q^{1/16}$, so the character becomes $$Tr_{R}[q^{L_{0}-c/24}]=\sqrt{2} q^{1/2}q^{-1/48}Tr_{R}[q^{L_{0}}].$$
Assuming that you know that $$Tr_{R}[q^{L_{0}}]=\prod_{r=1}^{\infty}(1+q^{r}),$$ we obtain that the 1-loop partition function for a free Ramond fermion is $$\sqrt{2}q^{1/16}q^{-1/48}\prod_{r=1}^{\infty}(1+q^{r}).$$
To finish our work notice that the only physical superstring degrees of freedom are the eight transverse (to the superstring) ones and this is also true for the oscillators of any worldsheet field operator; then we have one contribution of this kind for every fermion oscillator $$(\sqrt{2}q^{1/16}q^{-1/48}\prod_{r=1}^{\infty}(1+q^{r}))^{8}=2^{4}q^{1/3}\prod_{r=1}^{\infty}(1+q^{r})^{8},$$ a result that can be easily shown to be equivalent to what you want; It simply follows from elevating the identity $$\frac{\theta[1/2;0]}{\eta{ (\tau )}}=\sqrt{2}q^{1/12}\prod_{r=1}^{\infty}(1+q^{r})$$ to the fourth power $$(\frac{\theta[1/2;0]}{\eta{ (\tau )}})^{4}=2^{4}q^{1/3}\prod_{r=1}^{\infty}(1+q^{r})^{8},$$ exactly what we wanted to show.
What are the troubles with your computation?

*

*The first factor (the Casimir term) in your first equality must be $q^{-8/24}$ instead of $q^{8/24}$. Review how this term is obtained. The intuition is that if the hamiltonian for the string were something like $(L_{0}+c/24)$ instead of the correct $(L_{0}-c/24)$, then you wouldn't be able to get massless boson/fermion states and the conformal symmetry must be spontaneously broken.


*Your second line have a lot problems. I can't understand why you have a product that goes as $(1-q^{r})(1-q^{r})$ because terms such as $(1-q^{r})$ does not occur in the Ramond sector. Remember that each $\psi_{-r}^{I}$ for $r>0$ has a degeneracy that goes as $(1+q^{r}),$ namely one ground state and the single fermion state $\psi_{-r}^{I}$. Also you are missing an important $\sqrt{2}$ factor.


*You are not taking into account the conformal factor $q^{1/16}$. If you don't understand why this term is needed, I urge you to go to the wonderful "String Theory in a Nutshell" Kiritsis textbook and work out the exercises 4.36 and 4.37 in the page 122 of its first edition.  If you have any trouble, don't hesitate to post your doubts or solutions. Good luck.
