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In BBS's string theory book, in the equation (2.143), it says that $$\text{tr} \omega ^N=\prod _{n=1}^{\infty } \left(\prod _{i=1}^{24} \text{tr} \omega ^{\alpha _{-n}^i \:\alpha _n^i}\right)=\prod _{n=1}^{\infty } \frac{1}{\left(1-\omega ^n\right)^{24}}.$$ I agree with the first equality, but not the second. My calculation is as follows, $$tr\omega^{N}= \omega ^{\sum _{\phi } \langle \phi |N|\phi \rangle }=\omega ^{\sum _{\phi } \langle \phi |\sum_{n,i}\:\alpha_{-n}^i\;\alpha_{n}^i\;|\phi \rangle }=\omega ^{\sum_{n,i}\sum _{\phi } \langle \phi |\:\alpha_{-n}^i\;\alpha_{n}^i\;|\phi \rangle }=\prod _{n=1}^{\infty } \left(\prod _{i=1}^{24} \text{tr} \omega ^{\alpha _{-n}^i \:\alpha _n^i}\right),$$ my problem is that $$|\phi\rangle$$ is the basis of the whole Hilbert space, not the Hilbert space created by acting a specific creating operator $$\alpha_{-n}^i,$$ so $$ tr\omega^{\alpha_{-n}^i\;\alpha_{n}^i}\neq\sum_{i=0}^{\infty}\omega^{\;i\;n}=\frac{1}{1-\omega^{n}},$$instead I think that $$tr\omega^{\alpha_{-n}^i\;\alpha_{n}^i}=\sum_{i=0}^{\infty}\omega^{\;i\;m\;n}=\frac{1}{1-\omega^{m\;n}},$$where m is some infinity large constant. I can not find the problem in my calculation, so can some one point out my mistake? And show me how to organize the Hilbert space to get BBS's conclusion.enter image description here

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BBS omit to indicate which space they are tracing over in each step. Let me instead sketch why tracing over only the states created by the $\alpha_{-n}$'s in the last step is the correct thing to do. For simplicity I'll just do the $D=1$ case.

First an important lemma says that if our Hilbert space is of the form $V = V_1 \otimes V_2$ and we have an operator $A_1 $ which acts on $V$ by acting via some well defined action on $V_1$ trivially on $V_2$ and vice versa for an operator $A_2$, then we have $$\text{tr}_V(A_1 A_2) = \text{tr}_{V_1}A_1 \text{tr}_{V_2}A_2.$$ This is easy enough to show and I encourage you to prove it. Now we must apply this carefully to the present case. For the case of closed strings our Hilbert space is the following:

Since the Hamiltonian takes the following form $$H = \frac{1}{2}p_0^2 + \sum_{n=1}^{\infty} \alpha_{-n}\alpha_{n} + \widetilde{\alpha}_{-n}\widetilde{\alpha}_{n} -\frac{1}{12} = (L_{0} - \frac{1}{24}) + (\tilde{L}_0 - \frac{1}{24}) $$ our Hilbert space takes the form $$\mathcal{H} = \mathcal{H}_{\text{free}} \bigotimes (\otimes_{n=1}^{\infty}\mathcal{H_n}) \bigotimes (\otimes_{n=1}^{\infty}\widetilde{\mathcal{H}}_n) .$$ Here $\mathcal{H}_{\text{free}}$ corresponds to the free particle space which is due to the presence of the ${p_0}^2$, $\mathcal{H}_n$ is the space isomorphic to the one-oscillator Hilbert space created by $\alpha_{-n}$ and similarly for $\widetilde{\mathcal{H}}_n$. The partition function, which allows you to calculate how many states there are of a given energy is then computed by the following trace $$Z(q,\bar{q}) = \text{tr}_{\mathcal{H}}(q^{L_0 - 1/24} \bar{q}^{\tilde{L}_0 - 1/24}).$$

Now noting that $q^{\alpha_{-n} \alpha_n}$ acts non-trivially only on $\mathcal{\mathcal{H}_n}$ we can apply (a generalization of) the above Lemma to write $$Z(q,\bar{q}) = q\bar{q}^{-1/24} \text{tr}_{\mathcal{H}_{\text{free}}}(q\bar{q}^{p_{0}^2/4})\prod_{n=1}^{\infty}\text{tr}_{\mathcal{H}_n}(q^{\alpha_{-n}\alpha_{n}}) \prod_{n=1}^{\infty}\text{tr}_{\widetilde{\mathcal{H}}_n}(\bar{q}^{\widetilde{\alpha}_{-n} \widetilde{\alpha}_n}).$$ Finally, it's easy to see that $$\text{tr}_{\mathcal{H}_n}(q^{\alpha_{-n}\alpha_{n}}) = \frac{1}{1-q^{n}}.$$

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  • $\begingroup$ Thanks for your explanation, it helps me found my mistake, i thought $$w^{\sum_{\phi}}<\phi|a_{-1}a_1 +a_{-2}a_2 |\phi>=w^{\sum_{\phi}<\phi|a_{-1}a_1|\phi> * w^{\sum_{\phi}<\phi|a_{-2}a_2|\phi> $$ but it is obvious not true, Since these two phi are correlated $\endgroup$ – xjtan Jun 2 '15 at 13:19

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