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This questions concerns Exercise 2.11 in Polchinski. We are asked to compute the commutator $$L_{m}(L_{-m}|0;0\rangle) - L_{-m}(L_{m} |0;0\rangle)$$ By plugging the mode expansions, we use the definition from 2.7.6 $$L_{m}\sim \frac{1}{2}\sum_{n =-\infty}^{\infty} \alpha^{\mu}_{m-n} \alpha_{\mu n}. \tag{2.7.6}$$ Now, from the solution given here http://arxiv.org/abs/0812.4408, it is stated that we may consider only the summation $$L_{-m} |0;0\rangle = \frac{1}{2}\sum_{n=1}^{m-1}\alpha^{\mu}_{n-m}\alpha_{\mu (-n)}|0;0\rangle. \tag{36}$$ Now, I understand that for $n<0$, the operator will annihilate the state, but why do we cut it off at $m-1$? Why would the state be annihilated, at for example, the value $n=m+1$?

In addition, the calculation is carried out, and I was able to obtain the given result $$\frac{1}{4}\sum_{n=1}^{m-1}\sum_{n'=-\infty}^{\infty}\alpha^{\nu}_{m-n'}\alpha_{\nu n'}\alpha^{\mu}_{n-m}\alpha_{\mu(-n)}|0;0\rangle$$

However, in the next line, he proceeds the calculation with the equation $$\frac{1}{4}\sum_{n=1}^{m-1}\sum_{n'=-\infty}^{\infty}((m-n')n'\eta^{\nu \mu}\eta_{\nu \mu}\delta_{n' n}+(m-n')n'\delta^{\nu}_{\mu}\delta^{\mu}_{\nu}\delta_{m-n',n})|0;0\rangle$$

I can deduce that this is given by the commutator $$\frac{1}{4}\sum_{n=1}^{m-1}\sum_{n'=-\infty}^{\infty}[\alpha^{\nu}_{m-n'},\alpha^{\mu}_{n-m}][\alpha_{\nu n'},\alpha_{\mu(-n)}] + [\alpha^{\nu}_{m-n'},\alpha_{mu(-n)}][\alpha_{\nu n'},\alpha^{\mu}_{n-m}]$$

where we are given the commutator relations $$[\alpha^{\mu}_{m},\alpha^{\nu}_{n}] = m\delta_{m,-n}\eta^{\mu \nu}\tag{2.7.5a}$$ from equation (2.7.5a).

I can see why we could convert the equation to one in terms of commutators if taking the product of operators the other way annihilates the state, but I don't see why, for example, we could say $\alpha_{\nu n'}$ annihilates the state.

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