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For the commutation relation in quantising the bosonic string


we can then calculate this for $m=-n$ in between the vacuum state i.e.

$$\boxed{\left<0,0|\left[L_n,L_{-n}\right]|0,0\right>=\frac{D}{2}\sum_{m=1}^{n-1}m(n-m)\left<0,0|0,0\right>=\frac{D}{12}n(n^2-1)\left<0,0|0,0\right>} \, .$$

The second expression I can't seem to show explicitly although I know we can can write


and then it's surely something to do with how these $\alpha$'s act on the vacuum state. However, why the $m(n-m)$ and where's the $\frac{D}{2}$ come from?

Finally, the last term in the boxed calculation is obviously found from the commutation relation of $\left[L_n,L_{m}\right]$, but therefore $\left<0,0|L_0|0,0\right>=0$. Why is this?

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To show: $$\boxed{\left<0,0|\left[L_n,L_{-n}\right]|0,0\right>=<0,0|n^2L_0+\frac{D}{2}\sum_{m=1}^{n-1}m(n-m)\left|0,0\right>=<0,0|\frac{D}{2}\sum_{m=1}^{n-1}m(n-m)\left|0,0\right>= \frac{D}{2}\sum_{m=1}^{n-1}m(n-m)\left<0,0|0,0\right> = \frac{D}{12}n(n^2-1).}$$

Why is the first term vanishing?

To see that $<0,0|n^2L_0\left|0,0\right> =0$ one can use the algebra commutator relation and write $$ 2 L_0 |0,0 \rangle = (L_1 L_{-1} - L_{-1} L_1) |0,0\rangle =0 $$ as done in Polchinski, p 59. The expression is zero, because the $L_{\pm 1}$ can be viewed as normal ordered $$L_n=\frac{1}{2}\sum_{n=1}^{\infty} : \alpha_{n} \alpha_{-n} :$$ with all annihilation operators to the right destroying the $|0,0 \rangle$.

Why is $\frac{D}{2}\sum_{m=1}^{n-1}m(n-m)=\frac{D}{12}n(n^2-1)$ ?

Let's decompose the sum in 2 terms $$\frac{D}{2}\sum_{m=1}^{n-1}m(n-m) = \frac{D}{2} n \sum_{m=1}^{n-1}m -\frac{D}{2}\sum_{m=1}^{n-1}m^2 $$ The first sum can be identified with one known finite series $\sum_{k=1}^l k =\frac{l(l+1)}{2}$ giving $$\frac{D}{2} n \sum_{m=1}^{n-1}m = \frac{D}{2}n\frac{(n-1)n}{2}.$$ The second sum can be identified with another finite series from the same list $ \sum_{k=1}^l k^2 = \frac{l(l+1)(2l+1)}{6}$ giving $$\frac{D}{2}\sum_{m=1}^{n-1}m^2= \frac{D}{2} \frac{(n-1)n(2n-1)}{6}.$$ Subtracting the second sum from the first one arrives at desired result $$\boxed{\frac{D}{2}\sum_{m=1}^{n-1}m(n-m)=\frac{D}{12}n(n^2-1)}.$$

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