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In a classical theory, the functions $$l_n = \frac{\eta_{\mu\nu}}{2} \sum_{m} a_m^\mu a_{n-m}^\nu$$ form a Virasoro Algebra, i.e. they satisfy $$\{l_n, l_m\} = i (n-m) \delta_{n+m,0}.$$ When we quantise the theory, we have to apply normal ordering, so we define the operators $$L_n = \frac{\eta_{\mu\nu}}{2} \sum_m : a_m^\mu a_{n-m}^\nu:$$ These form a Virasoro Algebra with a central charge $$[L_n, L_m] = (n-m) L_{n+m} + \frac{c}{12} n(n^2-1)\delta_{n+m,0}$$ where the additional term appears because of the normal ordering. I have seen derivations of this where people split up the sums and deal with the normal ordering that way. However, the commutator of 2 creation/annihilation operators is just $$[a_n^\mu, a_m^\nu] = n\: \delta_{n+m,0}\:\eta^{\mu\nu}$$ which for each individual component is just a number (i.e. a multiple of the identity operator). If 2 operators satisfy $$[a,b] = z \cdot \operatorname{Id}$$ with some number $z$, then we should have $$[ab,c] = [ba,c] + z [\operatorname{Id}, c] = [ba, c]$$ for any operator $c$, since everything commutes with the identity. From this, I conclude that the normal ordering should not change the commutator at all: $$[:ab:, c] = [ab, c]$$ so that I can write $$[L_n, L_m] = \frac{\eta_{\mu\nu}\eta_{\alpha\beta}}{4}\: \sum_{l,k}\: [a_k, a_{n-k}, a_l a_{m-l}]$$ without any normal ordering. However, when I carry through with this calculation, I'm not getting the correct form of the Virasoro Algebra. What I get is $$[L_n, L_m] = (n-m)L_{n+m} + 2\:n\: D\: \delta_{n+m,0}\: \sum_{k>0}k $$ where $D$ is the dimension of the space. What am I missing here?

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The commutator $[a_m^\mu,a_n^\nu]=n\delta_{m+n,0}\eta^{\mu\nu}$, or better rewritten as $[a_m^\mu,a_{n-m}^\nu]=n\delta_{n,0}\eta^{\mu\nu}$ using $n\mapsto n-m$ for more convenience with the Kronecker symbol, indeed allows for simplification. As $\alpha_m^\mu\alpha_{n-m}^\nu=\alpha_{n-m}^\nu\alpha_m^\mu$ for $n\neq 0$, normal ordering won't have any effect at all. That means, that $L_n=:l_n:\;=l_n$ for $n\neq 0$, and only $l_0$ is affected by normal ordering.

If we take $n=0$, then we have $a_m^\mu a_{-m}^\nu=a_{-m}^\nu a_m^\mu+n\eta^{\mu\nu}$ and therefore $a_m a_{-m}=a_{-m} a_m+nD$ when contracting with $\eta_{\mu\nu}$, which yields: \begin{align*} L_0=\;:l_0: \;&=\frac{1}{2}\sum_{m\in\mathbb{Z}}:\alpha_{-m}\alpha_m: \;=\frac{1}{2}\sum_{m\in\mathbb{Z}, m\geq 0}\alpha_{-m}\alpha_m +\frac{1}{2}\sum_{m\in\mathbb{Z}, m<0}\alpha_m\alpha_{-m} \\ &=\frac{1}{2}\sum_{m\in\mathbb{Z}}\alpha_{-m}\alpha_m -\frac{D}{2}\sum_{n=1}^\infty n =l_0+\frac{D}{24}, \end{align*} where exactly the same divergent sum as in your final calculation arises. As it is the value of the Dirichlet series for $s=-1$, which describes the Riemann zeta function for $\operatorname{Re}(s)>1$, the divergence is averted by replacing the sum with $\zeta(-1)=-1/12$. But concerning the commutator, this can be avoided: The relation $a_m a_{-m}=a_{-m} a_m+nD$, which lead to divergence in the first place and adds a constant which would not change the commutator as you mentioned, is not used to compute an expression for $L_0$ before computing the commutator, but $a_ma_n=a_na_n+nD\delta_{m+n,0}$ is used after computing the commutator. Because why use the last line in the upper equation containing divergence, when there is the possibility, that certain terms will just cancel out? When quantizing, so going from $l_n$ to $L_n=l_n$ for $n\neq 0$ and $l_0$ to the different $L_0\neq l_0$, in the Witt algebra $\{l_m,l_n\}=(m-n)l_{m+n}$ both sides can be affected iff $m=0$, $n=0$ or $n=-m$. When computing the commutator, you indeed have to use normal ordering inside it, so: $$[L_m,L_n] =\frac{1}{4}\eta_{\kappa\lambda}\eta_{\mu\nu}\sum_{k,l\in\mathbb{Z}}\left[:\alpha_k^\kappa\alpha_{m-k}^\lambda:,:\alpha_l^\mu\alpha_{n-l}^\nu:\right].$$ Keep in mind, that you can easily make obvious mistakes with normal ordering. (For example, when applying it to a constant creation-anticreation commutator of the form $[a,a^\dagger]=1$ to concluding $0=1$.) Therefore, the calculations are usually broken into two pieces, the commutator $[L_m,\alpha_k^\xi]$ (with normal ordering used to the left) and then the commutator $[L_m,L_n]$ (with normal ordering used to the right). You can find it for example in this script in section 4.1.1 on Virasoro algebras.

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  • $\begingroup$ I guess, my main question is, Is it true that $[L_n, L_m] = \frac{\eta_{\mu\nu}\eta_{\alpha\beta}}{4}\: \sum_{l,k}\: [a_k a_{n-k}, a_l a_{m-l}]$ without any normal ordering? If not, what am I missing? $\endgroup$ Commented May 23, 2023 at 19:53
  • $\begingroup$ I have now updated the rest as promised and added an answer to that question. $\endgroup$ Commented May 26, 2023 at 14:23
  • $\begingroup$ I actually talked to my professor about this, and he said that removing the normal ordering inside the commutator should work. I tried to explain my reasoning why changing the order of two ladder operators inside the commutator shouldn't change the result. Where do you see a flaw in my reasoning? $\endgroup$ Commented May 27, 2023 at 21:43
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    $\begingroup$ I have thought about it for a while. Unfortunately, I don't know enough about the exact mathematical definition of normal ordering, which involves more than just reordering the operators - but this is how it is usually defined in literature. You also get a conflict with the example I mentioned, normal ordering being applied to $[a,a^\dagger]=1$ would give $0=1$. So there have to be further restrictions and rules on how to handle it properly. I have asked about this myself and also linked two other posts therein, you could find it helpful: physics.stackexchange.com/q/736394 $\endgroup$ Commented Jun 3, 2023 at 18:50

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