- The problem statement, all variables and given/known data
(I have dropped the hats on the $\alpha_{n}^{u}$ operators and $L_{m}$)
$[\alpha_{n}^u, \alpha_m^v]=n\delta_{n+m}\eta^{uv}$ $L_m=\frac{1}{2}\sum\limits_{n=-\infty}^{\infty} : \alpha_{m-n}^u\alpha_{n}^v: \eta_{uv}-\delta_{m,0}$
where " : " denotes normal-ordered.
Show that : $[\alpha_{m}^u,L_n]=m\alpha_{m+n}^u$
- Relevant equations
see above
- The attempt at a solution
For a given $n$ we are looking at the following commutator: $[\alpha_m,\alpha_{n-m}\alpha_m]$
to use commutator relation:
$[a,bc]=-a[b,c]-[a,c]b$
$a= \alpha_m, b= \alpha_{n-m}, c= \alpha_m $
$[a,c]=0$, $[b,c]=(n-m)\delta_{n=0}\eta^{uv}$ using (1)
$\implies [\alpha_{m}^u,L_n]=\alpha_m^u(m)\eta^{uv} $ which is wrong...
thanks in advance
First we can expand $L_n$ into its normally ordered form: \begin{align*} L_n = \frac{1}{2}\sum_{i=-\infty}^\infty:\alpha_{n-i}^u\alpha_i^v:\eta_{uv}-\delta_n = \frac{1}{2}\sum_{i=-\infty}^{-1}\alpha_i^v\alpha_{n-i}^u\eta_{uv}+\frac{1}{2}\sum_{i=0}^{\infty}\alpha_{n-i}^u\alpha_i^v\eta_{uv}-\delta_n \end{align*}
We want to find $[\alpha^u_m, L_n]$. In the derivation below I will use the fact that scalars commute, $[A,B+C] = [A,B]+[A,C]$, and a slightly rearranged version of the multiplication rule: $[A,BC]=[A,B]C+B[A,C]$. \begin{align*} [\alpha^u_m, L_n] &= [\alpha^u_m, \frac{1}{2}\sum_{i=-\infty}^{-1}\alpha_i^v\alpha_{n-i}^u\eta_{uv}+\frac{1}{2}\sum_{i=0}^{\infty}\alpha_{n-i}^u\alpha_i^v\eta_{uv}-\delta_n]\\ &= [\alpha^u_m, \frac{1}{2}\sum_{i=-\infty}^{-1}\alpha_i^v\alpha_{n-i}^u\eta_{uv}+\frac{1}{2}\sum_{i=0}^{\infty}\alpha_{n-i}^u\alpha_i^v\eta_{uv}]\\ &=\frac{1}{2}\sum_{i=-\infty}^{-1}[\alpha_m^u, \alpha_i^v\alpha_{n-i}^u\eta_{uv}]+\frac{1}{2}\sum_{i=0}^{\infty}[\alpha_m^u, \alpha_{n-i}^u\alpha_{i}^v\eta_{uv}]\\ &=\frac{1}{2}\sum_{i=-\infty}^{-1}([\alpha_m^u,\alpha_{n-i}^u\eta_{uv}]\alpha_i^v+\alpha_{n-i}^u\eta_{uv}[\alpha_m^u,\alpha_i^v])\\ &\phantom{==} + \frac{1}{2}\sum_{i=0}^{\infty}([\alpha_m^u,\alpha_{i}^v\eta_{uv}]\alpha_{n-i}^u+\alpha_i^v\eta_{uv}[\alpha_m^u,\alpha_{n-i}^u])\\ &= \frac{1}{2}\sum_{i=-\infty}^{-1}(m\delta_{m+(n-i)}\eta^{uu}\eta_{uv}\alpha_i^v+\alpha_{n-i}^u\eta_{uv}m\delta_{m+i}\eta^{uv})\\ &\phantom{==} + \frac{1}{2}\sum_{i=0}^{\infty}(m\delta_{m+i}\eta^{uv}\eta_{uv}\alpha_{n-i}^u+\alpha_i^v\eta_{uv}m\delta_{m+(n-i)}\eta^{uu})\\ \end{align*} Without loss of generality suppose $m>0$. (If $m = 0$, the whole thing is $0 = m\alpha_{m+n}^u$ as expected. By symmetry of the equation, you can see that $m<0$ will yield the same result.) Now the sum simplifies to the expression below, as the kronecker delta is always zero for the other components of the sum. \begin{align*} [\alpha^u_m, L_n] &= \frac{1}{2}\sum_{i=-\infty}^{\infty}(\alpha_{n-i}^u\eta_{uv}m\delta_{m+i}\eta^{uv}+\alpha_i^v\eta_{uv}m\delta_{m+(n-i)}\eta^{uu}) \end{align*} The first component in the expression is non-zero exactly when $i=-m$, and the second component is non-zero when $i=m+n$. Hence you get: \begin{align*} [\alpha^u_m, L_n] &= \frac{1}{2}(m\alpha_{m+n}^u\eta_{uv}\eta^{uv}+m\alpha_{m+n}^v\eta_{uv}\eta^{uu})\\ &= \frac{1}{2}(m\alpha_{m+n}^u+m\alpha_{m+n}^u)\\ &= m\alpha_{m+n}^u \end{align*} I'm still not confident on what the exact actions of $\eta^{uv}$ and $\eta_{uv}$ are, so I'm not sure if saying $\alpha^v\eta_{uv}\eta^{uu} = \alpha^u$ is correct. Apart from that my derivation should be okay.