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  Null State Level 2 in CFT

+ 3 like - 0 dislike

I'm reading Cardy's notes on CFT. He states the following in section 4.3:

$$\hat L_n\left(\hat L_{-2}|\phi_j\rangle-(1/g)\hat{L^2}_{-1}|\phi_j\rangle\right)=0.$$

I tried to work this out explicitly and I managed to prove it for $n=1$ and $n=2$, but I can't figure this out for general $n$. After some manipulations I get stuck here:

$$((n+2)\hat L_{n-2}-(n+1)(1/g)\hat{L}_{n-1})\hat L_{-1}|\phi_j\rangle+(n+1)\hat L_{-1}\hat L_{n-1}|\phi_j\rangle.$$

How is this zero?

This post imported from StackExchange Physics at 2014-06-15 18:00 (UCT), posted by SE-user user50473
asked Jun 14, 2014 in Theoretical Physics by user50473 (15 points) [ no revision ]
Thanks for the edit, I'm new here

This post imported from StackExchange Physics at 2014-06-15 18:00 (UCT), posted by SE-user user50473
Without knowing anything about the subject or equations, I still smell a classic induction problem a mile away.

This post imported from StackExchange Physics at 2014-06-15 18:00 (UCT), posted by SE-user Chris White

2 Answers

+ 2 like - 0 dislike

Since $|\phi_j\rangle$ is a highest weight state, i.e., $L_n\ |\phi_j\rangle=0$ for all $n>0$. it is easy to see that for $n>2$, $L_{n-1}|\phi_j\rangle=0$. Thus two of the three terms that you have go away. The remaining one needs you to use $[L_{n-1},L_{-1}]\propto L_{n-2}$. This lets you show that it also vanishes since for $n>2$, $L_{n-2}|\phi_j\rangle=0$.

answered Jun 16, 2014 by suresh (1,545 points) [ no revision ]
+ 2 like - 0 dislike

Although it is easy to prove it for general $n$, as shown by @suresh1 in the previous answer, it is in fact enough to show it only for $n=1$ and $n=2$, because $L_1$ and $L_2$ generate the positive degree part of the Virasoro algebra; e.g., $L_3 = [L_2,L_1]$, $L_4 = \tfrac12 [L_3,L_1] = \tfrac12 [[L_2,L_1],L_1]$, et cetera.

answered Jun 17, 2014 by José Figueroa-O'Farrill (2,315 points) [ revision history ]

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