A calculation problem on Conformal Field Theory

Question

This problem is on Di Francesco's book I. It's exercise 7.1: Calculate the norm of the following vector, where $\lvert h\rangle$ is the state of highest weight.

$$L_{-1}^n\lvert h\rangle$$

I have just tried to use the commutation relations of the $L$ operators and the fact that $L_1$ acts on $\lvert h\rangle$ is $0$. But as the calculation goes on, things began to be troublesome. I just found them too complicated.

Commutation relations: $$[L_n,L_m]=(n-m)L_{n+m}+\frac{c}{12}\delta_{n+m,0}n(n^2-1) $$
It is in fact asking us to calculate: $\langle h|L_1^nL_{-1}^n|h\rangle $. And we have the following relations: $$\langle h\rvert L_{-1}=0 \quad\mbox{and}\quad L_1\lvert h\rangle=0. $$

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Li Xinghe

Comments

This is a legitimate and very high-level technical questions, and I would bet quite a large amount of money that the closevoter(s) here have no clue what the OP is talking about. And just to say, even Qmechanic deems it worth an answer, so voting to close on such questions is quite preposterous ...

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Dilaton

---

I'll try again later this week, given you said that.

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Li Xinghe

---

This problem really intrigues me. I won't see the answer or hint before another try.

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Li Xinghe

Answer 1

I will outline the steps for you and you can fill in the details:

i)Calculate $[L_1, L_{-1}]=\cdots=2L_0$.

ii) Using the fact that $L_1 L_{-1}=[L_1, L_{-1}]+L_{-1} L_{1}$ and that $L_0|h\rangle=h|h\rangle $, try to calculate $\langle h|L_1 L_{-1}|h\rangle =\cdots\stackrel{?}{=}2h$.

iii) Calculate the following quantity that will (probably) be useful later: $$\langle h|(L_1 L_{-1})^n|h\rangle =\cdots\stackrel{?}{=}(2h)^n$$

iv) (The hard part) We have done the case for $n=1$ for the quantity $\langle h |L_1^n L_{-1}^n |h\rangle$ in i), but you will also need to do the case for $n=2$ and $n=3$ by hand, using the formulae in i) and ii) above and from that, try to deduce an inductive formula for the general case. I will have to admit though that deducing the general form (which then can be proved by induction) from a few cases might be quite hard!

Hope this helps!

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Heterotic

Comments

désolé, but that's not correct.

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Li Xinghe

---

You can take the n=2 case and try out your answer.

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Li Xinghe

---

What you have calculated is not what i am asking.

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Li Xinghe

Answer 2

Hint:

$$\langle h|L_1^n L_{-1}^n|h\rangle ~=~ \sum_{i=0}^{n-1} \langle h|L_1^{n-1} L_{-1}^{n-1-i}[L_1,L_{-1}]L_{-1}^i|h\rangle ~=~ \ldots $$ $$~=~ 2\sum_{i=0}^{n-1} \langle h|L_1^{n-1} L_{-1}^{n-1}(L_0-i)|h\rangle ~=~\ldots $$ $$~=~ n(2h-(n-1)) \langle h|L_1^{n-1} L_{-1}^{n-1}|h\rangle ~=~\ldots ~=~n!\prod_{i=0}^{n-1} (2h-i).$$

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Qmechanic

Comments

Nice answer Qmechanic! I will just add for clarification that the identity used in the first step is a simple generalization of the identities shown in eg en.wikipedia.org/wiki/Commutator#Identities_.28ring_theory.29

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Heterotic