# A calculation problem on Conformal Field Theory

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 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 ... Commented Feb 23, 2014 at 18:19
• I'll try again later this week, given you said that. Commented Feb 24, 2014 at 14:54
• This problem really intrigues me. I won't see the answer or hint before another try. Commented Feb 24, 2014 at 14:55

Hint:

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

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!

• désolé, but that's not correct. Commented Feb 23, 2014 at 11:40
• You can take the n=2 case and try out your answer. Commented Feb 23, 2014 at 11:41
• What you have calculated is not what i am asking. Commented Feb 23, 2014 at 11:42

Let $$f(n) = \langle h | L_{1}^n L_{-1}^n |h\rangle$$. From the virasoro algebra, $$[L_{1},L_{-1}] = 2L_{0}$$ Also, $$L_{-1}^n|h\rangle$$ is an eigenvector of $$L_0$$ with eigenvalue $$h+n$$ (eq. (7.8) of the book). Hence, \begin{align*} f(n)=\langle h|L_{1}^{n}L_{-1}^{n}|h\rangle & =\langle h|L_{1}^{n-1}\left(\left[L_{1},L_{-1}\right]L_{-1}^{n-1}+L_{-1}\left[L_{1},L_{-1}\right]L_{-1}^{n-2}+...\right)|h\rangle\\ & =2\langle h|L_{1}^{n-1}\left(L_{0}L_{-1}^{n-1}+L_{-1}L_{0}L_{-1}^{n-2}+...\right)|h\rangle\\ & =2\langle h|L_{1}^{n-1}\left(\left(h+\left(n-1\right)\right)L_{-1}^{n-1}+L_{-1}\left(h+\left(n-2\right)\right)L_{-1}^{n-2}+...\right)|h\rangle\\ & =2\left(nh+\frac{n\left(n-1\right)}{2}\right)\langle h|L_{1}^{n-1}L_{-1}^{n-1}|h\rangle\\ & =2n\left(h+\frac{\left(n-1\right)}{2}\right)f\left(n-1\right) \end{align*} So the final formula is: \begin{align*} f(n) & =\prod_{i=1}^{n}2i\left(h+\frac{\left(i-1\right)}{2}\right)\\ & =2^{n}n!\prod_{i=1}^{n}\left(h+\frac{\left(i-1\right)}{2}\right) \end{align*} The answer is with a + sign and not a minus sign. The correction was clarified in the errata of the first printing with can be found here: https://www.physique.usherbrooke.ca/pages/en/node/3449