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: $$\langle h\rvert L_{-1}=0 \quad\mbox{and}\quad L_1\lvert h\rangle=0. $$

Answer 1

Hint:

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

Answer 2

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!

Answer 3

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