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I'm trying to calculate the following quantity:

$ \left<(L_{-1}\phi)(w_1)(L_{-1}\phi)(w_2) \ldots (L_{-1}\phi)(w_N) \right>$

where $\phi(w_i)$ is a primary operator and $L_{-1}$ is the Virasoro generator of translations. Guided by equations (6.149), (6.155) and (6.161) in Di Francesco, Mathieu and Senechal, and after calculating a simple contour integral, the result came out to be:

$\sum_{i=1}^{i=N} \partial^2_{i} \left< \phi(w_1) \dots \phi(w_N) \right>$

I was specially guided by the OPE between the EM tensor and $\partial \phi$:
$T(z)\partial \phi(w)= \frac{2h\phi(w)}{(z-w)^3} +\frac{(h+1) \partial\phi(w)}{(z-w)^2} + \frac{\partial^2\phi(w)}{(z-w)}$
from which I was able to calculate the $\left< T(z)(L_{-1}\phi)(w_1)(L_{-1}\phi)(w_2) \ldots (L_{-1}\phi)(w_N)\right>$ and use it to obtain the result mentioned above. I'm not quite confident, however, this is 100% correct.

The reason I'm not quite confident of the validity of my result is that if I attempt to compute the correlation function as given above, I get mixed derivative terms, specifically $\partial_{w_i}\partial_{w_{i+j}}$ acting on the correlation function of the primary fields. I'm not sure, however, if I should get those terms in the final result. I only want to be left with $\sum_{i=1}^{i=N} \partial^2_{i}$ operator at the end.

Is there anything basic I'm missing?

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  • $\begingroup$ Generally, check my work type questions are off-topic here. Having said that, this is the most bodacious check my work question that I have seen posted here. $\endgroup$ – Alfred Centauri May 1 at 2:04
  • $\begingroup$ I was really not aware of this rule and was certainly NOT my intention. I sincerely apologize. I will perhaps make it very clear in the question what my source of confusion is. $\endgroup$ – Amro May 1 at 2:14
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Already by dimensional analysis you can see that your answer is wrong. If $\phi$ has weight $h$, then the correlator without $L_{-1}$ operators has total weight $N\cdot h$, and the one with $L_{-1}$ operators should have total weight $N \cdot (h+1)$. Yet your formula has total weight $N \cdot (h+2)$. In fact you're way way overcomplicating this. Just look at formula (6.136) of the Yellow Book and you'll get the correct result straight away.

Edit

I think you're confused at a basic level about the notation

$$(L_{-1} \phi)(w).$$

You seem to think that in a correlation function $$ \langle (L_{-1} \phi)(w) \psi(z) \ldots \rangle $$ you somehow get many crossterms that involve $L_{-1}$ acting on $\psi$ etc. However, by definition we just have

$$(L_{-1} \phi)(w) = [L_{-1},\phi(w)]$$

and likewise

$$(L_{n} L_{m} \phi)(w) = [L_{n},[L_{m},\phi(w)]]$$ and so forth. If you wish, you can rephrase the above definition in terms of integrating $T(z)$ around a small contour circulating $\phi(w)$. But it's not necessary, if you just use the formula from Di Francesco expressing $[L_{-1},\phi(w)]= \pm \partial_w \phi(w)$ (I forget about the sign).

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  • $\begingroup$ Thank you for your answer. You are absolutely right. Perhaps what I'm really after is more of inserting $L_{-1}$ operator at the left and computing something along the lines of equations 6.150 and 6.151 in the Yellow Book such that, in the end, I get the sum of second derivative differential operators acting on the $\left<\phi(w_1)\phi(w_2) \ldots \phi(w_N) \right>$. I'm not sure if this makes my question more clear. $\endgroup$ – Amro May 2 at 17:00
  • $\begingroup$ I mean I should get a total weight of $N.(h+2)$ for the following expression $\left< (L_{-1})(L_{-1}\phi)(w_1)(L_{-1}\phi)(w_2) \ldots (L_{-1}\phi)(w_N) \right>$. Correct? $\endgroup$ – Amro May 2 at 17:27
  • $\begingroup$ Wait a second. Eqs. (6.151) and (6.152) are completely explicit. Why don't you just set $n=1$ in the formula for $\mathcal{L}_{-n}$? And repeat that for all $N$ insertions. $\endgroup$ – Hans Moleman May 3 at 19:53
  • $\begingroup$ I have already tried doing that but this still wouldn't give me the answer I'm looking for. I think I was trying to calculate the wrong quantity since using equations (6.151) an (6.152) in the Yellow Book would directly yield a product of first derivative operators not a sum of 2nd-derivative ones. $\endgroup$ – Amro May 5 at 18:13
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I think I figured it out. The $n$-point correlation function of level-2 null field obeys a 2nd-order differential equation:
$\frac{3}{2(2h+1)} \partial^2_{w_1} \left<\phi(w_1) \dots \phi(w_N) \right> = \left<L_{-2}\phi(w_1) \dots \phi(w_N) \right> = \sum^N_{j=1} \left(\frac{h_j}{(z-z_j)^2} + \frac{\partial_j}{z-z_j} \right) \left<\phi(w_1) \dots \phi(w_N) \right>$.
If you then sum over ALL null fields, you get:
$\sum^N_{i=1}\partial^2_{w_i} \left< \phi(w_1) \dots \phi(w_N) \right> = \frac{2(2h+1)}{3} \sum^N_{i=1} \left( \sum^N_{j\neq i} \left(\frac{h_j}{(z_i-z_j)^2} + \frac{\partial_j}{z_i-z_j} \right)\right) \left< \phi(w_1) \dots \phi(w_N) \right>$
which is the operator I was looking for to start with.

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  • $\begingroup$ Sorry, but it's just not right. The right answer is simply $\sum_i \partial_{w_i} \langle \phi(w_1) \ldots \phi(w_N) \rangle$ up to an overall sign/factor of $i$. $\endgroup$ – Hans Moleman May 6 at 4:43
  • $\begingroup$ Let us back up a step. Three points I want to make: (1) The answer I gave to my question is NOT an answer to the original question I asked: What is $ \left<(L_{-1}\phi)(w_1)(L_{-1}\phi)(w_2) \ldots (L_{-1}\phi)(w_N) \right>$? As I have indicated in my last comment, this is not the right quantity to calculate since it clearly wouldn't give me what I want which is: $\sum^N_{i=1}\partial^2_{w_i}$. $\endgroup$ – Amro May 6 at 13:53
  • $\begingroup$ I believe I should edit my original question to say: What is the operator $\hat{O}$ inserted in $ \left< (\hat{O}) (L_{-1}\phi)(w_1)(L_{-1}\phi)(w_2) \ldots (L_{-1}\phi)(w_N) \right>$ that yields $\sum^N_{i=1}\partial^2_{w_i}\left<\phi(w_1) \dots \phi(w_N) \right>$? After this discussion with you, I really think it is $ \left< (L_{-1}) (L_{-1}\phi)(w_1)(L_{-1}\phi)(w_2) \ldots (L_{-1}\phi)(w_N) \right>$ since as you have indicated in your edited comment above $$(L_{n} L_{m} \phi)(w) = [L_{n},[L_{m},\phi(w)]]$$. $\endgroup$ – Amro May 6 at 13:53
  • $\begingroup$ (3) Your answer to my original question gives zero since $$\sum_i \partial_{w_i} \langle \phi(w_1) \ldots \phi(w_N) \rangle = 0$$ by virtue of the conformal Ward identity corresponding to translational symmetry. Don't you think? $\endgroup$ – Amro May 6 at 15:28

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