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commutation of operator product expansion

+ 3 like - 0 dislike
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In CFT, when we have an OPE: $$O_1(z)O_2(w)=\frac{O_2(w)}{(z-w)^2}+\frac{\partial O_2(w)}{(z-w)}+...$$ this holds inside a time-ordered correlation function, so $O_1(z)O_2(w)=O_2(w)O_1(z)$. Does it mean that $$O_1(z)O_2(w)=\frac{O_1(z)}{(w-z)^2}+\frac{\partial O_1(z)}{(w-z)}+...$$ ?

This post imported from StackExchange Physics at 2015-03-30 13:49 (UTC), posted by SE-user Barefeg
asked Mar 7, 2013 in Mathematics by Barefeg (15 points) [ no revision ]
If I recall correctly, OPEs in CFT aren't usually for time-ordered products.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user user1504
@user1504 I'm studying from Tong's notes so I guess when I said CFT It was in the context of string theory

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user Barefeg
I'm not familiar with Tong's notes. But neither Polchinski nor diFrancesco use time ordered products.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user user1504

1 Answer

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I am not an expert in 2d CFT. However I hope following manipulations are valid.

Assume that your second equation follows from first one.

Then on RHS of your first equation Taylor expansion of $O_2(w)$ at point $z$ gives :

$O_2(w)=O_2(z)+(w-z)\partial_z O_2(z)+ ...$

taking derivative wrt $w$ on both sides we get

$\partial_wO_2(w)=\partial_zO_2(z)+...$

Using these two results in your first equation we get

$O_1(z)O_2(w)= \displaystyle\frac{O_2(z)}{(z-w)^2}+regular\:terms$

Subtracting it from your second equation, multiplying with $(z-w)^2$ and taking limit $w\rightarrow z$ we conclude that $O_2$ and $O_1$ should be equal. Since to begin with we didn't assume any such thing regarding fields $O_2$ and $O_1$ so in general your second equation shouldn't follow from the first one.

I think equality of $O_2(w)O_1(z)$ and $O_1(z)O_2(w)$ (assuming fields are 'bosonic') within time ordered product only implies that their OPE should be symmetric under exchange of z and w. So if your first equation for OPE can be realized for some (bosonic) fields, then by exchanging z with w on RHS you should get the same result within a regular term.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user user10001
answered Mar 9, 2013 by user10001 (635 points) [ no revision ]
Thanks, that's what I thought. It only implies symmetry of w and z. I asked because I was trying to solve an exercise that involved the commutation of the operators but wasn't getting the answer so I thought that perhaps the second equation was valid. But at the end I found the solution without expanding the ope.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user Barefeg
@Barefeg I made some changes. Actually you can conclude from two equations that O_2 and O_1 are equal.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user user10001

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