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 $$O_1(z)O_2(w)=\frac{O_1(z)}{(w-z)^2}+\frac{\partial O_1(z)}{(w-z)}+...$$ ?

  • $\begingroup$ If I recall correctly, OPEs in CFT aren't usually for time-ordered products. $\endgroup$
    – user1504
    Mar 8, 2013 at 4:01
  • $\begingroup$ @user1504 I'm studying from Tong's notes so I guess when I said CFT It was in the context of string theory $\endgroup$
    – Prastt
    Mar 8, 2013 at 8:04
  • $\begingroup$ I'm not familiar with Tong's notes. But neither Polchinski nor diFrancesco use time ordered products. $\endgroup$
    – user1504
    Mar 8, 2013 at 13:14

1 Answer 1


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


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.

  • $\begingroup$ 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. $\endgroup$
    – Prastt
    Mar 9, 2013 at 21:26
  • $\begingroup$ @Barefeg I made some changes. Actually you can conclude from two equations that O_2 and O_1 are equal. $\endgroup$
    – user10001
    Mar 10, 2013 at 0:46

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