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In my String theory lecture radial ordering was introduced and I don't understand what it is. My first guess was $$R(A(z)B(w)) = A(z)B(w)\Theta(|z|-|w|) + B(w)A(z)\Theta(|w|-|z|).$$ But then we have proven the following standard formula by changing the contour ($C_a$ is a contour around $a$): $$\int_{C_0} \frac{\mathrm{d}z}{2\pi i} [A(z),B(w)] = \int_{C_w} \frac{\mathrm{d}z}{2\pi i} R(A(z)B(w)).$$

This formula is obviously wrong. For example set $w = 0$. Then on the right side there is an integral over $A(z)B(0)$ and on the left side we have the same integral over $[A(z),B(0)]$ which is not the same (ok, maybe $w = 0$ is excluded).

But in general the right hand side is not well defined since the integral over $C_w$ is path dependent (since $R(A(z)B(w)$) has a branch cut at $|z| = |w|$). I checked this directly for $A(z) = \frac{A}{z}$ and $B(w) = B = const.$ (I can add the calculation if needed).

After all I think that I misunderstood radial ordering. We were never given an explicit definition. So how is radial ordering defined? And if my definition is wrong why can we use radial ordering in the same way as time ordering?

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  • $\begingroup$ Are you following a textbook? $\endgroup$
    – Qmechanic
    Commented Dec 15, 2018 at 17:21
  • $\begingroup$ no. But I looked it up in some textbooks and I always find the same derivation with changing the contour (Should I add the derivation here?). $\endgroup$
    – toaster
    Commented Dec 15, 2018 at 18:14
  • $\begingroup$ Related: physics.stackexchange.com/q/405763/2451 $\endgroup$
    – Qmechanic
    Commented Dec 15, 2018 at 19:30
  • $\begingroup$ But what is radial ordering then exactly? I mean it must be defined in some way? $\endgroup$
    – toaster
    Commented Dec 15, 2018 at 20:55
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    $\begingroup$ Ok, but what is it then? And what does it change to the path dependence on the right side? No matter what you put for |z| = |w| you will always have a branch cut there. $\endgroup$
    – toaster
    Commented Dec 17, 2018 at 20:07

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