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How can we expand product of two currents into series?

by ndung200790
Tags: currents, expand, product, series
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ndung200790
#1
Feb28-12, 08:40 PM
P: 520
Please teach me this:
In QFT book of Peskin&Schroeder write:
J[itex]_{\mu}[/itex](x)J[itex]_{\nu}[/itex](0)~C[itex]_{\mu\nu}[/itex][itex]^{1}[/itex](x).1+C[itex]_{\mu\nu}[/itex][itex]^{q^{-}q}[/itex]q[itex]^{-}[/itex]q(0)+C[itex]_{\mu\nu}[/itex][itex]^{F^{2}}[/itex](x)(F[itex]_{\alpha\beta}[/itex][itex]^{a}[/itex])[itex]^{2}[/itex](0)...
I do not know how to expand the product into this series.
Thank you very much for your kind helping.
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TriTertButoxy
#2
Feb28-12, 11:10 PM
P: 194
Oh dear... I would love to know the answer to this one.

The best answer I have been given to date is: think of all the operators that transforms in the same way as the product of two currents, and those will be the ones appearing in the expansion.

It would be so awesome if someone could give a constructive algorithm for generating the series expansion for the product of two currents (or for that matter, for the non-local product of any two local operators).
ndung200790
#3
Feb29-12, 05:35 AM
P: 520
Given the series,they say that the coefficients in the series can be calculate by Feynman diagrams.Then I do not know how to give the Feynman diagrams for the coefficients.

ndung200790
#4
Mar1-12, 12:02 AM
P: 520
How can we expand product of two currents into series?

Why is the coefficient of operator 1 being the sum of diagrams with no external legs other than the current insertions?
ndung200790
#5
Mar1-12, 09:22 PM
P: 520
It seems that there is ''some arbitrary'' in OPE.We can ''adjust'' the coefficients in the series according with ''situation''.Is that correct?
ndung200790
#6
Mar3-12, 12:57 AM
P: 520
It seem to me that we can expand product of operators by Wick theorem and Taylor series?


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