- #1
binbagsss
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- 11
Homework Statement
I am trying to express ##T(\phi(x1)\Phi(x2)\phi(x3)\Phi(x4)\Phi(x5)\Phi(x6))## in terms of the Feynman propagators ##G_F^{\phi}(x-y)## and ##G_F^{\Phi}(x-y)##
where ##G_F^{\phi}(x-y) =\int \frac{d^{4}k}{(2\pi)^{4}}e^{ik(x-y)} \frac{ih}{-k.k - m^2 -i\epsilon} ##
and ##G_F^{\Phi}(x-y) =\int \frac{d^{4}k}{(2\pi)^{4}}e^{ik(x-y)} \frac{ih}{-k.k - M^2 -i\epsilon} ##
Homework Equations
[/B]
##<0|T(\phi(x1)\Phi(x2)\phi(x3)\phi(x4)\phi(x5)\phi(x6)) |0> = : non-fully contracted terms : + fully contracted terms = fully contracted terms ##
where ##T## is the time ordering operator
##<0| ## being the vacuum state,
since non-fully contracted terms, i.e. where every field is not involved in a contraction will vanish.
The Attempt at a Solution
My question is how/ can you contract over two different fields?
I am fine with contractions over two of the same field, but can't find any notes/examples on what to do in the case of two different fields - here ##\phi(x)## and ##\Phi(x)##
In particular, if you can't, the question ask to express the final answer in terms of ##G_F^{\phi}(x-y)## and ##G_F^{\Phi}(x-y)##, and I can't see how you could write the contraction over two different fields in terms of this.
I know ##G^{\phi} (x1-x3) = contraction of the fields \phi(x1) , \phi(x3)##
and similarly that ##G^{\Phi} (x4-x5) = contraction of the fields \Phi(x4) , \Phi(x5)##
but no idea how you would write e.g contraction over the fields ## \phi(x1), \Phi(x5)## in terms of ##G_F^{\phi}(x-y)## and ##G_F^{\Phi}(x-y)##?
Can you contract over two different fields?
Help really appreciated,
(Or a point to some notes on this)
Thanks