QFT Wicks theorem contraction -- different fields terms of propagation

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Homework Help Overview

The discussion revolves around the expression of time-ordered products of different quantum fields, specifically ##T(\phi(x1)\Phi(x2)\phi(x3)\Phi(x4)\Phi(x5)\Phi(x6))##, in terms of their respective Feynman propagators. The original poster is exploring the concept of contractions between different fields, ##\phi## and ##\Phi##, and how to express these contractions in terms of the given propagators.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster questions whether it is possible to contract fields of different types and seeks clarification on how to express such contractions in terms of the Feynman propagators. Some participants affirm that contractions between different fields cannot occur, while others suggest revisiting the foundational concepts of contractions.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of contractions and their implications in quantum field theory. There is a recognition of the original poster's confusion regarding the treatment of different fields, and some guidance has been offered to revisit the basic principles of field contractions.

Contextual Notes

Participants are navigating the complexities of quantum field theory, particularly the implications of commutation relations between different fields, which may influence the understanding of contractions.

binbagsss
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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
 
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binbagsss said:
Can you contract over two different fields?
No.
 
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Orodruin said:
No.

many thanks.

a possible explanation or a point towards one?
 
I can't type much (on my phone). Go back to how contractions were constructed, i.e., what does a contraction entail?
 
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Orodruin said:
I can't type much (on my phone). Go back to how contractions were constructed, i.e., what does a contraction entail?

oh right, no worries at all, thanks for your help,
so i looked at a derivation of time ordering of a field by splitting a field ##\phi## into its annihilation and creation operator components, and then get normal ordered + commutator terms, the commutator terms being the contraction...oh so commutators of different fields is zero- is this why?
 
binbagsss said:
oh so commutators of different fields is zero- is this why?
Right.
 

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