# Homework Help: QFT Wicks theorem contraction -- different fields terms of propagation

1. Nov 30, 2016

### binbagsss

1. The problem statement, all variables and given/known data

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 $G_F^{\Phi}(x-y) =\int \frac{d^{4}k}{(2\pi)^{4}}e^{ik(x-y)} \frac{ih}{-k.k - M^2 -i\epsilon}$

2. Relevant equations

$<0|T(\phi(x1)\Phi(x2)\phi(x3)\phi(x4)\phi(x5)\phi(x6)) |0> = : non-fully contracted terms : + fully contracted terms = fully contracted terms$

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.

3. 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

2. Nov 30, 2016

### Orodruin

Staff Emeritus
No.

3. Nov 30, 2016

### binbagsss

many thanks.

a possible explanation or a point towards one?

4. Nov 30, 2016

### Orodruin

Staff Emeritus
I can't type much (on my phone). Go back to how contractions were constructed, i.e., what does a contraction entail?

5. Nov 30, 2016

### binbagsss

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?

6. Nov 30, 2016

### Orodruin

Staff Emeritus
Right.