# Wick theorem proof 4 point correlator 2 different fields

1. Jul 21, 2017

### binbagsss

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

Attached:

2. Relevant equations

I've just changed the notation a tad to make things quicker for me :

$\phi_1=\phi_1(x_1)$ and $\Phi_2=\phi_2(x_2)$

: denotes normal product. i.e annhilator operators are on the RHS, so acting on a vacuum state will give zero.

I can split a field $\phi_1 = \phi_1^a + \phi_1^c$ , where a denotes the annihilating component of the field and c the creation operator component of the field.

$[\phi(x),\Phi(y)]=0$ , different fields commute, important here.

3. The attempt at a solution

I expect the result $T( \phi_1 \Phi_2 \phi_3 \Phi_4 ) = : \phi_1 \Phi_2 \phi_3 \Phi_4 : + [\phi_1^a,\phi_3^{c}] [\Phi_2^a,\Phi_4^{c}]$

The last term being fully contracted, as if any field is not involved in a contraction it will be zero when hitting a vacuum state.

If I assume that $x_1^0 > x_2^0>x_3^0>x_4^0$ then:
$T( \phi_1 \Phi_2 \phi_3 \Phi_4 ) = (\phi_1^a+\phi_1^c)(\Phi_2^a+\Phi_1^2)(\phi_3^a+\phi_3^c)(\Phi_4^a+\Phi_4^c) = (\phi_1^a\Phi_2^a+\phi_1^a\Phi_2^c +\phi_1^c\Phi_2^a +\phi_1^c\Phi_2^c) ((\phi_3^a+\phi_3^c)(\Phi_4^a+\Phi_4^c))= (\phi_1^a\Phi_2^a+\phi_1^a\Phi_2^c +\phi_1^c\Phi_2^a +\phi_1^c\Phi_2^c) (\phi_3^a\Phi_4^a+\phi_3^a\Phi_4^c +\phi_3^c\Phi_4^a +\phi_3^c\Phi_4^c)$
Now I will simplify this first expression by noting that anything in the first brackets with creation component on the left hand side will result in a normal ordered term as will anything in the second bracket with a annihilation component on the right hand side, so the only terms I need to consider are:

$= ( \phi_1^a\Phi_2^a +\phi_1^a\Phi_2^c ) (\phi_3^a\Phi_4^c +\phi_3^c\Phi_4^c) = \phi_1^a\Phi_2^a \phi_3^c\Phi_4^c + \phi_1^a\Phi_2^c \phi_3^a\Phi_4^c + \phi_1^a\Phi_2^a \phi_3^a\Phi_4^c + \phi_1^a\Phi_2^c\phi_3^c\Phi_4^c$
Where here I expect since different fields commute I expect the only term that will not be normal ordered or zero to be coming from the $\phi_1^a\Phi_2^a \phi_3^c\Phi_4^c$ after two iterations, since commutator of different fields is zero.
$= : : + \phi_1^a\Phi_2^c [\phi_3^a,\Phi_4^c ] + \phi_1^a\Phi_2^a [\phi_3^a,\Phi_4^c] + [\phi_1^a,\Phi_2^c]\phi_3^c\Phi_4^c +$
$= : : +0+0+0 + \phi_1^a \phi_3^c\Phi_2^a\Phi_4^c+ \phi_1^a[\Phi_2^a, \phi_3^c] \Phi_4^c$
$= = : : +0+0+0 + : : \phi_1^a \phi_3^c [\Phi_2^a,\Phi_4^c] + \phi_1^a[\Phi_2^a, \phi_3^c] \Phi_4^c$

QUESTION
So I have one commutator that is non-zero $[\Phi_2^a,\Phi_4^c]$, however there is no fully contracted term since $\phi_1^a , \phi_3^c$ are not contracted. Where has my proof gone wrong to not reveal a fully contracted term?

2. Jul 26, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Jan 6, 2018

### binbagsss

i've never met anyone quite as courteous as you bot <3.