Wick theorem proof 4 point correlator 2 different fields

[binbagsss]

1. Homework Statement

Attached:

Homework 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.

The Attempt at a Solution

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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?

Many thanks in advance !

 

[mark726]

Hi there,

Thank you for your post and for sharing your attempt at finding a solution to the problem. After reviewing your work, I believe there may be a few errors in your approach.

First, it seems like you are using the notation for annihilation and creation operators in a different way than what is typically used in quantum field theory. In most cases, the annihilation operator is denoted as "a" and the creation operator as "a†". Also, the notation for the normal ordered product is usually denoted as " : : " instead of just " : ". This may not affect the final result, but it can cause confusion when trying to follow your calculations.

Second, in your attempt at simplifying the expression, you have only considered terms in the first and second brackets that have either a creation or annihilation operator on the left or right side, respectively. However, there are other terms that should be considered, such as those with two creation or two annihilation operators. These terms may not result in a fully contracted term, but they should still be included in your calculations.

Lastly, I believe there may be some mistakes in your final steps. For example, in the line where you have $= : : +0+0+0 + \phi_1^a \phi_3^c [\Phi_2^a,\Phi_4^c] + \phi_1^a[\Phi_2^a, \phi_3^c] \Phi_4^c$, it seems like you have used the commutator relation for different fields incorrectly. The commutator of two different fields should be zero, so this term should not appear in your final result.

I hope this helps you in finding the correct solution to the problem. Keep in mind that there may be other approaches or techniques that can be used to solve this problem, so don't get discouraged if your initial attempt did not yield the desired result. Keep exploring and trying different approaches, and don't hesitate to seek help from your peers or instructors if you get stuck.

Best of luck with your work!