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

Question attached:

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  1. Relevant equations

Using the result from two fields that

$ T(\phi(x) \phi(y))= : \phi(x) \phi(y) : + G(x-y)$

Where $G(x-y) = [\phi(x)^+,\phi(y)^-] $

$ : $ denotes normal ordered

and $\phi(x)^+ $ is the annihilation operator part , and $ \phi(x)^- $ is the creation operator part.

  1. The attempt at a solution

Assume non-trivially that $ z^0 > x^0 > y^0 $

Then $ T(\phi(z),\phi(x),\phi(y)) = \phi(z) T(\phi(x) \phi(y)) $

$=(\phi(z)^+ + \phi(z)^-) T (\phi(x),\phi(y)) $

Since $\phi(z)^-$ is already normal ordered, look at the term multiplied by $\phi(z)^+$:

$=\phi(z)^+G(x-y) + \phi(z)^+:\phi(x)\phi(y): $ (1)

The term to be concerned with from

$\phi(z)^+:\phi(x)\phi(y):$ is $\phi(z)^+\phi(x)^-\phi(y)^-=\phi(x)^-\phi(z)^+\phi(y)^- +[\phi(z)^+,\phi(x)^-]\phi(y)^-= \phi(x)^-\phi(y)^-\phi(z)^+ +\phi(x)^-[\phi(z)^+,\phi(y)^-] + [ \phi(z)^+,\phi(x)^-]\phi(y)^-$

So putting this with (1) I have

$ T(\phi(z),\phi(x),\phi(y)) = : \phi(z) (\phi(x) \phi(y)): + [ \phi(z)^+,\phi(x)^-]\phi(y)^- +\phi(x)^-[\phi(z)^+,\phi(y)^-] +\phi(z)(G(x-y))= : \phi(z) (\phi(x) \phi(y)): + G(z-x) \phi(y)^- +\phi(x)^-G(z-y) +\phi(z)^+(G(x-y)) $

So compared to the model solution : - I should have a factor of both the creation and annihilation parts of the relevant field multiplying the relevant propagator? - and the fields should be on the RHS of the propagator rather than the LHS?

I'm not sure what I have done wrong.

  • $\begingroup$ Link to source? $\endgroup$ – Qmechanic Jul 23 '18 at 17:11

Your mistake comes from saying that $\phi(z):\phi(x)\phi(y):=\phi(z)\phi^-(x)\phi^-(y)$.
If we use the slightly easier notation $\phi(x_i):=\phi_i=(\phi_i^++\phi_i^-)$, with $x=x_1,\ y=x_2,\ z=x_3$, the above is actually given by: $$\phi_3:\phi_1\phi_2:=\phi_3:\left( \phi_1^-\phi_2^-+\phi_1^+\phi_2^++\phi_1^-\phi_2^++\phi_1^+\phi_2^- \right):\\=\phi_3\left( \phi_1^-\phi_2^-+\phi_1^+\phi_2^++\phi_2^+\phi_1^-+\phi_1^+\phi_2^- \right)$$ so that all the annihilation operators are pushed to the right.
If you now write $\phi_3=\phi_3^-+\phi_3^+$ and follow the same steps that you mention in you question, you should reach the correct answer.


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