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I'm trying to derive the relation:

$\phi(x)\phi(y)=:\phi(x)\phi(y):+\langle 0|\phi(x)\phi(y)|0 \rangle$

but struggling to see the first few steps I need to make. I've made the substitutions

$\phi(x)=\phi^+(x)+\phi^-(x)$

and the same for $\phi(y)$, but it hasn't got me very far.

I'm then unsure what the it means when it says to compute

$\langle 0|\phi(x)\phi(y)|\bf{k_1},\bf{k_2} \rangle$.

Any help would be great!

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1 Answer 1

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It might be a good idea to start with the Fourier expansion of the fields:

$$\phi(x)=\int \frac{d^3k}{\sqrt{2\omega}(2\pi)^{3/2}}a^\dagger e^{ikx}+ae^{-ikx}$$

Normal ordering then means putting every $a^ \dagger$ to the left of any $a$ by which it is multiplied. The identity is then fairly trivial because $\langle 0|a^\dagger=a|0\rangle=0$ so we see that $\langle 0|\phi(x)\phi(y)| 0\rangle$ is going to contain exactly everything but the normal-ordered terms.

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  • $\begingroup$ Is there not a more elegant way just using $\phi^+$ and $\phi^-$ and the commutation relations? I understand what you're saying, but I didn't think I need to resort to explicitly writing that expression in my calculations $\endgroup$
    – Phibert
    May 4, 2014 at 8:57
  • $\begingroup$ @user13223423 It's really just two lines... Plus, at least for me, it is good to try and get into the habit of writing many things out explicitly. $\endgroup$
    – Danu
    May 4, 2014 at 9:00
  • $\begingroup$ Yes I'll still give the right answer so I'll go with that. But how can the LHS not have anything to with the vacuum but the RHS have terms acting on the vacuum? $\endgroup$
    – Phibert
    May 4, 2014 at 9:09
  • $\begingroup$ @user13223423 Taking the vacuum expectation value simply 'kills' some of the terms. By the way, if you found this answer useful, please consider accepting it. $\endgroup$
    – Danu
    May 4, 2014 at 10:09
  • $\begingroup$ It's very helpful; but doesn't address the second half of my question $\endgroup$
    – Phibert
    May 4, 2014 at 10:31

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