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I have the following expectation value in a free theory for $N$ real massless scalar fields $ \phi^i $:

$$ \langle \phi^i\phi^i(x)\phi^j\phi^j(y)\rangle = 2N[G(x-y)]^2, \tag{2.11}$$

cf. arXiv:1404.1094. Can you explain me this equality? Using Wick's theorem I am not able to recover it and also this notation is not that clear to me.

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    $\begingroup$ are you missing some :...: for Wick ordering? $\endgroup$ – Abdelmalek Abdesselam Nov 9 '16 at 14:38
  • $\begingroup$ ehm, actually no, that's how the equality appears in the paper $\endgroup$ – cek Nov 9 '16 at 14:40
  • $\begingroup$ More than saying that the expectation value must depend on the difference $x-y$, it doesn't seem there's much going on. $\endgroup$ – gented Nov 9 '16 at 17:08
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    $\begingroup$ @GennaroTedesco - What he's missing in the question is that $G(x-y)$ is the free scalar propagator. $\endgroup$ – Prahar Nov 10 '16 at 19:02
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What is meant in Eq. (2.11) of the paper you mentioned is indeed for Wick ordered quantities. The formula computes $$ A=\sum_{i=1}^{N}\sum_{j=1}^{N}\langle \ :\phi_i(x)\phi_i(x):\ :\phi_j(y)\phi_j(y):\ \rangle_0\ . $$ I lowered the indices so there is no confusion with exponents. In principle for four fields there are 3 Wick contraction schemes but the colons forbid contractions within them. The two allowed contraction schemes give the same contribution, so one has $$ A=\sum_{i=1}^{N}\sum_{j=1}^{N}\ 2\ \langle \ \phi_i(x)\phi_j(y)\ \rangle_0 \langle\ \phi_i(x)\phi_j(y)\ \rangle_0\ . $$ Finally one has $$ \langle\ \phi_i(x)\phi_j(y)\ \rangle_0=\delta_{ij} G(x-y) $$ and summing over the indices gives the desired result $$ A=2N\ G(x-y)^2\ . $$

Note that the quadratic local fields involved, e.g $\phi^i\phi^i(x)$, are composite fields which need to be renormalized for $A$ to even make sense. In the free theory, renormalizing these local operators simply means putting the colons. The authors did not bother saying this explicitly because it's well known to their intended readership.

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