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Taking the example of two bosonic fields, Wick's theorem is \begin{equation} T\{\phi(x_1)\phi^\dagger(x_2)\} = N\{\phi\phi^\dagger\} + N\{(\phi\phi^\dagger)_c\} \end{equation} where the subscript $c$ denotes a contraction of the fields. Since $(\phi\phi^\dagger)_c$ is a commutator (more specifically propagator), this reduces to \begin{equation} T\{\phi(x_1)\phi^\dagger(x_2)\} = N\{\phi\phi^\dagger\} + (\phi\phi^\dagger)_c. \end{equation} My question is that since we know normal ordering of a commutator is zero, why don't we have $$T\{\phi(x_1)\phi^\dagger(x_2)\} = N\{\phi\phi^\dagger\}$$ only instead of the above? In QED, by expanding the $S$ operator, we also get many terms involving normal ordering of contractions between fields (and other uncontracted fields) but these are not zero. Why? Is the normal ordering of a commutator and contraction different from each other? What am I missing?

Qmechanic
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Kaitou1412
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  • Related: https://physics.stackexchange.com/q/323801/2451 , https://physics.stackexchange.com/q/18078/2451 and links therein. – Qmechanic May 15 '18 at 11:28
  • Hi I have checked that answer but it does not address the normal ordering of a contraction which I see is usually factored out of the $N$ operator. This is what I don't understand. – Kaitou1412 May 15 '18 at 11:41

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