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Let's have Wick's theorem in following form: for fields $$ A_{i}(x) ~=~ A_{i}^{+}(x) + A_{i}^{-}(x), $$ where first summand contains creation operator and the second contains destruction one, is a fair to say $$ \tag{1} T(A_{1}...A_{n}) ~=~ \sum_{k}(-1)^{\sigma}\overline{A_{i_{1}}A_{i_{2}}}~\overline{A_{i_{3}}A_{i_{4}}}\ldots \overline{A_{i_{2k - 1}}A_{i_{2k}}}: A_{i_{2k+1}}...A_{i_{n}}:, $$ where contractions are denoted with an overline $$ \overline{A_{i_{1}}(x)A_{i_{2}}(y)} ~=~ \langle | T(A_{i_{1}}(x)A_{i_{2}}(y))|\rangle ~=~ -iD^{c}(x - y), $$

and normal order is

$$ : A_{i_{k}}(x)A_{i_{k + 1}}(y): ~=~ A^{+}_{i_{k}}(x)A^{+}_{i_{k + 1}}(y) + A^{+}_{i_{k}}(x)A^{-}_{i_{k + 1}}(y) + A^{-}_{i_{k}}(x)A^{-}_{i_{k + 1}}(y) + A^{+}_{i_{k + 1}}(x)A^{-}_{i_{k}}(y). $$

In the beginning of proof of this theorem one says that permutation of fields in the left and right side of $(1)$ doesn't change it. Why this statement is correct?

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Andrew McAddams
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1 Answers1

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OP wrote (v1):

In the beginning of proof of this theorem one says that permutation of fields in the left and right side of (1) doesn't change it.

Answer: This is essentially due to the fact that operator ordering prescriptions (such as e.g. time ordering $T$ or normal ordering $::$) are (graded) symmetric $$ T(\hat{A}_{\pi(i_1)}\ldots\hat{A}_{\pi(i_n)}) ~=~\pm T(\hat{A}_{i_1}\ldots\hat{A}_{i_n} ),$$ $$ :\hat{A}_{\pi(i_1)}\ldots\hat{A}_{\pi(i_n)}: ~=~\pm :\hat{A}_{i_1}\ldots\hat{A}_{i_n} :,$$ where $\pi\in S_n$ is a permutation of $n$ elements. See also this Phys.SE answer. So more generally, the statement reads:

Permutation of fields in the left and right side of (1) doesn't change it, up to possibly an overall sign on both sides in the case of Grassmann-odd operators.

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