Let $T\{...\}$ denote time-ordering, $N\{...\}$ normal-ordering and $\left<ab\right>$ be the propagator.

Wick's theorem states that

$$ T\{ab\} = N\{ab\} + \left<ab\right>. $$

I now apply time-ordering to both sides of this equation. Because $T$ is idempotent, $T\{T\{ab\}\}=T\{ab\}$. Also $T\{N\{ab\}\}=T\{ab\}$ because we re-order operators inside the $T\{...\}$ anyway. $\left<ab\right>$ is not an operator, so time-ordering acts on it trivially. We obtain:

$$ T\{ab\} = T\{ab\} + \left<ab\right>$$

$$ \left<ab\right> = 0.$$

This result is obviously incorrect. What am I missing?


1 Answer 1


I) It is true that operator ordering procedures are idempotent operations

$$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots). $$

But it is not true that the outermost ordering cancels the effect of the innermost ordering

$$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$

In fact, the opposite is true

$$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$

as a special case of a nested${}^1$ Wick's Theorem, cf. Section II below.

Example. If for two operators $a$ and $b$, we have the relation

$$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$

where the contraction $\langle ab\rangle$ is a $c$-number, then

$$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linearity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$

II) More generally, if we want to bring a nested expression of the form

$$ \tag{6} T\left(N(\ldots) \ldots N(\ldots)\right) $$

on normal ordered form, there is a nested${}^1$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols.

Example. In OP's case (5), this means for the lhs. that

$$ \tag{7} T(N(ab))~=~N(ab), $$

since $a$ and $b$ belong to the same normal order symbol $N(ab)$, while the rhs. is

$$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$

[Note that $N(a)=a$ and $N(b)=b$.] See also e.g. this and this Phys.SE posts.


${}^1$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.