# Why is Wick contraction a $c$-number?

It is mentioned in Fetter's Quantum Theory of Many-Particle Systems (in contraction part of section 8 Wick's Theorem), that:

contractions are c numbers in the occupation-number Hilbert space, not operators.

(c-numbers are just complex numbers, right?)

I am confused because Fetter defines contractions as (time ordered operator) - (normal ordered operator). How does subtraction of two operator become a number? Is it implied that we surround it with $<\Psi_0|...|\Psi_0>$?

Update 2: See the links in the answer by Qmechanic for a further intuition on defining Wick contractions.

• The difference between the ordering of the operators can only give rise to zero or something proportional to the identity as we're dealing with creation and/or annihilation operators. – Count Iblis May 30 '16 at 5:09
• @CountIblis: That's an answer, you should add it as one. – ACuriousMind May 30 '16 at 10:14

1. The main fields assumption that goes into the proof of the Wick's theorem for fields $\hat{\phi}^i\in{\cal A}$ is that their (super)commutators $$[\hat{\phi}^i,\hat{\phi}^j]~\in~ Z({\cal A}) \tag{1}$$ are central elements of the operator algebra ${\cal A}$, cf. e.g. this Phys.SE post.
2. For free fields $\hat{\phi}^i\in{\cal A}$, their (super)commutators $$[\hat{\phi}^i,\hat{\phi}^j]~= (c~{\rm number}) \times \hat{\bf 1} \tag{2}$$ are proportional to the identity operator $\hat{\bf 1}$. Under mild assumptions one may prove that the contractions $$\tag{1} \hat{C}^{ij}~=~T(\hat{\phi}^i\hat{\phi}^j)~-~:\hat{\phi}^i\hat{\phi}^j: ~=~c^{ij}~ \hat{\bf 1}$$ are $c$-numbers times the identity operator $\hat{\bf 1}$, cf. e.g. this Phys.SE post. Be aware that in the physics literature, the word contraction sometimes refers to the operator $\hat{C}^{ij}$ and sometimes it refers to the $c$-number $c^{ij}$.