# Peskin & Schroeder's proof of Wick's Theorem

1. Oct 19, 2005

### emob2p

Hi,
I am stuck on a step Peskin & Schroeder give in their proof of Wick's Theorem (Intro to Quantum Field Theory, p 90). In the middle of the page when they consider the term with no contraction, it seems like in between the 1st and 2nd lines they somehow factor out the normal ordering operator. How is this legal? I've attached a bmp of what I'm talking about. Thanks

*Typo...There shouldn't be an equal sign after the last commutator in the bmp.

#### Attached Files:

• ###### ps.bmp
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2. Oct 20, 2005

### vanesch

Staff Emeritus
I think the step is far more complicated than it seems ; I think you have to work out first the commutator for each term of the kind:
$$| \phi_1^+, \phi_2^- \phi_3^-...\phi_m^+ |$$
which, by working out the distributivity of the commutator, equals:
$$| \phi_1^+, \phi_2^-| \phi_3^-... + \phi_2^-|\phi_1^+,\phi_3^-|\phi_4^- ...$$
Recognizing that the commutators are C-numbers, the remaining factors can be recognized to be in normal order, so you can go back to the N() notation, and you end up with the second term in the second line.
Sorry, I used | instead of brackets because it screwed up my inline latex
cheers,
Patrick.

Last edited: Oct 20, 2005
3. Oct 23, 2005

### emob2p

That was my thought too, except let's say $$\phi_n = \phi_n^+ + \phi_n^-$$. Then $$\phi_2\phi_3$$ won't simply be $$\phi_2^-\phi_3^- + \phi_2^+\phi_3^+$$ because you'll have the cross terms.

Last edited: Oct 24, 2005