
#1
Jan2911, 07:21 AM

P: 679

Notes: http://www.damtp.cam.ac.uk/user/tong/qft/three.pdf
On page 59, in deriving 3.48, from the first step to the second step, why can we just insert 0><0 like that? 



#2
Jan2911, 05:49 PM

P: 194

He didn't just insert 0><0. He's applying Wick's theorem, and his contractions correlations in the 0> state.




#3
Jan3011, 08:48 AM

P: 679

It's already normal ordered, so I don't think he applied Wick's theorem. And I still don't see why we can insert 0><0.




#4
Jan3011, 01:35 PM

P: 640

A confusion from David tong's notes on QFT
I just took a quick look and I'm not sure about his conventions, but from the conventions I'm used to here is my reasoning:
The expression [tex] : \psi_1^\dagger \psi_1 \psi_2^\dagger \psi_2 : [/tex] is normal ordered. So in any terms in the creation/annihilationoperator expansion of this, all annihilation operators act on the state to the right before any creation operators do. Since the incoming state on the right contains two particles and no antiparticles, the only contribution from this comes from terms where there are two particle annihilation operators (no antiparticle annihilation operators). This comes only from the [tex] \psi [/tex], not the [tex] \psi^\dagger[/tex]. The only possible result from the particle annihilation operators acting on the right is the vacuum state. Therefore he can factorize as he does, and put in 0><0 in the middle. 


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