A confusion from David tong's notes on QFT

  • Context: Graduate 
  • Thread starter Thread starter kof9595995
  • Start date Start date
  • Tags Tags
    Confusion Notes Qft
Click For Summary
SUMMARY

The discussion centers on the application of Wick's theorem in quantum field theory (QFT), specifically regarding the insertion of the vacuum state |0><0| in the derivation of equation 3.48 from David Tong's notes. Participants clarify that the expression :ψ₁†ψ₁ψ₂†ψ₂: is already normal ordered, allowing for the insertion of the vacuum state without violating the normal ordering rules. The reasoning hinges on the fact that annihilation operators acting on the right yield only the vacuum state, justifying the factorization in the equation.

PREREQUISITES
  • Understanding of Wick's theorem in quantum field theory
  • Familiarity with normal ordering of operators
  • Knowledge of creation and annihilation operators
  • Basic concepts of quantum states, particularly the vacuum state |0>
NEXT STEPS
  • Study the derivation of Wick's theorem in detail
  • Explore normal ordering techniques in quantum field theory
  • Review the properties of creation and annihilation operators
  • Examine examples of vacuum state contributions in QFT calculations
USEFUL FOR

Students and researchers in theoretical physics, particularly those focusing on quantum field theory and operator algebra, will benefit from this discussion.

kof9595995
Messages
676
Reaction score
2
Physics news on Phys.org
He didn't just insert |0><0|. He's applying Wick's theorem, and his contractions correlations in the |0> state.
 
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|.
 
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 : \psi_1^\dagger \psi_1 \psi_2^\dagger \psi_2 : is normal ordered. So in any terms in the creation/annihilation-operator 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 anti-particles, the only contribution from this comes from terms where there are two particle annihilation operators (no anti-particle annihilation operators). This comes only from the \psi, not the \psi^\dagger.

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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
988
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K