How to prove Wick's Theorem (Zee's eq. I.2 (16)) via Gaussian integration?
Answers
Answered by
0
'm working through Zee's QFT in a Nutshell but there's an integral [I.2 (16)] I couldn't quite derive. The problem is to find
⟨xixj...xkxl⟩=∫...∫dx1...dxnxixj...xkxlexp(−12x⃗ ⋅A⋅x⃗ )∫...∫dx1...dxnexp(−12x⃗ ⋅A⋅x⃗ )
using
∫...∫dx1...dxnexp(−12x⃗ ⋅A⋅x⃗ +J⃗ ⋅x⃗ )=((2π)ndet[A])−−−−−−−−√exp(12J⃗ ⋅A−1⋅J⃗ )
with A a real symmetric matrix. Zee suggests taking repeated derivatives with respect to J⃗ and then letting J⃗ →0. Here's my attempt at a solution:
The top of the numerator in the first equation can be obtained by differentiating the the above equation by ∂∂Ja for a=i,j,...,k,l. Dropping the constant for now, we find that:
∫...∫dx1...dxnxiexp(−12x⃗ ⋅A⋅x⃗ +J⃗ ⋅x⃗ )=(∑mnA−1imJm)exp(12J⃗ ⋅A−1⋅J⃗ ).
This process of taking derivatives can be repeated via the product rule to get messy stuff, but what worries me is that when one takes J⃗ →0 the integral will vanish, will it not? Because the sums over m that pop out will all vanish. The answer, Zee says, is not zero, but instead
⟨xixj...xkxl⟩=∑Wick(A−1ab)...(A−1cd)
"where the set of indices {a,b,...,c,d} represent a permutation of the set of indices {i,j,...,k,l}. The sum is over all such permutations or Wick contractions." How exactly does one reach this result?
Hope this helps you ✌️ ✌️ ✌️.....
Pls mark as brainliest answer.......
If it helps you.....
⟨xixj...xkxl⟩=∫...∫dx1...dxnxixj...xkxlexp(−12x⃗ ⋅A⋅x⃗ )∫...∫dx1...dxnexp(−12x⃗ ⋅A⋅x⃗ )
using
∫...∫dx1...dxnexp(−12x⃗ ⋅A⋅x⃗ +J⃗ ⋅x⃗ )=((2π)ndet[A])−−−−−−−−√exp(12J⃗ ⋅A−1⋅J⃗ )
with A a real symmetric matrix. Zee suggests taking repeated derivatives with respect to J⃗ and then letting J⃗ →0. Here's my attempt at a solution:
The top of the numerator in the first equation can be obtained by differentiating the the above equation by ∂∂Ja for a=i,j,...,k,l. Dropping the constant for now, we find that:
∫...∫dx1...dxnxiexp(−12x⃗ ⋅A⋅x⃗ +J⃗ ⋅x⃗ )=(∑mnA−1imJm)exp(12J⃗ ⋅A−1⋅J⃗ ).
This process of taking derivatives can be repeated via the product rule to get messy stuff, but what worries me is that when one takes J⃗ →0 the integral will vanish, will it not? Because the sums over m that pop out will all vanish. The answer, Zee says, is not zero, but instead
⟨xixj...xkxl⟩=∑Wick(A−1ab)...(A−1cd)
"where the set of indices {a,b,...,c,d} represent a permutation of the set of indices {i,j,...,k,l}. The sum is over all such permutations or Wick contractions." How exactly does one reach this result?
Hope this helps you ✌️ ✌️ ✌️.....
Pls mark as brainliest answer.......
If it helps you.....
Similar questions