How to Compute the Little Group Matrix in Weinberg's QFT Trilogy?

  • Thread starter Thread starter merius
  • Start date Start date
  • Tags Tags
    Qft Text
merius
Messages
1
Reaction score
0
Hi. I am new to this discussion group, so forgive me if this is not the appropriate forum for this question. A friend of mine and I are going through the first volume of Weinberg's QFT trilogy. My question regards the first problem at the end of chapter two. I think I completely understand the procedure for computing a Lorentz transformed positive definite spin state, but I can't seem to compute the little group element corresponding to the Lorentz transformation given in this problem. The matrix multiplication is very lengthy and I really get caught up in the algebra. Is there an easier way to compute the little group matrix than his formula (W = L^(-1)(lambda*p)*Lambda*L(p))? I would like to hear from someone who has done this problem already and, if possible to see their solution. The little group for mass positive definite spin states is the 3-d rotation group and I know the rotation for this particular problem is about the x1 axis, but the algebra is ridiculous. I have spent a lot of time on this problem and would greatly appreciate some help. Thanks.
Merius
 
Physics news on Phys.org
Originally posted by merius
I can't seem to compute the little group element corresponding to the Lorentz transformation given in this problem. The matrix multiplication is very lengthy and I really get caught up in the algebra. Is there an easier way to compute the little group matrix than his formula (W = L^(-1)(lambda*p)*Lambda*L(p))?

It's not so bad. Since Λ is a boost along the z-axis, it's only non-vanishing elements are

Λ11 = Λ22
Λ33 = Λ00
Λ03 = Λ30.

Since p is in the y-direction, the only non-vanishing elements of L(p) are

L11 = L33
L22
L02 = L20
L00.

One way to start off is by writing the little group element as

W(Λ,p) = L-1(Λp)Q(p)

where Q(p) ≡ ΛL(p) and note that the above immediately yields

Q1ν = Λ11L1ν
Q2ν = Λ22L2ν
Q0ν = Λ00L0ν + Λ03L3ν
Q3ν = Λ30L0ν + Λ33L3ν.

Noting that Λ11 = Λ22 = L11 = L33 = 1 simplifies things even further. Finally, L-1(Λp) = L(p') in which p'i = -Λiνpν.
 


Hi Merius, welcome to the discussion group! The first volume of Weinberg's QFT trilogy can be quite challenging, so it's great that you have a friend to work through it with. As for your question about the little group element, I understand how the algebra can get overwhelming. I would suggest trying to break down the matrix multiplication into smaller steps and using properties of matrix multiplication to simplify the calculations. Another helpful resource would be to look for online forums or communities specifically for QFT students where you can ask for guidance from others who have also worked through Weinberg's problems. I'm sure there are many people who have encountered the same difficulties and would be willing to share their solutions with you. Best of luck with your studies!
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top