Weinberg Vol 1: Understanding Index Arrangement in (2.4.8) LT Transformations

  • Thread starter Thread starter shehry1
  • Start date Start date
  • Tags Tags
    Weinberg
shehry1
Messages
39
Reaction score
0
Can anyone explain to me why in going from (2.4.7) to (2.4.8) the indices on the LT are arranged in the way they are. Why is mu the first index (lower) and rho the second (upper)?

Could they have been arranged in any other way? From the rules that I know, they can.
 
Physics news on Phys.org
Recall that the components on row \mu, column \nu of the matrices

\Lambda, \Lambda^T, \eta, \eta^{-1}, \omega

are written as

\Lambda^\mu{}_\nu, \Lambda^\nu{}_\mu, \eta_{\mu\nu}, \eta^{\mu\nu}, \omega^\mu{}_\nu

and that \eta^{-1} and \eta and used to raise and lower indices. The components of \Lambda^{-1} are

(\Lambda^{-1})^\mu{}_\nu=(\eta^{-1}\Lambda^T\eta)^\mu{}_\nu=\eta^{\mu\rho}\Lambda^\sigma{}_\rho\eta_{\sigma\nu}=\Lambda_\nu{}^\mu.

Let's use all of the above to evaluate the first term on the right-hand side of (2.4.7).

(\Lambda\omega\Lambda^{-1})_{\mu\nu}=\eta_{\mu\rho}(\Lambda\omega\Lambda^{-1})^\rho{}_\nu =\eta_{\mu\rho}\Lambda^\rho{}_\sigma\omega^\sigma{}_\lambda(\Lambda^{-1})^\lambda{}_\nu =\Lambda_{\mu\sigma}\omega^\sigma{}_\lambda\Lambda_\nu{}^\lambda=\Lambda_{\mu\rho}\delta^\rho_\kappa\omega^\kappa{}_\lambda\Lambda_\nu{}^\lambda

=\Lambda_{\mu\rho}\eta^{\rho\tau}\eta_{\tau\kappa}\omega^\kappa{}_\lambda\Lambda_\nu{}^\lambda =\Lambda_\mu{}^\tau\omega_{\tau\lambda}\Lambda_\nu{}^\lambda
 
Thanks a lot. Now I have just two short questions:

(1) In the last expression with the string of equalities, could you have expanded the bracket differently. Meaning that instead of \eta_{\mu\rho}\Lambda^\rho{}_\sigma\omega^\sigma{ }_\lambda(\Lambda^{-1})^\lambda{}_\nu , would it had been correct to put that rho on the Lambda inverse and the nu on the Lambda? I tried it and it didn't turn out correct.

(2) After posting nearly 2300 messages, does one become more natural at the notation (like adding or subtracting integers) or do you still have to think about all the indices. :)
 
shehry1 said:
(1) In the last expression with the string of equalities, could you have expanded the bracket differently. Meaning that instead of \eta_{\mu\rho}\Lambda^\rho{}_\sigma\omega^\sigma{ }_\lambda(\Lambda^{-1})^\lambda{}_\nu , would it had been correct to put that rho on the Lambda inverse and the nu on the Lambda? I tried it and it didn't turn out correct.
That doesn't work. Note that the only thing I'm using in this step is the definition of matrix multiplication:

(\Lambda\omega\Lambda^{-1})^\rho{}_\nu =\Lambda^\rho{}_\sigma\omega^\sigma{ }_\lambda(\Lambda^{-1})^\lambda{}_\nu

shehry1 said:
(2) After posting nearly 2300 messages, does one become more natural at the notation (like adding or subtracting integers) or do you still have to think about all the indices. :)
I still have to think about it. Probably took half an hour to remind myself about the things I needed to know before the actual calculation seemed trivial. Once I had written down the first equality in the last two lines (and knew why I was doing it), the rest was like adding integers. I try to avoid this notation when I can. I prefer an index free notation (e.g. \mbox{Tr}(AB) instead of A^i{}_j B^j{}_i), and my second choice is to write all the indices downstairs (Example).
 
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