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

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The discussion centers on the arrangement of indices in LT transformations from equations (2.4.7) to (2.4.8), specifically why the lower index mu precedes the upper index rho. Participants clarify that while the indices can theoretically be arranged differently, the specific arrangement is crucial for maintaining correct matrix multiplication. A participant attempts to rearrange indices in a calculation but finds it does not yield the correct result, reinforcing the importance of proper index placement. Additionally, there is a conversation about the familiarity with notation after extensive practice, with some still finding it necessary to think carefully about indices. Overall, the thread emphasizes the significance of index arrangement in tensor operations and the learning curve associated with mastering the notation.
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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.
 
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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).
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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