Step on Weinberg's QM Book (pp. 154-155)

1. Jan 20, 2014

carlosbgois

Hello all!

On the problem of taking elements of different (degenerated-)state vectors that do not vanish on the perturbation matrix, Weinberg uses the following approach, when dealing with the Zeeman effect:

In this way, he goes from the first to the second equation shown as attachments.

My main source of confusion arises by the fact that I can't see how can you align a (single-direction?) B vector with all the three (supposedly not parallel) axis of the coordinate system used. But I am surely completely missing the point in here.

Can someone help me?

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2. Jan 20, 2014

vanhees71

Weinberg just puts the $z$-axis in direction of the magnetic field and chooses the joint eigenbasis of the undisturbed hamiltonian, $\vec{L}^2$, and $l_z$.

Note that due to the rotationinvariance of the undisturbed hamiltonian you are allowed to choose any direction as the $z$-axis. For the perturbation it's just convenient to take it in the direction of $\vec{B}$.

Last edited: Jan 20, 2014