- #1

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## Homework Statement

Question: http://imgur.com/YzexPKl

**(only part c)**

## Homework Equations

From part a) I got |psi> = (1/sqrt(2))(e^(iwt/2)|z-up> + ie^(-iwt/2)|z-down>

I expanded this wave function in the x-spin basis and got,

|psi> = (1/(2))[(e^(iwt/2)+ie^(-iwt/2))|x-up> + (e^(iwt/2)-ie^(-iwt/2))|x-down>

## The Attempt at a Solution

I know the probability of |psi> being in some sate |q> is P = |<q|psi>|^2 but how does this change after the wavefunction has already collapsed once? Since we're making a measurement should it be P = |<x-up|Sx|psi>|^2? where Sx is the x-spin operator? I just don't get that if that's the case, since we know the first measurement is |x-up> we'd get Sx|psi> = Sx|x-up>, then |<x-up|Sx|psi>|^2 = |<x-up|Sx|x-up>|^2 =|sx|^2 where sx is the Sx operator's eigenvalue?