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Find probability of spin state after certain time.

  • Thread starter baouba
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  • #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?
 

Answers and Replies

  • #2
blue_leaf77
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Since we're making a measurement should it be P = |<x-up|Sx|psi>|^2?
No, that's not a probability.
In part c, your starting state is ##|x;+\rangle## because as you have realized the previous state collapsed upon measurement to this state. Then this state is let evolve for a time interval ##\Delta t = \pi/\omega## in a Hamiltonian ##H = -\omega S_z##. At the end of that interval, you are asked to calculate the probability for the state.##|x;+\rangle##.
 

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