Quantum version of Larmor precession

  • #1
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Homework Statement

gawdbdm.png


Homework Equations


I didn't get what this actually means.

The Attempt at a Solution


DtD4BYE.png

DpSH9Y6.png

I'm not sure whether it is correct. Could you take a look?
Regards!
 

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Answers and Replies

  • #2
TSny
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Hello. I haven't gone through all of the details of your calculations. But, I don't see where you have taken into account the evolution of the system between ##t = 0## and ##t = T##. It appears to me that you are assuming that the system is in state ##\chi_+^{(x)}## at time ##t = T##.

upload_2018-6-21_14-3-53.png
 

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  • #3
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Do you see any mistakes at the final answer? And yes, I'm assuming that the system in state ##\chi_+^{(x)}## at time ##t=T##. Staying tuned for your sincerely reply!
 
  • #4
TSny
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Maybe we are interpreting the statement of the problem differently. The way I read it, at time ##t = 0## the system is in state ##|\psi(0) \rangle = \chi_+^{(x)}##. It evolves for time ##T## with ##\mathbf B## in the z direction. Then, in the time interval ##T \leq t \leq 2T## it evolves with ##\mathbf B## in the y direction. So, you will need to find ##|\psi(T) \rangle## before you can determine the state for ##t > T##.

Your final result for the probability appears to be a complex number. But probabilities are real numbers. Also, shouldn't the answer depend on the magnetic field strength B?
 
  • #5
Chandra Prayaga
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It looks like this post has appeared twice.
Do you see any mistakes at the final answer? And yes, I'm assuming that the system in state at time ##t=T##. Staying tuned for your sincerely reply!
The systaem will not be in state ##\chi_+^{(x)}## at t = T. As TSny points out in the above post, the evolution between t = 0 and t = T is dictated by the magnetic field only in the z direction. You should calculate |ψ(T)> from that, and then let that evolve from t = T to t = 2T with the magnetic field only in the y direction.
 
  • #6
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I found the probability as ##\dfrac{B^2}{2}## is that correct?
 
  • #7
vela
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I found the probability as ##\dfrac{B^2}{2}## is that correct?
That answer can't possibly be correct. Probabilities are unitless.
 
  • #8
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Finally, I've found the probability as ##0## from ##P = \dfrac{1}{4}exp \biggr( \dfrac{-2t\mu BB}{\hbar}\biggr ) \biggr | (1-1) \biggr |^2 = 0## Does that seem correct now? Thanks in advance.
 
  • #9
Chandra Prayaga
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What steps did you follow to arrive at that answer?
 
  • #10
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Is that correct? Or what would be the correct probability we're looking for? I'm receiving too many different answers.
 
  • #11
vela
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No. The argument of the exponential should be unitless.
 

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