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Time dependent expectation value problems

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-3-21_20-44-19.png

    2. Relevant equations


    3. The attempt at a solution
    I tried to solve (a), but i don't know which approach is right ((1) or (2)) and how to solve (b).

    upload_2017-3-21_20-57-18.png
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2017 #2

    DrClaude

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    You don't start the problem correctly. The ##|S_z = + \hbar/2 \rangle## state is not an eigenstate of the Hamiltonian, so after a time ##T_2##, the state of the system is not ##e^{-i E T_2 / \hbar} |S_z = + \hbar/2 \rangle##.
     
  4. Mar 21, 2017 #3
    --------------------------------------------------------------------------------------------------------------------------
    H=-(e/mc)S * B , Sz and H differ just by a multiplicative constant, so they commute. The Sz eigenstates are also energy eigenstates.
    (* I tried to solve it Sz not Sx by mistake)
     
    Last edited: Mar 21, 2017
  5. Mar 21, 2017 #4

    DrClaude

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    In which direction is the magnetic field?
     
  6. Mar 21, 2017 #5
    In the problem, megnetic field is Bx direction.
     
  7. Mar 21, 2017 #6

    DrClaude

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    So how can you rewrite ##\mathbf{S} \cdot \mathbf{B}##?
     
  8. Mar 22, 2017 #7
    Sx * B
     
  9. Mar 22, 2017 #8

    BvU

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    So no commuting with ##{\bf S}_z## !
     
  10. Mar 22, 2017 #9
    Then how can i solve this problem. i would appreciate it if you can help me
     
  11. Mar 22, 2017 #10

    DrClaude

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    You have to express the initial state in terms of the eigenstates of the Hamiltonian.
     
  12. Mar 22, 2017 #11
    upload_2017-3-22_22-9-13.png

    I think it is not. Sx and H have same eigenstates, so the initial state l+> is lSx+> + lSx->
     
  13. Mar 22, 2017 #12

    DrClaude

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    Don't forget the normalization factor.

    But now that the state is expressed in terms of the eigenstates of the Hamiltonian, it should be easy to calculate the time evolution.
     
  14. Mar 22, 2017 #13
    bpGR.dn.gif
    I think the answer is weird. These are the probability for Sz + , and 1-cos(T2/h)(E1-E2) for Sz -
    So, the expectation value h/2(1+cos(T2/h)(E1-E2)) and -h/2(1-cos(T2/h)(E1-E2)) respectively
     
    Last edited: Mar 22, 2017
  15. Mar 22, 2017 #14

    DrClaude

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    Why? This is a very standard problem in QM and illustrates nicely the concept of precession.
     
  16. Mar 22, 2017 #15
    I replied the answer that i tried
     
  17. Mar 22, 2017 #16

    DrClaude

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    Ok, I had replied before you posted the solution.

    There is a problem, as you can clearly see that the prbability you get ranges from 0 to 2. Check again how you calculate the absolute value squared.
     
  18. Mar 22, 2017 #17
    I got it, then how can i approach problem(b) i think the result of first measurement is same with (a)
     
    Last edited: Mar 22, 2017
  19. Mar 23, 2017 #18

    DrClaude

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    The probability of measuring Sz = +ħ/2 is indeed given by the same equation, replacing T2 by T1. But you already know the measurement result, so you need to figure out what you get at T2.

    By the way, you can simplify the result you get using explicit values for E1 and E2.
     
  20. Mar 23, 2017 #19
    Result of first measurement at T1 is the same with (a). I should have to get a probability of that state before i get a expectation value as i did in (a).
    I think total result probability is probability(T1) * probability(T2)
    bq69.dn.gif

    Answer is quite complex, is it right??
     
  21. Mar 23, 2017 #20
    I changed it again.
    bq9e.dn.gif
     
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