1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spin time development

  1. Mar 18, 2015 #1
    1. The problem statement, all variables and given/known data
    I have a B field initially in the x-direction, and its constant:
    [itex] \widehat{H}= - (\dfrac{e}{mc})\overrightarrow{B} \dot{} \overrightarrow{S} [/itex]

    At t=0 it was prepared so that Sz has an eigenvalue of + hbar/2
    I want to show the time evolution of the initial state.

    2. Relevant equations
    [itex] \widehat{H}= - (\dfrac{e}{mc})\overrightarrow{B} \dot{} \overrightarrow{S} [/itex]
    [tex] \widehat{U} = e^{\dfrac{-i H}{\hbar}t} [/tex] Time evolution
    3. The attempt at a solution
    It would evolve as so:
    [tex] |\alpha, t> = c_{1} e^{\dfrac{-i H}{\hbar}t} |+> +c_{2} e^{\dfrac{-i H}{\hbar}t} |-> [/tex]

    But we know at t = 0 it should give us a spin up of hbar/2..
    So evaluate at t=0:

    [tex] |\alpha, 0> = c_{1} e^{\dfrac{-i H}{\hbar}0} |+> +c_{2} e^{\dfrac{-i H}{\hbar}0} |-> [/tex]
    [tex] |\alpha, 0> = c_{1} |+>+c_{2} |-> = 1|+> [/tex]
    Then that means that C1 = 1 and C2 = 0..?

    But intuitively I know that it precesses in the ZY plane since B is in the X-direction.. If I just got rid of the C2 coefficient, then i will never be able to show how it precesses into the spin down? Ahh i feel like I'm missing something here.. Any tips on what I am doing wrong?
     
  2. jcsd
  3. Mar 18, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    If ##| + \rangle## and ##| - \rangle## are eigenstates of ##S_z##, then they are not eigenstates of the Hamiltonian. This equation is therefore not correct.
     
  4. Mar 18, 2015 #3
    Then how can I express the state? I need to find the development of the state up to some arbitrary time T when the b field switches into the z direction...
     
  5. Mar 18, 2015 #4

    DrClaude

    User Avatar

    Staff: Mentor

    You have to write it in terms of the eigenstates of ##S_x##.
     
  6. Mar 18, 2015 #5
    AH. well I'm confused now.
    [tex] \vert\alpha, t \rangle = \dfrac{a}{\sqrt{2}} e^{i t\dfrac{H}{\hbar}} \vert +\rangle\pm\dfrac{b}{\sqrt{2}} e^{i t\dfrac{H}{\hbar}} \vert -\rangle [/tex]
    the summed state represents spin up in z and difference terms represent spin down?
     
  7. Mar 19, 2015 #6

    DrClaude

    User Avatar

    Staff: Mentor

    Let me introduce a notation to make things clearer:
    $$
    \hat{S}_z | \pm \rangle_z = \pm \frac{\hbar}{2} | \pm \rangle_z \\
    \hat{S}_x | \pm \rangle_x = \pm \frac{\hbar}{2} | \pm \rangle_x
    $$
    What you are writing is
    $$
    |\alpha, t \rangle = c_1 e^{-i \hat{H} t / \hbar} | + \rangle_z + c_2 e^{-i \hat{H} t / \hbar} | - \rangle_z
    $$
    The problem is that ##| + \rangle_z## and ##| - \rangle_z## are not eigenstates of ##\hat{H}## (you should show that). That means that ##e^{-i \hat{H} t / \hbar}## mixes the two spin states, and you can't treat it as a simple complex number. You also cannot easily calculate the exponential of a matrix that is not diagonal.

    You therefore need to go to a basis where ##\hat{H}## is diagonal, i.e., a basis of eigenstates of ##\hat{H}##, namely ##| + \rangle_x## and ##| - \rangle_x## (you can show that also). Then you need to take the correct initial state in this basis, look at its time evolution, and if you calculate for instance ##\langle \hat{S}_z \rangle##, you will find that it is time-dependent, hence you will indeed see a precession in the yz plane.
     
  8. Mar 21, 2015 #7
    So I started over and found the EigenVectors of the hamiltonian at t<T (Interval where B points in X-direction)..
    [tex]
    \mid\alpha\rangle = \dfrac{1}{\sqrt{2}} \mid +\rangle_{x} e^{\dfrac{\imath e B_{x} t}{2mc}}+ \dfrac{1}{\sqrt{2}} \mid -\rangle_{x} e^{\dfrac{-\imath e B_{x} t}{2mc}}\\
    Where: \\
    \mid +\rangle_{x} =\binom{1}{1}

    \mid -\rangle_{x} =\binom{1}{-1}
    [/tex]


    But I don't know what X is initially, I only know where it is in Z. When I evaluate it for t=0 I get:
    [tex]

    \mid\alpha, t=0 \rangle = \dfrac{1}{\sqrt{2}} \mid +\rangle_{x} + \dfrac{1}{\sqrt{2}} \mid -\rangle_{x}


    [/tex]
    Which gives us: //
    [tex]
    \mid\alpha, t=0 \rangle = \binom{1}{0}
    [/tex]

    Which is the spin up Z ket correct?
    So when t -> T
    My hamiltonian then changes to B in the Z direction
     
    Last edited: Mar 21, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted