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Particle's spin when subject to a constant magnetic field

  1. Jan 31, 2014 #1
    1. The problem statement, all variables and given/known data

    An alkali atom, on it's fundamental state, passes through a Stern-Gerlach apparatum, which will only transmit atoms with their spins aligned along the +z direction. After that the atoms travel, during a finite time τ, through a region of constant magnetic field [itex]\vec{B}=B\vec{e_{x}}[/itex].

    After that time τ the atoms pass through a new Stern-Gerlach apparatum, which only allows atoms with spin along -z to pass. What's the probability that they will pass?

    2. Relevant equations

    Pauli matrices
    [itex]\widehat{H}=-\vec{u_{B}}.\widehat{S}[/itex]
    [itex]u_{B}=\frac{q}{2m_{e}}[/itex]


    3. The attempt at a solution

    From the problem it's easy to see that the state of the system at the instant t=0 is:

    [itex]|\psi>(t=0)=|+>_{z}[/itex]

    Then I assumed that, while being under the influence of the constant magnetic field along the x direction, the state is:

    [itex]|\psi>=\alpha (t)|+>_{z} + \beta (t) |->_{z}[/itex]

    Next I applied the hamiltonian to my state [itex]|\psi>[/itex]. Since [itex]\vec{B} = B\vec{e_{x}}[/itex]:

    [itex]\widehat{H}|\psi>=-\frac{qB}{2m}\widehat{S_{x}}|\psi>[/itex]
    [itex]\widehat{H}|\psi>=-\frac{qB}{2m}\widehat{S_{x}}(\alpha (t)|+>_{z}+\beta (t) |->_{z})[/itex]

    Since [itex]\widehat{S_{x}}=\hbar \sigma _{x}[/itex], applying it to [itex]|\psi>[/itex] returns:
    [itex]\widehat{H}|\psi>=-\frac{qB\hbar}{4m}(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]

    Defining [itex]\omega=\frac{qB}{4m}[/itex]:

    [itex]\widehat{H}|\psi>=-w\hbar(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]

    Now, using Schrodinger's equation we get:

    [itex]\widehat{H}|\psi>=i\hbar \frac{d}{dt}|\psi>[/itex]
    [itex]\frac{d}{dt}(\alpha (t)|+>_{z} + \beta (t) |->_{z}) = iw(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]

    Separating this we get:

    [itex]\frac{d}{dt}\alpha (t) = iw\beta (t)[/itex]
    [itex]\frac{d}{dt}\beta (t) = iw\alpha (t)[/itex]

    Applying another derivative to the first differential equation we get:
    [itex]\frac{d^{2}}{dt^{2}}\alpha (t) = iw\frac{d}{dt}\beta (t)[/itex]
    [itex]\frac{d^{2}}{dt^{2}}\alpha (t) = -w^{2}\alpha[/itex]

    Doing the same to the second achieves a similar result:

    [itex]\frac{d^{2}}{dt^{2}}\beta (t) = -w^{2}\beta[/itex]

    Solving both I got:

    [itex]\alpha (t) = Ae^{iwt} + Be^{-iwt}[/itex]
    [itex]\alpha (t) = Ce^{iwt} + De^{-iwt}[/itex]

    Now, since I know that [itex]|\psi>(t=0) = |+>_{z}[/itex], I know that:
    [itex]A+B=1[/itex]
    [itex]C = -D[/itex]

    From this I can conclude that:
    [itex]\beta (t) = Fsin(wt)[/itex]

    This is as far as I can get.. I don't know what to do from here. Am I approaching the problem wrongly? Any help would be appreciated.

    Thanks.
    Daniel
     
  2. jcsd
  3. Feb 1, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, Daniel.

    First, there appears to be a problem with a factor of 2 in your expression
    I don't believe the 4 in the denominator is correct. Did you include the "g-factor" of the electron? Anyway, you might see if you can track down the error. (Hope I'm not the one in error, I always get a headache with the factors of 2 in this type of problem!)

    Later, you have


    In the second equation I think you meant to type [itex]\beta (t)[/itex] instead of [itex]\alpha (t)[/itex].

    This looks good (but your ##\omega## will need to be corrected for the factor of 2 problem mentioned above).

    But note, C and D are not independent of A and B. Remember you have

    So, you can determine [itex]\beta (t)[/itex] from [itex]\alpha (t)[/itex].

    You should be able to find A and B from the initial conditions and from the fact that you want your state to be normalized.
     
  4. Feb 2, 2014 #3
    I forgot that I was only interested in the [itex]|->_{z}[/itex] component.
    Regarding the factor of 2, I'll correct it. However my major issue has been resolved, so thanks!
     
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