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Flip the spin

  1. Dec 14, 2008 #1

    KFC

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    1. The problem statement, all variables and given/known data
    Assuming a spin 1/2 is put in a magnetic filed along z direction [tex]B_z = B*cos(w_0 t)[/tex]. At the beginning, the spin is in spin-up along x direction, i.e.

    [tex]\psi(0) = \frac{1}{\sqrt{2}}\left(
    \begin{matrix}
    1 \\ 1
    \end{matrix}\right)[/tex]

    Try to find out the minimum [tex]B[/tex] such that [tex]S_x[/tex] is flip.

    2. The attempt at a solution
    First of all, I write than the Hamiltonian of the system

    [tex]H \propto
    \left(
    \begin{matrix}
    B\cos w_0t & 0\\
    0 & -B\cos w_0t
    \end{matrix}
    \right)
    [/tex]

    From that, in any time t>0, the state will evolute as [tex]\psi(t)[/tex] (I already solved that). From the time-dependent solution, I can figure out the probability to find the spin-down when measuring [tex]S_x[/tex], which is of the following form

    [tex]P = \sin^2(\gamma \sin(w_0 t))[/tex]

    where [tex]\gamma[/tex] is a constant containing B. Hence, to make the system flip, I have to let P=1, i.e.

    [tex]\gamma \sin(w_0 t)=\pi/2[/tex]

    and solve for B gives

    [tex] B = \frac{k}{\sin w_0 t}[/tex]

    where k is another constant. For finding the minimum B, I just take [tex]sin w_0t =1[/tex]. I don't know if my solution is correct or not. Any comment?
     
    Last edited: Dec 14, 2008
  2. jcsd
  3. Dec 14, 2008 #2

    olgranpappy

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    Homework Helper

    it seems like something may be missing from the question... is that the exact statement of the question on the homework?
     
  4. Dec 16, 2008 #3

    KFC

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    Thanks for reply. This is not actually an hw. This is an exercise given by my instructor and that's all statement :)
     
  5. Dec 16, 2008 #4

    olgranpappy

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    Homework Helper

    then what you have looks good to me.
     
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