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Half spin.(Q in QM).

  1. May 23, 2008 #1
    a half spin has an eigenstate of the opertaor S_x (which is defined by the multiplication of half h bar times pauli sigma x matrix) of eigen value + half h bar at time t=0.
    the spin is at a magnetic field (0,0,B) which correspond to the hamiltonian [tex]H=w(B)\hbar*\sigma_z[/tex], at time T they change the direction of the magnetic field to the y direction: (0,B,0), after another time T a measurement of S_x was done, what is the probability that the value measured is the one we started with?

    My answer:
    now from 0<t<T we have that [tex]|\psi(t)>=e^{-iHt/\hbar}|\psi(0)>[/tex]
    which equals: [tex]|\psi(t)>=\frac{1}{\sqrt 2}(cos(wt)-isin(wt),cos(wt)+isin(wt))[/tex]
    now from T to 2T we have a magnetic field working in the y direction, does it mean we should act the above operator on |psi(T)> but with the appropiate change i.e should it be soemthing like this, at time t=T, [tex]|psi(T)>=\frac{1}{\sqrt 2}(cos(wT)-isin(wT),cos(wT)+isin(wT))[/tex], now in order to find |psi(t)> at [T,2T] should it be:
    [tex]|\psi(t)>=e^{-iw(B)\hbar \sigma_y(t-T)/\hbar}|\psi(T)>[/tex] or something else?
    from there in order to calculate the wanted probability i need to compute:
    is my approach correct or does it have loopholes?

    any input?
    thanks in advance.
  2. jcsd
  3. May 24, 2008 #2
    if someone already asked such a question please do point me to his thread.
  4. May 25, 2008 #3
    okay with my approach i got that that the probability is cos^2(2wT), is correct or not, i don't know, do you?
  5. May 26, 2008 #4


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    looks ok.
  6. May 26, 2008 #5
    Well I think I miss a factor of 0.25, there.
  7. May 26, 2008 #6
    my bad, I got through my calculuations and i don't think there's missing a factor.
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