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Spin 1 Particle in a time dependant magnetic field

  1. Dec 7, 2013 #1
    1. The problem statement, all variables and given/known data

    https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-prn2/1483071_10151813160569290_1331827985_n.jpg

    2. Relevant equations

    [tex]\left|\psi ,t\right\rangle=\sum_i c_i(t)e^{-\frac{i E_n t}{\hbar}}\left|n\right\rangle[/tex]

    3. The attempt at a solution

    I'm quite lost on how to even start this. I can work out that the hamiltonian is[tex]H(t)=\gamma S_z B(t)[/tex]Then I tried to put that into the time-dependant shrodinger equation. [tex]-i\hbar \frac{d}{dt}\left|\psi(t)\right\rangle=H(t)\left|\psi(t)\right\rangle[/tex]but I am not really sure if this is correct or even how to begin solving this.
     
  2. jcsd
  3. Dec 7, 2013 #2

    TSny

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    I think you're on the right track. After substituting the given expression for |ψ(t)> into the Schrodinger equation, you should be able to get a differential equation for each coefficient ci(t).
     
  4. Dec 8, 2013 #3
    I'm still stuck on this one actually. I don't think I am substituting [itex]\left|\psi ,t \right\rangle[/itex] into the equation properly. How to I get [itex]\psi (t)[/itex] from [itex]\left|\psi ,t \right\rangle[/itex]?
     
  5. Dec 8, 2013 #4
    wait, can I say[tex]\left| \psi ,t \right\rangle=c_1(t)\left| 1,1 \right\rangle+c_2(t)\left| 1,0 \right\rangle+c_3(t)\left| 1,-1 \right\rangle[/tex][tex]\left| \psi ,t \right\rangle=c_1(t)\left(\begin{matrix}1 \\ 0 \\ 0\end{matrix}\right)+c_2(t)\left(\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right)+c_3(t)\left(\begin{matrix}0 \\ 0 \\ 1\end{matrix}\right)[/tex][tex]\left| \psi ,t \right\rangle=\left(\begin{matrix} c_1(t) \\ c_2(t) \\ c_3(t)\end{matrix}\right)[/tex]
     
  6. Dec 8, 2013 #5
    and then [tex]i\hbar\frac{d}{dt}\left| \psi ,t \right\rangle=\gamma B(t) S_z\left| \psi ,t \right\rangle[/tex]
     
  7. Dec 8, 2013 #6
    [tex]i\hbar \left(\begin{matrix} c'_1(t) \\ c'_2(t) \\ c'_3(t)\end{matrix}\right)=\hbar\gamma B(t)\left(\begin{matrix}1 & 0&0\\0&0&0\\0&0&-1\\\end{matrix}\right)\left(\begin{matrix} c_1(t) \\ c_2(t) \\ c_3(t)\end{matrix}\right)[/tex]
     
  8. Dec 8, 2013 #7
    and then get [tex]c'_1(t)=-i\gamma B(t) c_1(t)[/tex][tex]c'_2(t)=0[/tex][tex]c'_3(t)=i\gamma B(t) c_3(t)[/tex]
     
  9. Dec 8, 2013 #8

    TSny

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    Yes, that looks good. Just have to solve each of these equations.
     
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