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Spin 1 Particle in a time dependant magnetic field

  • Thread starter richyw
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Homework Statement



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

Homework Equations



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

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.
 

Answers and Replies

  • #2
TSny
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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.
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).
 
  • #3
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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]?
 
  • #4
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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]
 
  • #5
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and then [tex]i\hbar\frac{d}{dt}\left| \psi ,t \right\rangle=\gamma B(t) S_z\left| \psi ,t \right\rangle[/tex]
 
  • #6
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[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]
 
  • #7
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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]
 
  • #8
TSny
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Yes, that looks good. Just have to solve each of these equations.
 
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