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Homework Statement
Assuming a spin 1/2 is put in a magnetic filed along z direction B_z = B*cos(w_0 t). At the beginning, the spin is in spin-up along x direction, i.e.
\psi(0) = \frac{1}{\sqrt{2}}\left(<br /> \begin{matrix}<br /> 1 \\ 1<br /> \end{matrix}\right)
Try to find out the minimum B such that S_x is flip.
2. The attempt at a solution
First of all, I write than the Hamiltonian of the system
H \propto<br /> \left(<br /> \begin{matrix}<br /> B\cos w_0t & 0\\<br /> 0 & -B\cos w_0t <br /> \end{matrix}<br /> \right)<br />
From that, in any time t>0, the state will evolute as \psi(t) (I already solved that). From the time-dependent solution, I can figure out the probability to find the spin-down when measuring S_x, which is of the following form
P = \sin^2(\gamma \sin(w_0 t))
where \gamma is a constant containing B. Hence, to make the system flip, I have to let P=1, i.e.
\gamma \sin(w_0 t)=\pi/2
and solve for B gives
B = \frac{k}{\sin w_0 t}
where k is another constant. For finding the minimum B, I just take sin w_0t =1. I don't know if my solution is correct or not. Any comment?
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