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## Homework Statement

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?

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