# Flip the spin

1. Dec 14, 2008

### KFC

1. The problem statement, all variables and given/known data
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.

$$\psi(0) = \frac{1}{\sqrt{2}}\left( \begin{matrix} 1 \\ 1 \end{matrix}\right)$$

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

$$H \propto \left( \begin{matrix} B\cos w_0t & 0\\ 0 & -B\cos w_0t \end{matrix} \right)$$

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?

Last edited: Dec 14, 2008
2. Dec 14, 2008

### olgranpappy

it seems like something may be missing from the question... is that the exact statement of the question on the homework?

3. Dec 16, 2008

### KFC

Thanks for reply. This is not actually an hw. This is an exercise given by my instructor and that's all statement :)

4. Dec 16, 2008

### olgranpappy

then what you have looks good to me.