Find Min B to Flip Spin-Up in Magnetic Field

[KFC]

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

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

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?

 

[olgranpappy]

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

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

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?

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

 

[KFC]

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

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

 

[olgranpappy]

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

then what you have looks good to me.