# Stern-Gerlach experiment with a magnetic field inbetween

An experiment is set up so that a beam of spin-1/2 is prepared for $$S_{z} = \hbar/2$$, it then passes a constant magnetic field $$\textbf{B} = B_{0}\textbf{e}_{x}$$ with the velcity $$v_{0}$$ for a distance of $$L$$ before it passes an aditional Stern-Gerlach apparatus in which only beams in $$S_{z} = -\hbar/2$$ can pass.

I've made a quick sketch of the installation. Now I'm wondering if my thought process is correct.

We're searching for the percentage of the initial beam that passes through the last apparatus.

The first apparatus blocks 50% of the incoming beam. Inside the magnetic field, I get $$\textbf{H} = -\gamma B_{0}\textbf{S}_{x}$$ Now through the Schrödinger equation I get $$i\hbar \frac{\partial \chi}{\partial t} = \textbf{H} \chi$$ $$\chi(t) = \begin{bmatrix} a e^{i\gamma B_{0}t/2} \\ b e^{i\gamma B_{0}t/2} \\ \end{bmatrix}$$ Intuitively $$\chi(0) = \chi_{+}^{(z)}$$ since that's what we get after we pass the first apparatus, but this becomes a problem since the probability of getting a spin down beam after the magnetic field becomes 0. $$\chi(t) =\begin{bmatrix} e^{i\gamma B_{0}t/2} \\ 0 \\ \end{bmatrix}$$ $$c_{-}^{(z)} = \chi_{-}^{(z)}\chi(L/v_{0}) =[0 \:\: 1]\begin{bmatrix} e^{i\gamma B_{0}(L/v_{0})/2} \\ 0 \\ \end{bmatrix} = 0 \implies P = |c_{-}^{(z)}|^{2} = 0$$

I'm quite certain that there are errors in my calculations since I'm unfamiliar with this field and would find it very helpful if you could point those out for me.

HINT: your solution to Schrödinger's equation is wrong. Try it out in terms of its components, remembering that $\textbf{H}$ is a matrix.

As pointed out in another answer, your Hamiltonian is wrong. In the $$z$$-basis, the representation for the $$x$$-component of a spin $$1/2$$ can be written as:

$$S_x = \frac{\hbar}{2} \begin{bmatrix} 0 &1 \\ 1 &0 \end{bmatrix}$$

This leads to the following Schrödinger equation:

$$\left\lbrace \begin{matrix} \dot{a} = i \frac{\gamma B_0 }{2} b \\ \dot{b} = i \frac{\gamma B_0}{2} a \end{matrix} \right. ,$$

where $$\chi(t) = \begin{bmatrix} a(t) \\ b(t) \end{bmatrix}$$.

By integrating this equation from $$t=0$$ to $$T$$, you should be able to answer what happens to the spin when it comes to the second Stern-Gerlach apparatus.