# Stern Gerlach with spin in opposite directions

So for the Stern-Gerlach apparatus, we assume that we either have a particle spin up or spin down. We also have the varying field, $\partial B/\partial z$. This initial configuration results in the particle wither going to plus $\hbar$ or minus $\hbar$.

Suppose instead of having spin up/down in the z direction, I sent it through with an initial spin aligned in the x direction (same exact configuration)? The Hamiltonian is given (for a linear B) as $$H=\frac{1}{2m}(p_x^2+p_z^2)-\mu \sigma_z(B_0+B'z)$$ So my equations of motion for the z direction would just give me $p_0t/m+z_0$ and $\dot p_x=0$. Do I need to account for the spin x now instead, or will the particle go undeflected?

• I noticed you posted a few questions about similar topics, you may want to check out Quantum Mechanics by Robert Scherrer, it essentially covers all 3 of your last questions in chapter 8. – Julien Mar 29 '14 at 22:44
• The first chapter of Townsend's book A Modern Approach to Quantum Mechanics deals with the S-G experiment in detail, including this particular situation. – Robin Ekman Mar 29 '14 at 23:17
• At the heart of the answer is $\left|+x\right\rangle = A\left|+z\right\rangle + B\left|-z\right\rangle.$ – BMS Mar 30 '14 at 2:05
• @BMS I am trying that. I know I have spin up in x, so I have the state $\frac {1}{\sqrt{2}}\begin{array}{c} 1\\ 1\\ \end{array}$. But when I try to find the initial z dependent spin, the get just 0. – yankeefan11 Mar 30 '14 at 2:12
