# Quantum mechanics, spin half particle in a magnetic field probability question.(schrodinger)

My understanding of how to solve this is a little fuzzy around certain areas.

Consider a spin-1/2 particle with a magnetic moment. At time $t = 0$, the state of the particle is$|\psi(t =0)\rangle = |+\rangle n$, with direction $n = (x - y)/\sqrt{2}$. The system is allowed to evolve in a uniform magnetic field $B = (B_0/\sqrt{2})(x+z)$. What is the probability that the particle will be measured to have spin up in the $y$-direction after time $t$?

I found the Hamiltonian to be $\hbar\omega/2\sqrt{2} (1 1;1 -1)$. Then I found my eigenvales to be $\pm \hbar\omega/2$. Then I used that to solve the equation $H(a;b) = \pm \hbar\omega/2 (a;b)$. But I'm not too sure what to do with my eigenvectors, and im not sure how the intial direction $n$ affects what basis I should be using. If anyone could shed some light on this and tell me step by step what I should be doing in general to solve these problems it would be much appreciated.

The stages to solve this generally is

1. Find the eigenvectors and eigenvalues
2. Decompose the original state ($t=0$) to the eigenvectors.
3. Now write the time dependent state for each eigenvector: $\lvert\psi_1\rangle e^{E_1t/\hbar}$ (where $E_1$ is the eigenvalue corresponding to the eigenvector $\psi_1$)
4. Take the inner product of the initial state with the time dependent state.
5. Square the result to obtain the probability of returning to the initial state as a function of $t$

note: the idea of almost every exercise of this sort is to transform the problem to the base of the eigenvectors of the Hamiltonian (where $H$ is diagonal), because there we know how each state evolves with time.