# 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.

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 $\psi_1$)
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.