# State of Spin 1/2 particle in a magnetic field

I am trying to solve this. But I got confused in part b because the particle is now in the up state so I was wondering what effect a magnetic field will have on the state? In other words, will the past state have an effect or I just compute the Hamiltonian and write the new state as a superposition of stationary states without any reference to its past state?

Consider a spin 1=2 particle. At time t = 0, the particle is in the state |+>

(a) At time t = 0, we measure Sx and find a value 1/2 h bar. What is the state vector immediately after the measurement?

(b) At the same instant of the measurement, we apply a constant magnetic field B = B_0k ^ on the particle and allow the particle to precess for a time T. What is the state of the system at t = T?

(c) At t = T, the magnetic field is very rapidly rotated so that it is now B = B_0^ j. After another time interval T, a measurement of Sx is carried out once more. What is the probability that a value 1/2 h bar is found?

The magnetic field $B_0 \hat{k}$ couples to the spin operator $\hat{S_z}$ and the system evolves with the hamiltonian $H \propto B_0 \hat{S_z}$ for time $T$. Now calculate what the unitarty evolution operator looks like at time $T$ and apply it to the state you get right after the measurement.