# Quantum state after change of magnetic field

I have the following conditions:

$\lvert\psi(0)\rangle=\lvert+\rangle_x=\frac{1}{\sqrt{2}}\lvert+\rangle+\frac{1}{\sqrt{2}}\lvert-\rangle$.

So the state at $t=T$ is $\lvert\psi(t)\rangle=\frac{1}{\sqrt{2}}e^{-\omega_0 t/2}\lvert+\rangle+\frac{1}{\sqrt{2}}e^{+\omega_0 t/2}\lvert-\rangle$. This was with respect to a magnetic field $\vec{B}=B_0\hat{z}$.

Now the field is very rapidly changed to $\vec{B}=B_0\hat{y}$. Then after some time $T$ a measurement of $S_x$ is made. I then need to find the probability of getting $\frac{\hslash}{2}$. However, the problem is that I don't know what state the particle should switch to after the change of the magnetic field. How is it determined? Would appreciate some clarification.

The state is still $\psi (t=T)$ right after the rapid change of B field to y-direction because the system doesn't have enough time to response the change.
• But then what happens with the state when an $S_x$ measurement is made after some time $T$ passed after the rapid change of the magnetic field? – sequence Jun 23 '16 at 16:41
• You need to work out the time evolution of the wavefunction using a new Hamiltonian (recall $H = B \cdot S$) , and express your wavefunction at $t = 2T$ in the basis of $S_{x}$, The modulus square of coefficient tells you the probability. – K_inverse Jun 24 '16 at 2:27
• Where is $t=2T$ coming from? – sequence Jun 24 '16 at 3:34
• At $t = T$, the wave function is given by $\psi (t = T)$ (as you mentioned). Then measurement of $S_{x}$ is performed after some time $T$. so now $t = 2T$. – K_inverse Jun 24 '16 at 6:25