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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – sequence Jun 23 '16 at 16:41
  • $\begingroup$ 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. $\endgroup$ – K_inverse Jun 24 '16 at 2:27
  • $\begingroup$ Where is $t=2T$ coming from? $\endgroup$ – sequence Jun 24 '16 at 3:34
  • $\begingroup$ 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 $t = 2T$. $\endgroup$ – K_inverse Jun 24 '16 at 6:25

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