# Half spin.(Q in QM).

Gold Member
Problem:
a half spin has an eigenstate of the opertaor S_x (which is defined by the multiplication of half h bar times pauli sigma x matrix) of eigen value + half h bar at time t=0.
the spin is at a magnetic field (0,0,B) which correspond to the hamiltonian $$H=w(B)\hbar*\sigma_z$$, at time T they change the direction of the magnetic field to the y direction: (0,B,0), after another time T a measurement of S_x was done, what is the probability that the value measured is the one we started with?

now from 0<t<T we have that $$|\psi(t)>=e^{-iHt/\hbar}|\psi(0)>$$
which equals: $$|\psi(t)>=\frac{1}{\sqrt 2}(cos(wt)-isin(wt),cos(wt)+isin(wt))$$
now from T to 2T we have a magnetic field working in the y direction, does it mean we should act the above operator on |psi(T)> but with the appropiate change i.e should it be soemthing like this, at time t=T, $$|psi(T)>=\frac{1}{\sqrt 2}(cos(wT)-isin(wT),cos(wT)+isin(wT))$$, now in order to find |psi(t)> at [T,2T] should it be:
$$|\psi(t)>=e^{-iw(B)\hbar \sigma_y(t-T)/\hbar}|\psi(T)>$$ or something else?
from there in order to calculate the wanted probability i need to compute:
$$||<\psi(0)|\psi(2T)>|^2$$
is my approach correct or does it have loopholes?

any input?

Gold Member
anyone?

Gold Member
okay with my approach i got that that the probability is cos^2(2wT), is correct or not, i don't know, do you?

malawi_glenn
Homework Helper
looks ok.

Gold Member
Well I think I miss a factor of 0.25, there.

Gold Member
my bad, I got through my calculuations and i don't think there's missing a factor.