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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 [tex]H=w(B)\hbar*\sigma_z[/tex], 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?

My answer:

now from 0<t<T we have that [tex]|\psi(t)>=e^{-iHt/\hbar}|\psi(0)>[/tex]

which equals: [tex]|\psi(t)>=\frac{1}{\sqrt 2}(cos(wt)-isin(wt),cos(wt)+isin(wt))[/tex]

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, [tex]|psi(T)>=\frac{1}{\sqrt 2}(cos(wT)-isin(wT),cos(wT)+isin(wT))[/tex], now in order to find |psi(t)> at [T,2T] should it be:

[tex]|\psi(t)>=e^{-iw(B)\hbar \sigma_y(t-T)/\hbar}|\psi(T)>[/tex] or something else?

from there in order to calculate the wanted probability i need to compute:

[tex]||<\psi(0)|\psi(2T)>|^2[/tex]

is my approach correct or does it have loopholes?

any input?

thanks in advance.

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# Homework Help: Half spin.(Q in QM).

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