How Does a Fluctuating Hamiltonian Affect the Expectation Value of Sx?

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Niles
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Homework Statement


Hi

Say I have a Hamiltonian given by H = δSz acting on my system, where δ is a random variable controlled by some fluctuations in my environment. I have to show that if I start out with <Sx>=½, then the Hamiltonian will reduce <Sx> to

<Sx> = ½<cos(δt)>

where the <> around the cosine means averaged over all values of δ. What I would do is to use

<eiHtSx(0)e-iHt> = <Sx(t)>

but this seems very tedious. Am I on the right path here?


Niles.
 
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Isn't the easiest way to just get the eigenfunctions of Sz and then project Sx on that?
 
Niles said:
where the <> around the cosine means averaged over all values of δ. What I would do is to use

<eiHtSx(0)e-iHt> = <Sx(t)>

but this seems very tedious. Am I on the right path here?

Yes, but it shouldn't be a tedious computation. The expectation value of that time-evolved operator is just equal to x^H*Sx*x, where x is the time-evolved state vector. x is a 2x1 vector (written using the eigenvectors of H as a basis, because that's convenient) and Sx is a 2x2 matrix, so the computation shouldn't be too hard.