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

  • Thread starter Thread starter Niles
  • Start date Start date
  • Tags Tags
    Hamiltonian Qm
Niles
Messages
1,834
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top