The time-dependence of the expectation values of spin operators

  • #1
Rayan
16
1
Homework Statement
Assume a spin s= 1/2 is subjected to an external magnetic field B. The Hamiltonian is then given by H, and that at t= 0, the spin of the particle is in the eigenstate of the S_x operator with the eigenvalue:
Relevant Equations
$$ \hat{H} = -\frac{eB}{mc} \hat{S}_z = w\hat{S_z} $$
$$ \hat{S}_x|\psi (t= 0)⟩= \frac{\hbar}{2}|\psi(t= 0)⟩$$
So first I derived the expressions for the dynamics of the spin operators and got:
$$ \frac{d\hat{S}_y}{dt} = w\hat{S}_x^H $$
$$ \frac{d\hat{S}_x}{dt} = w\hat{S}_y^H $$
$$ \frac{d\hat{S}_z}{dt} = 0 $$

Now I want to calculate the time-dependence of the expectation values of the spin operators, To do that I used Ehrenfest theorem (for an arbitrary $S_i$):

$$ \frac{d}{dt} ⟨ S_i ⟩_H = \frac{1}{i\hbar} ⟨ [ \hat{S}_i , \hat{H} ] ⟩ + ⟨ \frac{\partial S_i }{dt} ⟩ $$

Starting with the first term:
$$ ⟨ [ \hat{S}_i , \hat{H} ] ⟩ = ⟨ {S}_i \hat{H} ⟩ - ⟨ \hat{H} \hat{S}_i ⟩ = w ( ⟨ \hat{U}^{\dagger} \hat{S}_i \hat{U} \hat{U}^{\dagger} \hat{S}_z \hat{U} ⟩ - ⟨ \hat{U}^{\dagger} \hat{S}_z \hat{U} \hat{U}^{\dagger} \hat{S}_i \hat{U} ⟩ ) =$$
$$ = w ( ⟨ \hat{U}^{\dagger} \hat{S}_i \hat{S}_z \hat{U} ⟩ - ⟨ \hat{U}^{\dagger} \hat{S}_z \hat{S}_i \hat{U} ⟩ ) $$

So my question is what is the best/easiest way to go now? I tried using the definition of expectation value and the fact that the state at t=0 is ( changing to z-basis ):

$$ |\psi (t= 0)⟩ = |+⟩_x = \frac{1}{\sqrt{2}} ( |+⟩_z + |-⟩_z ) $$

So that

$$ ⟨ [ \hat{S}_i , \hat{H} ] ⟩ = \frac{w}{2} \Bigl( ⟨ ⟨+|_z + ⟨-|_z | e^{-iw\hat{S}_zt} \hat{S}_i \hat{S}_z e^{iw\hat{S}_zt} | |+⟩_z + |-⟩_z ⟩ \Bigr) $$

But I don't really know how to continue here to find the expectation value of the exponential term with t-dependence! Any advice appreciated:)
 
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  • #2
Rayan said:
Homework Statement: Assume a spin s= 1/2 is subjected to an external magnetic field B. The Hamiltonian is then given by H, and that at t= 0, the spin of the particle is in the eigenstate of the S_x operator with the eigenvalue:
The problem statement is incomplete. What is it that you are supposed to calculate?

Rayan said:
So first I derived the expressions for the dynamics of the spin operators and got:
$$ \frac{d\hat{S}_y}{dt} = w\hat{S}_x^H $$
$$ \frac{d\hat{S}_x}{dt} = w\hat{S}_y^H $$
$$ \frac{d\hat{S}_z}{dt} = 0 $$
Are you sure about those?

Rayan said:
Starting with the first term:
$$ ⟨ [ \hat{S}_i , \hat{H} ] ⟩ = ⟨ {S}_i \hat{H} ⟩ - ⟨ \hat{H} \hat{S}_i ⟩ = w ( ⟨ \hat{U}^{\dagger} \hat{S}_i \hat{U} \hat{U}^{\dagger} \hat{S}_z \hat{U} ⟩ - ⟨ \hat{U}^{\dagger} \hat{S}_z \hat{U} \hat{U}^{\dagger} \hat{S}_i \hat{U} ⟩ ) =$$
$$ = w ( ⟨ \hat{U}^{\dagger} \hat{S}_i \hat{S}_z \hat{U} ⟩ - ⟨ \hat{U}^{\dagger} \hat{S}_z \hat{S}_i \hat{U} ⟩ ) $$
Stat by finding a simplified equation for ## [ \hat{S}_i , \hat{H} ] ##.

And do you have to use the Heisenberg picture?
 

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