The time-dependence of the expectation values of spin operators

Question

Assume a spin s= 1/2 is subjected to an external magnetic field $B=Be_z$. The Hamiltonian is then given by $$ \hat{H} = -\frac{eB}{mc} \hat{S}_z = w\hat{S_z} $$ and that at t= 0, the spin of the particle is in the eigenstate of the $S_x$ operator with the eigenvalue $ \hbar/2 $, e.g. $$ \hat{S}_x|\psi (t= 0)⟩= \frac{\hbar}{2}|\psi(t= 0)⟩.$$

So first I derived the expressions for the dynamics of the spin operators and got: $$ \frac{d\hat{S}_y^H}{dt} = iw\hat{S}_x^H $$ $$ \frac{d\hat{S}_x^H}{dt} = {-iw} \hat{S}_y^H $$ $$ \frac{d\hat{S}_z^H}{dt} = 0 $$

Now I want to calculate the time-dependence of the expectation values $ ⟨ \hat{S}_x ⟩ $, $ ⟨ \hat{S}_y ⟩ $ and $ ⟨ \hat{S}_z ⟩ $. 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:)

Answer 1

In the Schrödinger picture, you can write the Hamiltonian operator in the basis of the eigenstates of $\hat{S}_z$: $|+\rangle_z=\left(\begin{array}{c}1\\0\end{array}\right)$ and $|-\rangle_z=\left(\begin{array}{c}0\\1\end{array}\right)$ as $$ H = \frac{\hbar w}{2} \left(\begin{array}{cc}1 & 0\\0 & -1\end{array}\right) $$ and the initial state as $$ |\psi(t=0)\rangle = \frac{1}{\sqrt{2}} \left( |+\rangle_z + |-\rangle_z \right) = \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\1\end{array}\right). $$

Now from the Schrödinger equation we get the formula for the time evolution of a state: $$ i\hbar \partial_t \psi = H \psi \rightarrow |\psi(t)\rangle = \exp{\left[-iHt/\hbar\right]} |\psi(t=0)\rangle. $$ The matrix exponential of a diagonal matrix is trivial: you just exponentiate every diagonal term, so: $$ |\psi(t)\rangle = \left(\begin{array}{cc} e^{-iwt/2} & 0 \\ 0 & e^{iwt/2} \end{array}\right) \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\1\end{array}\right) = \frac{1}{\sqrt{2}} \left(\begin{array}{c}e^{-iwt/2}\\e^{iwt/2}\end{array}\right). $$ Knowing the time evolved state, you can compute expectation values of any operator, including for instance $$\hat{S}_x = \frac{\hbar}{2} \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \;\;\;\;\;\;\;\; \hat{S}_y = \frac{\hbar}{2} \left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right) :$$ $$ \langle \psi(t) | \hat{S}_x | \psi(t) \rangle = \frac{\hbar}{2} \cos{(wt)} \;\;\;\;\;\;\; \langle \psi(t) | \hat{S}_y | \psi(t) \rangle = \frac{\hbar}{2} \sin{(wt)} $$

Hope this helps! :)

Answer 2

Since you are working in the Heisenberg picture where states are time-independent, you may automatically take expectation values of your operator equations, resulting in the same uniform rotation structure, $$ \frac{d \langle \hat{S}_y^H\rangle }{dt} = iw \langle \hat{S}_x^H\rangle , \\ \frac{d \langle \hat{S}_x^H\rangle }{dt} = - iw \langle \hat{S}_y^H\rangle ,\\ \frac{d \langle \hat{S}_z^H \rangle }{dt} = 0 .$$

In your notation, their evident solution is $$ \langle \hat{S}_y^H (t)\rangle = \cos( wt ) \langle \hat{S}_y^H(0)\rangle + \sin(wt ) \langle \hat{S}_x^H(0)\rangle , \\ \langle \hat{S}_x^H (t)\rangle = \cos(wt ) \langle \hat{S}_x^H(0)\rangle - \sin( wt ) \langle \hat{S}_y^H(0)\rangle , \\ \langle \hat{S}_z^H (t)\rangle = \langle \hat{S}_z^H (0)\rangle .$$