Spin expectation values in x and y direction

[bobred]

I have found what I think is the correct answer I just want to check an assumption. The magnetic field points in the +ve z-direction. We are given the initial state vector

\left| A \right\rangle_{initial}=\frac{1}{5}\left[ \begin{array}{c}3\\4\end{array} \right]

Am I right in thinking that this is in the z-direction?
So that

\left| A \right\rangle_{initial}= a_1 \left| \uparrow_z \right\rangle + a_2 \left| \downarrow_z \right\rangle = \frac{3}{5}\left[ \begin{array}{c}1\\0\end{array} \right] + \frac{4}{5}\left[ \begin{array}{c}0\\1\end{array} \right]

 

[tiny-tim]

hi bobred!

Am I right in thinking that this is in the z-direction?

if you mean, do the notations (1 0) and (0 1) mean the +ve and -ve z directions, then yes

(that is the general convention, and you can assume it unless stated otherwise)

 

[bobred]

Hi
Yes that's what I meant. We are then asked to find the expectation values at a later time which I have.

We are then asked using the generalized Ehrenfest theorem to find the rate of change of \left\langle S_x \right\rangle and \left\langle S_y \right\rangle in terms of \left\langle S_x \right\rangle and \left\langle S_y \right\rangle and ω. Using the generalized Ehrenfest theorem I find them in terms of \hat{S_x} and \hat{S_y} and ω. Any ideas?

 

[TSny]

We are then asked using the generalized Ehrenfest theorem to find the rate of change of \left\langle S_x \right\rangle and \left\langle S_y \right\rangle in terms of \left\langle S_x \right\rangle and \left\langle S_y \right\rangle and ω. Using the generalized Ehrenfest theorem I find them in terms of \hat{S_x} and \hat{S_y} and ω. Any ideas?

How far have you gotten? Are you able to get an expression for [S_x, H] in terms of S_x and/or S_y?

 

[bobred]

Hi, I get (we are told \omega = -\gamma_s B)

\left[ \hat{S_x},\hat{H} \right]=-\frac{\gamma_s B \hbar^2}{2}\left[ \begin{array}{cc}0&-1\\1& 0 \end{array} \right]=\frac{\omega \hbar^2}{2}\left[ \begin{array}{cc}0&-1\\1& 0 \end{array} \right]

and

\displaystyle\dfrac{d\left\langle S_x \right\rangle}{dt}=-\frac{\omega \hbar}{2}\left[ \begin{array}{cc}0&-i\\i& 0 \end{array} \right]=-\omega \hat{S_y}

 

[TSny]

Hi, I get (we are told \omega = -\gamma_s B)

\left[ \hat{S_x},\hat{H} \right]=-\frac{\gamma_s B \hbar^2}{2}\left[ \begin{array}{cc}0&-1\\1& 0 \end{array} \right]=\frac{\omega \hbar^2}{2}\left[ \begin{array}{cc}0&-1\\1& 0 \end{array} \right]

and

\displaystyle\dfrac{d\left\langle S_x \right\rangle}{dt}=-\frac{\omega \hbar}{2}\left[ \begin{array}{cc}0&-i\\i& 0 \end{array} \right]=-\omega \hat{S_y}

Looks pretty good. But note that Ehrenfest's theorem states

\displaystyle\dfrac{d\left\langle S_x \right\rangle}{dt}= \frac{1}{i\hbar}\langle\left[ \hat{S_x},\hat{H} \right]\rangle

The right hand side contains an expectation value of $\left[ \hat{S_x},\hat{H} \right]$.

 

[bobred]

From the commutator I get the expectation value of the spin y operator... can't seem to see where to go from there.

 

[TSny]

From the commutator I get the expectation value of the spin y operator... can't seem to see where to go from there.

Doesn't that provide the answer to the question? That is, you now have the rate of change of $\langle S_x \rangle$ expressed in terms of $\omega$ and $\langle S_y \rangle$. Or have I misunderstood what you are trying to find?

 

[bobred]

Hi, my fault I had confused myself. Thanks