Finding Spin Expectation Values At Any Time t > 0

[Leechie]

Homework Statement

Write down a spinor that represents the spin state of the particle at any time t > 0. Use the expression to find the expectation values of ##S_x## and ##S_y##

Homework Equations

The particle is a spin-##\frac 1 2## particle, the gyromagnetic ratio is ##\gamma_s \lt 0##, and the magnetic field points in the ##z## direction.

The initial spin state is: ##| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}##

The Attempt at a Solution

This is where I've got so far:
$$| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}=a_u|\uparrow_n \rangle+a_d|\downarrow_n \rangle$$
Finding ##|\uparrow_n \rangle## and ##|\downarrow_n \rangle## using ##|\uparrow_n \rangle=\begin{bmatrix}\cos(\theta / 2)\\e^{i\phi}\sin(\theta / 2)\end{bmatrix}## and ##|\downarrow_n \rangle=\begin{bmatrix}-e^{-i\phi}\sin(\theta / 2)\\\cos(\theta / 2)\end{bmatrix}##. The magnetic field points in the ##z## direction so ##\theta = 0## and ##\phi = 0##:
$$|\uparrow_z \rangle=\begin{bmatrix}\cos(0)\\e^0\sin(0)\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix} \\ |\downarrow_z \rangle=\begin{bmatrix}-e^0\sin(0)\\\cos(0)\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$
Finding the coeffecients ##a_u## and ##a_d##:
$$a_u=\langle \uparrow_z | A \rangle=\frac 1 5 \begin{bmatrix}1&0\end{bmatrix} \begin{bmatrix}3\\4\end{bmatrix}=\frac 3 5 \\ a_d=\langle \downarrow_z | A \rangle=\frac 1 5 \begin{bmatrix}0&1\end{bmatrix} \begin{bmatrix}3\\4\end{bmatrix}=\frac 4 5$$
So:
$$| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}=\frac 3 5 \begin{bmatrix}1\\0\end{bmatrix} + \frac 4 5 \begin{bmatrix}0\\1\end{bmatrix}$$
Using the the equation for spin at any time ##| A \rangle=a_u e^{-iE_ut/\hbar}|\uparrow_n\rangle + a_d e^{-iE_dt/\hbar}|\downarrow_n\rangle## and since ##\gamma_s \lt 0## the energy eigenvalues are ##E_u=+\frac {\hbar \omega} 2## and ##E_d=-\frac {\hbar \omega} 2## I get:
$$| A \rangle=\frac 3 5 e^{-i\omega t/2}\begin{bmatrix}1\\0\end{bmatrix} + \frac 4 5 e^{+i\omega t/2}\begin{bmatrix}0\\1\end{bmatrix}$$
And so the spinor I get to is:
$$| A \rangle=\frac 1 5\begin{bmatrix}3 e^{-i\omega t/2}\\4 e^{+i\omega t/2}\end{bmatrix}$$
For the expectation value of ##S_x## I get:
$$\begin{align} \langle S_x \rangle & =\langle A | \hat {\mathrm S}_x | A \rangle \nonumber \\ & =\frac 1 5 \begin{bmatrix}3 e^{+i\omega t/2} & 4 e^{-i\omega t/2}\end{bmatrix} \frac \hbar 2 \begin{bmatrix}0&1\\1&0\end{bmatrix} \frac 1 5 \begin{bmatrix}3 e^{-i\omega t/2} \\ 4 e^{+i\omega t/2}\end{bmatrix} \nonumber \\ & = \frac 1 5 \begin{bmatrix}3 e^{+i\omega t/2} & 4 e^{-i\omega t/2}\end{bmatrix} \frac \hbar {10} \begin{bmatrix}4 e^{+i\omega t/2} \\ 3 e^{-i\omega t/2} \end{bmatrix} \nonumber \\ & = \frac \hbar {50} \left( 12e^{i\omega t} + 12e^{i\omega t} \right) \nonumber \\ & = \frac {12\hbar} {25} e^{iwt} \nonumber \end{align}$$
Could someone tell me if I'm along the right lines with this. I've been working on this for so long now I'm starting to lose sight of how this should workout.
Thanks

 

[DrClaude]

$$\begin{align} \langle S_x \rangle & =\langle A | \hat {\mathrm S}_x | A \rangle \nonumber \\ & =\frac 1 5 \begin{bmatrix}3 e^{+i\omega t/2} & 4 e^{-i\omega t/2}\end{bmatrix} \frac \hbar 2 \begin{bmatrix}0&1\\1&0\end{bmatrix} \frac 1 5 \begin{bmatrix}3 e^{-i\omega t/2} \\ 4 e^{+i\omega t/2}\end{bmatrix} \nonumber \\ & = \frac 1 5 \begin{bmatrix}3 e^{+i\omega t/2} & 4 e^{-i\omega t/2}\end{bmatrix} \frac \hbar {10} \begin{bmatrix}4 e^{+i\omega t/2} \\ 3 e^{-i\omega t/2} \end{bmatrix} \nonumber \\ & = \frac \hbar {50} \left( 12e^{i\omega t} + 12e^{i\omega t} \right) \nonumber \\ & = \frac {12\hbar} {25} e^{iwt} \nonumber \end{align}$$

You made a mistake going from line 4 to line 5 line 3 to 4 in there, so the final result is not correct. Otherwise, it looks fine.

 

[PeroK]

Finding ##|\uparrow_n \rangle## and ##|\downarrow_n \rangle## using ##|\uparrow_n \rangle=\begin{bmatrix}\cos(\theta / 2)\\e^{i\phi}\sin(\theta / 2)\end{bmatrix}## and ##|\downarrow_n \rangle=\begin{bmatrix}-e^{-i\phi}\sin(\theta / 2)\\\cos(\theta / 2)\end{bmatrix}##. The magnetic field points in the ##z## direction so ##\theta = 0## and ##\phi = 0##:
$$|\uparrow_z \rangle=\begin{bmatrix}\cos(0)\\e^0\sin(0)\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix} \\ |\downarrow_z \rangle=\begin{bmatrix}-e^0\sin(0)\\\cos(0)\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$

Just an observation. The formulas you are using are valid for the "z-basis" where ##\begin{bmatrix}1\\0\end{bmatrix}## is the z-spin-up and ##\begin{bmatrix}0\\1\end{bmatrix}## is the z-spin-down. So, ##a_u = \frac35## and ##a_d = \frac45## immediately.

In other words, by definition:$$| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}=\frac35 |\uparrow_z \rangle + \frac45 |\downarrow_z \rangle$$

 

[Leechie]

I had a feeling there was something wrong somewhere, I'll take another look. Thanks for your help.

 

[Leechie]

Just an observation. The formulas you are using are valid for the "z-basis" where ##\begin{bmatrix}1\\0\end{bmatrix}## is the z-spin-up and ##\begin{bmatrix}0\\1\end{bmatrix}## is the z-spin-down. So, ##a_u = \frac35## and ##a_d = \frac45## immediately.

In other words, by definition:$$| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}=\frac35 |\uparrow_z \rangle + \frac45 |\downarrow_z \rangle$$

Thanks PeroK. When I was working through that I realized it just led back to the initial spin state because of the z-basis. Is the method I used a general way of calculating a spinor in any direction ##n##?

 

[PeroK]

Thanks PeroK. When I was working through that I realized it just led back to the initial spin state because of the z-basis. Is the method I used a general way of calculating a spinor in any direction ##n##?

Yes, the formulas you used would give you (expressed in the z-basis) the eigenspinors in the direction ##n##.

 

[Leechie]

Yes, the formulas you used would give you (expressed in the z-basis) the eigenspinors in the direction ##n##.

Thanks. I think I'm finally starting to get my head round this now.