- #1

Leechie

- 19

- 2

## 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