- #1

fyzikapan

- 12

- 0

## Homework Statement

An electron is an eigenstate of s

_{z}at time t = 0 (spin up). It is in a magnetic field [tex]\vec B = (B \sin \theta, 0,B\cos\theta)[/tex]. Find the probability of finding the electron with spin down at time t.

## Homework Equations

[tex]

U(t) = \exp \left( -i \mathcal{H} t/\hbar\right)\\\\

P(t) = \left| \left \langle \downarrow | \chi(t) \right \rangle \right|

[/tex]

## The Attempt at a Solution

[tex]

\begin{align*}

U(t) &= \exp \left( -i \mathcal{H} t/\hbar\right)\\

&= \exp \left( -i \frac{et}{mc\hbar} \vec S \cdot \vec B\right)\\

&= \exp \left( -i \frac{\omega_0 t}{2} \left(\sigma_x \sin \theta + \sigma_z \cos \theta\right)\right)\\

&= \exp \left(-i \frac{\omega_0 t}{2} \left[ \begin{array}{cc}\cos \theta & \sin \theta \\

\sin \theta & \cos \theta \end{array}\right]\right)\\

&= \left( \begin{array}{cc}

\exp (- i \frac{\omega_0 t}{2} \cos \theta) & \exp (- i \frac{\omega_0 t}{2} \sin \theta)\\

\exp (- i \frac{\omega_0 t}{2} \sin \theta) & \exp ( i \frac{\omega_0 t}{2} \cos \theta)

\end{array}\right)

\end{align*}[/tex]

Then at time t,

[tex]

\begin{align*}

\chi(t) &= U(t) \left| \chi(0) \right \rangle\\

&= \left( \begin{array}{cc}

\exp (- i \frac{\omega_0 t}{2} \cos \theta) & \exp (- i \frac{\omega_0 t}{2} \sin \theta)\\

\exp (- i \frac{\omega_0 t}{2} \sin \theta) & \exp ( i \frac{\omega_0 t}{2} \cos \theta)

\end{array}\right)\left( \begin{array}{c}1\\0\end{array}\right)\\

&= \left( \begin{array}{cc}

\exp (- i \frac{\omega_0 t}{2} \cos \theta)\\

\exp (- i \frac{\omega_0 t}{2} \sin \theta)\end{array}\right)

\end{align*}

[/tex]

But this gives [tex]P(t) = \left| (\begin{array}{cc} 0 & 1 \end{array})\left( \begin{array}{cc}

\exp (- i \frac{\omega_0 t}{2} \cos \theta)\\

\exp (- i \frac{\omega_0 t}{2} \sin \theta)\end{array}\right)\right|^2 = 1[/tex], which is obviously wrong.

I've looked over my math and don't see any obvious mistakes, so I'm not sure what I'm doing wrong.