fyzikapan
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Homework Statement
An electron is an eigenstate of sz at time t = 0 (spin up). It is in a magnetic field \vec B = (B \sin \theta, 0,B\cos\theta). Find the probability of finding the electron with spin down at time t.
Homework Equations
<br /> U(t) = \exp \left( -i \mathcal{H} t/\hbar\right)\\\\<br /> <br /> P(t) = \left| \left \langle \downarrow | \chi(t) \right \rangle \right|<br />
The Attempt at a Solution
<br /> \begin{align*}<br /> U(t) &= \exp \left( -i \mathcal{H} t/\hbar\right)\\<br /> &= \exp \left( -i \frac{et}{mc\hbar} \vec S \cdot \vec B\right)\\<br /> &= \exp \left( -i \frac{\omega_0 t}{2} \left(\sigma_x \sin \theta + \sigma_z \cos \theta\right)\right)\\<br /> &= \exp \left(-i \frac{\omega_0 t}{2} \left[ \begin{array}{cc}\cos \theta & \sin \theta \\ <br /> \sin \theta & \cos \theta \end{array}\right]\right)\\<br /> &= \left( \begin{array}{cc}<br /> \exp (- i \frac{\omega_0 t}{2} \cos \theta) & \exp (- i \frac{\omega_0 t}{2} \sin \theta)\\<br /> \exp (- i \frac{\omega_0 t}{2} \sin \theta) & \exp ( i \frac{\omega_0 t}{2} \cos \theta)<br /> \end{array}\right)<br /> \end{align*}
Then at time t,
<br /> \begin{align*}<br /> \chi(t) &= U(t) \left| \chi(0) \right \rangle\\<br /> &= \left( \begin{array}{cc}<br /> \exp (- i \frac{\omega_0 t}{2} \cos \theta) & \exp (- i \frac{\omega_0 t}{2} \sin \theta)\\<br /> \exp (- i \frac{\omega_0 t}{2} \sin \theta) & \exp ( i \frac{\omega_0 t}{2} \cos \theta)<br /> \end{array}\right)\left( \begin{array}{c}1\\0\end{array}\right)\\<br /> &= \left( \begin{array}{cc}<br /> \exp (- i \frac{\omega_0 t}{2} \cos \theta)\\<br /> \exp (- i \frac{\omega_0 t}{2} \sin \theta)\end{array}\right)<br /> \end{align*}<br />
But this gives P(t) = \left| (\begin{array}{cc} 0 & 1 \end{array})\left( \begin{array}{cc}<br /> \exp (- i \frac{\omega_0 t}{2} \cos \theta)\\<br /> \exp (- i \frac{\omega_0 t}{2} \sin \theta)\end{array}\right)\right|^2 = 1, 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.