# Question on measuring expectation value of spin with time variation

I have a particle with the following wave function:

$$\psi(t) = \frac12 |\uparrow \rangle e^{-i(\omega_1+\omega_2)t/\hbar} +\frac12 |\uparrow \rangle e^{-i(\omega_1-\omega_2)t/\hbar} +\frac{1}{\sqrt{2}} |\downarrow \rangle e^{-i(-\omega_1-\omega_2)t/\hbar}$$

I am trying to calculate the expectation value of the spin. How should I go about doing that? My best guess is to calculate $\psi ^* \psi$ to find the time-dependent probability of both spin up and spin down and then multiply each probability by the appropriate eigenvalue ($\pm \hbar/2$) and add them. However, when I did this I got a time-dependent probability for spin up but a time-independent probability for spin down, which seems impossible to me.

$$P(\uparrow)=\frac12 + \frac14 e^{i2\omega_2t/\hbar} + \frac14 e^{-i2\omega_2t/\hbar}$$

$$P(\downarrow)=\frac12$$

I don't know if I messed up or if the answer is starting me in the face and I can't see it.

Note: the two omega values of the time-dependent terms of the wavefunction come from two different particles; my question is only about the expectation value of spin for the first particle so I have not included other non-necessary information on the second particle.

The reason you are getting a strange result is because you present a strange wavefunction. The wavefunction you give is not normalized: \begin{align} \left\langle \psi | \psi \right\rangle & = \bigg( \frac{1}{2} \left( e ^{ i ( \omega _1 + \omega _2 ) t / \hbar } + e ^{ i ( \omega _1 - \omega _2 ) t / \hbar } \right) \left\langle \uparrow \right| + \frac{1}{ \sqrt{2} } e ^{ i ( - \omega _1 - \omega _2 ) t / \hbar } \left\langle \downarrow \right| \bigg) \notag \\ & \times \bigg( \frac{1}{2} \left( e ^{ - i ( \omega _1 + \omega _2 ) t / \hbar } + e ^{ - i ( \omega _1 - \omega _2 ) t / \hbar } \right) \left| \uparrow \right\rangle + \frac{1}{ \sqrt{2} } e ^{- i ( - \omega _1 - \omega _2 ) t / \hbar } \left| \downarrow \right\rangle \bigg) \\ & = \frac{1}{2} \left( 1 + \cos \frac{ 2\omega _2 t }{ \hbar } \right) + \frac{1}{2} \\ & = 1 + \frac{1}{2} \cos \frac{ 2 \omega _2 t }{ \hbar } \end{align} which is only normalized if $\omega_2 = 0$.