Transitions in Hydrogen Atom
Suppose we have a Hydrogen atom in a superposition of the 1s ground state, $\psi_1 = R_{1,0}(r)Y_{0,0}(\theta,\phi)$, and the $m_l = 0$ state of the 2p configuration, $\psi_2 = R_{2,1}(r)Y_{1,0}(\theta,\phi)$. The time-dependent wave function will therefore be
$$ \Psi(\vec{r},t) = A\psi_1(\vec{r})e^{-i\omega_1t}+B\psi_2(\vec{r})e^{-iw_2t},$$ where $\omega \equiv E/\hbar$. Concerning this system, I have a couple of questions that I am stuck with:
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In my book, it says for any superposition of stationary states, you can write the electronic charge distribution as
\begin{equation} -e|\Psi(t)|^2=-e{|A\psi_1|^2+|B\psi_2|^2+|2A^B\psi_1^\psi_2|\cos(\omega_{12}t-\phi)}, \end{equation}
where $\omega_{12} \equiv \omega_2-\omega_1$. Essentially, this boils down to writing out the amplitude squared, from which I got
\begin{align} |\Psi|^2 &\equiv \Psi^\Psi = (A^\psi_1^e^{i\omega_1t}+B^\psi_2^e^{i\omega_2t})(A\psi_1e^{-\omega_1t}+B\psi_2e^{-i\omega_2t})\ &= |A\psi_1|^2 +|B\psi_2|^2+A^B\psi_1^\psi_2e^{-i\omega_{12}t}+AB^\psi_1\psi_2^e^{i\omega_{12}t} \end{align*} From there, I don't see a clear way to simplify any further, which concerns me because the amplitude squared should be positive and real.
UPDATE: Note that the last term is simply the complex conjugate of the second to last term, so we have something like $z + z^* = (a+bi)+(a-bi) = 2a$, which is exactly the expression from the book.
1. Next order of business, from the electronic charge distribution, we see that there is some oscillation happening over time, from which you can think of having an oscillating electric dipole moment $-e\vec{D}(t)$. Quantum mechanically, this corresponds with the dipole moment being
\begin{equation} -e\vec{D}(t) = -e<\vec{r}>_{\Psi}, \end{equation} the expectation value of the position in the state $\Psi$. 649