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:
- 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.
- 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 $\Psi$. To find an expression for this electric dipole moment, I tried plugging in the wave function by definition, since
\begin{align} <\vec{r}>_\Psi &\equiv \int \Psi^*\vec{r}\Psi d^3\vec{r} = \int \vec{r}|\Psi|^2 d^3r\\ &= \int \vec{r} \bigg[|A\psi_1|^2+|B\psi_2|^2+|2A^*B\psi_1^*\psi_2|\cos(\omega_{12}t)\bigg]d^3r\\ &= |A^2|\int\vec{r}|\psi_1|^2d^3r + |B^2|\int\vec{r}|\psi_2|^2d^3r+|2A^*B|\cos(\omega_{12}t)\int\vec{r}\psi_1^*\psi_2d^3r\\ &= |2A^*B|\cos(\omega_{12}t)\int\vec{r}\psi_1^*\psi_2d^3r\\ &= |2A^*B|\cos(\omega_{12}t)\int\vec{r}(Y_{0,0}Y_{1,0})(R_{1,0}R_{2,1})d^3r\\ &= |2A^*B|\cos(\omega_{12}t)\Bigg\{\hat{e}_x\int_{\Omega}\bigg[\int_0^{\infty}rR_{1,0}R_{2,1}r^2dr\bigg]\sin\theta\cos\phi Y_{0,0}Y_{1,0}\sin\theta d\theta d\phi\\ & \ \ + \hat{e}_y\int_{\Omega}\bigg[\int_0^{\infty}rR_{1,0}R_{2,1}r^2dr\bigg]\sin\theta\sin\phi Y_{0,0}Y_{1,0}\sin\theta d\theta d\phi\\ & \ \ + \hat{e}_z\int_{\Omega}\bigg[\int_0^{\infty}rR_{1,0}R_{2,1}r^2dr\bigg]\cos\theta Y_{0,0}Y_{1,0}\sin\theta d\theta d\phi \Bigg\}\\ &= |2A^*B|\mathcal{I}_{\text{ang}}\bigg[\int_0^{\infty}rR_{1,0}R_{2,1}r^2dr\bigg]\cos(\omega_{12}t)\hat{e}_z \end{align} That was a lot, but it simplified the problem. In the above simplifications, I used the book's equation for $|\Psi|^2$, and noted that $\int\vec{r}|\psi_1|^2d^3r = \int\vec{r}|\psi_2|^2d^3r = 0$, and the later integrals in the $x$ and $y$ directions are zero due to symmetry. In addition, $\mathcal{I}_{\text{ang}}$ is the angular integral from above. So this is the form of the equation that the book has for the electric dipole moment. However, I am having some trouble understanding this result. In the book, they defined this angular integral as $$ \mathcal{I}_{\text{ang}} = \int_{\Omega}Y_{l_2,m_2}^*(\theta,\phi)\hat{r}\cdot\hat{e}_{\text{rad}}Y_{l_1,m_1}(\theta,\phi)d\Omega, $$ where $\hat{e}_{\text{rad}}$ is the unit vector in the direction of the oscillating radiation field. From the analysis above, this would mean that the EM radiation field would be in the $z$ direction, wouldn't it? If so, is this a general result, or only valid for these two states? Thanks!
