Intuition behind the following calculation for hydrogen atom

I calculated the following the integrals: $$\langle 100|\hat{\overline{r}}|100 \rangle=0$$ $$\langle100|\hat{\overline{r}}|200\rangle=0$$ $$\langle100|\hat{\overline{r}}|210\rangle=\frac{128\sqrt{2}a_{0}}{243}\hat{k}$$ $$\langle100|\hat{\overline{r}}|211\rangle=\frac{-128a_{0}}{243}\hat{i}-\frac{i128a_{0}}{243}\hat{j}$$ where $$\hat{\overline{r}}= r \sin\theta \cos\phi \, \hat{i}+r \sin\theta \sin\phi \, \hat{j}+r \cos\theta \, \hat{k}$$ and $$|nlm_{l}\rangle$$ are the eigen functions of the hydrogen atom. I have no issue with the mathematics, but I am not getting any physical understanding of such values, especially the last one in which we have imaginary number in the $$j^{th}$$-component. I am hoping to get some intuition behind these values. My class teacher said that these are very important in spectroscopy.

• Use \langle and \rangle instead of < and > Nov 20 '20 at 12:30
• we have imaginary number Matrix elements are often imaginary. They aren’t directly measurable quantities so this is not a problem. If you compute a measurable quantity it will be a real number. Nov 20 '20 at 18:16

Your $$\vec{r}$$ is just a position vector pointing from the origin. Thus, what you are calculating is the matrix elements of the dipole moment (omitting the charge), which characterize the intensity of coupling between the atom and the electromagnetic field. Where the dipole moment is zero, such a coupling is absent, whereas the non-zero values characterize the field polarization that could cause transitions between the two states.
• You do understand that your $\vec{r}$ is just the position vector from the center of the atom? Nov 20 '20 at 12:44
• An electromagnetic plane wave can be written as $\vec{E}(\vec{r},t) = \vec{E}_0e^{i\vec{k}\vec{r}} + c.c.\approx \vec{E}_0(1+i\vec{k}\vec{r}) + c.c.$ The first term is constant, while the second is proportional to $\vec{r}$. Thus, when you calculate the interaction matrix element, it is essentially the matrix element of the position vector. Btw, have you studied QM before? If this is your first pass, on you are still on the hydrogen atom, then it will be hard to explain... Nov 20 '20 at 12:53