# Position operator in spherical basis

In a set of notes it states that we can define spherical basis vectors in terms of the Cartesian basis vectors $\hat{x}, \hat{y}$ and $\hat{z}$ by $$\hat{e}_{\pm 1} := \mp \frac{1}{\sqrt{2}}(\hat{x} \pm i \hat{y})~~\text{and}~~\hat{e_{0}} = \hat{z}~~~~~~~~~~~~~~(*)$$

In terms of the components of a vector $\mathbf{A} = A_x \hat{x} + A_{y}\hat{y} + A_{z}\hat{z}$ we have $$\hat{A}_{\pm 1} := \mp \frac{1}{\sqrt{2}}(A_x \pm i A_y)~~\text{and}~~A_{0} = A_z~~~~~~~~(**)$$

It then states the following regarding the components of the position vector: "the components of the position vector $\mathbf{r}$ can be written $$r_{\pm 1} = \mp \frac{r}{\sqrt{2}} \sin \theta e^{\pm i \phi}~~~\text{and}~~~r_0 = r \cos \theta~~~~\text{(position operator in spherical basis)}$$ or more compactly $$r_{q} = r \sqrt{\frac{4 \pi}{3}}Y_{1}^{q}(\theta, \phi)~~~(\text{position operator as spherical harmonics})"$$

Question: So it is clear that $r_q$ is obtained from $(**)$ by using spherical coordinate substitution $$x = r sin \theta cos \phi,~~~y = r \sin \theta \sin \phi,~~~z = r \cos \theta$$ What is the reason for writing it in this form and how does it relate to the position operator in spherical basis as is stated in brackets?

## 1 Answer

Imagine a charge $q$ coupled with a classical electrical field, the transition between states $|i \rangle$ and $| f \rangle$ in the dipole approximation is proportional to

$$p_{if} \sim \langle i | q{\bf r} | f\rangle = q\langle i | {\bf r} | f\rangle$$

If you think of both $|f\rangle$ and $| i \rangle$ as bound states of, say, the Hydrogen atom, you can obtain the selection rules that govern the transition between energy levels when the atom is coupled with light.

I am not going to give you the whole math, but that results in the famous $\Delta l = \pm 1$ and $\Delta m = 0, \pm 1$