# Transforming Cartesian Position Operators into Spherical Coordinates

I'm attempting to show $$\langle n,l',m'|\hat z|n,l,m \rangle = 0$$ for $$m\neq m'$$ using the explicit form of $$Y_{l,m}(\theta,\phi)$$.

Question:

I wasn't sure how to handle the position operator $$\hat z$$, so my idea was to express it in spherical coordinates.

It is not clear to me if I'm allowed to do something like $$\hat z = \hat r \cos(\theta)$$

Do position operators work like this?

• This would suggest that the radial unit vector $\hat{r}$ is always in the $z$-direction, which can't be true. See here for how to go between Cartesian and spherical unit vectors. – march Feb 20 at 1:48
• You can also try to express the spherical harmonics in Cartesian. – Superfast Jellyfish Feb 20 at 3:46
• @march I may be confused but I was under the impression that the position operator $/hat x$ was different than the unit vector $/hat x$ – s0ggyj0hns0n Feb 20 at 4:24
• I see. I misunderstood the hat to be indicating a unit vector rather than an operator. (That always gets me). Yes, then you are correct. – march Feb 20 at 4:57

You can write $$z$$ as

$$z = r\left(\frac{4\pi}{3}\right)^{1/2} Y_{1,0}(\theta,\phi)$$

This amounts to evaluating the matrix element in the following way

$$\langle n\;l^\prime m^\prime | z | nlm\rangle = \sqrt{\frac{4\pi}{3}}\int R_{n^\prime l^\prime}(r)R_{nl}(r)r^3\,dr\int Y_{l^\prime m^\prime}(\theta, \phi)Y_{10}(\theta, \phi)Y_{lm}(\theta, \phi)\,d\Omega$$

I carried out the full calculation just to give an idea of how to evaluate this integral in the case of linear polarization on the $$z$$ axis.

Since we are only interested into the projection of the angular momentum, we just have to consider the part of the spherical harmonics which depends upon $$m$$. This part, in the integral, is given by the product of the three exponential factors of the three spherical harmonics which depend on $$\phi$$, so $$Y_{l^\prime m^\prime}Y_{1q}Y_{lm} \propto e^{-im^\prime \phi}e^{iq\phi}e^{im\phi}$$ in our case $$q=0$$. The integral in $$\phi$$ is non-vanishing only when $$m^\prime = m+q$$ which in our case amounts as saying that the integral is non-vanishing only is $$m^\prime = m$$

• There's too much detail here. We don't provide complete answers to homework-like questions. They don't do the OP much good. – garyp Feb 20 at 11:31
• @garyp Yes, i understand, but this calculations can easily be found everywhere, even in many books. I thought that giving an example to the question asked by the OP would help him understand the basics so that he could do the other calculations in a similar manner. I'll edit my answer anyway – Davide Morgante Feb 20 at 11:35