# Spherical harmonics of Hydrogen-like atoms

1. May 11, 2014

### Nikitin

Hi.

http://www.nt.ntnu.no/users/jensoa/E-FY1006-31mai2012.pdf

It appears, after all, I understand nothing when it comes to the wave function of Hydrogen like atoms. So I kindly ask you to answer some questions I got:

1) "A selection of these atoms will then be left in an ensemble described by the wave function $\psi(r,\theta,\phi)$ " - can somebody please translate this? What, precisely, is an ensemble?

2) What does it mean when the angular function $Y(\phi,\theta)$ is a linear combination of other $Y_{lm}(\phi,\theta)$ functions? Like, in the problem, "$Y(\phi,\theta) = Y_{p_x} n_x + Y_{p_y} n_y + Y_{p_z} n_z$", meaning $Y$ is a combination of a bunch of p-orbital states. How can this be? Surely an electron can only be in one state, one orbital, after measurement?

Or does this have something to do with the ensemble stuff?

3) "$Y = \sqrt{\frac{3}{4 \pi}} \hat{n} \cdot \hat{r}$" - Why are they multiplying the direction vector with a normal vector? How does this become the angular function?

Thank you.

Last edited: May 11, 2014
2. May 11, 2014

### Nikitin

Oh it seems one of the reasons I was struggling on this is that the details concerning orbitals aren't part of the curriculum this year (so we didn't go through it in detail). What a waste of time.

But can somebody still please answer the 3 questions? I am curious

3. May 12, 2014

### king vitamin

1. The "ensemble" is a reference to the fact that "a large number of such atoms" were measured. They measured the properties of a bunch of cesium ions, and then they consider a subset of those that they measured in the 3p state with zero $\hat{n}\cdot \hat{L}$. All of these have definite values of these three quantum numbers, but if you want to know the probability distribution of another observable, you should make a bunch of independent observations of that observable on the individual ions in this ensemble. So they represent a statistical ensemble in the following sense: http://en.wikipedia.org/wiki/Statistical_ensemble_(mathematical_physics)

2. The question is poorly worded for something presented to an intro QM student. The $Y$ on the left-hand side is in an eigenstate of $\hat{n} \cdot \hat{L}$, while the $Y_{lm}$ on the right hand side are eigenstates of $L_z$. For $\hat{n} \neq \hat{z}$, these will not coincide. If you now measure $L_z$, the state will collapse into one of its eigenstates.

3. Recall that $\vec{r} = r \cos(\theta) \sin(\theta) \hat{x} + r \sin(\theta) \sin(\theta) \hat{y} + r \cos(\theta) \hat{z} = x \hat{x} + y \hat{y} + z \hat{z}$. So by looking at the spherical harmonics, $z = CrY_{10}$, $x pm iy = \mp KrY_{1\pm1}$ with constants C and K. Does this help with understanding why one could have $Y = \sqrt{\frac{3}{4\pi}}\hat{n} \cdot \hat{r}$?

4. May 13, 2014

### Nikitin

OK. So from the ensemble you can measure the probability of finding the electron in the 3px,3py or 3pz given it is in the 3p orbital?

oh it's not a problem from the text, it's just something I don't understand about the text. I didn't ask for help with solving the problems, just with understanding them.

Why?

Well OK, fair enough. So $n_x$, $n_y$ and $n_z$ are normalization coefficients or something?

5. May 13, 2014

### king vitamin

Since the spin is initially measured along the $\hat{n} \cdot \hat{r}$ direction, the wave function you're given is in an eigenstate of $\hat{n} \cdot \vec{L}$. That's why Y takes the form is does. See if you can convince yourself that, if $\hat{n} = \hat{z}$, everything still holds. No matter which direction you measure, you can always rewrite it in another basis. The problem seems to be rewriting in terms of 3px, 3py, 3pz.

The components of $\hat{n}$ parametrize which direction the spin was originally measured along. They should be normalized (it's a unit vector), and they characterize the probability distribution the wave function is current in.