Find the possible outcomes of ]##L^2## and ##L_{z}##

[keyzan]

TL;DR Summary: Find the possible outcomes of ]$L^2$ and $L_{z}$ and their respective probabilities of an electron of an idrogen athom with function:
$\psi(r) = ze^{-\alpha r}$

Hi guys, I have a problem with this exercise.

The electron of a hydrogen atom is found with direct spin along the z axis in a state with an orbital wave function:

$\psi(r) = ze^{-\alpha r}$

with alpha greater than zero. The first exercise asks: find the possible outcomes of ]$L^2$ and $L_{z}$ and their respective probabilities.

My solution:
First of all I have to write the function as the product of a radius function multiplied by a spherical harmonic. So I have:

$\psi(r) = r \cos{\theta} \text{ } e^{-\alpha r} = \sqrt{\frac{8 \pi}{3}} \Upsilon^{0}_{1} \text{ } \frac{1}{2\sqrt{\alpha^{3}}} \varphi_{2,1}$

Here I deduced that the radial part has quantum numbers $n=2$ and $l=1$, because the coefficient is a power of $1$ and starts from $r^{1}$ and therefore $n=2$ and $l=1$, but in reality this would be if we had a stationary state, but it is not since we have an $\alpha$ in the argument of the exponential. Can I still write this function with these quantum numbers or is it an error and should I consider a linear combination of radial functions? Any kind of help will be appreciated

 

[PeterDonis]

Moderator's note: Thread moved to advanced physics homework forum.

 

[PeterDonis]

this would be if we had a stationary state, but it is not since we have an $\alpha$ in the argument of the exponential

Why would that make the state not stationary?

 

[vela]

It is already a function of $r$ only so what you have is a function of $r$ multiplied by a spherical harmonic function of $1$ (which is not the spherical harmonic you wrote down). What quantum numbers does that correspond to? (They're not the ones you wrote down.)

The original wave function has a factor of $z$.

 

[PeterDonis]

The original wave function has a factor of $z$.

What is $z$?

 

[vela]

What is $z$?

Given the context, I would assume it's the standard cartesian coordinate, which in terms of spherical coordinates is $z=r\cos\theta$. What did you take it to mean?

 

[PeterDonis]

Given the context, I would assume it's the standard cartesian coordinate, which in terms of spherical coordinates is $z=r\cos\theta$.

Ah, ok. I had been thinking of atomic number. Normally spherical coordinates are used in these kinds of problems, not cylindrical coordinates, so I wasn't expecting $z$ to be a coordinate.

 

[keyzan]

Why would that make the state not stationary?

It is a steady state just in case $\alpha = \frac{1}{n a}$ Where $a$ is the Bohr radius and in this case $n=2$. So is a steady state only if $\alpha = \frac{1}{2a}$ In all other cases it is a combination of stationary states.

 

[vela]

Would that affect your answers to the questions asked?

 

[keyzan]

Would that affect your answers to the questions asked?

Yes, but now how do I find the outcomes of $L^2$ and the probabilities? That is, if I have to write as a linear combination, then I should have as outcomes $\hbar^2 l (l+1)$ for the values $l=1,2,3,4,..$. But at this point the probabilities are impossible (or very difficult) to find