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

• keyzan
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

keyzan said:
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?

PeterDonis said:
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##.

vela said:
The original wave function has a factor of ##z##.
What is ##z##?

PeterDonis said:
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?

vela said:
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.

PeterDonis said:
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 said:
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

Does anyone know how to solve the problem?

keyzan said:
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
Why would the probabilities be difficult to find? Do you know what it means to write a state as a linear combination of eigenstates of an operator?

keyzan
According to my intuition I would write the function in terms of the radial eigenfunctions:

##\psi(r) = \frac{\sqrt{8\pi}}{3}\Upsilon^{0}_{1} \hspace{0.5cm} 2\sqrt{\alpha^{3}} \sum_{n=2}^{\infty} \sum_{l=-n}^{l=n} \varphi_{n,l} (r)##

where I first normalized the radial part. The summation starts from ##n=2##, because each eigenstate is multiplied by a coefficient that minimum starts from ##r## due to the multiplication factor ##r## in the starting ##\psi(r)##. So since the eigenstates are multiplied by polynomials of degree ##n-1## starting from the power ##r^l##, ##n## must start from ##2##.
At this point the stationary states corresponding to the different values of ##L^2## and ##L_{2}## are:

##\phi(r)_{n,l,m} = \Upsilon_{l}^{m} \varphi_{n,l}##

To know the probabilities we should project:

##|\langle \phi_{n,l,m} | \psi(r) \rangle |^2##

From here I deduce that only the terms with ##l=1## and ##m=0## remain due to the spherical harmonic which leads to the probability for all other values of l and m being canceled out. So we have that the only outcomes are ##2\hbar^2## for ##L^2## and ##0## for ##L_{z}## both with probability ##1##. The wave function becomes:

##\psi(r) = \frac{\sqrt{8\pi}}{3}\Upsilon^{0}_{1} \hspace{0.5cm} 2\sqrt{\alpha^{3}} \sum_{n=2}^{\infty} \varphi_{n,1} (r)##

The spherical harmonica saved my life ahahahaah. Does this seem like correct reasoning to you?

keyzan said:
According to my intuition I would write the function in terms of the radial eigenfunctions
What are you measuring? If you're measuring ##L^2##, you need to write the state in terms of eigenfunctions of ##L^2##. Not just "radial" eigenfunctions, but overall eigenfunctions of that operator. In general those won't just be functions of ##r##.

PeterDonis said:
What are you measuring? If you're measuring ##L^2##, you need to write the state in terms of eigenfunctions of ##L^2##. Not just "radial" eigenfunctions, but overall eigenfunctions of that operator. In general those won't just be functions of ##r##.
ye it was a writing error, in fact as you can see I wrote the ##\psi(r)## in terms of both the angular and radial parts.

keyzan said:
From here I deduce that only the terms with ##l=1## and ##m=0## remain
This is inconsistent with your earlier post that, except for certain special values of ##\alpha##, the state will be a linear combination of stationary states.

keyzan said:
ye it was a writing error, in fact as you can see I wrote the ##\psi(r)## in terms of both the angular and radial parts.
I think that thanks to the presence of the spherical harmonic it was simple to find the probabilities of the outcomes of ##L^2## and ##L_{z}##, but imagine if he had asked me the probabilities of the energy outcomes, it would have been a mess. Right?

keyzan said:
I think that thanks to the presence of the spherical harmonic it was simple to find the probabilities of the outcomes of ##L^2## and ##L_{z}##, but imagine if he had asked me the probabilities of the energy outcomes, it would have been a mess. Right?
No. Stationary states are also eigenstates of ##L^2## and ##L_z## since those operators commute with the Hamiltonian. So a linear combination of stationary states should also be a linear combination of eigenstates of ##L^2## and ##L_z##.

keyzan
PeterDonis said:
This is inconsistent with your earlier post that, except for certain special values of ##\alpha##, the state will be a linear combination of stationary states.
I know, but this result is not due to ##\alpha##, but is due to the presence of that spherical harmonic. Despite this, the radial part can still be a linear combination of eigenfunctions with ##l=1## but different values of ##n##, right?

PeterDonis said:
No. Stationary states are also eigenstates of ##L^2## and ##L_z## since those operators commute with the Hamiltonian. So a linear combination of stationary states should also be a linear combination of eigenstates of ##L^2## and ##L_z##.
Yes but, we have a degeneracy, so we can have different values of n, where the angular part does not change

So i guess i'm right lol

keyzan said:
the radial part can still be a linear combination of eigenfunctions with ##l=1## but different values of ##n##, right?
Yes, in principle it can. The question is whether the ##l = 1##, ##m = 0## spherical harmonic is really the only one that can produce a factor ##\cos \theta##--in other words, that there is no way to form a linear combination of other spherical harmonics with different values of ##l## and/or ##m## that results in ##\cos \theta##. If you can establish that, then I think your solution is correct. Otherwise you might need to reconsider.

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