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

 

[keyzan]

Does anyone know how to solve the problem?

 

[PeterDonis]

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?

 

[PeterDonis]

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 $r$.

 

[keyzan]

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.

 

[PeterDonis]

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]

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?

 

[PeterDonis]

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]

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?

 

[keyzan]

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

 

[keyzan]

So i guess i'm right lol

 

[PeterDonis]

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.