Recent content by keyzan

I would like to thank those who have helped me in recent months to understand something about theoretical physics. My 25th goes to you too. Thanks guys. I owe at least 5 points to you. :heart::heart::heart:

So i guess i'm right lol

Yes but, we have a degeneracy, so we can have different values of n, where the angular part does not change

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?

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?

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.

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...

Does anyone know how to solve the problem?

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

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.

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 professor more or less solved it this way, but there are things I don't understand. Let's consider the state: ##|2, 1, 0, +\rangle = |n=2, l=1, m=0, m_{s}=1/2\rangle## Then we have to consider the matrix: ##\hspace{3cm}m=-1\hspace{1cm}m=0\hspace{1cm}m=1## ##m_{s} = \frac{1}{2}...

Hi guys, I have a problem with point 2 of this exercise: The electron of a hydrogen atom is initially found in the state: having considered the quantum numbers n,l,m and epsilon related to the operators H, L^2, Lz and Sz. I am asked: determine the possible outcomes of a measurement of J^2...

Is everything ok? Can I continue with the exercise? Although this zero contribution to the first order of perturbation theory seems a bit strange to me

2. Determine the shift of the energy of the ground state to the first order of the theory of perturbations in ##\lambda##. Solution: At the first order of perturbation theory we have that the energies will be: ##E_n (\lambda) = E_n^{(0)} + \langle \phi_n| \beta W|\phi_n \rangle +...