hokhani's latest activity

Since ##[H,L^2]=0## we have ##L^2H|l,m\rangle=HL^2|l,m\rangle=l(l+1)H|l,m\rangle##. So, ##H|l,m\rangle## is an eigenstate of ##L^2##...

##|l,m\rangle## are the common eigenstates of ##L^2## and ##L_z##. Also, ##H## is the free particle Hamiltonian (or any Hamiltonian with...

Not the one you wrote down, no. That describes a particle in a bound state. The angular momentum operators of course can be applied to...

Thanks, I think since ##[H,L^2]=0## we have ##H |lm\rangle = \sum_m {\alpha_m |l,m\rangle}## and since ##[H,L_z]=0## we have ##H...

Right, I didn't notice the degeneracy. Thanks.

It is not true that ##\psi## is a free-particle wavefunction just because ##\hat{H},\hat{L}^2,\hat{L}_z## commute when acting on...

For the free particle the three operators Hamiltonian, ##L^2## and ##L_z## commute togethers. So, it seems that ##\gamma_{3,3}## is also...

let's look at the problem quite quantum mechanically. The wave function ##\gamma_{3,3}## describes a free particle and so we expect the...

You know that ##L^2 \gamma_{l,m}(\theta, \phi)=l(l+1)\hbar^2 \gamma_{l,m}(\theta, \phi)## and ##L_z \gamma_{l,m}(\theta, \phi)=m...

Nothing. When QM allows something that is classically forbidden, the explanation is that it isn't really forbidden and the classical...

There is no such thing. Quantum mechanics is not classical mechanics. The fact that a state has definite ##L^2## and ##L_z##, which is...

I think the OP is confused about models and is using the term "classically forbidden" not to refer to any actually valid physics model...

A far as I know, the root of difference between QM and classical mechanics is in the wave nature of quantum particles. So, to deal with...

I would like to know what causes in the wave picture that particle to penetrate in the classically forbidden regions?

Take as an example the quantum state ##\gamma_{3,3}## of angular momentum where ##\vec{L}## makes the angle ##\frac{\pi}{6}## with...