Recent content by hokhani

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## with eigenvalue ##l(l+1)## and so it is linear combination of ##|l,m\rangle## as ##H|l,m\rangle=\sum_m \alpha_m |l,m\rangle##. Similarly, since ##[H,L_z]=0## we...

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

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 |lm\rangle = \sum_l {\beta_l |l,m\rangle}##. Isn't it?

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

For the free particle the three operators Hamiltonian, ##L^2## and ##L_z## commute togethers. So, it seems that ##\gamma_{3,3}## is also an eigenfunction of the free particle. Could you please explain more?

let's look at the problem quite quantum mechanically. The wave function ##\gamma_{3,3}## describes a free particle and so we expect the probability of the presence of particle to be identical everywhere while this wave function depends on the polar angle ##\theta##!

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 \hbar\gamma_{l,m}(\theta, \phi)## which means that at each ##(\theta, \phi)## the wave ##\gamma_{l,m}(\theta, \phi)## is so that we have ##L^2=l(l+1) \hbar^2## and...

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 quantities, we have to use operators which act on all the positions that a particle may be present. In this view, when a particle has specific ##L^2## and...

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 ##z-## axis. This angular momentum state in classical model, describes a circular motion in a plane perpendicular to ##\vec{L}##. So, in all the possible motions...

Ok, and I told since ##[H,\hat{A}=\hat{S_z}]=0##, in the Stern-Gerlach experiment after the time ##t_1## the system is always in the specific spin, say up, and this way my problem in post #13 is resolved. I hadn't considered there the commutation of ##[\hat{A} , H]=0##.

No, I approved your comment about S-G experiment by taking ##\hat{A}=\hat{S_z}## and ##A## in the Hamiltonian is not operator but the vector potential for magnetic field.

You are right, I edited the post #15.

Right, so for example in the S-G experiment, before the first measurement we have the Hamiltonian ##\frac{\hat{P}^2}{2m}##, at first measurement at ##t_1## we have Hamiltonian in the form ##\frac{(\hat{P}-eA/c)^2}{2m}+S_zB## and between ##t_1## and ##t_2## we have again ##\frac{\hat{P}^2}{2m}##...

Thank you and Matterwave. We are at a good point to review my problem again. We would like to measure the quantity ##\hat{A}## and ##[\hat{H}, \hat{A}]\ne0##. By the first measurement which is done at time ##t_1## the system collapses to one of the eigenfunctions of ##\hat{A}##, say...