Recent content by hokhani

I'm terribly sorry for my serious error. I disregarded the non-commutation of position and Hamiltonian. So, it seems that by time evolution, at least this phase change of the components, results in ##\langle r \rangle =0##. Many Thanks again.

I would appreciate if you could please explain what really occurs in this case, with considerations beyond the classical electromagnetism?

I got to this problem when I was thinking about the electronic transition among atomic states. I am also interested in possibly making correspondence between quantum and classic physics.

Right, but to analyse the problem we solve Schrodinger equation, considering the initial conditions. Although the relative phases vary, ##\langle \hat{r} \rangle ## is always constant. It seems that this problem also persists for all the other QM systems which have stationary states.

Ok, thanks. but the problem persists for a proton instead.

If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we...

As you clearly outlined, ##\gamma_l^m## has complex values on a sphere. In the case of ##p_z## there is no problem because ##\gamma## is real. But generally, ##\gamma=Re+iIm## and so the complex number ##\gamma_1^{+1(-1)}## cannot be represented as dumbbell-shaped ##p_{x(y)}##. Following...

The Spherical harmonics are in the form ##\gamma_l^{m}(\theta, \phi) \propto P_l^{m}(cos \theta) e^{im\phi}## (https://en.wikipedia.org/wiki/Spherical_harmonics).

I think you agree that the p-orbitals are spherical harmonics ##\gamma_l^{m}(\theta, \phi) \propto P_l^{m}(cos \theta) e^{im\phi}## with ##l=1## (https://en.wikipedia.org/wiki/Spherical_harmonics). For ##m=1##, the orbital shape is a dumbbell along the x-axis (please see...

Sorry, I don't know which example do you mean? As far as I understand, the spherical harmonics ##\gamma_l^{m}## are atomic orbitals. For ##l=1## they present p orbitals which are introduced as dumbbell shaped in the literatures. In fact, I don't know how the atomic orbitals are represented in...

The wavefunction of an atomic orbital like ##p_x##-orbital is generally in the form ##f(\theta)e^{i\phi}## so the probability of the presence of particle is identical at all the directional angles ##\phi##. However, it is dumbbell-shape along the x direction which shows ##\phi##-dependence!

Also in classical view, when a particle is rotating about z-axis, the x,y components of angular momentum are not zero. They are zero in the specific case which the ring is in the xy-plane, otherwise the angular momentum ##L## is rotating with the particle.

Please see "Solid State Physics", by Giuseppe Grosso, Giuseppe Pastori Parravicini, chapter 1, Eq. (1.71). It is interesting that in the solids, we have ##\langle \psi_k|\frac{\hat{p}}{m}|\psi_k\rangle = \frac{1}{\hbar} \frac{dE}{dk}## which is compatible with the definition of ##v_g## as ##v_g=...

Could you please see the post #7? It approves more the discrepancy said in the post #1.

However, in the post #1, I sent the function ##\psi=f(r,\theta) e^{im\phi}## which is always the eigenfunction of ##\hat{\mathbf{L_z}}## but not always the eigenfunction of ##\hat{\mathbf{L^2}}##.