Recent content by hokhani

I look for the error in the term ##\langle x |P| x \rangle=-i\hbar \frac{\partial}{\partial x} \delta(0)## which seems undefined. Thanks for regarding the preview check, but I don't know why my preview sometimes doesn't work!

Right, it is undefined. However, still another problem remains: The eigenfunctions of the particle in box, ##\psi_m(x)##, should form a complete basis and any function, ##\psi(x)##, seems to be expanded as ##\psi(x)=\sum_m a_m \psi_m(x)## while ##\psi_m(x)## is zero outside the box!

For infinite walls I did that in the post #8.

Very nice separation of the Hamiltonian. If we take ##\psi(x)=\sum_m a_m \psi_m## where ##\psi_m## are the eigenfunction of the Hamiltonian, then it seems that we have ##[V(x),P]\psi(x)=0## since inside the box (##|x|<L##) the potential is zero ##V(x)=0## and for ##|x| \geq L## we have ##\psi_m=0##!

I would like to know how to calculate the ##[\hat{H}, \hat{P}]## for a particle in a 1D box? At the first glance it seems that they commute but they don't get diagonalized in identical basis. How to calculate this commutation?

Definitely. I think I got the answer. A distinguished feature of QM is the quantization of quantities but usually this quantization does not practically manifest itself, unless in the case of a single measurement which system collapses into one of the eigen-kets. Otherwise, the average observed...

Thanks, the goal of raising this question was to know, step by step, how quantization manifests itself in practice while the system is not necessarily in an eigenstate and can be in a superposition of eigenstates. Also, the statistical measurement of an observable gives the expectation value...

If we have ##\psi (0)= \sum_m c_m \psi_m## at ##t=0##, then ##\psi(t)=\sum_m c_m e^{\frac{-iE_m t }{\hbar}} \psi_m## so ##\langle H \rangle=\sum_m |c_m|^2 E_m## is constant.

I don't know why the electrons in atoms are considered in the orbitals while they could be in sates which are superpositions of these orbitals? If electrons are in the superposition of these orbitals their energy expectation value is also constant, and the atom seems to be stable!

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...