If this is indeed the case, I cease to participate in this forum. I shall inquire about by the mentor.weirdoguy said:The prefix is about your level of knowledge, so that everyone can adjust their responses to that.
Is the statement of weirdoguy correct? If this is indeed the case, I cease to participate in this forum. My opinion about its todays level is desastrouos.berkeman said:Mentor Note -- Changed thread prefix "A" Advanced (graduate school level) to "I" Intermediate (undergraduate level)
Unfortunately we cannot continue the dialogue, since I am leaving this desastrous forum after the insult of the Mentor.Laci said:I am sorry, You are wrong. You should be more careful with mathematical formulations. " A hydrogen atom's states are not (or are not) plane waves." is not a meaningful statement.
If a Hamiltonian is separable, then its eigenstates are products of the eigenfunctions of the separated degrees of freedom. This is trivial and You must have learned it already on the example of the motion in a fixed Coulomb potential. Here, due to the spherical symmetry of the Hamiltonian one may separate the radial and the angular motions and the eigenfunctions are ##R_{nl}(r)Y_{lm}(\theta, \phi)## , while the energy in this peculiar case has an ##n^2## degeneracy . If the potential is not Coulombian , the degeneracy in ##l## may be lifted.
In the case of a translational invariant Hamiltonian, like the one of Coulomb interacting charged particles in the absence of an external potential, one may separate the c.m. motion from the relative motion and the stationary states are products of the plane waves for the c.m. motion and the stationary states of the relative motion. The relative motion may be bounded i.e. a true normed eigenstate. The energy of the stationary states are the sum of these stationary eigenenergies. (Plane waves are however not eigenstates, since they are not normed.)
In the case of the hydrogen atom (bound state of electron and proton) the stationary states are the product of the above discussed Eigenfunctions depending on the relative coordinates (##r, \theta, \phi ##) of the electron to the proton (however with the reduced mass ##\frac{1}{\mu}=\frac{1}{m_e}+\frac{1}{m_p}## ) multiplied with the plane wave of the c.m. cooordinate satisfying the Schrödinger equation with the total mass ##M=m_e+m_p##. The total energy is the sum of the kinetic energy of the c.m. plus the eigenenergies of the relative motion. All this You may find in any textbook on QM or You may perform the trivial calculus.
Of course , You may construct normed states (wave packets) of the c.m. motion out of plane waves, but they are not stationary. Any of these atomic wave packets have still a conserved average energy. This energy decreases with the initial width of the packet.
Therefore ona may find hydrogen atoms in any of these states with stationary plane waves or broadening wave packets of the c.m. The motion of the c.m. does not affect the internal (relative) motion of the electron and proton.
After You swallowed the above explanations we may continue the discussion.
Frabjous said:It‘s actually a comment about the level of discussion of the thread, not your abilities.
Frabjous said:the discussions here can get ruder than many would prefer
If someone thinks they are at postgraduate level but they can't understand why a car does not exhibit fundamental QM behaviour, then it's a not a thick skin that's needed.Frabjous said:It‘s actually a comment about the level of discussion of the thread, not your abilities. For this thread, no one is breaking out the formal mathematics.
That being said, the discussions here can get ruder than many would prefer so one needs to develop a thick skin.
Then it is time to lock this thread.Laci said:Unfortunately we cannot continue the dialogue, since I am leaving this desastrous forum after the insult of the Mentor.