I Why no plane waves of macroscopic bodies? The micro-macro threshold...

[berkeman]

Mentor Note -- Changed thread prefix "A" Advanced (graduate school level) to "I" Intermediate (undergraduate level)

 

[Laci]

I fully agree with this decision. Hopefully, young students still know the basics of Quantum Mechanics.

 

[PeterDonis]

the motion of the c.m. is separable and has no influence on the energy of the relative motion

The Hamiltonian is separable. But that does not mean the state is separable; it doesn't mean you can just ignore the "internal" part of the Hamiltonian when looking at the stationary states of the system. A hydrogen atom's stationary states are not plane waves, even though its Hamiltonian is separable; the separability of the Hamiltonian is most commonly made use of to set the energy of the center of mass motion to zero, i.e., to justify working in the center of mass frame. But even if you don't work in the center of mass frame, the atom's stationary states are not plane waves.

 

[Laci]

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.

 

[weirdoguy]

I fully agree with this decision. Hopefully, young students still know the basics of Quantum Mechanics.

The prefix is about your level of knowledge, so that everyone can adjust their responses to that.

 

[Laci]

The prefix is about your level of knowledge, so that everyone can adjust their responses to that.

If this is indeed the case, I cease to participate in this forum. I shall inquire about by the mentor.

Mentor Note -- Changed thread prefix "A" Advanced (graduate school level) to "I" Intermediate (undergraduate level)

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.

 

[Laci]

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.

Unfortunately we cannot continue the dialogue, since I am leaving this desastrous forum after the insult of the Mentor.

 

[Frabjous]

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.

 

[weirdoguy]

It‘s actually a comment about the level of discussion of the thread, not your abilities.

Actually both, but the abilities in the context of the subject of the thread. No one says anything about Laci's abilities in general, but this discussions shows that he lacks some of the basics in QM, even though he thinks otherwise.

the discussions here can get ruder than many would prefer

I wouldn't call it rude (there are some exceptions) but mainly straightforward. I know that some people do not like to hear that they lack knowledge in some aspects, but if they really lack it, why should we lie?

 

[PeroK]

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.

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.

 

[DrClaude]

Macroscopic objects are constantly in interaction with an environment (be it, at the very least, the cosmic microwave background), and this interaction is sufficient to localize macroscopic objects to the point where they have a given position FAPP.

 

[DrClaude]

Unfortunately we cannot continue the dialogue, since I am leaving this desastrous forum after the insult of the Mentor.

Then it is time to lock this thread.