Orbits from Orbitals: Maximally Commuting Operators & Coherent States

[arivero]

TL;DR Summary

Asking for optimal ways to make continuous/frequent measurements for a trajectory.

For a particle in a central potential, an orbital is the state determined by Energy, Angular Momentum, and Third Component of the Angular momentum, which constitute a maximal set of commuting operators, right?

Even in classical mechanics, this is not an orbit... it would be a set of orbits, perhaps with some statistics density of probability. But in classical physics, the measurement of position should reveal the exact orbit. In fact, as we now both E and J, it is enough the measurement of angular position.

Of course, in quantum mechanics a precise measurement of position will collapse the state towards a full indeterminacy of J and E. So we need a less precise measurement, so that they are still somehow determined, and then subsequents measurements would approximate/reveal an ellipse, would them? (a circle in the case l=n, a straight line across the center in the case l=0)

Question is, which is the best strategy to make this sequence of measurements? Do you know any article/textbook doing this exercise?

EDIT: I think the question is related to coherent states, but I do not see exactly how. Or perhaps some way to define a set of maximally "quasi-commuting" operators for some set of states.

 

[PeroK]

Summary:: Asking for optimal ways to make continuous/frequent measurements for a trajectory.

For a particle in a central potential, an orbital is the state determined by Energy, Angular Momentum, and Third Component of the Angular momentum, which constitute a maximal set of commuting operators, right?

Even in classical mechanics, this is not an orbit... it would be a set of orbits, perhaps with some statistics density of probability. But in classical physics, the measurement of position should reveal the exact orbit. In fact, as we now both E and J, it is enough the measurement of angular position.

Of course, in quantum mechanics a precise measurement of position will collapse the state towards a full indeterminacy of J and E. So we need a less precise measurement, so that they are still somehow determined, and then subsequents measurements would approximate/reveal an ellipse, would them? (a circle in the case l=n, a straight line across the center in the case l=0)

Question is, which is the best strategy to make this sequence of measurements? Do you know any article/textbook doing this exercise?

What you are asking is fundamentally un-Quantum mechanical. There is no trajectory. There is only the quantum state, fully defined by the three quantum numbers.

 

[arivero]

What you are asking is fundamentally un-Quantum mechanical. There is no trajectory. There is only the quantum state, fully defined by the three quantum numbers.

Yep it is un-quantum in the sense that semi-classicality is unquantum, I guess.

And I am asking for a sequence of quantum measurements of position, applied first to such state and then to the subsequent states. Of course for low n (low Energy) it is very possible that there is not an optimal strategy, but consider a state for n=1000000000. Or consider to start from a quantum state whose radial solution has the main peak about 1000000 km from the center. Do you think that successive measurements of position, with a small error, will be randomly dispersed across all the initial pdf?

 

[PeroK]

Yep it is un-quantum in the sense that semi-classicality is unquantum, I guess.

And I am asking for a sequence of quantum measurements of position, applied first to such state and then to the subsequent states. Of course for low n (low Energy) it is very possible that there is not an optimal strategy, but consider a state for n=1000000000. Or consider to start from a quantum state whose radial solution has the main peak about 1000000 km from the center. Do you think that successive measurements of position, with a small error, will be randomly dispersed across all the initial pdf?

Your question is partly answered here:

https://physics.stackexchange.com/q...ually-observe-the-position-of-the-electron-in

An electron that far from a nucleus, I would say, is experimentally indistiguishable from a free electron.

 

[arivero]

Your question is partly answered here:

https://physics.stackexchange.com/q...ually-observe-the-position-of-the-electron-in

An electron that far from a nucleus, I would say, is experimentally indistiguishable from a free electron.

Thanks. What I do not like of this solution is that it starts always from the same state, thus revealing the orbital. So the classical object to compare is a sample of a set of objects having all of them the same fixed E, J. Such set of objects of course has spherical symmetry, or axial if you also fix J_z

The point of destructivity is interesting. I guess that for large n what we have is so many energy states that transition from n to n+1 is unappreciated. In this sense yep is is pretty indistinguishable from a free electron, sure, but force is still there. Hey, a tiny neutral particle still orbits the sun even it it is as far as the Oort cloud, and electrostatic force is stronger than gravity.

 

[PeroK]

Thanks. What I do not like of this solution is that it starts always from the same state, thus revealing the orbital. So the classical object to compare is a sample of a set of objects having all of them the same fixed E, J. Such set of objects of course has spherical symmetry, or axial if you also fix J_z

The point of destructivity is interesting. I guess that for large n what we have is so many energy states that transition from n to n+1 is unappreciated. In this sense yep is is pretty indistinguishable from a free electron, sure, but force is still there. Hey, a tiny neutral particle still orbits the sun even it it is as far as the Oort cloud, and electrostatic force is stronger than gravity.

The Sun's gravity dominates the solar system. The electrostatic potential from a single proton does not. A better analogy is asking how the gravitational field of a football on Earth affects something in the Oort cloud.

 

[vanhees71]

Thanks. What I do not like of this solution is that it starts always from the same state, thus revealing the orbital. So the classical object to compare is a sample of a set of objects having all of them the same fixed E, J. Such set of objects of course has spherical symmetry, or axial if you also fix J_z

The point of destructivity is interesting. I guess that for large n what we have is so many energy states that transition from n to n+1 is unappreciated. In this sense yep is is pretty indistinguishable from a free electron, sure, but force is still there. Hey, a tiny neutral particle still orbits the sun even it it is as far as the Oort cloud, and electrostatic force is stronger than gravity.

If you want to measure a probability distribution you need of course well-defined conditions. If you want to find the position probability distribution of an electron around a proton in the hydrogen ground state you should by definition always prepare this state and then measure the electron's position. That's how the probabilities in QT defined.

It is utmost important for the right understanding of QT to understand that in such a state the electron does not move (it's a energy-eigenstate, i.e., a stationary state) and that it doesn't make sense to think about the electron moving in some orbit. This was the great drawback of the old Bohr-Sommerfeld quantum theory of atoms. By chance it works for the energy levels of the hydrogen atom (which is due to the large symmetry group the Kepler problem has), but for all other electrons it doesn't work, and the ad hoc solution of inventing "non-radiating orbits" is contradicting in itself. The modern quantum theory a la Heisenberg, Born, Jordan, Schrödinger, and Dirac solved this problem by admitting static solutions for the electron's motion around a proton (and for all other atoms with more than 1 electron too).

You are right in saying that the Coulomb force is long-ranged (despite the fact that it tends to be screened by other charges, which is a great relief in many-body theory, where the Coulomb potential makes a lot of trouble, e.g., in plasma physics described with classical kinetic theory; here Debye screening is your friend), and that the asymptotic free states of point particles are tricky, but that's another story.

 

[arivero]

An issue I have with this kind of objections is that at the end of the way we want to explain classical mechanics as a high energy, or high angular momentum, limit of the quantum state, because h->0 is non existent in the real world.

 

[Nugatory]

An issue I have with this kind of objections is that at the end of the way we want to explain classical mechanics as a high energy, or high angular momentum, limit of the quantum state, because h->0 is non existent in the real world.

Then the Ehrenfest theorem is the explanation that you're looking for.

 

[arivero]

Then the Ehrenfest theorem is the explanation that you're looking for.

Hmm, yes, that is a good hint.

 

[vanhees71]

An issue I have with this kind of objections is that at the end of the way we want to explain classical mechanics as a high energy, or high angular momentum, limit of the quantum state, because h->0 is non existent in the real world.

Classical mechanics in this sense (WKB method or equivalently the saddle-point approximation of the path integral for the propagator) occurs, because the typical action is large compared to $\hbar$ and you can expand the full quantum solution of the Schrödinger equation in a formal series in powers of $\hbar$ (starting with $1/\hbar$). It's a socalled "singular perturbation", and what you get is usually an asymptotic series. Physically you "expand around the classical path", and in this case the quantum fluctuations are small compared to the typical average values of observables.

Then the equations for the averages are also well approximated by the classical equations of motion (the hint to Ehrenfest's theorem is indeed very valuable here!).

 

[arivero]

Note, I just read an interesting dead end to draw keplerian orbits: to start, instead of an eigenstate of angular momentum, from an eigenstate of the Runge-Lenz vector

https://physics.stackexchange.com/questions/89659/eigenfunctions-of-the-runge-lenz-vector

This is because classically one can fix an orbit by having the energy, the angular momentum and the runge-lenz vector, or alternatively the hamilton vector. But all the three quantities do not fix the position of the particle in the orbit (one could consider the dilation operator, xp+px, for it... Berry-Keating anyone :-) ?), so there is still some remnant of indeterminacy.

There is an attempt to use L and A to parametrize solutions here https://link.springer.com/article/10.1007/s002200050152 and here https://link.springer.com/chapter/10.1007/978-94-007-0196-0_9 but I have not finished the reading yet

 

[arivero]

Classical mechanics in this sense (WKB method or equivalently the saddle-point approximation of the path integral for the propagator) occurs, because the typical action is large compared to $\hbar$ and you can expand the full quantum solution of the Schrödinger equation in a formal series in powers of $\hbar$ (starting with $1/\hbar$).

Ah, I see, this paper does a very similar thing,

https://pdfs.semanticscholar.org/3528/8dfad1167b34bdc50ecd694477fbbb9c52ba.pdf