The position of a particle/electron question

[science_rules]

When scientists say the position of a particle or electron cannot be known at the same time as its speed, do they mean the particle's position in reference to jumping from energy levels in a single atom, the position in reference to traveling around the nucleus of an atom, or do they mean position in reference to moving from atom to atom? Or all three?

 

[PeterDonis]

They mean position, period. The restriction applies to any attempts whatsoever to measure a particle's position and momentum.

 

[drvrm]

When scientists say the position of a particle or electron cannot be known at the same time as its speed, do they mean the particle's position in reference to jumping from energy levels in a single atom, the position in reference to traveling around the nucleus of an atom, or do they mean position in reference to moving from atom to atom? Or all three?

position and momentum of a quantum mechanical particle obeys uncertainty principle(Heisenberg's Uncertainty Principle) that's why when you try to locate/define/measure one of them(the pair) exactly the conjugate member ( variable) gets a spread /uncertainty in their value...the principle does not restrict whether a particle is in any state of dynamical motion therefore its not relevant for the measurement process..

Moreover the quantum description is slightly different from our day to day classical experiences with bodies -our classical world is deterministic whereas the quantum discriptions are a bit 'hazy'.

 

[hilbert2]

When scientists say the position of a particle or electron cannot be known at the same time as its speed, do they mean the particle's position in reference to jumping from energy levels in a single atom, the position in reference to traveling around the nucleus of an atom, or do they mean position in reference to moving from atom to atom? Or all three?

If you describe a hydrogen atom's quantum state as a combination of its states of relative motion (motion of the electron and proton with respect to the common center of mass) and collective motion (the motion of the atom as a whole through space), the momentum operator for the collective motion commutes with the position operator of relative motion, and vice versa. Therefore you can in principle know the position of the whole atom's center of mass precisely while at the same time knowing the momentum components of relative motion precisely, too, but that would be a rather crazy quantum state because the electron density distribution around the nucleus would be spread evenly across an infinite volume of space.

 

[PeterDonis]

the momentum operator for the collective motion commutes with the position operator of relative motion, and vice versa.

This is true, but neither of these operators are position or momentum operators for the electron in the atom. They are operators for different subsystems of the complete system (the atom), whose dynamics are uncoupled, which is why they commute.

 

[vanhees71]

When scientists say the position of a particle or electron cannot be known at the same time as its speed, do they mean the particle's position in reference to jumping from energy levels in a single atom, the position in reference to traveling around the nucleus of an atom, or do they mean position in reference to moving from atom to atom? Or all three?

Scientists, when very careful thinking about the foundations of quantum theory, don't say that. They rather say, as the formalism of quantum theory tells them, that position and momentum cannot be determined accurately, i.e., if you prepare particles to have a very well determined momentum their position is very indetermined and vice versa. As with any continuous variable neither position nor momentum can be determined exactly, and this is not a matter of technical problems to do so but an inherent property of these observables. It's formalized in terms of the Heisenberg-Robertson uncertainty relation
$$\Delta x \Delta p \geq \frac{\hbar}{2},$$
where $\Delta x$ and $\Delta p$ are the standard deviations for the $x$ component and $p$ that of the momentum component in $x$ direction (of course it's valid for any direction) with respect to any (pure or mixed) quantum state of the particle.

A careful scientist also would never say that anything "jumps" in quantum theory. The equation of motion for the wave function, i.e., the probability amplitudes, is the Schrödinger equation, i.e., a partial differential equations and thus nothing abruptly jumps. In your example of atoms going from one to another energy state you only have a pretty rapid transition due to the interaction with some disturbance like an electromagnetic field (or a spontaneous emission process due to quantum fluctuations of the electromagnetic field, but that's more complicated since it needs quantum field theory to be correctly described).

Finally the electrons in an energy eigenstate of an atom don't move. Energy eigenstates are stationary states, i.e., the probability distribution for position, $|\psi|^2$, is time independent and so are all expectation values of observables.

 

[science_rules]

If you describe a hydrogen atom's quantum state as a combination of its states of relative motion (motion of the electron and proton with respect to the common center of mass) and collective motion (the motion of the atom as a whole through space), the momentum operator for the collective motion commutes with the position operator of relative motion, and vice versa. Therefore you can in principle know the position of the whole atom's center of mass precisely while at the same time knowing the momentum components of relative motion precisely, too, but that would be a rather crazy quantum state because the electron density distribution around the nucleus would be spread evenly across an infinite volume of space.

What is the "momentum operator" and what do you mean that it "commutes with the position operator of relative motion"?? Also, what is the atom's "center of mass"? And what do you mean that the "electron density distribution would be spread evenly across an infinite volume of space"?

 

[science_rules]

This is true, but neither of these operators are position or momentum operators for the electron in the atom. They are operators for different subsystems of the complete system (the atom), whose dynamics are uncoupled, which is why they commute.

Can you explain this in simpler words? What is an "operator" and what do you mean by "different subsystems" and "dynamics are uncoupled which is why they commute"? I don't know what an operator is, and I also don't know what a subsystem or uncoupled dynamics is. Why do they commute from uncoupled dynamics, whatever that is?

 

[PeterDonis]

Can you explain this in simpler words?

Not without giving you a course on quantum mechanics; these are all basic QM terms. I would suggest working your way through a QM textbook. I have found Ballentine's QM textbook (which I believe you can find PDFs of online) to be a fairly good one.

 

[science_rules]

okay, thanks

Not without giving you a course on quantum mechanics; these are all basic QM terms. I would suggest working your way through a QM textbook. I have found Ballentine's QM textbook (which I believe you can find PDFs of online) to be a fairly good one.