Particle in a potential Well interacts with the wall without touching it?

[Bob_for_short]

You may consider them as continuining outside. The zero vaue outside is due to the fact that apart from the incident wave there is an emitted with the wall material one and they come in the exactly opposite phases. This is how reflection happens in reality.

Their discretenes is due to considering a specific wave function - an eigenstate of the Hamiltonian which is discrete in the well.

We can consider a superposition of eigenstates - with no certain energy an momentum in it but only with average ones.

 

[sweet springs]

You may consider them as continuining outside.

Thanks. Continueing outside means that we have non-zero possibility to observe the particle outside the well. I feel it wrong.
Regards.

 

[Bob_for_short]

It is not alone outside. It comes together with the emitted wave and they cancel each other. So no need to consider something outside.

 

[I_am_learning]

From the moment you SEE the marble, it isn't in its ground state anymore. The act of "seeing it" has changed its state.

Thanks, it has enlightened me. I forgot that observing any particle would change its state.

Now, I am thinking of one thing.
Is this concept of pressure on walls same as that with the kinetic theory of gases?
What my train of reasoning is that
Since you can't peek inside the box, you don't know anything about the particles position; Weather its in the center, or near the boundary (about to hit the wall) etc. (thats why in QM we don't say the particle's position is somewhere now and its moving; instead we say that the particle is spread out in the volume of the box).
Since QM is involved with probability and Statistics,
it implies that if we take a lot of particles inside the box (provided they don't interfere), or equivalently take a lot of boxes with particles, then within a small time dt, there are finite no. of such particles hitting the walls. And this is what constitute the pressure on the walls.

Just like In particle in a box scenario, the probability distribution won't be apparent in just one experiment with one particle and one box, but it will be highly apparent if we do large no. of such experiments;
I think the pressure will also be not apparent with just one experiment.

I would be very happy to know if all of my above reasoning were correct?

 

[sweet springs]

It is not alone outside. It comes together with the emitted wave and they cancel each other. So no need to consider something outside.

Thanks. So we may agree that the more precise formula of the ground state wave function
is exp(iπx/L) - exp(-iπx/L) multiplied by the distribution
{θ(x)θ(L-x)} which is 1 inside and 0 oustide of the well,
Do you? The results of Fourier transform are different with {θ(x)θ(L-x)} multiplied or not.
Regards.

 

[Bob_for_short]

Yes, we agree about the WF and about sin(πx/L). Without cut-off factor the sine has only one Fourier component, with this factor we need much more harmonics to get the wave "localized".

 

[sweet springs]

Yes, we agree about the WF and about sin(πx/L). Without cut-off factor the sine has only one Fourier component, with this factor we need much more harmonics to get the wave "localized".

Much more harmonics corresponds with much more momentum to take.
Simple reflection between two walls of just two plus-minus momentum values is not accurate in this case.

Thank you for your suggestion.

 

[vanesch]

Thanks, it has enlightened me. I forgot that observing any particle would change its state.

Now, I am thinking of one thing.
Is this concept of pressure on walls same as that with the kinetic theory of gases?

Yes, although that's absolutely not the way people look at it in standard kinetic theory of gasses!

What my train of reasoning is that
Since you can't peek inside the box, you don't know anything about the particles position; Weather its in the center, or near the boundary (about to hit the wall) etc. (thats why in QM we don't say the particle's position is somewhere now and its moving; instead we say that the particle is spread out in the volume of the box).
Since QM is involved with probability and Statistics,
it implies that if we take a lot of particles inside the box (provided they don't interfere), or equivalently take a lot of boxes with particles, then within a small time dt, there are finite no. of such particles hitting the walls. And this is what constitute the pressure on the walls.

Be very careful: quantum mechanics is not a substitute for ignorance (although some people tend to think of it that way). It is not because you don't know something that that thing has necessarily a spread-out wavefunction. It is not because you don't know the results of the lottery tomorrow, that the little balls in the lottery machine have a wavefunction that is spread out over the entire machine for instance.

There is a fundamental difference in quantum theory between a wavefunction which has, say, spatial extension, and an ensemble of particles that have a statistical spread of spatial positions, but each of them has a sharp "position" wavefunction. That said, for several measurements, both systems (the one with a broad wavefunction, and the one with a statistical ensemble) will give similar results, but there are experiments in principle that can show the difference between both. Indeed, the "spread out wavefunction" system can show (quantum) interference, while the "ensemble" system can only show "statistical evolution".

In fact, "the act of measurement" converts the "spread out wavefunction" system into a "statistical ensemble", or at least in a system that behaves as if it were correctly described by a statistical ensemble ; it "kills" the interference properties of the spread-out wavefunction. Decoherence theory is a way of looking at this process.

But, to come back to the original question: in as much as you could consider the particles in a gas as being described by wavefunctions that "fill the box", you would indeed find that the pressure on the walls is exactly the same as in the usual description of "little balls that bounce off the wall". However, particles that have wavefunctions that fill the box would only be a correct description with very cold gasses and with low pressures I would guess.

But for pressure, it is one of those experiments that DOESN'T make any difference between the "spread out wavefunction" and the "ensemble". However, there are other possible experiments that would make a difference, and then for most gasses, they would show you that there' s no such wavefunction.

(I'm thinking of making two pinholes in the box on the same surface, at a certain distance one from another, and look at interference patterns: I think you would be disappointed with most gasses).



Just like In particle in a box scenario, the probability distribution won't be apparent in just one experiment with one particle and one box, but it will be highly apparent if we do large no. of such experiments;
I think the pressure will also be not apparent with just one experiment.

I would be very happy to know if all of my above reasoning were correct?[/QUOTE]

 

[I_am_learning]

So, vanesch, is it safe to conclude that-->
Both "Statistical Ensemble" And "QM" can explain quantum mechanical phenomenons, like the pressure on the walls, probability Distribution and so on. However, interference and Diffraction of the wave function is the key experiment that establishes QM's validity over "statistical ensemble".
?

 

[I_am_learning]

If what I said above is true, why not use the concept of statistical Ensemble to explain quantum mechanical phenomenon except Interference and Diffraction.