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

[I_am_learning]

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

The ground state Energy for a particle in an infinite potential well is given by
E = h² / 8mL²

Consider a particle with high mass, say 1 kg. Then the ground state Energy will be very very low. So the velocity of the particle is very very low. It takes thousands of years for such particle to hit the walls of a say, 1m³ box that confined it.

Suppose I was able to actually setup such condition and put the particle to its lowest Energy level.
Now, If I ask a Quantum Mechanical student, What is the Energy of this particle?
He will calculate E = h² / 8mL² and tell me some value x.
Now, without doing anything to the particle, I decrease the volume (hence L) of confinement by say 10%.
Now, If I ask the same student about the Energy, he will use the same formula but give me another answer, say y. But y > x.

Now I ask him, Why are you supposed to say that the Particles Energy has increased, For I have never touched it, even the slightly?

What will he say now?

 

[Bob_for_short]

It's a mystery of the quantum mechanics!

 

[I_am_learning]

It's a mystery of the quantum mechanics!

If so, can't we say that its fallacy of QM.

 

[Count Iblis]

If so, can't we say that its fallacy of QM.

There is no fallacy. The mass always interacts with the walls (it is not localized). When you decrease the volume, you perform work.

 

[sokrates]

What is "touching" at the quantum level anyway? Why is this even mysterious?

With your numbers L=1 m, m= 1 kg, the energy difference is on the orders of h^2 ~ 1e-68 J ! ...which is zero, even exactly. Your student will say to you that he really can't measure any difference. To get a sensible shift in energy you have to confine a 1 kg ball into a nanometer cube or so, which is physically impossible, considering sizes and weights of atoms, unless you want to put a tiny black hole into the box!

Count Iblis's argument also is very convincing, I can't even feel "a mystery" here.

 

[I_am_learning]

What is "touching" at the quantum level anyway? Why is this even mysterious?

With your numbers L=1 m, m= 1 kg, the energy difference is on the orders of h^2 ~ 1e-68 J ! ...which is zero, even exactly. Your student will say to you that he really can't measure any difference. To get a sensible shift in energy you have to confine a 1 kg ball into a nanometer cube or so, which is physically impossible, considering sizes and weights of atoms, unless you want to put a tiny black hole into the box!
Count Iblis's argument also is very convincing, I can't even feel "a mystery" here.

First thing is that I am not discussing here about whether the Energy Differences of the order 1e-68 J is practically measurable or not. Its measurable, at least theoretically, or say mathematically .

Now the mystery is the result of the thought experiment I already mentioned.
Suppose that we have got an instrument that's able to detect Energies as low as--1e-68 J.
So, my question is as soon as I decrease the volume of confinement, do the Energy Reading of the Instrument Increase? If Yes, then I was able to increase the Energy of a Distant Particle, instantly (thus sending Information faster than light??)

 

[xepma]

You're working with non-relativistic QM, so the speed of information is taken to be infinity anyway.

If you take into account relativistic effects (i.e. a finite speed of light) you will have to deal with the fact that the response of the wavefunction to your change is not instantanious. There is a retardation.

 

[Count Iblis]

Also: The particle will only remain in the ground state with certainty if you change the volume infinitely slowly. This is the Adiabatic Theorem of quantum mechanics.

In the opposite limit you change the volume infinitely fast and then the wavefunction of the particle stays the same . In this case this so-called "sudden approximation" does not work if you decrease the volume, only if you increase the volume.

So, if you increase the volume fast enough, the wavefunction andthus the expectation value of the energy will stay the same. The particle will thus not be in the new ground state.

 

[f95toli]

The effect of time varying potential can be handled even in non-relativisti QM (for "v<<c"). If you suddently start changing the potential the particle will undero a Landau-Zener transition to a new state; the probability of this transition scales with the rate of change.

 

[I_am_learning]

Whatever be the case, (relativistic / non-relativistic),
Does QM implies that--,
if I put a small marble in a big box near the center at almost rest, the marble is always interacting with the wall, though the wall may be kilometers away, So that, any changes in the potential energy or distance between wall results in the change of marble's wave function.

If so the wall should be constantly sending Signal to the marble. What type of signal is it.
I mean, it can't surely be EM signal?

 

[Count Iblis]

The crucial point is this:

Suppose I was able to actually setup such condition and put the particle to its lowest Energy level.

In practice this means that that the whole system has to be cooled to a very low temperature (Exercise: estimate this temperature). Only then will the system be in the ground state and become "de-localized".

 

[I_am_learning]

The crucial point is this:
In practice this means that that the whole system has to be cooled to a very low temperature (Exercise: estimate this temperature). Only then will the system be in the ground state and become "de-localized".

Are you trying to tell me that we can't setup such condition practically or are you trying to tell that there is no way to do that even in theoretical basis.

 

[Dale]

So that, any changes in the potential energy or distance between wall results in the change of marble's wave function.

If so the wall should be constantly sending Signal to the marble.

I don't think it is correct to think of it as "the marble's wave function". The wave function describes the state of the system, and the system includes both the marble and the box. As you change the box you are changing the system so of course the state and therefore the wavefunction must also change. In relativistic QM it does so in a way that respects relativity.

 

[sweet springs]

Now, without doing anything to the particle, I decrease the volume (hence L) of confinement by say 10%.

Hi.
Ｐressure ∂E/∂Ｌ　workｓ　on the wall. You cannot but change energy by moving the wall.
Regards.

 

[Count Iblis]

Are you trying to tell me that we can't setup such condition practically or are you trying to tell that there is no way to do that even in theoretical basis.

No, what I'm saying is that in the given problem, you cannot picture the marble as being in the middle of the box. If you just put the marble in the middle of the box, it will definitely not be in the ground state.

You can write down a wavefunction for the marble's center of mass motion that does describe what will go on in reality if you just put it in the middle of the box. Then you can show that changing the volume does not have any effects.

 

[Bob_for_short]

Let us consider a light QM particle in an infinitely high well of some finite size L. The wave function is zero at the well walls. Does it mean the particle does not touch the walls?

 

[xepma]

The wavefunction goes to zero near the walls. This means the derivative, and also the second derivative of the wavefunction is not zero. The kinetic energy of the wavefunction is determined through the momentum operator squared, which is precisely the second derivative of the wavefunction: $\nabla^2/2m$. So the fact that the wavefunction tends to zero gives the particle some momentum and kinetic energy.

If you move the wall, you force the wavefunction to go faster to zero. This is a consequence of the fact that wavefunction stays normalized. The second derivative increases, and so does the kinetic energy.

 

[Bob_for_short]

What a use of the kinetic energy if the particle does not touch the walls?

 

[xepma]

If you're going to say that the wavefunction or particle doesn't touch the walls, then the only way to realize this is by making the wall an infinitely large and infinitely steep barrier. The typical textbook problem. It's not a very realistic model, but fine, let's just move on from here. In this case there is no overlap with the barrier since the wavefunction cannot have some value in the regime of infinite potential, so indeed, there is no "interaction energy" present.

That doesn't mean the barrier doesn't do anything. On the contrary, the barrier imposes very strict boundary conditions on the wavefunction since it needs to go to zero at the boundary, it needs to stay normalized and it needs to stay continious. In effect, the kinetic energy term goes up.

 

[Bob_for_short]

I am joking. The stationary wave function is zero because it is a superposition of two running waves - the incident and reflected. Of course, the walls participate and experience some pressure.

 

[I_am_learning]

The more simple I wan't this thread to be the more complicated it goes.
Now, My only question is this real life scenario
I make a cardboard box of say 100000000m³. volume. I then roughly place a small marble near the center at roughly stationary with respect to cardboard (Of course you may start saying, that there is nothing like placing near center and roughly stationary, in QM, its just wave function of the particle spread out in space-time. But I am telling these scenario with respect to what I perceive, what I see). To discard complications (if any) I suppose doing this in far outer space.

Now my question is--
If I now move the walls does it change the particles behavior?

 

[Vanadium 50]

Count Iblis answered your question correctly. There is pressure on the walls, and you do work in pushing them in.

Furthermore, when you say "I have a particle in the ground state" and "I have a particle in the middle of the box", you are making two simultaneous statements- you are claiming the particle is simultaneously in an energy eigenstate and it's in a position eigenstate. This is not possible in QM. Asking that it behave classically ("But I am telling these scenario with respect to what I perceive, what I see") will surely not help you understand QM.

 

[sweet springs]

I am joking. The stationary wave function is zero because it is a superposition of two running waves - the incident and reflected. Of course, the walls participate and experience some pressure.

I have a question relating to the waves.
The stationary wave function is wave train of fineite length which is zero outside.

We can represent this wave as superposition of waves of infinite legth by Fourier transformation.

As wave images of this system, which is more appropriate?

Regards.

 

[I_am_learning]

There is pressure on the walls, and you do work in pushing them in.

Do you know of some experiments to detect the pressure on the walls?

Furthermore, when you say "I have a particle in the ground state" and "I have a particle in the middle of the box", you are making two simultaneous statements- you are claiming the particle is simultaneously in an energy eigenstate and it's in a position eigenstate.

I made a big mistake there by using the word middle. I meant to say, I am sure about the particle being in ground state, but not very sure of its position.

"But I am telling these scenario with respect to what I perceive, what I see" will surely not help you understand QM.

It may not, but your help will certainly help me
Is perception BANNED in QM?
In the scenario I mentioned, What do we see then? Do we see the marbles's wave function instead of seeing the marble? Do we see the marble to be spread out in the 100000m³ box, instead of seeing it somewhere?

 

[Bob_for_short]

I have a question relating to the waves.
The stationary wave function is wave train of finite length which is zero outside.
We can represent this wave as superposition of waves of infinite length by Fourier transformation. As wave images of this system, which is more appropriate?

You mean, more practical? It is the first one because it contains only one term. A Fourier series is difficult to sum up exactly and to use in some numerical calculations (iterations) due to numerical instability.

The best basis for any wave in a well is the well proper basis. It may converge quickly.

 

[vanesch]

thecritic,

people have already correctly answered your question, but I think you didn't realize it - which is normal when you just started studying quantum mechanics, BTW. It is extremely counter-intuitive in the beginning.

When you say that you put your particle *in the ground state*, that means that you put it in a state where the wavefunction is non-zero everywhere in the potential well, except at the walls. That, in turn, means that the particle is "everywhere at the same time". (you can discuss about that, I prefer to say it that way, probably because my preferred mental picture is Many Worlds - it is my preferred picture exactly because it helps me explain things that way - interpretation doesn't matter for physics, but it can help to get a mental picture, an explanation). You cannot say that you "put the particle in the middle of the box" and at the same time that you "put it in the ground state". These states correspond to two different quantum states.

Now, accepting that the particle is everywhere at the same time in its ground state, if you move the walls over a finite distance, OF COURSE you have to interact with the particle, because you have to "push" it such that it isn't any more in those places where it was before the wall moved, but which are now outside the well after moving the walls.

So you had to push the particle. It is this pushing which gives it some extra energy. You will also have done work on the walls to push them - which corresponds to the pressure on the wall exerted by the particle in the ground state.

The difficulty is that you are probably still thinking of the particle to be at one place at the same time. In quantum theory, you can't see it that way.

 

[vanesch]

Do we see the marbles's wave function instead of seeing the marble? Do we see the marble to be spread out in the 100000m³ box, instead of seeing it somewhere?

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

 

[sweet springs]

The best basis for any wave in a well is the well proper basis. It may converge quickly.

Thank you for your reply. I agree that the former is easier to calculate.
The benefit of the latter is to see momentum of the system, because the sinusoidal wave of iinfinite lemgth is eigenfunction of momentum. Am I right?

Regards.

 

[Bob_for_short]

...The benefit of the latter is to see momentum of the system, because the sinusoidal wave of iinfinite lemgth is eigenfunction of momentum. Am I right?

Yes, you are rigth as long as you apply it within the well premises. In fact, it is not really necessary to write the running wave outside the well:

sin(π*x/L) = [exp(iπx/L) - exp(-iπx/L)]/2i

Each exponential is already a proper function of the particle momentum. From here you see why the particle momentum is not certain: we have a superposition of different momentum eigenfunctions. Their physical sense is clear: they are running waves (incident and reflected) corresponding to a certain quantized (kinetic in this simple case) energy in the well. You see, everything is already here, no additional Fourier expansion is actually necessary.

 

[sweet springs]

:

sin(π*x/L) = [exp(iπx/L) - exp(-iπx/L)]/2i

Each exponential is already a proper function of the particle momentum.

Thanks for your quick reply. I think these two would be proper functions if they continue outside the well, but really they vanish outside. Momentum distribution is continuous not two plus-minus values. The ground state momentum wave function has only one peak at p=0.
applet in www.falstad.com/qm1d was helpful.
Regards.

 

[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.