How Does Quantum Mechanics Combine Particle and Wave Descriptions?

[flowerew]

"The particle and the wave picture are both simplified forms of the wave packet description, a localized wave consisting of a combination of plane waves with different wavelength."

it's confusing. Can somebody explain it? particle is an object, it seems wave is also an object.How to combine the two thing together?

 

[masudr]

flowerew: How to combine the two things together?

The quantum state vector combines wave and particle properties; one representation (in particular, the position representation) of the state vector is the wavefunction.

 

[FunkyDwarf]

i don't think that's what he's asking. your (masudr) are referring to the wavefunction ie shrodingers equation or wavefunction for position and probility whereas i think flowerew is asking how perpendicular EM and electrical fields can be construted as a particle, which is a good question.

 

[XVX]

I think of it as a localized wave.

Imagine the waves in a pool combining in such a way that there is a big bump in the middle of the pool that smooths out to the boundaries of the pool. A bump, that is a localized standing wave.

Pool waves never do this, because the wavenumbers are pretty much random. But for particles, there is an average wavelength, a Gaussian distribution (for a free particle) that the wavenumbers mostly hang around. Because of this, when you add together an infinite set of waves then, you get a localized standing wave "envelope" that is the particle.

Its comprised of waves, so it has wave properties, but the waves form an overall localized wave that gives it its particle properties then.

As far as EM fields and quantization, you'll need to read some Quantum Field Theory and search some "second quantization" stuff. In short, one can quantitize any normalizable field, which then becomes a QFT. The EM field is normalizable, which is why QED is so successful. But the Einstein Field Equations are not normalizable and hence the trouble.

 

[lightarrow]

XVX,
Many people on this forum agrees on the fact that the wavefunction is just a mathematical representation and not a "real" wave.

 

[actionintegral]

it's confusing. Can somebody explain it? particle is an object, it seems wave is also an object.How to combine the two thing together?

How dare you even speak such words! You obviously are not smart enough to comprehend quantum mechanics. You should be placed under house arrest and go back to studying inclined planes.

 

[quetzalcoatl9]

there are no particles, there are only waves

 

[lightarrow]

there are no particles, there are only waves

How would you prove it?

 

[vanesch]

XVX,
Many people on this forum agrees on the fact that the wavefunction is just a mathematical representation and not a "real" wave.

The wavefunction of 2 particles already doesn't "live" in ordinary space, but in a 6-dimensional configuration space...

 

[masudr]

...how perpendicular EM and electrical fields can be construted as a particle, which is a good question.

The question is asked the wrong way round. It's not that photons (or any excitations of a fundamental quantum field) are constructed out of EM fields. Instead, it's that EM fields are constructed from many zillions of photons.

 

[quetzalcoatl9]

How would you prove it?

where in the Schrödinger equation is there anything but a wavefunction?

what we call a "particle" is just the localized wavepacket in the classical limit

 

[jtbell]

there are no particles, there are only waves

I prefer to say that there are no (pure) particles, nor (pure) waves. Quantum-mechanical objects have both particle-like and wave-like aspects.

 

[ueit]

where in the Schrödinger equation is there anything but a wavefunction?

what we call a "particle" is just the localized wavepacket in the classical limit

AFAIK you need to assume point particles in order to write down the Schrödinger's equation. So, it seems strange to deny the existence of those point particles, isn't it?

 

[masudr]

No you don't.

You need to assume a classical system that possesses a Hamiltonian (or equivalently a Lagrangian).

 

[ueit]

No you don't.

You need to assume a classical system that possesses a Hamiltonian (or equivalently a Lagrangian).

For a hydrogen atom we assume the electron and nucleus to be point charges (we know this is not true for the nucleus and there are deviations for the calculated energy because of that) so the force between them is given by the Coulomb law.

The squared amplitude of the wave function gives the probability of detecting those point charges in a certain place. So, we have particles and their motion is statistically described by a wave.

The motion of water molecules can be described by a wave but this doesn't mean that a water molecule is a wave.

 

[masudr]

For a hydrogen atom we assume the electron and nucleus to be point charges

Well, technically speaking, the Hydrogen atom is treated as a system with Hamiltonian (I'll write it in Cartesians and their conjugate momenta since they are the usual canonical variables in the electrostatic case, although it'd be more concise in spherical polar co-ordinates):

$$\hat{H}(<br /> \hat{x}_1, <br /> \hat{p_x}_1, <br /> \hat{y}_1, <br /> \hat{p_y}_1, <br /> \hat{z}_1, <br /> \hat{p_z}_1, <br /> \hat{x}_2, <br /> \hat{p_x}_2, <br /> \hat{y}_2, <br /> \hat{p_y}_2, <br /> \hat{z}_2, <br /> \hat{p_z}_2) =$$

$$\frac{\hat{p_x}_1^2+<br /> \hat{p_y}_1^2+<br /> \hat{p_z}_1^2}<br /> {2m_1}+<br /> \frac{\hat{p_x}_2^2+<br /> \hat{p_y}_2^2+<br /> \hat{p_z}_2^2}<br /> {2m_2}-<br /> \frac{1}{4\pi\epsilon_0}<br /> \frac{e^2}<br /> {\sqrt{<br /> (\hat{x}_1 - \hat{x}_2)^2+<br /> (\hat{y}_1 - \hat{y}_2)^2+<br /> (\hat{z}_1 - \hat{z}_2)^2}}$$

You haven't assumed anything of the system apart from the fact that it has the above Hamiltonian. To get to that expression, you might use some classical physics of point particles and the Coulombic interaction between charged particles. However, note that the quantum mechanics of the system starts here, at the Hamiltonian.

You haven't assumed anything more, specifically, you haven't talked about particles - you've talked about systems. I'm not being pedantic: in QFT, you don't deal with particles directly, you deal with oscillating fields, where particles happen to emerge.

P.S. You usually eliminate 6 of the variables by transforming to so-called "centre-of-mass" and "relative" co-ordinates, and while this has physical content in this case, it didn't have to.

 

[FunkyDwarf]

there are no particles, there are only waves

particle physics would disagree

 

[quetzalcoatl9]

particle physics would disagree

really? i don't think so

tell me what happens to your particles when their average separation is on the order of the de broglie thermal wavelength? are they still "particles"?

 

[ueit]

Well, technically speaking, the Hydrogen atom is treated as a system with Hamiltonian (I'll write it in Cartesians and their conjugate momenta since they are the usual canonical variables in the electrostatic case, although it'd be more concise in spherical polar co-ordinates):

$$\hat{H}(<br /> \hat{x}_1, <br /> \hat{p_x}_1, <br /> \hat{y}_1, <br /> \hat{p_y}_1, <br /> \hat{z}_1, <br /> \hat{p_z}_1, <br /> \hat{x}_2, <br /> \hat{p_x}_2, <br /> \hat{y}_2, <br /> \hat{p_y}_2, <br /> \hat{z}_2, <br /> \hat{p_z}_2) =$$

$$\frac{\hat{p_x}_1^2+<br /> \hat{p_y}_1^2+<br /> \hat{p_z}_1^2}<br /> {2m_1}+<br /> \frac{\hat{p_x}_2^2+<br /> \hat{p_y}_2^2+<br /> \hat{p_z}_2^2}<br /> {2m_2}-<br /> \frac{1}{4\pi\epsilon_0}<br /> \frac{e^2}<br /> {\sqrt{<br /> (\hat{x}_1 - \hat{x}_2)^2+<br /> (\hat{y}_1 - \hat{y}_2)^2+<br /> (\hat{z}_1 - \hat{z}_2)^2}}$$

You haven't assumed anything of the system apart from the fact that it has the above Hamiltonian.

Then please tell me what is the meaning of x1, px1 and m1? What is their physical significance?

To get to that expression, you might use some classical physics of point particles and the Coulombic interaction between charged particles.

Is there any other way?

However, note that the quantum mechanics of the system starts here, at the Hamiltonian.

Indeed.

I'm not being pedantic: in QFT, you don't deal with particles directly, you deal with oscillating fields, where particles happen to emerge.

I have no training in QFT so I cannot contradict you here, but my understanding was that QFT treats fields as particles and not the other way around.

 

[rlduncan]

The SE can't be solved until you specify the potential energy function, U. For the H-atom, U(r) = -e^2/r in which you assume point charges. Is this not a contradiction to current quantum interpretations?

 

[zbyszek]

Or perhaps you misinterpret wave functions? Does a wave function correspond
to a single quantum object or to an ensamble of single quantum objects?

If the second is true then QM has nothing to say about particle or wave character
of a single quantum object, a single electron for example. QM describes ensambles.

If the first is true then how come that electrons appear as points on the screen
in Young-like experiment although the their wave functions spread across the
whole screen? In other words if QM applies to single quantum objects then one
should be able to predict WITH CERTAINTY (not just probability) an outcome of a
single run of an experiment with one electron!

Cheers!

 

[Careful]

Or perhaps you misinterpret wave functions? Does a wave function correspond
to a single quantum object or to an ensamble of single quantum objects?

Is this a polish local realist speaking ?

 

[vanesch]

Or perhaps you misinterpret wave functions? Does a wave function correspond
to a single quantum object or to an ensamble of single quantum objects?

What is sure, is that the wave function generates correct probabilities for an ensemble. Now, as whether or not one can give a meaning to the wavefunction of a single quantum object depends on what philosophical preferences one has about ontology.
Personally, I think it is impossible not to consider that a wavefunction describes individual systems, given that there aren't always ensembles.

If the second is true then QM has nothing to say about particle or wave character
of a single quantum object, a single electron for example. QM describes ensambles.

This also has foundational problems of course. Where does the ensemble come from if you only do a thing once ? It gives problems in cosmological considerations. Now of course, without "ensemble", probabilities don't make sense either and hence the "single-event" description is only a description, without predictive value, but at least it is a description.

The question is: is the ensemble, the ensemble of objects, or the ensemble of observers ? If you do a thing only once, but this generates an ensemble of observers, then you have your probabilities there...

If the first is true then how come that electrons appear as points on the screen
in Young-like experiment although the their wave functions spread across the
whole screen? In other words if QM applies to single quantum objects then one
should be able to predict WITH CERTAINTY (not just probability) an outcome of a
single run of an experiment with one electron!

This is because one has taken as a hidden ontological assumption that determinism of observation should be true, in other words, that it is not possible to generate an ensemble of observers.

 

[masudr]

The question is: is the ensemble, the ensemble of objects, or the ensemble of observers ? If you do a thing only once, but this generates an ensemble of observers, then you have your probabilities there...

Hehe, the true vanesch style!

 

[zbyszek]

Is this a polish local realist speaking ?

Yes. Not sure if realist, though.

Cheers!

 

[zbyszek]

What is sure, is that the wave function generates correct probabilities for an ensemble.

We agree so far.

Now, as whether or not one can give a meaning to the wavefunction of a single quantum object depends on what philosophical preferences one has about ontology.
Personally, I think it is impossible not to consider that a wavefunction describes individual systems, given that there aren't always ensembles.

I don't care what the philosophical preferences are. The simple question is:
Does QM apply to individual quantum objects or not?

The current state of knowledge is that QM certainly applies to ensembles. If you
claim that it also applies to single objects, justify it.
However, the evidence is against you. Even if a quantum state of a single
object is known perfectly, no one can predict the outcome of a measurement. With
the exeption of several special cases.

The question is: is the ensemble, the ensemble of objects, or the ensemble of observers ? If you do a thing only once, but this generates an ensemble of observers, then you have your probabilities there...

This is because one has taken as a hidden ontological assumption that determinism of observation should be true, in other words, that it is not possible to generate an ensemble of observers.

See, you know so many philosophical concepts and still wasn't able to answer the simple
plain-English question. You just restated it with use of more difficult words.

 

[vanesch]

However, the evidence is against you. Even if a quantum state of a single
object is known perfectly, no one can predict the outcome of a measurement. With
the exeption of several special cases.

Why should I have to predict exactly the outcome of a measurement, in order for my description to be "physical" and not just "statistical" ?

If I know that my wife is in the state "she's doing fine", that doesn't mean that I can predict exactly what she's going to do this evening. But does that mean that the state "she's doing fine" only describes an ensemble of wives of me ? Do you really think that there is a description, so perfect, of my wife (or, of any woman, for that matter), that it tells me what she's going to do, and that "she's doing fine" is not related to herself, as a physical description of her individual state ?

See, the requirement of determinism and perfect predictability is your requirement, and it doesn't need to be so. It is something we got used to in classical mechanics, and it is a special property of that theory, but I don't see why the argument "you can't tell exactly what's going to happen" is necessarily a proof of the statement that the state description you have is only a description of ensemble, and why it cannot have any physical significance for the single object you consider.

If nature, for one reason or another, is fundamentally non-deterministic, your "current state of knowledge" will remain for ever that you don't have a description of the state of nature, but only of ensembles.

 

[zbyszek]

Why should I have to predict exactly the outcome of a measurement, in order for my description to be "physical" and not just "statistical" ?

If I know that my wife is in the state "she's doing fine", that doesn't mean that I can predict exactly what she's going to do this evening. But does that mean that the state "she's doing fine" only describes an ensemble of wives of me ? Do you really think that there is a description, so perfect, of my wife (or, of any woman, for that matter), that it tells me what she's going to do, and that "she's doing fine" is not related to herself, as a physical description of her individual state ?

See, the requirement of determinism and perfect predictability is your requirement, and it doesn't need to be so. It is something we got used to in classical mechanics, and it is a special property of that theory, but I don't see why the argument "you can't tell exactly what's going to happen" is necessarily a proof of the statement that the state description you have is only a description of ensemble, and why it cannot have any physical significance for the single object you consider.

If nature, for one reason or another, is fundamentally non-deterministic, your "current state of knowledge" will remain for ever that you don't have a description of the state of nature, but only of ensembles.

You are joking, right?

In a single run of an electron through slits it ends up in a spot on a screen. Do you
have a theory predicting the spot? Don't do exactly, do it with an arbitrary precision.
If you cannot do that (despite the fact that you know all the states exactly) I say you have no theory of single objects.

What you actually can do, you can tell me the distribution of spots that can be recovered after many repetition of the same experiment with an arbitrary precission, a statistical average over an ensemble. In my dictionary that constitutes a statistical theory.

It very well might be that in Nature there is no theory of single objects. It still does not give
us freedom to put monsters in the uncharted area or stretch known maps to hide them.

But here is a quick sanity quiz:

1. Do you agree that QM does not reveal the spot?

2. How would you call a hypothetical theory that does?

3. If such a theory appears, would you still consider QM applicable to individual objects?


Cheers!

 

[lightarrow]

If the first is true then how come that electrons appear as points on the screen in Young-like experiment although the their wave functions spread across the whole screen?

In which other way could it appear on the screen? The screen is made of many single revelators, and each of them can only make a "click" or not making it.

If many revelators, let's say 1000, on a spread area on the screen, would click at the same time, this wouldn't be interpreted as "the electron has been revelated on a spread area of the screen" but, instead: 1000 electrons have arrived at the same time.

IMO, the problem is to find a physical meaning for the word "electron" (or "particle") because, unless we are talking about an energetic "thing" (and, so, well spatially localized) the "electron" is not a well defined concept and we can only think of it as the "click" of the revelator.

Another, probably similar, problem, is, for example, which precise physical meaning we could give to the word "position" in the case of a wave packet, even for a real wave of any kind. The physical meaning is made by the act of measuring in some way. If the measuring process means to make the wave interact with revelators that can "click" in any instant of time when the wave hits them, then, by definition the "position" of the wave cannot have a more precise value.

I mean, IMO, it's not because of some strangeness of physical word, but just because we have given that meaning to that physical property/concept.

 

[ueit]

But here is a quick sanity quiz:

1. Do you agree that QM does not reveal the spot?

2. How would you call a hypothetical theory that does?

3. If such a theory appears, would you still consider QM applicable to individual objects?

I tend to agree with you that QM is statistics, however, I have an objection to your argument.

The double slit experiment is explained using not pure QM but a semiclassical approximation of it. The electron is treated as a quantum, but the wall, source and detector not. In order to find out what QM really predicts for such an experiment we need a detailed description of the system, that is, the quantum state of the whole system. The result of such a ridiculously complex calculation could give you a much better prediction about the experimental outcome.