How Does Quantum Mechanics Combine Particle and Wave Descriptions?

• flowerew
In summary, 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.
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?

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.

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.

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.

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

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.

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there are no particles, there are only waves

quetzalcoatl9 said:
there are no particles, there are only waves
How would you prove it?

lightarrow said:
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...

FunkyDwarf said:
...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.

lightarrow said:
How would you prove it?

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

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

quetzalcoatl9 said:
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.

quetzalcoatl9 said:
where in the schrodinger 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 Schrodinger's equation. So, it seems strange to deny the existence of those point particles, isn't it?

No you don't.

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

masudr said:
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.

ueit said:
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):

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

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

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.

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quetzalcoatl9 said:
there are no particles, there are only waves
particle physics would disagree

FunkyDwarf said:
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"?

masudr said:
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):

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

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

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.

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?

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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!

zbyszek said:
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 ?

zbyszek said:
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.

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vanesch said:
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!

Careful said:
Is this a polish local realist speaking ?
Yes. Not sure if realist, though.

Cheers!

vanesch said:
What is sure, is that the wave function generates correct probabilities for an ensemble.
We agree so far.

vanesch said:
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.

vanesch said:
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.

zbyszek said:
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.

vanesch said:
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!

zbyszek said:
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.

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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.

lightarrow said:
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.

Why not? One can control how many electrons enter a double slit experiment. It can
be just one. If 1000 revelators click simultaneously that would mean that the electron is apporprieatelly spread (like its wave function ? .
No one has seen such wonder yet.

lightarrow said:
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.
Does, for instance, an 'event horizon' have any physical meaning? If it does, who measured it?

lightarrow said:
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.

Our set of concepts evolves. There is still hope :).

Cheers!

zbyszek said:
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.

As I said elsewhere, in an MWI setting, the single electron, after interaction with the screen, the photodetector, the electronics and all that, generates an *ensemble of observers* for that single event (aka "branches" for that observer). Because "you" are only one of them, you can't predict what you'll observe, because you only have the law for the ensemble of observers. So yes, quantum theory is not a theory of a single observer, but only of an ensemble of observers. Nevertheless, it can perfectly well describe a single *object*.

Ok, you don't have to adhere to that view, but to me it is a perfectly sensible one, for two reasons. The main reason is that this is what comes out of the formalism itself, if you consider that everything in the universe is described by QM (and hence has a quantum state, including human bodies and all that) and that you consider that there is only one genuine dynamical law. The second reason is that it allows you to give some ontological status to the formal elements of quantum theory.

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.

I find it more difficult to conceive that *there is no theory of single objects* (and hence that there is no theory of the universe as a whole), than to make a few assumptions that fit perfectly well with the formalism that we have at hand.

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

Yes.

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

If it works, a better theory of course.

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

That will depend on the new theory, with its new ontology and how it relates to the older QM (how the older QM is emergent from the newer one). Maybe the new theory will tell you *how to pick the observer* from its ensemble!

If we change theory, we change of course ontology, and the entire world picture changes completely.

ueit said:
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.

That "ridiculously complex calculation" is nothing else but an MWI view on things. The "problem" with it is the following: if *everything* is described by quantum theory (ie, has a quantum state, a state in Hilbert space), then there's no "outside" which can make an "observation", and the only thing that's left is applying the Schroedinger equation to that quantum state, in which the Hamiltonian includes all the physical interactions between the electron, the screen, the photodetector, the computer, the computerscreen, your eyes, your brain and everything. Indeed at first sight ridiculously complex.
But we know that the Schroedinger equation, no matter how complicated the hamiltonian, is a linear equation.

That means that superpositions of solutions are also solutions.

Now, "electron through the left slit" is the starting state |left>, and if we take this as the starting situation, and solve this tremendiously complicated equation, we will find that the screen, the computer, your eyes, your brain ... will be in a certain quantum state, which we will symbolically represent by |observed-left>.

Similarly for "electron through the right slit", |right> ... which will result in the final quantum state |observed-right>

Well, what will now be the result when we allow for interference ?
Due to the linearity of the Schroedinger equation:

from |left> + |right> will simply follow:

|observed-left> + |observed-right>

In other words, "you" now appear TWO TIMES in the final state, with two *different* observations. This is when, in MWI, we say that "the observer has branched", which means, he now appears in two different states, with each a different outcome.
Or, we say that we now have "two copies" of the observer.
Or, we can say that we have now an "ensemble of observers" with two possibilities.

We didn't do anything special here. We didn't introduce any extra formal elements, we didn't change any equation... we simply assumed the axioms of quantum theory valid "all the way", and applied the Schroedinger equation, which contained all physical interactions (at a ridiculously detailled level). That's why I say that the "multiplication of observers" or the "appearance of an observer-ensemble" appears naturally in the quantum formalism, if you use it rigorously all the way.

Now, it turns out that we, subjectively, don't experience this "multitude of copies". So where does the "statistical aspect" of quantum mechanics seem to reside ? In the fact that we have to repeat the experiment somehow ? That doesn't appear nowhere in the formalism: we did everything for one single incident electron. We didn't start out with a hilbert space of states of several electrons.
No, we saw that, upon this single-electron event, there appeared an ensemble of observers "out of the single one that was present", through the use of the Schroedinger equation.

It hence seems *more natural* to me, as an interpretation of the quantum formalism, that in as much there is an ensemble (which must appear somehow, given that there are probabilities), that the ensemble is on the observer.

I don't want to ram this through people's throats (although it sometimes may sound like that :-), everybody is free to have his/her own ideas on the matter. But I don't find this view as "evidently untenable" as it is often suggested.

vanesch said:
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 ?

my god man, an ensemble of wives?! do you not know that one is quite enough? furthermore, a complete specification of the wife state is an unknowable observable - the exact state can only be known by guessing, and god help you if that guess is wrong.

:)

sorry, i couldn't resist

vanesch said:
As I said elsewhere, in an MWI setting, the single electron, after interaction with the screen, the photodetector, the electronics and all that, generates an *ensemble of observers* for that single event (aka "branches" for that observer). Because "you" are only one of them, you can't predict what you'll observe, because you only have the law for the ensemble of observers. So yes, quantum theory is not a theory of a single observer, but only of an ensemble of observers. Nevertheless, it can perfectly well describe a single *object*.

Oh! Now I get it! You are an MWI guy. I always wanted to understand this branching thing.
Could you help me a little here?

1. Whether a state is a superposition or not depends on the choice of basis. A gaussian
packet can be viewed as a base vector or as a superposition of position eigenvectors.
So, when the branching occures? For one gaussian is there as many worlds as position
eigenstates or just a one, or some other number?

2. Since nobody was able to quantize gravitation it is concivable that it is classical. If it
does not follow the rules of QM then there is just one gravitational field for all the worlds.
Change in one world affects the field so it affects all other worlds.Perhaps we should
be able to detect all other worlds? If the spliting generated enough worlds the gravitation
would be considerable, right?

3. What advantages does MWI provide over other interpretations of QM?

vanesch said:
I find it more difficult to conceive that *there is no theory of single objects* (and hence that there is no theory of the universe as a whole), than to make a few assumptions that fit perfectly well with the formalism that we have at hand.

Or you are too modest! You can easily conceive concepts of ontology and, I bet, epistemology and last but not least infinite complexity of MWI. The lack of theory
for single objects is no match for such a flexible imagination.

Cheers!

P.S. If your do not referee for PRL yet, please ask the editor. You will.

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