• What do we actually mean when we say that matter is a wave?

  • What does the wavelength of this matter wave indicate? The idea of a particle behaving like a wave is kinda incomprehensible to me.

  • Further, why is the wavelength inversely proportional to the momentum?

Please help me out.


5 Answers 5


Though @Christoph and @poorsod cover the mathematical concepts, the basic meaning of attributing a wave nature to matter is not emphasized enough.

It is not a matter wave in space time, it is a probability wave that is described by quantum mechanics.

A probability tells me what are my chances to find the particle at a particular (x,y,z,t) and nothing more than that. That the probability has a wave solution due to the nature of quantum mechanical equations, does not make it into a mystical field or entity. It just says that potentially the behavior of matter in a measurement can have the attributes of a wave.

That is the nature of probability functions: when we say that the probability of finding a classical particle with energy E follows a gaussian distribution about E, we do not mean that the particle is really distributed in increments of E. We just estimate the probability of finding the value E when we measure the energy.

Further, why is the wavelength inversely proportional to the momentum?

Because it is a conjecture to start with consistent with the Heisenberg Uncertainty Principle which is a lynch pin of Quantum Mechanics which arises from its basic equations. There is ample experimental verification of this relation.

The answer is because the statement is consistent with experiments.


What do we actually mean when we say that matter is a wave?

We mean that particles like electrons and photons exhibit wave-like phenomena, such as superposition and diffraction. This is more properly known as wave-particle duality. The thought experiment that best illustrates this is the double-slit experiment, in which electrons behave as waves while propagating, and particles when they arrive at the detector. The explanation is that our ideas of particle and wave are in fact ill-defined - every physical object is both a particle and a wave.

Edit 2 to expand upon Anna V's comment below:

This idea is mathematically built-in to the modern formulation of quantum mechanics, which considers the wavefunction $\psi$ to be the fundamental mathematical object describing the particle. The wavefunction interacts with its environment similarly to how a light or water wave interacts with its environment. The wave-particle duality manifests itself when you measure it: the square of the wavefunction $|\psi|^2$ gives you the probability of finding your particle at a given position.

What does the wavelength of this matter wave indicate? The idea of a particle behaving like a wave is kinda incomprehensible to me.

You're in good company - when wave-particle duality was first demonstrated in Planck's calculation of the black-body spectrum it caused quite a stir, as you can imagine.

The wavelength of a particle has a precisely analogous role to the wavelength of the waves you are used to (like sound, light, etc) - for example it is the characteristic length over which diffraction happens.

Why is the wavelength inversely proportional to the momentum?

This is another idea that is built-in to the mathematical formulation of quantum mechanics. As the uncertainty principle tells us, states in which momentum and position both have definite values don't exist. However, states in which momentum has a definite value do exist, and as you might expect in these states position has a totally indefinite value.

I won't go into the detail because I guess it's above your level of knowledge, but it turns out that these states of well-defined momentum are plane waves $\psi = e^{-i k x}$, where the momentum $p = \hbar k = 2 \pi \hbar/\lambda$, and $\lambda$ is the wavelength.

Edit: OK, here is (some of) the detail.

In the formalism of quantum mechanics, states in which an observable quantity (like momentum) has a well-defined value are the eigenstates of the operator associated with that observable. Then the allowed values of the observable are the eigenvalues of that operator.

We're interested in the momentum operator, which is defined as $\hat{p} = - i \hbar \frac{\partial}{\partial x}$. So let's use the eigenvalue equation to find out what the $p$-states are: $$ -i \hbar \frac{\partial \psi}{\partial x} = p \psi, $$ where $p$ here is a number (not an operator). This is a differential equation in $x$, which is solved by the plane-wave solution given above (substitute it in and try it!).

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    $\begingroup$ I would also stress that the wave amplitude expresses the probability of finding the particle at that (x,y,z,t) coordinate. In that it is different than a sound wave, for example, whose energy is spread all over the wave front. $\endgroup$
    – anna v
    Commented Aug 2, 2012 at 18:20
  • $\begingroup$ Good point, I'd neglected to mention that. I'll add it in an edit. $\endgroup$ Commented Aug 2, 2012 at 21:30
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    $\begingroup$ Thanks for the answer!!its very helpful.But, after posting this question I did a little bit of research on the net and got an excellent article from one of the forums or blogs whose names I don't remember . I have posted it as an answer here to ensure that I get it right and don't end up having misconceptions.I know I'm answering my own question but please consider having a look at it. $\endgroup$ Commented Aug 4, 2012 at 19:08

I know you think you've got an answer, but let's give it another try.

First of all, you should get familiar with the double-slit experiment.

Don't just sort of memorize it, but puzzle over it until you can see the ramifications.

What it's saying is that there is something that acts like a probability wave. Normal waves, like in water, work by interfering, that is, by reinforcing and/or canceling each other.

We call the height of a wave its amplitude, which can be positive or negative at any place and point in time. When two amplitudes hit each other and have the same sign, they add together to make a big wave. When they hit each other and have opposite sign, they subtract and make a small wave, or no wave at all (at that place and time). You knew all that.

A wave also has energy, which is proportional to amplitude squared, and it is always positive, right? If you square a positive number, the result is positive. If you square a negative number, the result is positive. That way, whether a wave at a particular place is "high", or "low", its energy is always positive.

OK, now here's the leap into quantum-land:

There is a kind of wave whose energy at a place and time is equal to the probability that a particle exists at that place and time. That is the particle's wave function.

Back to the double-slit. A single particle is shot from the gun. It continually exists at a place and time, which is to say its wave function has a big lump of energy there. But, the wave spreads out. Part of it goes through slit A, and part through slit B, and part gets lost. On the other side, the remaining two components of the wave travel forward and interfere, as waves do. So finally, at the screen, there are places where the waves cancel out, so there's no energy there, so there's no probability there, so there's no particle there. There are other places where the waves reinforce, so there's plenty energy there, so there's plenty probability there, so the particle is likely to be found there.

That's the key idea. You have waves of a "substance" where the energy in the wave is the same as the probability of something. That should make you think, because it means we don't actually know what's so. We only know possibilities, and those possibilities conspire among themselves.

  • $\begingroup$ Thank you for this excellent answer. "Wave whose energy at place and time ~ probability that that particle exists at that place and time" (paraphrased); does this mean that every individual particle has its own wavefunction? And does your statement also carry the implication that since the maximum probability can not exceed 1, there is an upper limit to the energy of the wave (function?) for that particle? $\endgroup$
    – Dunois
    Commented Aug 2, 2020 at 10:49
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    $\begingroup$ @Dunois: Every individual particle has its own wave function? Think of the wave function as existing in a multi-dimensional space. So one dimension for particle 1, another for particle 2, and two dimensions for the pair of particles. Except that over-simplifies it. Maximum energy? My explanation is not great. Feynman's could be the best. I'm a little rusty, but look at a photon as a particle and a wave $exp[i \omega(t-x)]$. Since $\omega$ has no limit, there is no limit to energy. $\endgroup$ Commented Aug 3, 2020 at 0:34
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    $\begingroup$ @Dunois: Now it's coming back to me. Consider an electron as a wave where its time frequency is its energy, and its space frequency is its momentum. What uncertainty means is there could be two or more of each kind of frequency, so you have multiple wave functions. They all interfere with each other (produce a "beat") resulting in the electron being more or less localized in space, and having more or less certain momentum. $\endgroup$ Commented Aug 3, 2020 at 0:45
  • $\begingroup$ I've never heard of this explanation before, but it really gets the intuition across (I don't have any background in physics). Is the "temporal frequency ~ energy, spatial frequency ~ momentum" more hand-waving or is it more ground truth? Is your explanation using the uncertainty principle and interference for localization also true for photons? (I see that you mentioned this in the context of electrons.) $\endgroup$
    – Dunois
    Commented Aug 3, 2020 at 12:18
  • $\begingroup$ @Dunois; I apologize, it's been 30 years since I programmed this stuff, but the basic form of a wave function is $exp( i (w_1 t + w_2 x))$ where $w_1$ is the time-frequency (corresponding to energy) and $w_2$ is the space-frequency (corresponding to momentum). (If I remember it correctly.) Uncertainty is when $w_1$ or $w_2$ or both are not known precisely but follow distributions. In that case you get many closely related wave functions that interfere with each other, producing a "beat" pattern, where the probability is concentrated in a lump that may or may not have velocity. $\endgroup$ Commented Aug 4, 2020 at 19:27

In quantum field theory, elementary particles are the normal modes of fundamental fields, so according to the standard model, matter can indeed be considered a wave.

However, the so-called wave-particle duality is a property of any quantum system, and you can do double-slit experiments not only with photons or electrons, but also $C_{60}$ buckyball molecules. Should we therefore consider buckyballs as excitations of a buckyball-field filling the whole of spacetime? Probably not.

It's just that quantum system exhibit properties not present in classical mechanics, but similar to ones familiar from classical wave optics. The wave-function is not a physical field like the electromagnetic field, though, but rather an object which lives in abstract phase space and related to Hamilton's principal function of the Hamilton-Jacobi formalism.

Goldstein's book on classical mechanics has a chapter on that (9-8 in the 2nd edition): The time-independent Hamilton-Jacobi equation is formally equivalent to the eikonal equation of geometrical optics, where Hamilton's characteristic function plays the role of phase and the energy the role of frequency. The relationship between momentum and wavelength follows from that. This wave length is not the length of a physical wave, but a wave in phase space, where the wave fronts are surfaces of constant action.

  • $\begingroup$ There's nothing wrong with considering a Buckyball field, except that it ceases to be elementary already at rather low temperatures, when the photons can excite the electrons, and turn it into a quantum of an excited buckyball field. $\endgroup$
    – Ron Maimon
    Commented Aug 3, 2012 at 5:18

Thank you for your excellent answers they were so helpful. But, after posting this question I did a little bit of research on the net and got an excellent article from one of the forums or blogs whose names I don't remember . I thought of posting it here to ensure that I get it right and don't end up having misconceptions.I know I'm answering my own question but please consider having a look at it.

In QM, a "wave" isn't what we normally imagine: something that moves up and down and moves in one direction, like water. It's just a function that evolves with time and has a (in general) different value at every point in space. The wave does not "exist" per se in physical space. It can be drawn (superimposed) on physical space, but that just means that it has a value at every point there.

The wave associated with an electron shows the probability of finding it at a particular point in space. If an electron is moving, it will have a "hump" in its vicinity, which shows it's probability at every point in time. This hump will move just like the electron does. When you observe the electron, you collapse the hump to a peak. This peak is still a wave, just narrowly confined so it looks like a particle.

Your issue is that you're trying to look at the "electron" and "wave" simultaneously. This isn't exactly possible. The wave is the particle. You can look at it as if you exploded the electron into millions of fragments and spread it out over the hump. There is a fraction of an electron at every point. The fraction corresponds to the probability of finding it there. At this point, there is no electron-particle. So there's nothing that's "waving". Of course, we never see a fraction of an electron, so these fellows clump together the minute you try to make an observation.

Quantum mechanics has a nice concept called wave particle duality. Any particle can be expressed as a wave. In fact, both are equivalent. Exactly what sort of wave is this? Its a probability wave. By this, I mean that it tracks probabilities.

I'll give an example. Lets say you have a friend, A. Now at this moment, you don't know where A is. He could be at home or at work. Alternatively, he could be somewhere else, but with lesser probability. So, you draw a 3D graph. The x and y axes correspond to location (So you can draw a map on the x-y plane), and the z axis corresponds to probability. Your graph will be a smooth surface, that looks sort of like sand dunes in a desert. You'll have "humps" or dunes at A's home and at A's workplace, as there's the maximum probability that he's there. You could have smaller humps on other places he frequents. There will be tiny, but finite probabilities, that he's elsewhere (say, a different country). Now, lets say you call him and ask him where he is. He says that he's on his way home from work. So, your graph will be reconfigured, so that it has "ridges" along all the roads he will most probably take. Now, he calls you when he reaches home. Now, since you know exactly where he is, there will be a "peak" with probability 1 at his house (assuming his house is point-size, otherwise ther'll be a tall hump). Five minutes later, you decide to redraw the graph. Now you're almost certain that he's at home, but he may have gone out. He can't go far in 5 minutes, so you draw a hump centered at his house, with slopes outside. As time progresses, this hump will gradually flatten.

So what have I described here? It's a wavefunction, or the "wave" nature of a particle. The wavefunction can reconfigure and also "collapse" to a "peak", depending on what data you receive.

Now, everything has a wavefunction. You, me, a house, and particles. You and me have a very restricted wavefunction (due to tiny wavelength, but let's not go into that), and we rarely (read:never) have to take wave nature into account at normal scales. But, for particles, wave nature becomes an integral part of their behavior

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    $\begingroup$ It is not a favorite view for me, since this collapse business is disorienting for me and imo misleading. Also it is a gross exaggeration to say that you and me have "a" wavefunction. Only if we assume we are an elementary particle with the mass of a human body, which is nonsense. We are composed by a decohered, i.e. classical for all intents and purposes,conglomerate of atoms and molecules. $\endgroup$
    – anna v
    Commented Aug 5, 2012 at 6:39
  • $\begingroup$ I agree with @annav. I think this is a rather sloppy description of the wavefunction, and it does not account for the really important features of QM like interference. The Wikipedia pages I linked to in my answer, particularly the one for the double-slit experiment, are much more illuminating. $\endgroup$ Commented Aug 10, 2012 at 22:40

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