I heard from my lecturer that electron has dual nature. For that instance in young's double slit experiment electron exhibits as a particle at ends but it acts as a wave in between the ends. It under goes diffraction and bends. But we don't see a rise in energy. It has to produce 500kev of energy (please correct if my approximation is wrong) according to mass energy equivalence relation. But wave is a form of pure energy and doesn't show properties of having mass as of experimental diffraction. So where is the mass gone?

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    $\begingroup$ phdcomics.com/comics/archive.php?comicid=1684 $\endgroup$
    – Ajasja
    Commented Feb 24, 2014 at 8:19
  • $\begingroup$ Well, if one uses the Einstein relation between mass and energy then both are essentially the same. The problem is that we tend to think (classically) as "mass$\leftrightarrow $ localized structure". So, if you like, energy is more fundamental in nature because it applies both to particles and waves (even in the classical limit). $\endgroup$ Commented Nov 25, 2022 at 11:50

7 Answers 7


I don't really like the whole wave-particle duality business because it obscures the more startling truth about particles: they aren't sometimes waves and sometimes particles, and they also don't transform into waves sometimes before reforming as particles, they are something completely different.

It's like the story of the blind men and the elephant: a group of blind men are trying to describe an elephant by touch, but each man is touching a different part of the elephant. The man who touches the elephant's side says it's like a rough wall, the man who touches its leg says it's like a pillar, the man who touches its tail says its like a rope, and so on. All the men are right, of course, but they simply have incomplete pictures of the elephant because they cannot observe its full character.

Similarly, when we observe the behavior of things like electrons or photons, we sometimes think they are acting as particles, and other times that they are acting as waves. But really, they are neither waves nor particles, but something new that has properties of both; it is only the case that in many cases, only their particle-like or wave-like behavior happens to be relevant to their behavior, and so we treat them as such. So while the phrase "wave-particle duality" might make it seem as if particles can become waves (hence your question), what actually exists is a strange sort of object that always has properties of both particles and waves, only one of which may be easily observable in some cases, and the question of where an electron's mass goes when it becomes a wave isn't really applicable, since it doesn't become a wave at all.

EDIT: It's worth noting, as Anna pointed out, that the wave-like character of objects isn't quite the same as a normal wave, as there isn't really any physical substance that's waving in any real sense. Instead, the "wave" is a probability function that assigns a probability to each point in space, representing the likelihood of finding the particle there; it just so happens that this function takes the mathematical form of a wave. This is a deep subject, so I'll refer you to Anna's answer for further information.

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    $\begingroup$ +1 Duality is severely botched in introductory courses and books, and doesn't get fixed until the student gets a taste of quantum field theory (assuming even then that the student recognizes the fix), in other words: never for almost everyone. $\endgroup$
    – garyp
    Commented Feb 23, 2014 at 17:18
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    $\begingroup$ You should clarify that the "wave" part does not mean a variation of energy/mass in space, but it is a probability wave . The questioner seems to think that the mass is waving. $\endgroup$
    – anna v
    Commented Feb 23, 2014 at 19:45
  • $\begingroup$ A very well written answer. +1 $\endgroup$
    – Steeven
    Commented Feb 26, 2015 at 18:39

The waves of quantum mechanics are probability waves. The solutions of quantum mechanical equations are the wave functions and the square of the wave function gives the probability of finding the particle at $(x,y,z,t)$.

That is why the solutions for the electrons in the field of a nucleus are not orbits, but orbitals, i.e. probability distributions.

The wave particle duality for particles with mass says that they behave like classical particles when detected macroscopically, as in this bubble chamber picture, or as probability waves when looked at small distances, commensurate with $\hbar$. For photons the particle nature appears in the photoelectric effect, and the probability wave formalism builds up the classical electromagnetic wave with the same frequency.

  • $\begingroup$ no distance can be commensurate with ℏ $\endgroup$
    – hyportnex
    Commented Feb 23, 2014 at 19:05
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    $\begingroup$ @user31748 well, one has to look at the variables of the problem and see where the h_bar enters with the Heisenberg Uncertainty Principle. If delta(x) is very small then delta(p) gets large; if the indeterminacy of the momentum is important then one is in the probability QM range. I guess I used "commensurate" as an approximation to dragging the HUP in. $\endgroup$
    – anna v
    Commented Feb 23, 2014 at 19:37

It seems that you have misundrstood the wave-particle duality.

What happens in the double slit experiment is that the electrons impact at the screen as they were particles. But they also interfere, just as waves. So you can see a wave-particle behaviour.

But it doesn't say that the electron is destroyed, becomes a wave and then a particle again (as you seem to think).

But wave is a form of pure energy and doesn't show properties of having mass as of experimental diffraction.

NO, waves are not pure energy (for example, sound waves have mass). Maybe you understood that it was supposed to be an electromagnetic wave?

As a sidenote, I think that the wave-particle duality gives a wrong impression, since it's not a classical wave and it can be explained using just particles . See my answer here.

Basically, the wave effect is due to the probabilistic behaviour in quantum mechanics, why can be explained using the wavefunction. But we can predict everything using particles and no waves.

You just have to consider that the electrons passes through both slits and sum the contributions of each path:

enter image description here

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    $\begingroup$ In what sense do sound waves have mass? Do you mean relativistic energy-mass relation? Or maybe some sort of effective mass like that of electron in crystal? $\endgroup$
    – Ruslan
    Commented Feb 24, 2014 at 7:27
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    $\begingroup$ @Ruslan I mean, air has mass. $\endgroup$
    – jinawee
    Commented Feb 24, 2014 at 9:10
  • $\begingroup$ Air has mass and since sound is a pressure wave there is more preasure so more air so more mass in the peaks. Right? $\endgroup$
    – aalku
    Commented Feb 24, 2014 at 10:27

The wave-particle duality thing becomes important when you are dealing in a microscopic scale where quantum mechanics becomes relevant and you have to discard your ordinary notion of particle and wave. So don't expect to relate "particle" or/and "wave" notion that you usually get from picturing a marble or water wave from classical world surrounding you.

Point Particle is a concept of structureless, dimensionless entity which only has its existence but no internal structure, so you can't expect an electron to be pictured as a marble ball. And wave is a concept that anything which is a function of space and time whether it is something physical or purely mathematical. So when we talk about "wave-particle duality" we actually mean that in the microscopic world scale - say for election, it's not actually wave or particle at the same time or neither, rather we mean it behaves like particle in some experiment and behaves like wave in some other experiments depending on conditions. Being a particle or wave is fundamentally very different than behaving like particle or wave. moreover we really don't know what a electron really is, for details see The Feynman Lectures on Physics.

So we demanded a wave function to the best describe a electron's nature in quantum mechanics. But when we work in quantum field theory we describe fundamental particles as a quantum field, neither the wave function or quantum field is a physical thing that you can measure or picture or relate to anything to your classical analog. These are the mathematical abstraction that helps the physicist to describe fundamental particles nature in subatomic scales and their interactions. Hope this helps.

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    $\begingroup$ removed "the answers above" as you suggested, actually it is my first post in physics.stackexchange, so I didn't notice the change in order of answeres.thanks @DavidRicherby $\endgroup$
    – Remon
    Commented Feb 24, 2014 at 16:03

Electron is accompained by waves, so there still exists electron which has mass. This solves your problem I hope.

Look here at what de Broglie says in his Nobel lecture of 1929 (this is an extracted portion):

I thus arrived at the following overall concept which guided my studies: for both matter and radiations, light in particular, it is necessary to introduce the corpuscle concept and the wave concept at the same time. In other words the existence of corpuscles accompanied by waves has to be assumed in all cases. However, since corpuscles and waves cannot be independent because, according to Bohr’s expression, they constitute two complementary forces of reality, it must be possible to establish a certain parallelism between the motion of a corpuscle and the propagation of the associated wave.

We shall content ourselves here by considering the general significance of the results obtained. To sum up the meaning of wave mechanics it can be stated that: "A wave must be associated with each corpuscle and only the study of the wave’s propagation will yield information to us on the successive positions of the corpuscle in space". In conventional large-scale mechanical phenomena the anticipated positions lie along a curve which is the trajectory in the conventional meaning of the word. But what happens if the wave does not propagate according to the laws of optical geometry, if, say, there are interferences and diffraction? Then it is no longer possible to assign to the corpuscle a motion complying with classical dynamics, that much is certain. Is it even still possible to assume that at each moment the corpuscle occupies a well-defined position in the wave and that the wave in its propagation carries the corpuscle along in the same way as a wave would carry along a cork? These are difficult questions and to discuss them would take us too far and even to the confines of philosophy. All that I shall say about them here is that nowadays the tendency in general is to assume that it is not constantly possible to assign to the corpuscle a well-defined position in the wave. I must restrict myself to the assertion that when an observation is carried out enabling the localization of the corpuscle, the observer is invariably induced to assign to the corpuscle a position in the interior of the wave and the probability of it being at a particular point M of the wave is proportional to the square of the amplitude, that is to say the intensity at M.

Hear from Broglie about interference:

If we consider a cloud of corpuscles associated with the same wave, the intensity of the wave at each point is proportional to the cloud density at that point (i.e. to the number of corpuscles per unit volume around that point). This hypothesis is necessary to explain how, in the case of light interferences, the light energy is concentrated at the points where the wave intensity is maximum: if in fact it is assumed that the light energy is carried by light corpuscles, photons, then the photon density in the wave must be proportional to the intensity.

  • $\begingroup$ You can read the full noble lecture here $\endgroup$
    – Sensebe
    Commented Feb 23, 2014 at 19:02
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    $\begingroup$ While the lecture may have been noble, it is a Nobel lecture! $\endgroup$ Commented Feb 23, 2014 at 20:30
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    $\begingroup$ @Godparticle: Electron is accompanied by a wave. . .. This is not Copenhagen interpretation or Born's interpretation; you are actually talking of pilot-wave theory. $\endgroup$
    – user36790
    Commented Feb 26, 2015 at 18:20

The so called Copenhagen Interpretation avoids the question about whether the electron is a particle or a wave. This question is directly not allowed. In fact, the wave function is an instrument of the theory with not physical meaning. Acording to CI, the goal of the theory is only to make predictions about the results of a specific experiment. In the case of the double slit experiment, we can ask: what is the probability that a given detector at screen will be activated? The theory has a precise answer to this question. But the question: how does an electron behave? does not relate with a specific experimental situation.

  1. mass=energy(actually mass is a form of energy) so everything who has energy(wave) has mass also
  2. if you squeeze all wave in a very little place you get a solid item(like you).
  3. if you squeeze a solid(like you again) in a very little place you will get a black hole
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    $\begingroup$ Photons have energy but no mass. I don't get the squeezing part. $\endgroup$
    – jinawee
    Commented Feb 23, 2014 at 21:21
  • $\begingroup$ photons have mass.sometimes we just say photons has no mass and when it comes to E=mc2 we say photons have mass. We can say both ways in different conditions just how @EtaZetaTheta has thought. squeezing part is easy. $\endgroup$ Commented Feb 23, 2014 at 21:27
  • $\begingroup$ Almost every physicists talks about mass at rest. Solid objects are not at all related to the spread of the wavefunction. And I not sure how can you describe a black hole with a wavefunction. $\endgroup$
    – jinawee
    Commented Feb 23, 2014 at 21:31
  • $\begingroup$ A boson might have a concentrated wavefunction but it won't be a solid. $\endgroup$
    – jinawee
    Commented Feb 23, 2014 at 21:44
  • $\begingroup$ @jinawee that site has no special message section!! now that' s really interesting. we are bullying on the good question maybe we can continue in another place. $\endgroup$ Commented Feb 23, 2014 at 22:22

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