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  Why do people still talk about "wave-particle" duality?

+ 2 like - 0 dislike

It comes to me that I have never used "wave-particle duality" in any calculation or interpretation for any physics problem, scattering problem, bound state problem etc. In fact all I have used are just the "wave" part, and the closest I could get to a "particle" picture is just a highly localized wave packet.  It seems the early usage of "particle behavior" was all about some unexpected discreteness showed up in some experiments(photo-electric effect, black-body radiation etc.), but now we know such discreteness can be demonstrated just from waves. To me in these contexts, "particle" seems to be just a synonym for "discreteness", which is a generic attribute of waves under certain circumstances. 

My question is why people still talk about "wave-particle duality" as if there were still mysteries left? Or do they?  Or is it just a semantic game that people like to play with?

asked Aug 5, 2014 in Theoretical Physics by Jia Yiyang (2,640 points) [ no revision ]

3 Answers

+ 6 like - 0 dislike

Maybe an experimentalist's point of view is not useless in this context. All my professional life I have worked with elementary "particles" , i.e. entities localized in space and time enough to leave tracks in bubble chambers and hits in ( x,y,z,t) in electronic detectors. The same electron passing through a double slit and registering as a point on the screen , a point in the interference pattern, will make a beautiful circle turning in the magnetic field of the bubble chamber. It would have been fun to see the interference pattern in  circles in a bubble chamber :).

The way I, and I expect most experimentalists, reconcile this is by keeping very good track  in our point of view of when we are talking of probability distributions  for particles which eventually pertain to the small dimensions where $\hbar$ limits phenomena, and where the dimensions are macroscopic enough to talk about classical type of particles. Probability distributions follow quantum mechanical equations, which after all are wave equations, with boundary conditions and all the paraphernalia. Particles follow classical solutions for macroscopic trajectories in detectors.  And it is particles, with their mass and quantum numbers in the symmetries  of SU(3)xSU(2)xU(1) that led us forward to the standard model, using probability waves to define their cross-sections and widths.

So it is not really a duality  in experimentalist's POV. It is like the proverbial description of an elephant  by four blind men. At the moment we have gotten hold of two attributes  :).

answered Aug 6, 2014 by anna v (2,005 points) [ revision history ]
edited Aug 6, 2014 by Arnold Neumaier

Thanks, I like your answer, +1. I take your answer as an expansion on what Arnold said, "Wave-particle duality naturally and efficiently bridges the two modes of thinking about the microscopic world. "

I considerably prefer your Answer to Arnold's, ... but I would prefer your "Particles follow classical solutions for macroscopic trajectories in detectors" to be instead "sequences of detector events approximately follow classical solutions for macroscopic trajectories". If a number of detector events are obviously lined up, we can more-or-less assign all the events to a single trajectory, whereas more generally we cannot be sure which events are in which trajectory. We say that a set of lined up detector events is caused by a particle, but we could equally say that "particle" is a shorthand way to refer to a lined up set of detector events.

In quantum mechanics more generally, detector events are not the only way that discrete structure is manifest (e.g. line spectra, generated, approximately, using a prism or a grating, correspond to discrete energy eigenvalues, not to discrete detector events), which for me adds enough complexity that I was unwilling to Answer the Question.

+ 5 like - 0 dislike

There is no mystery left, since QM explains wave-particle duality nicely. Indeed, ``particle'' is just the name for a highly localized wave packet that in the (unphysical) classical limit $\hbar\to 0$ would become a classical point particle.

However, the terminology in physics is colored very much by one of the two views. We have research groups on elementary particle theory, not elementary wave theory, since much of the thinking is in terms of particles though the wave description is far more appropriate. Wave-particle duality naturally and efficiently bridges the two modes of thinking about the microscopic world. 

You may like my thermal interpretation of quantum mechanics, which differs from the mainstream interpretations in that it consistently takes the field point of view.

answered Aug 5, 2014 by Arnold Neumaier (15,787 points) [ no revision ]

Thanks and +1, so I guess it's really me getting the wrong impression.

Dear Dr Neumaier,

I appreciate the general description of wave-particle phenomena that you put forward in your thermal interpretation. It is the first convincing explanation of matter diffraction experiments that I could find on the net so far.

Right now I am trying to figure out how it could be applied it to the explanation of the tracks left in a collider detector after complex interactions, producing multiple tracks and secondary decay vertices.

I am familiar with Mott's explanation of alpha particle tracks in a bubble chamber in terms of spherical waves. What would be similar, and what different to that case?

I understand, following Anna, that the field excitations generated by the decays can be approximately described in terms of point particles in the relevant experimental conditions, but what would a full wave description of the phenomenon look like?

In general, has this problem ever been tackled in detail from a theoretical and/or simulation point of view?

Thank you for your time,


Please post this as a separate question, rather as a comment here, where it is a side issue.

Thank you.

I have rephrased the question in more general terms to avoid narrowing down the responses to a specific interpretation.


+ 1 like - 0 dislike

The problem is that the observed "discreteness" is not due to highly localized wave packet. In a double-slit experiment the wave is wide and the observed signals are discrete, at certain positions whose uncertainty is much smaller than the wave size. So the "wave picture" consists of (drawn by) many discrete points.

It looks like before the screen, particles propagate as a wave, they "know "about rather distant obstacles, and while registration they show discreteness, as if they were "point-like" entities. Thus we assign the wave to a probability amplitude rather than to the particle wave packet concentration.

If a particle is indeed point-like or small, how does it know about distant obstacles? Note, for determining the wave propagation, one should take into account some boundary conditions, which are "distant", but influence the wave form. Hence, people speak of this duality.

@Dilaton: As I have no right to reply to your comment and to comment the post, I am writing my disagreement in the answer body. A detector can reveal the effect of extendedness of a photon. For example, if a photon has a very narrow width in energy, i.e., if is is nearly monochromatic, in order to register it "completely" by absorption, the detector should not change its velocity (or state) during detecting. The period of detecting should be rather large for such a photon (a long wave-train, many-many wave-lengths). Otherwise the photon may escape detecting (I mean a resonance or a thershold method).

answered Aug 5, 2014 by Vladimir Kalitvianski (102 points) [ revision history ]
edited Aug 5, 2014
Most voted comments show all comments

I like the part about the wave being just a probability wave. But I think for being absorbed as a particle, it is not important if the particle is really point like or not. The detector can certainly not reveal effects of extendedness of elemetary particles such as photons ...

``The discrete points shown on the screen are just waves "collapse"(let's use the collapse picture for now, I don't want to go into interpretational issues) into position eigenstates?''

Photons do not have position eigenstates! See the entry Localization and position operators from my  theoretical physics FAQ.

Vladimir, your argument is multiply faulty:

1. A photon has no classical equation of motion.

2. For massive particles, Ehrenfest's theorem implies that the position $\langle x\rangle$ in the ensemble mean satisfies the classical equation only for a free particle.

3. the position $\langle x\rangle$ if considered as a mean over all points in space covered by an object is a point even for very extended objects like the Moon, but nobody considers the Moon to be pointlike. 

The Ehrenfest equation only superficially looks like a classical mechanics equation. The left hand side is a derivative of <q(t)> but the right hand side is not a function of <q(t)> but of the operator q(t). Moreover, <q(t)> is the ensemble mean, not the mean of the positions occupied by  the spot. In a double slit experiment, <q(t)> is therefore situated at the center of the interference diagram! For the Moon, the Ehrenfest equation is not a differential equation fir the center of mass, but has 10^28 components or so!

And geometric optics is not a theory of point particles - it was invented long before anyone dreamed of a photon. 

@ArnoldNeumaier: I just wanted to say that we often (if not always) call a "particle" something "big" by taking some sort of average for obtaining its "position".

Most recent comments show all comments

@ArnoldNeumaier, I know, and I got to know it first from your FAQ! As for my previous comment on double slit experiment, maybe what really matters is that the atoms making up the screen have well defined positions?

Arnold, to obtain a "point" from a "spot", it is not necessary to decrease the spot size (like $\hbar\to 0$). It is sufficient to take an average position of all spot points and you obtain a unique "position". It is this "position" who follows the classical trajectory.

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