I am not asking about wavefunction collapse. I do understand that QM is one of the most experimentally proven theories, but there are different interpretations. What I am asking about is whether the wavefunction travels like a real particle or a virtual particle (just a mathematical model).

I have read these questions:

"Reality" of EM waves vs. wavefunction of individual photons - why not treat the wave function as equally "Real"?

Is there a direct physical interpretation for the complex wavefunction?

Is the wavefunction a real physical wave or only a mathematical abstraction?

where Bob bee says:

Yes, it is physical enough. It is real enough. Eigenstates or projections or some other description of the state of the particle, such as a wave function, are equivalent for your purposes. And the fact is that they have an amplitude (sqrt of probability) and a phase. Both are real. So, whichever words, the property (and we call it all those like prob. amplitude, projection into eigenstates, wave function, etc) is real, is physical. Not just a math concept. As you said, otherwise they would not interfere.

On the nature of the collapse of the wave function

where John Rennie says:

The wave function is not an actual wave - like an electromagnetic wave. It is a collection of numbers that summarizes our knowledge about the physical system and that can be used to make predictions. Any attempt to "overinterpret" the wave function and "visualize" it as a real wave that objectively exists etc. is fundamentally flawed.

Do photons oscillate or not?

where ACuriousMind says:

The wavefunction that models a freely travelling particle is usually a Gaußian wavepacket. This moves, but it does not "oscillate".

If the wavefunction really moves, like a Gaussian wavepacket, then it could be a wave, and if it does have amplitude and phase, it could be a wave, and if it interferes (causes interference), then it could mean that it is a wave too.

Yet, wavefunctions has no physical significance at singularities, at the initial singularity, because the wavefunction is a probability density (its square modulus), not a probability, so the wavefunction started to gain physical significance later, the wavefunction itself might be a human creation, like virtual particles, just a mathematical model.

So basically the wavefunction could either be like a real particle (traveling as a wave) or a virtual particle (just a mathematical model).


  1. Does the wavefunction propagate as a Gaussian wave, or is it just a mathematical model, a virtual wavefunction?
  • 2
    $\begingroup$ Given that there is no agreement among physicists that a photon even has a wavefunction, why are you asking whether wavefunctions travel like photons? Couldn’t you have asked whether a wavefunction travels like an electron? $\endgroup$
    – G. Smith
    Dec 13, 2019 at 23:06
  • $\begingroup$ There are many wavefunctions that are not Gaussian (in position space I assume?). Additionally, you make it seem like propogation of a Gaussian and the wavefunction being a model are mutually exclusive possibilities. Why could it not be both (or neither)? I don't understand the question. What differentiates a wave from a model for you? $\endgroup$ Dec 13, 2019 at 23:26
  • $\begingroup$ @G.Smith you are right I will edit thank you. I only used the photon because there are virtual photons that we use regularly as a mathematical model. I did not hear of virtual electrons though. $\endgroup$ Dec 14, 2019 at 1:33
  • $\begingroup$ @AaronStevens as I understand we describe real photons (traveling as waves), but we use virtual photons only as mathematical models (to describe static fields). What I was trying to ask whether wavefunctions are more like a real photon or just a virtual one (just a mathematical model). Real photons are measurable entities, even as single units in an experiment. Virtual photons are just a mathematical model to describe a static field, and though experiments prove the existence of static fields, no experiment can prove the existence of a single virtual photon (it is just a mathematical model). $\endgroup$ Dec 14, 2019 at 1:37
  • 1
    $\begingroup$ You are still reasoning on the level of words without engaging with any of the math they imperfectly describe, which is a recipe for bad thinking. Witness: "Nothing is better than ice cream. Spinach is better than nothing. Therefore spinach is better than ice cream." This is what all your reasoning with "it must be a wave" sounds like. $\endgroup$
    – knzhou
    Dec 14, 2019 at 3:58

2 Answers 2


The wave function does not “travel” like photon. For one thing the wave function describing two particles in 3d lives in 6-dimensional space, whereas a photon would always travel in 3d space. For another the wavefunction can be complex so one should really concentrate on the time-dependence of $\vert \psi(x,t)\vert^2$ rather than $\psi(x,t)$ itself.

Moreover, a Gaussian wavepacket is just a convenient example because of the simple properties of Gaussian distributions. Since the Gaussian is non-negative, there is nothing oscillating here. In addition, there is nothing to prevent the wavepacket to have any particular other shape, have multiple local maxima (v.g a sum of two separated Gaussians): it basically depends on the initial conditions, i.e. on the initial shape of the wave packet. Moreover, one can craft wavepacket which do not deform as a function of time: the best example would be a particle described by a coherent state in a harmonic well.

  • $\begingroup$ thank you so much. Do you think a wavefunction is just a virtual thing, a mathematical model? Just like a virtual photon (describing a static field as a mathematical model)? $\endgroup$ Dec 14, 2019 at 1:40
  • $\begingroup$ it's not a visual thing. It's just a device to compute quantities. There is an interpretation in terms of state preparation: Peres, Asher. "What is a state vector?." American Journal of Physics 52.7 (1984): 644-650 but otherwise it's just math. $\endgroup$ Dec 14, 2019 at 2:49
  • $\begingroup$ so it is just a mathematical model (like a virtual particle). Thank you. $\endgroup$ Dec 14, 2019 at 2:50
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    $\begingroup$ It's not like a virtual particle or a model. It's just a function that encapsulates the information about the system. $\endgroup$ Dec 14, 2019 at 2:51
  • $\begingroup$ @ÁrpádSzendrei The wavefunction in QM is analogous to the configuration space coordinates in classical mechanics. Systems don't physically move through configuration space (i.e. we can't see this space), yet it is very useful to use configuration space to explain how the system behaves, it's properties, etc. $\endgroup$ Dec 14, 2019 at 12:03

The way I understood this question, I think it is something which humanity has no way to investigate. There is no way to guarantee that a description of what is there is the "actual way it is" and not just a mathematical model. There could always be something more fundamental underlying what is known, and it's not even knowable if math is the language nature would use to describe how it works in the end. You can rule out theories from being the "ultimately correct theory" if they contradict experiment, but you can't prove that they are the final answer.

In the case of Quantum Mechanics / Quantum Field Theory, we certainly don't have the final answer, because of various known problems in the theory, so it seems like there's a good chance that nature won't describe reality using a wave function in the end; the concept of wave function in position space has big issues in QFT anyway.

  • $\begingroup$ "the concept of wave function in position space has big issues in QFT anyway" can you please tell me more about this? $\endgroup$ Dec 14, 2019 at 2:51
  • $\begingroup$ Take a look at this question, for which the only answer is also not lorentz covariant: physics.stackexchange.com/questions/367740/… $\endgroup$ Dec 24, 2019 at 13:53

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