# Deny of wave nature of matter [closed]

I have encountered a particle physicist saying that the wave nature of electron doesn't exist and the wavefunction just gives the probability of finding the electron in a particular position.

Well, actually how can he/she explain the interference pattern in the double-slit experiment? When we don't turn on detectors between the gun of electrons and the result display we see the full interference pattern, but when we turn on the detectors, we see the same pattern to be created particle by particle.

• I think it depends on what you mean by "wave nature". – BioPhysicist Oct 31 '19 at 15:14
• This physicist sounds like a pilot-wave interpretation proponent. I.e. the particle just follows the pilot wave into an interference pattern (the particle and pilot wave being conceptually different things). I personally think it’s a silly position. – Gilbert Oct 31 '19 at 16:10
• You seem to be under the false assumptions that a single particle can create an entire interference pattern and that if we determine which slit a particle went through we would still end up with an interference pattern. Both of these assumptions are false. – BioPhysicist Oct 31 '19 at 20:27
• @writteninmymind Yeah the probability of measuring it at some position in space would still be described by the wavefunction determined by the double slit experiment. That doesn't mean a single particle produced an entire interference pattern though – BioPhysicist Oct 31 '19 at 20:36
• If the electron exists at all positions... then how is its mass distributed across all space? Do we have an infinite amount of mass existing in space, or is the mass spread out infinitesimally everywhere? If the electron exists at all positions then why don't I detect it to be everywhere? – BioPhysicist Oct 31 '19 at 20:40

I am not quite sure what the question is, but the denial of particle wave duality doesn't really have any ground. Of course the wave portion of this duality can be modeled in a probabilistic manner, but there are certainly cases where the physical propagation is a wave form. In the double slit experiment it is noted that single electrons were fired at a time, and despite there only being "one" electron, it interfered with itself causing the same interference pattern as if multiple electrons were fired at once. We aren't exactly sure why this happens, but nonetheless it does. I'd be happy to go into my own theories of why electrons exhibit particle like propagation when observed (if you'd like), but they are unfounded and irrelevant to the topic at hand :)

• "We aren't exactly sure why this happens" - Sure we do. It is called Quantum Electrodynamics known as the most exact theory ever created by the humankind that matches observations to one part in a trillion. – safesphere Oct 31 '19 at 17:48
• Sure, we have the mathematical models to describe HOW the system will behave, but is WHY the system adheres to this model not an entirely different issue? ;) – CuriousOne Oct 31 '19 at 18:04
• I really do not like the "interferes with itself" explanation. What does that even mean? Why does a free electron keep itself to itself, but then when it encounters two slits it says "you know what, I am going to interfere with myself". I just feel like it is an unnecessary explanation that brings along with it some unintended physical consequences. – BioPhysicist Oct 31 '19 at 18:11
• @writteninmymind Is the electron a wave, or is the probability distribution of where we might find it described by a wave? And to the others here, I have no issue with the idea that there is interference of the resulting wave function (as in some parts are amplified and others are decreased by superposition), but I feel like saying "the electron interferes with itself" gives the wavefunction more of a physical role than it ought to have. – BioPhysicist Oct 31 '19 at 18:16
• @CuriousOne I am not saying interference does not occur (please read my previous comment). I just think that specific description equates the electron with the wavefunction. – BioPhysicist Oct 31 '19 at 18:19

Let me say once more what for quantum particles wave-particle duality means.

When quantum mechanical particles interact , they give a footprint of one point, within the measurement errors and the Heisenberg uncertainty principle. That is why they are called "particles".

Here is the double slit one electron at a time experiment.

Electron buildup over time

Note the random pattern becoming an interference one, with the accumulation of the distribution of different electrons with the same energy and through the same boundary conditions. It is obvious that an interference pattern exists. This accumulation is a probability distribution for each electron to be found at the (x,y) of the screen. What is waving is the probability, i.e. the solutions of the quantum mechanical equation, the $$Ψ^*Ψ$$ for the experiment "electron scattering off two slits given distance and given width". That is why it is called a "wavefunction", it is a solution of a wave differential equation.

Again note, the individual electron is not spread all over the screen. The accumulation of electrons displays interference patterns expected by waves.

When "a which way detector" is put after the slits, one is changing the boundary conditions of the experiment and a different wavefunction solution applies. This is seen in this experiment

Overall, the results suggest that the type of scattering an electron undergoes determines the mark it leaves on the back wall, and that a detector at one of the slits can change the type of scattering. The physicists concluded that, while elastically scattered electrons can cause an interference pattern, the inelastically scattered electrons do not contribute to the interference process.

A rule of thumb for the so much abused word "duality" is that when quantum elementary particles interact, they interact as point particles, with a probability following the wave equation solutions for the particular experimental setup.

Here is a bubble chamber picture of an electron

Beam tracks are $$K^-$$ at $$4.2 GeV/c$$ and one of them hits a hydrogen atom with enough momentum to expel an energetic electron, seen to lose energy as it ionizes hydrogen atoms while turning in the magnetic field (B, perpendicular to the picture). All the dots making the tracks are the usual small energy transfers that lead to ionization and allows to see the charged tracks.

There is no spread of the $$K^-$$ all over the place, they behave like classical particles ( until they undergo a deep inelastice interaction with a proton, when a lot of tracks can be produced. See the link for more . It is the accumulation of $$K^-p$$ that allows to study the quantum mechanical behavior/probabilities .)

• anna v for electrons if we perform the double-slit experiment without any detectors the all the screen behind the double-slit will be coloured(just like a wave) with blank spaces due to interference.How do you explain that? – written in my mind Oct 31 '19 at 18:12
• You are implying that an electron physically travels through space as not a wave, but as a single point. If this is the case, then why is there a distinction in the double slit experiment when an observer is added at one of the slits? – CuriousOne Oct 31 '19 at 18:16
• @CuriousOne the last part, the which way experiment explains it, different boundary conditions mean different solutions. Yes, the electron positron beams at LEP were controled basically using classical equations. Electrons scattereed on positrons in a controlled region and all the fits to the data show point particles (ithin HUP) . Bubble chamber interactions also show beam particles coming in, interacting at a point and the accumulation gives the quantum mechanical crossection/rpbability – anna v Oct 31 '19 at 18:48
• @writteninmymind if you do the experiment one electron at a time, you will see the same behavior as above, randomness slowly building up to an interference pattern. – anna v Oct 31 '19 at 19:05
• I think many questions about two slit diffraction could be avoided by careful consideration of your phrase "one is changing the boundary conditions of the experiment and a different wavefunction solution applies". This important point is often missed. – garyp Oct 31 '19 at 19:36

Do you remember details of what he said? If by "wave nature" it's meant "wave-like kinematics", then the statement is wrong. If by "wave nature" its meant the electron is not a wave even though it has wave like kinematics, the statement has merit.

The position/momentum relationship expressed in the Heisenberg Uncertainty Principle applies to all wave-like phenomenon, including classical waves. That it applies to electrons suggest at least a kinematic wave nature. In the Bohr model of the atom, electrons are pictured as waves circling the nucleus, the waves have a wavelength corresponding with an angular momentum that is an integer multiple of $$\hbar$$. Here there is not only wave-like kinematics, the electron itself is considered a wave.

De Broglie–Bohm theory , a pilot wave theory, is one explanation of quantum phenomena that explains wave like kinematics without considering the electron to be a wave. The electron in the double split experiment doesn't go through both holes, nor does it interfere with itself as a wave. This might be what the particle physicist was getting at.

• But the Bohr model is not actual QM. In QM an electron is a vector in a Hilbert (rigged/Fock or whatever) space. No need to say the electron is a wave or not. – jinawee Oct 31 '19 at 17:26
• 'In the present day model for elementary particles, the standard model, the electron is a point particle with the given mass and other attributes shown in the table, which have been experimentally determined. It is not a wave. What has a wave behavior is the wavefunction Ψ determining the probability of finding the electron at (x,y,z,t) by the value of Ψ∗Ψ , dependent on the boundary conditions of the specific problem.' – written in my mind Oct 31 '19 at 17:28
• R.Romero the electron is a wave-particle . It is both particle and wave depending on if it is measured or not. – written in my mind Oct 31 '19 at 17:32
• The standard model predicts the particle properties of an electron. – written in my mind Oct 31 '19 at 17:33
• And he/she says that the wave function is a propability function for the position of the electron.If it was a propability function we wouldnt get the interference pattern in the double-slit experiment. – written in my mind Oct 31 '19 at 17:36

She didn't say "the wave nature of electron doesn't exist." She said an individual electron's position is a point, which is true. There is one position operator $$\hat{x}$$ (for each dimension). Expanding a particle's state in the basis of $$\hat{x}$$ is sufficient to obtain a wave-function.

The "wave nature" comes from the wave-function. And that's exactly what she said: "What has a wave behavior is the wavefunction."

Contrast this with a quantum field, which takes a value at every point in space, and whose wave-functional is obtained from field eigenfunctions rather than position eigenstates.

Edit: Based on your comments, you don't understand the difference between "having a wave nature" and "being a wave". If you want to find out, compare an individual electron in QM to a field in QFT.

• Hmm, actually she does say it. She says that the wavefunction is a mathematical model and the wave doesn't exist – written in my mind Oct 31 '19 at 20:06
• Exactly!She says it is not a wave which is not correct. – written in my mind Oct 31 '19 at 20:09
• @writteninmymind If you send just one electron through the double slit you won't see an interference pattern – BioPhysicist Oct 31 '19 at 20:17
• @writteninmymind "the electron until observed takes all points , it is wave" — Wrong. Having a wave-function does not make it a "wave." – alexchandel Oct 31 '19 at 20:18
• @writteninmymind "How is the interference pattern created on the screen after 1 time experiment was done huh?" — a "1 time experiment" doesn't have an interference pattern. However, the probability distribution is produced by evolution of the wave-function. But again, having a wave-function doesn't make it a wave. You fundamentally misunderstand what a "wave" and "wave-function" are. – alexchandel Oct 31 '19 at 20:20