I don't think the double slit should count as there is a good chance those two waves going through the slits are entangled. What do you know about unobserved quantum wave behavior?

I agree that the double slit experiment doesn't seem relevant as (in the quantum version) it involves a single particle rather than two independent ones.

You might find something here: https://iopscience.iop.org/article/10.1088/1367-2630/12/6/065029/meta

Or here: https://www.nature.com/articles/474586a

Or here: https://arxiv.org/search/?query="matter+wave+interference"&searchtype=all&abstracts=show&order=-announced_date_first

Or here: https://scholar.google.com/scholar?q=matter+wave+interference

Quote

Is there an experiment that verifies what happens when two unrelated matter waves collide?

If the two objects are orbital electrons belonging to different atoms, the two may combine to form new molecular orbitals, one a bonding orbital and the other an antibonding orbital. These new orbitals have new wavefunctions, different to but formed from the original wavefunctions.

Edited August 22, 20196 yr by studiot

You consider an electron a matter wave?

3 minutes ago, scifimath said:

You consider an electron a matter wave?

Why wouldn't I?

Or is your question "do I consider an electron to be exclusively a matter wave?"

Just now, scifimath said:

You consider an electron a matter wave?

Don't mix "matter wave" aka "Broglie wavelength"

https://en.wikiversity.org/wiki/De_Broglie_wavelength

with wave-functions of probability of finding electron in atom and/or molecule..

 

 

All particles are field excitations which is a wavefunction. All fermionic particles count as matter this includes electrons.

I guess I want something more substantial. Thinking electrons represent whole atoms seems dangerous.

5 minutes ago, studiot said:

If the two objects are orbital electrons belonging to different atoms, the two may combine to form new molecular orbitals, one a bonding orbital and the other an antibonding orbital. These new orbitals have new wavefunctions, different to but formed from the original wavefunctions.

..but he is talking about "matter wave" aka "de Broglie wavelength".. i.e. in diffraction or interference experiments with electrons you use de Broglie wavelength.

1 minute ago, scifimath said:

Thinking electrons represent whole atoms seems dangerous.

I didn't say they did, but how do you think an electron microscope works?

15 minutes ago, Mordred said:

All particles are field excitations which is a wavefunction. All fermionic particles count as matter this includes electrons.

This is great (guess?), but do have any results besides orbitals?

6 minutes ago, scifimath said:

This is great (guess?), but do have any results besides orbitals?

You specified unrelated 'matter waves'

If you are using particles (matter waves) generated by the same source, how do you guarantee that they are unrelated?

If they are already in separate orbitals, this is guaranteed.

I think orbitals go by a slightly different set or rules. :/

why would you generate them by the same source?

Just now, scifimath said:

I think orbitals go by a slightly different set or rules. :/

Well explain what your rules are and what your objectives are.

Interaction of matter waves is an everyday common or garden event in this universe.

 

The point is twofold

(1)

To have a wave you need a wave equation to be satisfied.
There are varying degrees of sophistication of wave equations.

(2)

To pick out the appropriate solution to said wave equation you also need to apply boundary conditions.

 

In both (1) and (2) we normally use  appropriate versions to the situation.

28 minutes ago, scifimath said:

This is great (guess?), but do have any results besides orbitals?

Consider how particles can decay into other particles when the original particle does not contain the other particles.

Consider how two protons can form  Higgs boson as one example.

This is just one example where the field excitation formalism of QFT makes more sense than thinking of particles as little billiard balls.

Well, the point I'm heading to is that they aren't physical.

I want two atom cannons pointed at each other. Shoot a million at random angles, one of them is bound to hit another one ..right? Or do the two waves just ghost eachother?

 

2 minutes ago, scifimath said:

Well, the point I'm heading to is that they aren't physical.

I want two atom cannons pointed at each other. Shoot a million at random angles, one of them is bound to hit another one ..right? Or do the two waves just ghost eachother?

 

If you must have atoms, rather than electrons, the gadget you want is called a field ion microscope.

An unobserved atom smasher is what I need.

8 minutes ago, scifimath said:

An unobserved atom smasher is what I need.

Why unobserved?

I want them to be quantum waves when they "hit" each other.

There are accelerators that use heavy ions. LHC and RHIC. But as these are high energy, they probably don’t show what you’re looking for. Cold atom experiments may be a better bet. (See below)

12 minutes ago, scifimath said:

I want them to be quantum waves when they "hit" each other.

deBroglie wave behavior isn’t wave function behavior, as Sensei already pointed out.

 

 

https://arxiv.org/pdf/1111.6196.pdf

“When the de Broglie wavelength is comparable to the interparticle spacing, the coherent matter waves associated with the various particles in the gas are forced to overlap, meaning (pictorially) that the number of independent quantum states in the gas becomes comparable to the number of gas particles. At this point, the quantum statistics of particles come into play in describing the nature of the gas.”

So, statistically, what happens?

How about two unobserved double slit experiments pointed at each other?

Edited August 22, 20196 yr by scifimath

Letter me ask you a question that you can use to help you.

What happens when two waves interact ? It doesn't have to be matter waves or force waves.

Here

https://en.m.wikipedia.org/wiki/Wave_interference

but it does have to be matter waves

because we, apparently, have never tested with them before.

Why would you think that considering the responses you have recieved ?