Understanding the Wave Nature of Photons in Double Slit Experiments

[jartsa]

If the shape of the wave envelope is such that there are sharp corners, then it's not possible to produce an interference pattern were there is complete cancellation somewhere.

Because the wavelength in uncertain.

I mean complete cancelleation of all the possible waves with different wavelengths is impossible.

Actually wavelength is always somewhat uncertain, so complete cancellation is never possible, I guess.

 

[DParlevliet]

I am not struggling to get my head around QM, but to construct questions and traps to get something out of here. But you don't know and are not interested in the answer. Probably it is too basic, to simple, you only go for answers which covers the whole universe (and outside). So this is not the right place.

The wave is clearly visible for everyone who looks to the double slit experiment. It is the interference pattern. That is the shape, the envelope. Most photons which are absorbed by the detector had this shape when they entered the slits (the rest are for you). With different geometry's you will find the same pattern. Of course there must a formula which describes this as an approximate. But I will not find that here.

But still thanks for all your answers and effort. I still learned from it (although less about QM).

 

[Cthugha]

I am not struggling to get my head around QM, but to construct questions and traps to get something out of here. But you don't know and are not interested in the answer.

You have been given the answer several times. You just do not like it. We cannot help with that. You do not even need to wrap your head around QM. The math is exactly the same as for the double slit in classical optics.

The wave is clearly visible for everyone who looks to the double slit experiment. It is the interference pattern. That is the shape, the envelope.

No. Not at all.

Most photons which are absorbed by the detector had this shape when they entered the slits (the rest are for you).

Also: no.

With different geometry's you will find the same pattern.

Third no. Change the slit distance or their width and you will get a different pattern. Move the position of your light source and you will get a different pattern. If you move too close, you will even get no pattern at all. Change slit orientation and you will get a different pattern. The interference pattern is nothing but a map of the phase difference between two paths leading from the same initial point to the same end point. That is all there is to it. All that matters is the wavelength of the photon used. Besides that, ALL of the double slit pattern just depends on geometry.

 

[craigi]

I am not struggling to get my head around QM, but to construct questions and traps to get something out of here. But you don't know and are not interested in the answer. Probably it is too basic, to simple, you only go for answers which covers the whole universe (and outside). So this is not the right place.

The wave is clearly visible for everyone who looks to the double slit experiment. It is the interference pattern. That is the shape, the envelope. Most photons which are absorbed by the detector had this shape when they entered the slits (the rest are for you). With different geometry's you will find the same pattern. Of course there must a formula which describes this as an approximate. But I will not find that here.

But still thanks for all your answers and effort. I still learned from it (although less about QM).

Sorry dude, you still misunderstand it. Are you still in education? Has this subject cropped up at school/college/university yet?

The double slit experiment is usually not taught until the age of 16 or 17, though a good understanding of it, isn't expected until undergraduate level. Perhaps you need to wait a little while, or perhaps just keep reading, until it falls into place for you.

Do you have a book that covers the subject? I'd recommend using one, if not. Piecing together information from the internet isn't the best way to learn this stuff.

I know for a fact that many of the people in the quantum mechanics section of this forum, have a very good understanding of this subject, beyond anything you'll find from a teacher in High School.

 

[Naty1]

The animation posted above by Dparl is non relativistic:

[source is here:]

http://en.wikipedia.org/wiki/Schrödinger_wave_equation#Time-dependent_equation


Can someone explain how it is appropriate to be referencing such an illustration for light waves,that is photons? If it is ok, where does such a representation breakdown...

Wikipedia says it this way:


http://en.wikipedia.org/wiki/Relati...#Early_1920s:_Classical_and_quantum_mechanics

...the Schrödinger and Heisenberg formulations are non-relativistic, so they can't be used in situations where the particles travel near the speed of light, or when the number of each type of particle changes (which happens in real particle interactions; the numerous forms of particle decays, annihilation, matter creation, pair production, and so on).

 

[Naty1]

What does the following mean in the context of all the explanations in this thread??

http://en.wikipedia.org/wiki/Quantum_field_theory#Unification_of_fields_and_particles

Sometimes, it is impossible to define such single-particle states, and one must proceed directly to quantum field theory. For example, a quantum theory of the electromagnetic field must be a quantum field theory, because it is impossible (for various reasons) to define a wavefunction for a single photon. In such situations, the quantum field theory can be constructed by examining the mechanical properties of the classical field and guessing the corresponding quantum theory.

 

[Bill_K]

What does the following mean in the context of all the explanations in this thread??

This is what VanHees was referring to in #18 above - the impossibility of defining a position operator for a photon.

 

[Naty1]

This is what VanHees was referring to in #18 above -

Ah, I skimmed that post but did not get the context!

Thanks...