Double slit: Human vs Machine Observer

[craigi]

In glass the forward licht also has interference with the back-reflected light (Feynman, QED). Reflection is based on absorbed-emitted photons. Therefore I expect it will be the same in my case.

The difference with reflection is that it's from electrons which are free to take on any vibrational state, so re-emit a photon that is coherent with the incident photon, leaving no record of it's reflection. The electrons in orbit around atoms can't do this.

 

[DParlevliet]

So there is a difference between a reflection back and "forward"?
If you reflect on a small angle, it is almost forward
Or photons which in the glass are absorbed and retransmit an photon forward
Let us forget the foil. The atoms are the "foil" (it is a thought experiment)

 

[craigi]

So there is a difference between a reflection back and "forward"?
If you reflect on a small angle, it is almost forward
Or photons which in the glass are absorbed and retransmit an photon forward
Let us forget the foil. The atoms are the "foil" (it is a thought experiment)

You can reflect on a small angle, no problem. Stand infront of a mirror, then keep moving to the side. It still reflects.

The point is that there's different ways that photons interact with materials. Some presevere coherence, others don't. Scattering is different to specular reflection. Firing things through foil, scatters them. Bouncing things off mirror like surfaces reflects them specularly.

It's not the atoms, as such, that determine specular reflection. It's the surface. The interactions between photons and atoms are different.

A thought experiment isn't supposed to ignore actual physics and make up new physics. It's supposed to be an experiment that you could theoretically perform with actual physics, given hypothetical technology, time and resources.

 

[DParlevliet]

Alright, so now we have a case (1) where photons are not detected (they are emitted by the atoms with the same energy), but no interference.
Then case 2: mirrors are placed at both slits. The light comes from the side and is reflected by the mirrors into the slits. No other light can enter the slits. There is no detection. I suppose there will be interference now?
Compare case 2 with detectors at the slits. In both cases the photon interferes with matter and passes a/the photon. Then the conclusion should be that a detector (and case 1) works in a way that it changes something in the photon, making it unable to interfere.
Could it be possible that a photon can only interfere with its own wave, not with another photon/wave with the same properties. In detectors (and case1) new photons are emitted. But that would mean that in a mirror not a new, but the same photon is re-emitted (or wave reflected).

 

[craigi]

Alright, so now we have a case (1) where photons are not detected (they are emitted by the atoms with the same energy), but no interference.
Then case 2: mirrors are placed at both slits. The light comes from the side and is reflected by the mirrors into the slits. No other light can enter the slits. There is no detection. I suppose there will be interference now?
Compare case 2 with detectors at the slits. In both cases the photon interferes with matter and passes a/the photon. Then the conclusion should be that a detector (and case 1) works in a way that it changes something in the photon, making it unable to interfere.
Could it be possible that a photon can only interfere with its own wave, not with another photon/wave with the same properties. In detectors (and case1) new photons are emitted. But that would mean that in a mirror not a new, but the same photon is re-emitted (or wave reflected).

Detectors end coherence, mirrors, lenses, electromagnetic and gravitational fields don't. You need coherence for interference patterns (but not for interference).

Photons can occupy the same position and state and do interfere with each other. Individual photons interfere with themselves also.

In the cases that you mention, I would say that the photon interacts with matter rather 'interferes with matter'.

Mirror-like surfaces do absorb and emit new photons. The important thing about these photons is that they preserve the phase relationship, frequency and leave no record of path information. This means that they remain coherent and interfere with the incident photon in a way that is consistent with the probabilty distribution of an interference pattern (and all other photons which originate from processes that preserve coherence). Some surface interactions reverse the phase but coherent interference still takes place, in analogy to classical wave mechanics.

 

[DParlevliet]

In case 1, supposing only a layer of atoms, does it leave a record, does it change frequency (you told it did change phase)

 

[craigi]

In case 1, supposing only a layer of atoms, does it leave a record, does it change frequency (you told it did change phase)

More importantly, it changes phase in a non-deterministic way, which prevents the emergence of an interference pattern. A classical analog of this would be destroying waves by waving your hand around eratically on the surface of water.

In your previous post you stipulate that the required process must emit a photon of the same energy. Two photons of the same energy must have the same frequency.

If you use light of the right frequency for a particular atom you can achieve this by exciting an electron to a higher quantum state. After a random time the electron will spontaneously emit a photon of the same frequency. This photon will be of uncorellated phase to the incident photon so you'll see no interference pattern.

It's worth noting that this isn't the only way that photons interact with atoms.

The question of the record of path information is more complex, but the process is not thermodynamically reversible. Information remains in the layer of atoms so that, in principle it could be known which slit the photon went through.

 

[DParlevliet]

Does "leaving a record" not always change frequency and/or phase?

Is the loss of interference not basicly caused by the loss of coherence of the photon, regardless if it leaves a record or not?

 

[craigi]

Does "leaving a record" not always change frequency and/or phase?

No

Is the loss of interference not basicly caused by the loss of coherence of the photon, regardless if it leaves a record or not?

Leaving a record of path information means that the particle takes an exact position so it can't take part in superposition.

Coherent superposition is required for the interference pattern to emerge, which requires the particle to maintain a coherent phase and frequency and have a distributed position. If you take one of these away then, the photon can't contribute to an interference pattern.

It's worth being clear that interfence still takes place when no interference pattern is present. It's just being averaged out.

 

[DParlevliet]

That is better. Now we have only physical properties. I don't like the impression (also in above topic) that the pattern disappers just because nature want to prevent us knowing what is happening.

But if the photon leaves the detector it is distibuted again (when you place a new 2-slit after that, it again gives a interference pattern). In a mirror also the photon is absorbed and emitted by one atom and becomes distributed again. Unless you suppose there is a relation between the distrubuted before and after the mirror. That looks not QM.

 

[craigi]

That is better. Now we have only physical properties. I don't like the impression (also in above topic) that the pattern disappers just because nature want to prevent us knowing what is happening.

But if the photon leaves the detector it is distibuted again (when you place a new 2-slit after that, it again gives a interference pattern). In a mirror also the photon is absorbed and emitted by one atom and becomes distributed again. Unless you suppose there is a relation between the distrubuted before and after the mirror. That looks not QM.

The apparent conspiracy that nature has against us, is just our the failure of our intuition from the macroscopic world. We try to force it on nature and we don't like it when it won't behave according to our prejudices, but it's just a passive set of mathematical rules. It's not out to get us.

I'm not sure what you're trying to say in your 2 new examples.

 

[DParlevliet]

That's right (and you pass nicely the remark about distribution :)

Back to case 1: now I understand what effect you mean. But X-ray diffraction is based on coherent scattering of photons by atoms according the Thomson effect. It radiates in all directions, so also forward. So I think case 1 could be made to have an interference pattern. For the discussion it does not matter.

 

[craigi]

That's right (and you pass nicely the remark about distribution :)

Back to case 1: now I understand what effect you mean. But X-ray diffraction is based on coherent scattering of photons by atoms according the Thomson effect. It radiates in all directions, so also forward. So I think case 1 could be made to have an interference pattern. For the discussion it does not matter.

Could you explain again your "remark about distribution". I didn't pass over it. I just don't understand what it is about "that looks not QM" to you.

Regarding your case 1. The scattering process involved in X-ray diffraction, that you refer to is probably Compton Scattering. Thomson Scattering is a type of Compton Scattering. Neither Thomson Scattering nor Compton Scattering are photon interactions with an atom. They occur when a photons interact with a free charged particles. The "Thomson Effect", is distinct from this and pertains to heat and electric currents. It might help if you could link the webpage that you got this from because it's getting difficult to understand what you mean.

 

[DParlevliet]

Regarding your case 1, Thomson scattering isn't a photon interaction with an atom. It occurs when a photon interacts with a free charged particle.

which is bound to an atom. I did mean everything what happens in a atom. It differs if with Thomson (it was called "effect" in my very very old learning book) the photon is not absorbed by the atom/free charged particle, but only sacttered.
A mirror is that also based on the Thomson effect?

 

[craigi]

which is bound to an atom. I did mean everything what happens in a atom. It differs if with Thomson (it was called "effect" in my very very old learning book) the photon is not absorbed by the atom/free charged particle, but only sacttered.
A mirror is that also based on the Thomson effect?

Nope. A free electron is the exact opposite of an electron bound to an atom.

In Thomson Scattering, a photon is absorbed by a free election (ie. one not bound to an atom). The electron emits another photon of the same frequency, plus a doppler shift, and returns to its initial energy state.

Can you give the name of the book and the author that you're getting this from?

 

[DParlevliet]

Perhaps it was Handbook of X-rays from Kaeble (1967). So:
on what principle is X-ray differaction based?
on what principle is a mirror based?
Then I can search furthre on that.

 

[craigi]

Perhaps it was Handbook of X-rays from Kaeble (1967). So:
on what principle is X-ray differaction based?
on what principle is a mirror based?
Then I can search furthre on that.

Just search for specular reflection and x-ray diffraction.

I suspect that you're looking for an explanation that involves an isolated interaction of only 2 particles. If so, I think you'll be disappointed. As we've already discussed, a mirror requires a surface and you can't make a surface from only one particle. X-ray diffraction typically, involves a lattice and you can't make a lattice with only one particle either.

 

[DParlevliet]

No, I want to know for sure if the principle is based on real absorption-emission or only scattering (probably both the first)

But on my travel home I realized that I forgot the second slit. For instance there is one detector on slit 1, slit 2 is open. I suppose there is no interference pattern.
- If a photon is detected in the slitdetector then is it absorped there (I suppose that happens in every detector) so the wave disappears. If the new photon is emitted it is in the detector, so its wave cannot reach the second slit. There is no interference because there is no wave though slit 2. Slit 1 acts like a new photon source.
- If the photon is detected only in the main detector, it went throught slit 2. If there is no interference pattern, then the conclusion should be that a wave cannot pass a detector, for whatever reason.

Or is now too simple?

 

[DParlevliet]

Just search for specular reflection and x-ray diffraction.

X-ray diffraction is based on Rayleigh scattering. But in articles of both never the physical background is explained. So this leaves open the possibility that it is not based on absorption-emision

 

[craigi]

X-ray diffraction is based on Rayleigh scattering. But in articles of both never the physical background is explained. So this leaves open the possibility that it is not based on absorption-emision

You'll have to tell me what "it" is that you think might "not based upon absorption-emission".

You'll also have to provide a referernce for your Rayleigh scattering X-ray diffraction technique. There are a number of X-ray diffraction techniquies, I'm not aware of any that use Rayleigh scattering.

 

[DParlevliet]

Coherent superposition is required for the interference pattern to emerge, which requires the particle to maintain a coherent phase and frequency and have a distributed position. If you take one of these away then, the photon can't contribute to an interference pattern.

After some thinking I am less satisfied with this.
- Phase is right (when uncorrellated, not fixed). With one photon there will be interference (there are positions on the detector where the photon will never arrive). But with multiple photons it is not possible to build up a measurable pattern.
- Frequency: Suppose a detector exists which emits a new photon with lower frequency but fixed phase. Then according classical wave still interference would be possible. But transferring higher to lower frequency with fixed phase is essential not possible. So the basic reason here is also uncorrellated phase.
- Distributed position: as mentioned before, after leaving the detector the photon is a wave again, so a distributed position. There is no difference with the photon before the detector.

 

[craigi]

After some thinking I am less satisfied with this.
- Phase is right (when uncorrellated, not fixed). With one photon there will be interference (there are positions on the detector where the photon will never arrive). But with multiple photons it is not possible to build up a measurable pattern.
- Frequency: Suppose a detector exists which emits a new photon with lower frequency but fixed phase. Then according classical wave still interference would be possible. But transferring higher to lower frequency with fixed phase is essential not possible. So the basic reason here is also uncorrellated phase.
- Distributed position: as mentioned before, after leaving the detector the photon is a wave again, so a distributed position. There is no difference with the photon before the detector.

Regarding frequency, as I've mentioned before, photons always intefere, but to see an interference pattern the interference must be from photons that have properties that are coherent across multiple paths. A coherent frequency shift could still result in an interference pattern. For example, if at each slit the frequency is shifted by an equal amount. A shift of a different amount on each path is unlikely to result in interference, exceptions to this would be very small shifts ie. wavelength shifts that are much smaller than than the scale of the experimental apparatus or shifts that involve exact or nearly exact wavelength shifts.

Regarding position detection. If a photon has been detected going through one slit, it can't have gone through the other. There is no wave from the other slit to interfere with, so no inteference is possible with this path, hence this photon can't contribute to an interference pattern. The important thing to remember here is the wave represents the probabilty of finding the particle at a location. If it is found at one location, it can't be at another. Once its position is known the wave spreads out again in future time, in fact the more accurately the position is known the wider the spread of the wave. This is the same as the process which causes the wave to spread out after passing through a slit.

 

[DParlevliet]

Position detector: I agree, as I also proposed in my answer #48. With detectors there is no interference pattern because the wave of the new emitted photon in the detector cannot reach the second slit and the old wave is gone.
Therefore my remark about phase is also not applicable anymore.