Double slit: Human vs Machine Observer

[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.