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When a quantum of light arrives at a double slit, it passes through both slits as a wave and arrives upon a second screen with the interference pattern of a single wave that has been split into two waves, that have then interfered with each other.

If a detector is placed at one of the slits and the duality is detected at either of the two slits the detected duality proceeds to and arrives at the second screen as a photon/particle that has 'emerged' from its electromagnetic wave.

Therefore can one assume that detection has 'caused' the collapse of the wave portion of the duality?

How has detection precisely influenced the duality? Can anyone clarify?

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    $\begingroup$ Hi and welcome to PSE. This answer is related to your question: physics.stackexchange.com/q/134849 $\endgroup$ – user179430 Dec 25 '17 at 23:52
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    $\begingroup$ arrives at the second screen as a photon/particle that has 'emerged' from its electromagnetic wave. Not true, where did you get this idea? $\endgroup$ – Ben Crowell Dec 26 '17 at 14:46
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    $\begingroup$ As I understand it the quantum duality may behave as either a particle, or a wave. No, a more accurate statement would be that it always behaves as both a particle and a wave. Outside of measurement it contains both. "Contains" isn't the right word. There isn't a photon that contains both a particle and wave. A photon just is both a particle and a wave. It's a particle because you can't have half of one. It's a wave because it obeys the principle of superposition. $\endgroup$ – Ben Crowell Dec 26 '17 at 23:08
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    $\begingroup$ My understanding of the double slit in the presence of detectors is that particulate probability is increased (emerges) and in the absence of detection wave probability is increased or is maintained. No, this is wrong. There is not a probability of being a particle and a probability of being a wave. Therefore I am assuming that a given quantum duality has the emergent properties of a particle and or those of a wave. This isn't what emergent means. It means that you have a theory that doesn't explicitly have some feature built in, but that feature emerges, e.g., snowflakes are hexagonal. $\endgroup$ – Ben Crowell Dec 26 '17 at 23:13
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    $\begingroup$ Paper reference for one realization of this experiment: Chapman MS, et al.*Phys Rev Lett.* 1995 Nov 20;75(21):3783-3787. doi.org/10.1007/978-1-4757-9742-8_18 (and available for free from chapmanlabs.gatech.edu/papers/scattering_ifm_prl95.pdf) $\endgroup$ – dmckee Dec 27 '17 at 21:25
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When a quantum of light arrives at a double slit, it passes through both slits as a wave and arrives upon a second screen with the interference pattern of a single wave that has been split into two waves, that have then interfered with each other.

This is not correct. The photons arrive one at a time whole, not split in space. In any case, in quantum mechanics what is waving is the probability of detecting the particle not the particle itself.

Here is the double slit experiment displaying one photon (quantum of light) at a time, and what happens when many photons are accumulated.

singlephotdouble slit

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

At the frame on the far left the footprints of the individual photons are seen. The photons do not leave a signal all over the place, they hit at a specific (x,y)at a distance z, according to the probability of the solution for the setup "photons scattering off two slits with specific width and distance". This probability is given by the $Ψ*Ψ$ of the specific wavefunction and it looks random in the first frame on the left.

The accumulation of photons shows the classical interference pattern, which for the quantum level means the probability distribution $Ψ*Ψ$.

A detector after one of the slits intercepting the photon, changes the boundary conditions to a different system, and thus a different $Ψ*Ψ$. It is no longer the same experimental setup. It should be obvious that if the detecting instrument after the slit , absorbs the photon like the screen does, only the untouched slit will give a signal on the far screen, which could not interfere with itself .( A sophisticated experiment with electrons which tries to minimally show the effect came to the conclusion that the detecting level acts as a point source for the electrons going through it, i.e. a different $Ψ*Ψ$ for the electron which is no longer coherent so as to show the interference pattern.)

Therefore can one assume that detection has 'caused' the collapse of the wave portion of the duality?

Detection at the screen has picked ("collapsed ")an instance of (x,y,z) of the original wavefunction and removed that photon from the final screen. In general after the detection of "which slit" the photons are in a different wave function with new boundary conditions.

How has detection precisely influenced the duality? Can anyone clarify?

The duality is not affected by detection, the mathematical model that describes the probabilities , $Ψ*Ψ$, has a different Ψ because the boundary conditions have changed and the coherence necessary to display interference is lost.(coherence in the phases describing the photons in spacetime). Again, the term wave particle duality has to do with the mathematics of the quantum mechanical probabilities. The probability is a wave, (a solution of a quantum mechanical system) the particle manifests as a point in (x,y,z,t) when interacting in a measurement, in accumulation of many particles with the same boundary conditions, the probability distribution is built up.(It is the same as throwing dice. The probability distribution versus the numbers 1-6 is seen in the accumulation of many throws).

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    $\begingroup$ A detector after one of the slits intercepting the photon, changes the boundary conditions to a different system, and thus a different Ψ∗ΨΨ∗Ψ. It is no longer the same experimental setup. $\endgroup$ – Marcus de Brun Dec 26 '17 at 22:10
  • $\begingroup$ "A detector after one of the slits intercepting the photon, changes the boundary conditions to a different system, and thus a different Ψ∗Ψ. It is no longer the same experimental setup." Anna: Can you clarify what experimental conditions have been changed by detection. When you refer to 'boundary conditions', what do you mean? My understanding is that a detector does not interact with the quantum in any material way that might affect its behaviour. Therefore can I assume that the "boundary conditions" that you refer to, lie outside of the experimental construct and are themselves hypothetical? $\endgroup$ – Marcus de Brun Dec 27 '17 at 7:20
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    $\begingroup$ The two slits solution of the Schrodinger equation exists, but is complicated and comes in power series. To get a prediction to check against data, you have to put down the distance of the slits, the width of the slits and the path length to the screen. If you insert a detecting element on the way, it is a new set of wavefunctions where the effect of the existence of a detector has to be taken into account mathematically. It is a different problem. This is true whether quantum mechanically or classically. Think of a wave leaving a point source disturbance. $\endgroup$ – anna v Dec 28 '17 at 19:26
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    $\begingroup$ The waves are a radial solution of the differential equation of fluids, and are uniform in theta and phi expanding in time. If you insert a pencil on the way , or a counter to count the flow, the original solution will no longer fit , it is different boundary conditions that generate new waves dependent on the position of the counter. It is the same with probability waves. New detectors means new boundary conditions on the general solution. the link shows how coherence is lost between the incoming beam and the beam passing the slits when detection happens, in the electron case. $\endgroup$ – anna v Dec 28 '17 at 19:30
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    $\begingroup$ in particle physics the photon is also a particle on par with the electron. $\endgroup$ – anna v Dec 28 '17 at 19:30
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A few weeks ago, I had a paper to do about why the interference patterns disappears when you place a detector to determine through which slit a photon passes through. This gave me a pretty good idea of what's going on, and answered some of these questions that I was asking myself actually. As this paper was in french, I'll try my best summing up what I understood, in english.

If we consider a beam of light directed to 2 slits and on the other side of the slits, a screen mounted on 2 oscillators (that's our detector), like so:

Here's what it looks like

First let's consider that the screen is stated. You don't make any measurements so what you see is a simple interference pattern with distinguishable rays, and without demonstrating it (it would take too long) we have:

This phenomenon is proper to waves.

Secondly, we make a measurements, which means the screen can now oscillate and inform us about a photon's momentum (px):

So, to be able to know from which slit a photon originated from, we must know if it corresponds to p1x or p2x:

And I'm sure you've heard about Heinsenberg uncertainty principle which gives us the following inequality:

We notice that delta X has the same order of magnitude then the distance between 2 rays! Which means we can't see an interference pattern distinctly anymore! Knowing that the interference pattern is proper to waves, we can say that light no longer has the property of a wave and behaves like a particule!

This is called the complementarity principle: we can't see light behaving like a wave and a particule at the same time.

I hope this helped don't hesitate if you have any questions :-)

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  • $\begingroup$ Gornemant: You write that P=h/wavelength, and this is the same for all photons emitted by S if S is a monochromatic source. Is this assumption not contrary to Special Relativity. If you are assuming that the unobserved or undetected photon has a momentum, you are giving it a mass. If it travels at the speed of light its mass becomes infinite. How can you assume the momentum of an unobserved photon, when an unobserved photon has never been observed? $\endgroup$ – Marcus de Brun Dec 27 '17 at 7:00
  • $\begingroup$ Sorry I may have made a mistake on this part, but I have no scientific background on special relativity... However, why does saying a photon has a momentum gives it a mass? $\endgroup$ – Gornemant Dec 27 '17 at 10:10
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    $\begingroup$ I agree with @MarcusdeBrun on that point. A photon is a photon. It being relativistic or not shouldn't influence its intrinsic characteristics such as its mass. $\endgroup$ – Gornemant Dec 28 '17 at 16:36
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    $\begingroup$ (A) Its not about your handwriting. It is about searchability, editability, and an expectation of professionalism. (B) The Compton effect shows that light carries momentum; the best limit on mass comes from experiments on Coulomb's Law. $\endgroup$ – dmckee Dec 28 '17 at 20:30
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    $\begingroup$ Gornemant: Don't let the down voting get you down, complete agreement produces nothing but silence. Science progresses through an antagonism of ideas! $\endgroup$ – Marcus de Brun Dec 29 '17 at 20:51
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"A detector after one of the slits intercepting the photon, changes the boundary conditions to a different system, and thus a different Ψ∗Ψ. It is no longer the same experimental setup."

Anna: You appear to be suggesting that a detector will interact with the photon in a classical sense, as it "intercepts" the photon. Can you clarify what experimental conditions have been changed by detection. When you refer to 'boundary conditions', what do you mean? My understanding is that a detector does not interact with the quantum in any material way that might affect its behaviour.

Therefore can I assume that the "boundary conditions" that you refer to, lie outside of the experimental construct and are themselves hypothetical?

double slit in presence of detector(s)

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  • $\begingroup$ Of course a detector interacts with a photon. The photon is absorbed. After that it’s can no longer contribute to a interference pattern on the screen. $\endgroup$ – Bill Alsept Dec 26 '17 at 23:28
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    $\begingroup$ As Holger was saying, When individual photons travel through a slit they are influenced left and right by the two edges. The photons may go straight, left or right And eventually form a single slit pattern on the screen. If there is a second slit it will do the same thing but the two patterns will interfere with each other. It is still all derived from single coherent photons. $\endgroup$ – Bill Alsept Dec 27 '17 at 0:03
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    $\begingroup$ Marcus I think the only way a photon can be detected is by absorption. $\endgroup$ – Bill Alsept Dec 27 '17 at 0:08
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    $\begingroup$ A photon can only be detected one time. Either it makes it to the detection screen and contributes to a pattern or it is blocked along the way by some snooping devise. Obviously if you close one of the slits it is blocked along the way. And yes the interference pattern on the screen can be derived from millions of single photon absorptions $\endgroup$ – Bill Alsept Dec 27 '17 at 0:30
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    $\begingroup$ the OP asked what effect the detector would have if placed at the slit. Any photons detected at the slit cannot continue on to the final detection screen. They are taken out of the experiment. If enough photons are absorbed at the first ( slit detector) then not enough will make it to the screen to form an interference pattern. $\endgroup$ – Bill Alsept Dec 27 '17 at 6:45
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When a quantum of light arrives at a double slit, it passes through both slits as a wave and arrives upon a second screen with the interference pattern of a single wave that has been split into two waves, that have then interfered with each other.

What could you observe and what should you interpret?

What could you observe

You could observe that a photon that passes a double slit impacts as a photon on an observer screen (for example a CCD chip). Repeating this setup for a while you could observe that the impacts have an intensity distribution broader the double slit and swelling between nearly no impacts and some maximum impacts (from Wikipedia):

enter image description here

But you are curious and remove one of the slits and later you replace the slit by a sharp edge. In all cases you could observe an intensity distribution (fringes) behind the obstacle:

enter image description here

What should you interpret?

Young concluded that a beam of light, going through two slits, works like a water wave interference (from Wikipedia):

enter image description here

That is a simplified interpretation because any water wave interference produces a moving interference pattern which is not the case for the intensity pattern from light or even electrons.

Furthermore behind an edge water waves get bended but not have interference pattern.

The influence of the edges

If even for single thrown photons after some time an intensity distribution occurs, haven’t we to ask about the influence of the edges? On the edges surface electrons are concentrated, do they interact with the photons?

If a detector is placed at one of the slits and the duality is detected at either of the two slits the detected duality proceeds to and arrives at the second screen as a photon/particle that has 'emerged' from its electromagnetic wave.

If a detector is placed at one slit, a particle could be detected in approximately 50% of the cases. Does this support Young’s point of view about the wave nature of light or does it support the point of view that photons still are quanta under the influence of the edges?

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  • $\begingroup$ The initial question is concerned with the cause 'detection' and effect 'quantum behaviour-change'. Might we consider what occurs at the edges of the slits after the initial question has been addressed? $\endgroup$ – Marcus de Brun Dec 26 '17 at 22:38
  • $\begingroup$ Holger, Glad to see you still think of light as photons. $\endgroup$ – Bill Alsept Dec 27 '17 at 1:03
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    $\begingroup$ @Bill Who will not think of light as photons if consider how EM radiation “starts his life”? ;-) If someone could give an example how light occurs without emission from subatomic particles, I’ll change my point of view. $\endgroup$ – HolgerFiedler Dec 27 '17 at 7:20
  • $\begingroup$ @Marcus It was my intention to show that other presumptions are possible without quantum behavior-change. Having an easier explanation for “what is not observable” (regarding the quantum mechanics sciences) both points of view have to be accepted until it is founded an inconsistency. ... Doesn’t my explanation give an answer? $\endgroup$ – HolgerFiedler Dec 27 '17 at 7:39
  • $\begingroup$ "Doesn’t my explanation give an answer?" Yes your explanation gives an answer. But I doubt that it contains THE answer. If there is a an occurrence at the edge of the slits, this should correlate or change in some manner proportional to the size or state of the atomic or poly-atomic duality that is sent through the slits. Furthermore the presence or absence of 'edge effect' has no influence upon the function: detection (all slits have edges), which appears to be the point change in experimental outcome. $\endgroup$ – Marcus de Brun Dec 27 '17 at 8:06

protected by Qmechanic Dec 31 '17 at 18:58

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