What Would Happen If We Set Up the Double Slit Experiment Differently?

[mfb]

stewart brands: Did you see my previous post? You can block, or even destroy the photon source if you like, before you do the double slit experiment. It does not matter.

 

[DrChinese]

I ponder the filament idea because...

Your filament is a photon. What you are asking is whether photons have (or not) existence independent of the source and detector. This has nothing to do whatsoever with the double slit experiment. As previously mentioned by several others, you can get interference from things other than photons.

Your speculative idea (which is not really appropriate) calls for the photon to act the same as usual, but there is also this previously unknown connection too. As mfb says, this cannot be something which in any ways is detectible. Ergo it is excess baggage.

 

[marksesl]

Is it not possible that the photon is a filament of oscillating energy from the source to its destination? Is it not possible that the oscillation of this filament of energy(at c "speed") would enter each slit at different times and subsequently interfere with itself. The distance between the slits is tiny,but the velocity huge therefore any time differences could not be measured. The time to transverse a mm at c for instance.
Is it not the case that the photon passing both slits simultaneously is an assumption in disregard other possible realities, designed to prove a theory?
Perhaps the appearance of an interference pattern does not prove that the photon did go through both slits but is used to prove the hypotheses that it did.
The assumption that the photon is distinct at time t from the source is also ,it would appear,to be an unfounded assumption.
I am unaware of any experiment whereby a single photon is contained. Perhaps a photon cannot be distinct because it is attached to the source until it entangles. Hence the proposal of a stretched out vibration"filament" of huge period relative to the collapsing wavelength("colour")

We know from the De Broglie equation that all particles should and do have a wave property. So, why should we need some special explanation for how a photon can manage to get through both slits at the same time? Even electrons and some molecules exhibit the same behavior. The maxima of the various particles tested correspond to their mass with the frequency going up and the fringes getting closer. It's case closed, small particles do go through both slits simultaneously, by virtue of their wave ccharacteristics.

 

[welcomeblack]

For my explanation, you need to know some elementary QM. The probability to go from initial state to final state is the probability amplitude squared. This probability amplitude is denoted <f|i>, so P(i→f)=|<f|i>|². The total probability amplitude is the sum of all indistinguishable paths between final state and intial state. If the paths are distinguishable, then it is the probabilities that are summed. Now, let's set up the double slit in the following way:

Let |i> represent the initial state at the source, and |f> represent the state at the detector. We denote the particle passing through slit one by the state |1>, and the particle passing through slit two by |2>.

The probability amplitude for the particle to be emitted by the source and to go through slit one is <1|i>. The probability amplitude for the particle to travel from slit one to the detector is <f|1>. Thus the probability amplitude for the particle to travel from the source, through slit to, then to the detector, is <f|1><1|i>. The same goes for slit two, just replace |1> with |2>.

With no way of telling the difference between the particle passing through slit one or two, the probability amplitude is <f|i>=<f|1><1|i>+<f|2><2|i>=A₁+A₂. Then the probability to go from source to detector is P(i→f)=|A₁+A₂|²=A₁²+A₂²+A₁A₂cos(Δψ), where Δψ is the relative phase between the two amplitudes. The last term, A₁A₂cos(Δψ), is the interference term.

Now, if you put some particle detection mechanism at slit one, you CAN distinguish between the two paths from intial state to final state. The means that we may no longer add probability amplitudes, but must rather add the probabilies, just like in classical mechanics. Explicitly then, P(i→f)=P(i→1→f)+P(i→2→f)=|<f|1><1|i>|²+|<f|2><2|i>|²=A₁²+A₂². No interference term. All because you can distinguish the path.

That's the formal/mathematical explanation. The physical explanation is given elsewhere in this thread.

 

[marksesl]

With no way of telling the difference between the particle passing through slit one or two, the probability amplitude is <f|i>=<f|1><1|i>+<f|2><2|i>=A₁+A₂. ..
Now, if you put some particle detection mechanism at slit one, you CAN distinguish between the two paths from intial state to final state.

That's all fine and well and very interesting, but very misleading, i.e., "With no way of telling the difference," vs. ".. you can distinguish between the two paths ... ." Surely it is the very act of placing the detector that causes decoherence and wave-function collapse. We can only know of what we caused ourselves.

 

[marksesl]

As mentioned, if you start with the idea that the photon is a wave: then interference makes perfect sense. But that same photon is quantized and appears at a single point only, giving it particle-like properties. So there are elements of both.

Your reasoning is circular about the importance of the experiment though. You can see that the behavior is fundamentally different when you cannot determine which of 2 slits the light traverses. Such matches the classical view of light as a wave, and what you make of that is up to you. I appreciate that all of this is "self-evident" to you but that is hardly the case for most. For example, the classical view of light as a wave is strictly ruled out in this experiment:

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

Not so obvious.

I appreciate your discussion, but there can be no doubt that light and even a single photon is a wave. Photons, as you know, are produced by the quantum fall of electrons in atoms. It is during that fall that a photon's frequency is determined, imparting the very color the photon will have. All photons have an inherent frequency and wavelength. They are waves, but of discrete energy packets which give a particle quality as well. The assumption I'm making is not that light (photons) are waves, that's a given, but that as waves, photons can travel through both slits at the same time. The experiment is flawed, not because I'm assuming that light is wave, but because it is assuming that light is only discrete particles that cannot go through both slits, and by making that false assumption it is thought the photons can be correctly determined as to which way they actually traveled. But, if they do actually go through both slits, the experiment automatically destroys that "both-way" information because waves at right-angles cannot interfere anyway. The experiment eliminates one of the three possibilities, left, right, or "both." And, in eliminating the "both" option, it indirectly destroys knowledge of left or right as well because if the photons do in fact go through both, then left or right knowledge must be something the experiment just makes appear to happen. I'll read the article you linked to though.

 

[harrylin]

[..] Surely it is the very act of placing the detector that causes decoherence and wave-function collapse. We can only know of what we caused ourselves.

[..] All photons have an inherent frequency and wavelength. They are waves, but of discrete energy packets which give a particle quality as well. [...] The experiment is flawed [..] because it is assuming that light is only discrete particles that cannot go through both slits, and by making that false assumption it is thought the photons can be correctly determined as to which way they actually traveled. [..]

I agree with that - to describe a propagating photon as a particle is just no good. There have been papers explaining the same, for example this recent paper by Hobson: http://ajp.aapt.org/resource/1/ajpias/v81/i3/p211_s1 .
Note that perhaps also not all his interpretations are correct, see my post https://www.physicsforums.com/showthread.php?p=4458704

 

[marksesl]

As mentioned, if you start with the idea that the photon is a wave: then interference makes perfect sense. But that same photon is quantized and appears at a single point only, giving it particle-like properties. So there are elements of both.

Your reasoning is circular about the importance of the experiment though. You can see that the behavior is fundamentally different when you cannot determine which of 2 slits the light traverses. Such matches the classical view of light as a wave, and what you make of that is up to you. I appreciate that all of this is "self-evident" to you but that is hardly the case for most. For example, the classical view of light as a wave is strictly ruled out in this experiment:

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

Not so obvious.

The experiment is clamed to be proof that photons exist, and the view of quantum physics is justified. It says light is not "classical" waves. So, Einstein was correct. No big news there; after all he won the Nobel Prized for being correct.

The wave aspect of a photon is contained within the photon (which is what I'm referring to). It is a wave packet. That wave packet is due to the drop of an electron from a higher state to a lower in the atom it is part of. The drop has a distinct beginning and ending place and literally squirts out a photon. Thus light is quantized into photons that have color, frequency, and wavelength. The photon is a wave in and of itself. A photon (wave packet) has an unbounded field that can pass through both slits at the same time. Upon striking something that it can interact with, it decoheres and collapses to a much more distinct place, such as to a point of the surface of any detector, giving the impression it only went through one slit, though it went through both.