High School Photon Continuity in Double-Slit Experiment

[jeremyfiennes]

TL;DR

Is there experimental evidence for photon conitinuity.

In the double-slit experiment, is there experimental evidence that a photon detected passing a slit always results in one and one only screen point?.

 

[vanhees71]

Well, with there's always some probability that a photon is not registered, but if it's registered then at one point and only one point of the screen.

 

[jeremyfiennes]

Thanks. So that a photon behaves as a 'particle' in the classical sense - that if at one point in time it is found at some point, then at a later point in time it will be found at some other (not necessarily predictable) point - is in fact not experiementally demonstrable.

 

[A. Neumaier]

In the double-slit experiment, is there experimental evidence that a photon detected passing a slit always results in one and one only screen point?.

No, because whenever a photon is detected, it ceases to exist.

 

[jeremyfiennes]

Right. So photon continuity is an unverifiable hypothesis. Which doesn't of course mean that it is untrue. But just that in practice it is at present experimentally unverifiable.

 

[A. Neumaier]

So photon continuity is an unverifiable hypothesis.

Nobody claims it anyway.

Photons are just a fancy name for detector clicks in response to faint light. In addition they are a a technical name for ingredients building up the state of electromagnetic radiation.

 

[jeremyfiennes]

But delayed eraser experiments, for instance, DO imply particle-like continuity: that an idler photon arriving in a detector corresponds to a signal photon point on the screen.

 

[A. Neumaier]

But delayed eraser experiments, for instance, DO imply particle-like continuity: that an idler photon arriving in a detector corresponds to a signal photon point on the screen.

This is just a figure of speech for a more complicated situation.

It allows one to reduce the discussion to the essentials but is misleading when taken literally.

 

[jeremyfiennes]

Ok. Thanks.

 

[jeremyfiennes]

In the Mach-Zender 'particle' setup, would it be true to say that a photon is detected either in one detector, or in the other, but never in both simultaneously?

 

[DrChinese]

In the Mach-Zender 'particle' setup, would it be true to say that a photon is detected either in one detector, or in the other, but never in both simultaneously?

Generally, that's the right idea. This experiment spells it out pretty well. 1 photon, 1 click.

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

 

[jeremyfiennes]

Thanks. So photons are at least more trustworthy than atoms.

 

[DarMM]

Note there is no state that will definitely cause $N$ clicks in an appropriate device.

In a sense a "one photon" state just means a state of the electromagnetic field very likely to cause one click in a photon detector, but not certain to.

 

[A. Neumaier]

Generally, that's the right idea. This experiment spells it out pretty well. 1 photon, 1 click.

No. There is always some loss due to detector inefficiency. The 1-1 correspondence is just a convenient simplification.

 

[f95toli]

But delayed eraser experiments, for instance, DO imply particle-like continuity: that an idler photon arriving in a detector corresponds to a signal photon point on the screen.

This is ultimately a philosophical question: what does "exist" mean?
The vast majority of people working with photons do think of them as being as "real" as e.g. atoms, but this does of course not mean that they are in any way classical objects,

 

[DrChinese]

No. ... The 1-1 correspondence is just a convenient simplification.

Well, I used the word "generally" (indicating a "convenient simplification") and cited the specific experiment itself in case anyone wanted to see more depth; I'm not sure how you can say NO to that as an answer to a simple question.

We agree that every experiment has some degree of inefficiency, so I don't see the value of your contrary conclusion. Yes, 1:1 is the relevant takeaway here - and you would get that with perfect efficiency (if there were such a thing). In fact, as mentioned in the reference: "a single photon can only be detected once!*"

IMHO: We do a disservice to some readers when we qualify things so much that the original question gets completely lost. *Originally Grangier, Roger and Aspect, 1986.

 

[A. Neumaier]

We agree that every experiment has some degree of inefficiency [...]

IMHO: We do a disservice to some readers when we qualify things so much that the original question gets completely lost.

Nevertheless, the only correct B-level answer to the original question

In the double-slit experiment, is there experimental evidence that a photon detected passing a slit always results in one and one only screen point?

is no.

The experimental evidence is overwhelming that a photon passing undetected a slit always results in zero or one screen point, never more. A zero count is due to detector inefficiencies.

On the other hand, if the photon is detected passing a slit it disappears and always results in no point on the screen.

 

[jeremyfiennes]

"The experimental evidence is overwhelming that a photon passing undetected a slit always results in zero or one screen point, never more."
In SPDC terms: "there is overwhelming experimental evidence that if an idler photon is detected, then in principle one and only signal photon will necessarily be detected. And if it isn't, this will be attributed to detector inefficiency" (correct?).

"No. ... The 1-1 correspondence is just a convenient simplification."
This is the essence of my query. Is the near 1:1 ratio due to a given wave amplitude generally resulting in a photon detection at both points (slit and screen)? Or is is a necessary condition due to the particle properties of photons. And that if a photon is detected at one point and not the other, then this must be due to detector inefficiency at the no-show point.

 

[A. Neumaier]

"The experimental evidence is overwhelming that a photon passing undetected a slit always results in zero or one screen point, never more."
In SPDC terms: "there is overwhelming experimental evidence that if an idler photon is detected, then in principle one and only signal photon will necessarily be detected. And if it isn't, this will be attributed to detector inefficiency" (correct?).

If only one photon is around at the time of measurement, there can be only one detection event.

(But you seem to be discussing here the case of an entangled pair of photons, an idler photon and a signal photon. For this, there are two possible detection events.)

"No. ... The 1-1 correspondence is just a convenient simplification."
This is the essence of my query. Is the near 1:1 ratio due to a given wave amplitude generally resulting in a photon detection at both points (slit and screen)? Or is is a necessary condition due to the particle properties of photons. And that if a photon is detected at one point and not the other, then this must be due to detector inefficiency at the no-show point.

No. One can detect a given photon only once, as it is destroyed by the detection process. Hence only at the slit or at the screen, not twice. But because of detector inefficiency, both detectors can possibly miss it.

 

[jeremyfiennes]

But you seem to be discussing here the case of an entangled pair of photons, an idler photon and a signal photon. For this, there are two possible detection events.)

But do these in principle (ie. ignoring detection inefficiency) always correlate: if one is detected, then the other necessarily will be; and if not then not?

 

[A. Neumaier]

But do these in principle (ie. ignoring detection inefficiency) always correlate: if one is detected, then the other necessarily will be; and if not then not?

The detection failures are independent. But assuming 100%efficiency, two photons imply two detection events. This has nothing to do with correlations.

 

[DrChinese]

But do these in principle (ie. ignoring detection inefficiency) always correlate: if one is detected, then the other necessarily will be; and if not then not?

You are talking about PDC pairs. So the answer is YES, despite everything being said that might indicate otherwise. When one of the pair goes through a double slit, and since it is a photon: it only has the opportunity to be detected once. That is what A. Neumaier is pointing out, and there is no argument about that. The key takeaways are:

a. There is 1 click at both detectors, or no clicks at either*. That is what you refer to as "correlated"; which is often used with PDC to mean something a bit different (again why A. Neumaier's answer is different than mine). But you are correct generally, that is the entire point of the reference I provided.

b. There are never cases of 1 click at one detector*, and 2 (or more) clicks at the other. That is because a PDC pair - sometimes referred to as a biphoton - is a special state with a known photon number (2 in this case). Such a state is called a Fock state. Most light does not appear in this state, as photon number is not usually a conserved quantity. In fact, PDC photon pairs are created from a single input photon (in those cases where a biphoton results).

c. Although often ignored with a double slit setup: it IS possible to determine which slit a photon goes through without destroying it. That's another subject though.

d. You should never make assumptions about what quantum particles are doing - at least not rigorous assumptions - when not being observed. They have a nasty habit of doing the impossible when you make an assumption which is reasonable yet wrong.

Now obviously, and as A. Neumaier correctly points out, there are many details being glossed over in our discussion. You can read more at the reference I provided. It is very good as it is intended for an undergrad lab. Note that this experiment soundly refutes one of the classical views of light. *Of course many photons that are part of PDC pairs never make it through the double slit in the first place (in your setup); so we are ignoring that for the purposes of our discussion.

 

[vanhees71]

Nevertheless, the only correct B-level answer to the original question

is no.

The experimental evidence is overwhelming that a photon passing undetected a slit always results in zero or one screen point, never more. A zero count is due to detector inefficiencies.

On the other hand, if the photon is detected passing a slit it disappears and always results in no point on the screen.

This is a bit strange formulation for a beginner.

Fact is that there is some probability for a single photon, which state can be prepared today easily via parametric downconversion, to be detected by a detector or not.

If photon is detected and if the detector resolves the position of the detection event a single photon state is registered at one "point" (a point has of course necessarily a finite size) and not as a smeared distribution. That's one of the most simple indications for the necessity of field quantization. It cannot be explained within classical electrodynamics.

 

[A. Neumaier]

This is a bit strange formulation for a beginner.

What is strange?? What you wrote does not contradict what I wrote.

 

[vanhees71]

No, it doesn't contradict anything, but I hope it makes it more clear to a beginner in QED.

 

[jeremyfiennes]

Dr Chinese: thanks. Ok, for simplicity let's restrict discussion to PDC pairs, and assume hypothetical 100% detection efficiency.

"There is 1 click at both detectors, or no clicks at either."
That is what I have been calling "correlated". Sorry, I didn't realize it could have another meaning.

So when Einstein said that
"A particle can only appear as a limited region in space where the field strength, or the energy density, is particularly high. We can consider as matter those regions of space when the field is extremely intense."
this doesn't exclude a particle (e.g. a photon) being detected at one point but not at a subsequent one, and so doesn't fit the present model. (Right ?)

"It IS possible to determine which slit a photon goes through without destroying it. That's another subject though."
Could you give a reference? Does it confirm photon continuity: that one detected passing a slit always corresponds to one and one only screen point?

"You can read more at the reference I provided. The experiment soundly refutes one of the classical views of light."
I read and liked it. Which classical view does it refute?

 

[vanhees71]

I don't know, what precisely you mean by "photon continuity". As I said before, if you have prepared a single photon and it hits a position resolving detector it leaves only one spot though the expectation value of the em-field energy density, which is by definition the intensity of the em. field, can be a broad distribution. That distinguishes a single-photon state from a classical em.-wave state (QFT-wise something like a coherent state or some other state, e.g., thermal radiation like a Planckian black body).

One very elegant way to gain which-way information in the double-slit experiment with photons is to start with linearly polarized photons (say horizontally polarized wrt. the $x$ direction) and put quarter-wave plates into the slits one ("slit 1") oriented with $\pi/4$ relative to the $x$ direction and the other (slit 2) with $-\pi/4$. Then a photon running through slit 1 gets left-circularly polarized and a photon running through slit 2 gets right-circularly polarized, and thus you can (in principle or really) determine behind the slits through which each photon came. Since the corresponding states are orthogonal to each other there's no double-slit interference pattern (but only an incoherent addition of the two single-slit interference patterns). As I said before, to get a distribution you need many photons. Each single photon leaves one point-like spot on the screen.

 

[DrChinese]

Dr Chinese: thanks. Ok, for simplicity let's restrict discussion to PDC pairs, and assume hypothetical 100% detection efficiency.

"There is 1 click at both detectors, or no clicks at either."
That is what I have been calling "correlated". Sorry, I didn't realize it could have another meaning.

So when Einstein said that
"A particle can only appear as a limited region in space where the field strength, or the energy density, is particularly high. We can consider as matter those regions of space when the field is extremely intense."
this doesn't exclude a particle (e.g. a photon) being detected at one point but not at a subsequent one, and so doesn't fit the present model. (Right ?)

"It IS possible to determine which slit a photon goes through without destroying it. That's another subject though."
Could you give a reference? Does it confirm photon continuity: that one detected passing a slit always corresponds to one and one only screen point?

"You can read more at the reference I provided. The experiment soundly refutes one of the classical views of light."
I read and liked it. Which classical view does it refute?

Correlated... I understood what you meant.

Einstein's comment is a little vague to be useful here. There was a lot of quantum optics theory/experiment developed after his death: Fock states and PDC for example. One good thing about PDC is that you can essentially get 1 photon to herald (announce) the impending presence/detection of another.

Most of the time, particles (photons) can be thought to move in a classical path. But you want to be careful about assumptions about any quantum particle's trajectory between detection events. The assumption goes only so far, and breaks down completely where there are interference effects, multiple indistinguishable paths, etc. So the specific setup is important.

Here is a reference to an experiment where which-slit can be determined using polarization information. Polarizers are placed by the slits and varied, causing interference to appear or disappear. This occurs because it is merely the possibility of obtaining the which-slit information which controls the interference - you do not need to actually learn it.

http://sciencedemonstrations.fas.ha...-demonstrations/files/single_photon_paper.pdf

 

[vanhees71]

It doesn't make sense to say a quantum particle moves along a classical path. For photons you don't even have a position observable to begin with!

 

[DrChinese]

It doesn't make sense to say a quantum particle moves along a classical path. For photons you don't even have a position observable to begin with!

Just wondering if you read entire sentences. Yes, MOST OF THE TIME, you can think of a photon as moving in a straight line at c (that would be a classical path, my friend). You can also predict its arrival time at a specified position within very narrow limits. You can also manipulate its path. OTHER TIMES: this is not possible with quantum particles, such as photons.

Really, why make every answer so complex that no one can discuss anything without referencing the entire body of physics? Any scientist doing an experiment on photons operates with basic assumptions that are useful for the job at hand. That including how a photon travels. They don't set up fiber, mirrors, optics and detectors at random places. They pair PDC entangled photons using precise estimates of path and travel time - and that in fact assumes a specific velocity. So I don't see the point of stating that there are exceptions to general rules when someone states to begin with (as I did)... it's a general rule. It is a general rule, and it does apply MOST OF THE TIME exactly as I said.

The idea that a photon lacks a position observable is laughable, as they are regularly observed at specific points in hundreds of experiments (as both detector clicks and in other manners) exactly as predicted. What you mean to say is that field theory does not allow for a photon position observable in the same sense as other quantum particle observables.