# Slit interference pattern depends or whether there is a clock in the room?

1. May 24, 2010

### m.e.t.a.

Below is a question (paraphrased) which I addressed to two physics lecturers on the subject of slit diffraction. The problem I have is that answer I received from them sounds so ridiculous and impossible that I find it difficult to accept. I asked the question independently to both lecturers, and their answers were in agreement with each other. Furthermore, both are accomplished, respected, and are expert on this subject; and so I have no reason to doubt the truth of what they say. I would be grateful to hear any thoughts that anybody has on this subject. Anyway, here is the question:

"Suppose that you pass single photons through a standard double slit apparatus one at a time. Over time, a diffraction pattern builds up on the screen, the screen being made up of an array of closely-packed photon detectors. Additionally, you have an accurate clock which is connected both to the photon emitter and to the detection array. When a photon is emitted, the clock starts; when that photon is detected, the clock stops. The two slits are positioned far apart from each other (see diagram), and hence a photon, in passing through one slit or the other, may take either a short path or a long path to the screen. In other words, the photon may spend either a short or a long period of time "in transit" between emission and detection. The purpose of the clock is to observe when a short transit time occurred, and when a long transit time occurred, and hence to provide unambiguously the (retrospective) knowledge of which slit the photon passed through.

Now, isn't it true that in all experiments designed to determine which slit the photon went through, the diffraction pattern is destroyed? That is, no matter how gently and sensitively you try to detect which slit the photon (or electron etc.) goes through, in carrying out this detection you must interfere with the photon slightly, and this interference is what destroys the diffraction pattern somehow? If it is a fact of Nature that one can never both get a diffraction pattern AND tell which slit the photon went through, then what happens in my clock experiment as described above? A clock is passive and does not physically interfere with the photon in any way at all. It merely times the photon's travel time and calculates which slit the photon must have gone through. My question is: By what mechanism does a clock destroy the interference pattern?"

The above is heavily paraphrased but my basic question remains the same. Here is the answer given me by both lecturers: If you switch off the clock, you get a diffraction pattern. If you switch on the clock and collect its timing data, you do not get a diffraction pattern. If you switch on the clock but throw away its timing data as soon as each is collected, you do not get a diffraction pattern. If the clock is switched off, but an advanced robot is present in the room (I began to ask quite desperate questions at this point), and the robot, possessing an accurate internal clock, is able to tell how much time passed between photon emission and detection merely by looking at the experiment, then you do not get a diffraction pattern.

Presumably, then, the diffraction pattern will appear and disappear if the robot opens and closes its eyes? Or if robots sound too far-fetched, presumably the diffraction pattern will appear and disappear as you switch the clock's power supply on and off? This all sounds too ridiculous to be true! What is going on?

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2. May 24, 2010

### alxm

The standard answer to that is: Decoherence, i.e. an interaction with the greater environment. The details of how decoherence occurs is however not entirely worked out though.

Note the underlying assumption here: Your detector (in this case a clock) is assumed to work classically, whereas the thing you're detecting is acting quantum-mechanically. But what if your 'detector' consisted of a single quantum-mechanical particle measuring another? Then the result of your 'measurement' would be an entangled state, where the 'subject' particle was both in the state of "being measured" and "not being measured", and the 'detector' would both be in the state of "having measured" and "not having measured". So your 'detector' doesn't 'collapse' the wave function in that case.

So decoherence is the answer to how you get from the latter state of affairs to the former.
To give a hand-waving answer, a macroscopic detector (say, a PMT tube) is in a metastable state. The tiny interaction triggers a much bigger one, sufficiently large to create entropy, which in turn irreversibly 'forces' the system into being in either one state or the other, to an extremely high degree of probability.

Well, it makes more sense if you rephrase what your teacher actually meant by 'turning the clock on and off', which is that you're turning the interaction with the environment (hence the decoherence) on and off.

3. May 24, 2010

### DrChinese

OK, clearly the clock has nothing itself to do with the interference pattern appearing or not. The relevant issue is HOW you choose to observe the photon. If you do it in a way which allows the path to be determined, at least in principle, then the interference disappears. Looking at a clock will not change anything. But changing the experimental setup does, and that is active, not passive.

4. May 25, 2010

### m.e.t.a.

alxm, I suspect that I need to learn a lot more quantum theory before I can fully appreciate your answer. Could I ask you to clarify your final point? You say that turning the clock on and off does cause the interference pattern to appear/disappear? And that this is because you are turning the "interaction with the environment" on and off? Could you explain what this interaction with the environment physically is? (As with much of QM, I am confused, but especially so in this case. I thought that a clock must be a totally passive, non-interacting element of the experiment, and as such does not in any way interact with the photon.)

DrChinese, do you mean that the physical presence or absence of a clock makes no difference at all to the appearance or non-appearance of an interference pattern? If that is true then the world makes more sense to me... And yet my instructors were adamant (and alxm seems to say the same) that the clock is crucial! Do I misunderstand them?

In my experiment there is a difference in length, $\Delta l$, between the two possible photon paths. Since photons travel at speed c there is an associated time difference, $\Delta t$, between the paths, and it is this $\Delta t$ which enables a clock to determine which slit the photon passed through. If I understand you correctly, it is the fact of whether a clock could be used to determine which slit the photon passes through that is important—not whether or not a clock was actually used in the laboratory.

Does this mean that there is a physical limit to how large one can make $\Delta l$ and still observe an interference pattern? The reason I ask this question is as follows. If you start off with a very tiny $\Delta l$ (i.e. the situation of 2 slits very closely spaced) then no physical clock, not even a clock of maximum theoretical accuracy, would be able to tell you with 100% certainty which slit the photon went through—$\Delta t$ would just be too small compared to the inherent uncertainty in the clock's measurements. An inteference pattern would be observed. If $\Delta l$ is slowly increased between experiments then won't there eventually come a point at which $\Delta t$ is sufficiently large that a physical clock can begin to tell you (at least some of the time) which slit the photon passed through? At this point would the interference pattern begin to wash out and eventually disappear?

Also: alxm and DrChinese, your answers seem to me partly contradictory. I could easily be misunderstanding, but alxm seems to suggest that the presence of a clock is important, while DrChinese seems to suggest that the presence of a clock is irrelevant. I apologise if I have failed to understand one or both of you in this respect. What I am curious to know is: has this experiment been done and the results published?

5. May 26, 2010

### Zarqon

I think this is the main problem with your arguement (and also what DrC tried to point out I think). Any measurement apparatus, clock or otherwise, unavoidably interacts with your system, i.e. cannot be passive.

For example, a single photon source is a quantum mechanical object. Any conceivable clock that you would try to attach to this system, to measure exactly at what time the photon is sent out and then detected, would have to be entangled with this single photon. Thus, at the point later when you choose to measure the value of your clock, you are unavoidably causing decoherence to the photon path, such that the interference pattern is destroyed.

6. May 26, 2010

### Cthugha

You have a rather wrong idea of how the photon emission process works. A photon is not a ball thrown at some well defined time. The exact time of emission is rather undefined and smeared out. If this uncertainty exceeds the difference in the times the photon would have to travel on the short and long path, you see an interference pattern, otherwise you do not.

Your idea of a stopwatch starting at the emission of a photon automatically assumes that you have some mechanism which causes the uncertainty in emission time to be short, so that you are able to detect a certain emission time. As a rather simplifying model you could imagine a very light emitter that shows significant recoil upon photon emission. Usually this system will be in a superposition of states due to conservation of momentum:
1: emitter is stationary, photon is not emitted
2: photon is emitted, emitter recoils and moves

The whole photon-emitter system is entangled. Now your idea of a stopwatch requires some measurement of the emitter recoil (or a similar measurement) which breaks the entanglement and thus necessarily results in no interference pattern.

7. May 26, 2010

### alxm

Whatever starts and stops the clock. The starting or stopping of the clock is a macroscopic, classical, event. The clock is either running or it's not - it doesn't behave quantum mechanically. Your detection of the photon, and the subsequent chain of events that lead to the macroscopic event of the clock getting its measurement, is the decoherence process.

DrChinese isn't contradicting me when he says the presence of the clock in itself doesn't change the result. The presence of a 'non-interacting' clock in the room won't change anything. It's the presence or absence of this interaction. Also, as mentioned, it can't be any old interaction, but one which (in priniciple) can allow you to determine which path the photon took.

8. May 26, 2010

### m.e.t.a.

Thank you, Zarqon. I had not considered what it means to detect/isolate the precise time of emission of a single photon. I can see now that this is not a passive process; and yes, you have helped clarify DrChinese's point to me! Could I ask: does the physical process of detecting a photon's emission time alone destroy any chance of an interference pattern, or is it the combination of detecting both its emission and detection times? Or are these just circumstantial examples of a more underlying reason why interference is sometimes destroyed and sometimes not?

Here is why I ask: I had assumed that when a photon hits the screen and is absorbed there then its wavefunction is isolated at that instant to a small pocket of space and time ("collapsed"?), and therefore the screen might as well be connected to a clock since an irreversible interaction has taken place there (as alxm said, entropy is created). E.g. when slit-diffracted electrons are detected one by one at a phosphor screen, they cause individual flashes of light which blatantly isolate each electron's position, but these flashes in no way destroy the interference pattern. (I could easily be wrong about this but hopefully this is right so far...?) If this is true, then is it also true that no property of the screen can affect the presence or absence of a diffraction pattern?

Hi, Cthugha. Oh yes, I should have mentioned this in my question, but after having asked two similar (but long) questions in the last few weeks, and receiving no replies, I felt that I ought to keep this new question as brief as possible. Anyway, yes, I am aware that there is a finite uncertainty in both the emission and detection times of the photon. In my previous post I asked: what if we use a clock of maximum theoretical accuracy? This would mean that when the photon is detected, we detect its arrival time to the greatest possible accuracy. (I am assuming that the fact that this detection causes wavefunction collapse is allowable and does not destroy the interference pattern. Please correct me if I'm wrong.) When the photon is emitted, however, won't we destroy the interference pattern if we try to detect its emission time via a physical interaction (e.g. detecting recoil)? If this is true then what about if we employ a subtler, more "gentle" way of isolating a photon's emission time which does not involve any physical interaction with the photon?

For example, if we only switch on our light source for a few moments then we know to a high probability that if a photon that was emitted then it was emitted during that brief window. Or, alternatively, we could place a camera shutter in front of the light source and open the shutter for a brief moment so as to let 0 or 1 photon pass through...and so on. There are many possible methods of pinning down the emission time of a photon to within a small time window. These example methods that I have mentioned are not very ingenious, but to me they certainly appear passive. I.e. at no point do they involve physically "looking at" the photon or interrupting its path, and neither do they involve measuring the recoil or other properties of the body which emitted the photon. If these methods are not truly passive then please correct me.

It seems reasonable to say, then, that it is possible to determine the emission time of a photon to within a finite uncertainty without destroying the interference pattern. With the best possible equipment, this finite uncertainty would reach a minimum value (call it t_min). Cthugha, you said:

When you say "If this uncertainty exceeds...", by "this uncertainty" do you mean the minimum possible uncertainty, t_min? If this is the case, then is t_min always the same constant, no matter the experimental setup? I.e. does t_min represent the best possible accuracy to which one can passively know the emission time of any photon under any circumstances, without "looking"? And if t_min is always the same constant, then can my question from earlier be answered: "Does this mean that there is a physical limit to how large one can make $\Delta l$ and still observe an interference pattern?"

My tutors, if I understood them correctly, seemed of the opinion that $\Delta l$ could be made arbitrarily large and an interference pattern would still be seen. This is why I am confused!

9. May 26, 2010

### DrChinese

My point exactly. If you could know, the interference disappears. So it is setup, the context. So that is what is relevant. And not the clock per se, and not whether someone looks at the clock.

10. May 26, 2010

### GeorgCantor

Quantum objects can be said to 'conspire' against aquiring prohibited knowledge, i.e violating the HUP. This is a curiousity of the quantum world.

11. May 27, 2010

### Cthugha

Yes, you will destroy the pattern.

Even these methods are not passive. The one thing you can say for sure about particle-like behavior is, that interactions happen in discrete quanta. Not more. All the classical notions of a fixed in-flight position and a well-defined time of emission lead to severe problems. The precise details of this quantized interaction is determined by wavelike properties (like everything in QM): probability amplitudes. Just as the intensity of a beam depends on the square of the total electric field, the probability to detect some photon somewhere depends on the square of the sum of all probability amplitudes leading from one initial state to one final state. If you introduce a shutter, all probability amplitudes for the events corresponding to emission times when the shutter was closed will vanish and you have changed the state of the system. Accordingly, if the difference in photon travel time distance to the two slits is longer than the interval during which the shutter was opened, there will be no interference pattern.

This depends on the light source used. In general one can say that a larger uncertainty in the photon emission time (or equivalently the coherence time), allows for larger values of $$\Delta l$$.

I am not talking about the minimal possible uncertainty t_min, but the uncertainty that is actually present. You can make this uncertainty as small as you like, but you want it to be large to see an interference pattern. So to make $$\Delta l$$ large, you would be more interested in t_max, the largest possible uncertainty determined by your source or your measurements concerning emission time.

12. May 27, 2010

### m.e.t.a.

Is the HUP something one can "work out" from other principles? Or is it too fundamental to be worked out from anything simpler?

Oh! So it is truly wrong to think about the "crest" of a probability wave as propagating outwards at speed c. Are you saying that you have to consider probability amplitudes for the photon travelling faster and slower than c while it is "in transit" between emission and detection, even though the photon's average speed for the whole journey must be c?

Cthugha, I'm sorry that you had to repeat yourself before I took notice, but I'm glad you said that! In a previous thread I started, "Time period measured in double slit experiment", I make the argument that the interference pattern should disappear if $\Delta l$ is made too large. I know I have asked a lot of questions already, but could I ask you to check over my reasoning below and tell me which parts do and do not make sense?

[Let T be the total travel time of a photon between its emission and subsequent detection; let $\delta T$ be the inherent (?) uncertainty on T. Let $\Delta l$ be the difference in length between the long and short paths, with negligible uncertainty on this measurement. (I'm using lower case delta for uncertainties, capital delta for differences.)]

Ok, so the slits/clock experiment is carried out as previously described. Photons are emitted one by one. When a photon chances to be absorbed at the detector, that photon's travel time, T, is recorded and logged. The experiment is run many times. Finally, the results are plotted in the form of a probability distribution: [T] vs. [frequency of T]. (For a single slit I would expect this probability distribution to be roughly Gaussian in shape.)

For each of the two paths, $l_1$ and $l_2$, there is an associated travel time for a photon traversing that path:

$${T_1} = \frac{{{l_1}}}{c}$$

$${T_2} = \frac{{{l_2}}}{c}$$

I would therefore expect the plot to show two (~ Gaussian) probability distributions, tails overlapping, with mean values centred around $T_1$ and $T_2$ (after correcting for skew). However, if two distinct probability distributions are visible in your data then this means that you can determine to a probability >50% which of the two slits a given photon went through. (I.e. you would look at a photon's T-value, and then by consulting your probability distrubtion plot you would be able to say, "Aha, there is a 72% probability that a photon with that particular T value went through slit number 1" (for example).) If $\Delta l$ is made larger and larger then the probability distributions centred around $T_1$ and $T_2$ will get further apart and overlap less and less. The confidence with which you know which slit a photon goes through then approaches 100%. Therefore, as $\Delta l \to \infty$ the interference pattern should wash out and ultimately disappear.

On the other hand, if $\Delta l$ is made smaller and smaller then $T_1$ and $T_2$ start to get very close to each other. Eventually, their probability distributions will overlap so completely that the two distributions will appear as one. Beyond this point it will be impossible to tell from a given photon's T value which slit it went through. In such cases the interference pattern ought to be very strong. I have tried to illustrate this with some diagrams (attached).

So, does the above make sense? If so then I have a related question—which I promise will be the final question!

How might you make t_max large?

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13. May 27, 2010

### Cthugha

Oh, sorry, Feynman's approach was not what I was thinking about. Although that is another way to put it. You have to consider probability amplitudes for the photon being emitted at different times. My point was rather to consider the width of this outspreading probability wave. The classical particle picture will produce some delta-like peak spreading out from the light source. In a situation which shows interference using a quantum particle picture, this peak will become very broad due to the coherence of the emitter-photon state or equivalently uncertainty in the emission time. If it is broad enough to be nonzero at both slits simultaneously, you get interference. Otherwise you do not.

The explanation is ok, I think.

Well, it is not really a good explanation for single photon emitters, but one of the best ways to make t_max large is stimulated emission used in a common laser. Some photon gets emitted into a laser cavity at time A. It bounces back and forth and causes stimulated emission when it hits some atom in an excited state at time B. These photons are completely indistinguishable and now also bounce back and forth. These same two photons also cause stimulated emission again at later times C, D and E. Even later one of these indistinguishable photons leaves the cavity. Was it emitted at A, B, C, D or E? you cannot know in principle. t_max gets very large.

14. May 27, 2010

### m.e.t.a.

Thank you, Cthugha, and thanks to everyone who replied, you've all been a great help. I was going to ask another question, but, given all this new information, I will need to think about it some more before asking. Thanks again.

- m.e.t.a.

15. May 28, 2010

### GeorgCantor

Yes, the HUP is the result of the relationship between momentum and position operators. As you get a more accuate reading of the position of a particle, it's corresponding wavelength becomes larger. If you knew with precision where a 'particle' is, it's momentum would be indefinite, meaning the 'particle' could be anywhere.

As for the fundamental part, i don't know what you mean by fundamental. If you believe QM is complete, you could say that the HUP is a fundamental postulate, as long as you realize the trap you're getting yourself into by using that word.

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16. May 28, 2010

### m.e.t.a.

I was probably using the word "fundamental" improperly. What I should have said was: which of the following is true?

1. The HUP is the product of more elementary physical laws. E.g. in the same way that Kepler's laws of planetary motion are the product of the more elementary laws of conservation of energy and momentum. The HUP can be worked out from so-called "first principles" without the need for one to observe the quantum world.

2. The HUP is itself an elementary law. It is observed to always hold, but the reason why it holds cannot be expressed in terms of more elementary laws. In other words, there is no satisfactory answer to the question of why it is true. When I say "no satisfactory answer", I mean that the answer would have to be "Because the Universe is just that way".

I am perhaps wrongly thinking of the words "elementary" and "fundamental" as being roughly synonymous. A better word might be "empirical".

17. May 28, 2010

### DaTario

I think it is passive, but this is not the important point. Passive instruments couple with the system being studied. And this creates interactions which produce effects on interference patterns.

Best Wishes

DaTario

18. May 28, 2010

### GeorgCantor

I'd choose 1, but i'd say that 2 is also true because of the wave-like nature of matter. If the universe/reality/call-it-whatever didn't consist of waves, there'd be no HUP.

This how you derive the HUP from more basic concepts:

https://www.physicsforums.com/library.php?do=view_item&itemid=207

19. May 28, 2010

### m.e.t.a.

Thank you for the link, but I'm afraid that those "more basic concepts" are much too advanced for me! Perhaps in a year or two when I have picked up some knowledge I will think to re-read this thread with new eyes.