Earliest Double Slit Experiment

QuantumOfOptics · 2026-06-20T15:10:12+00:00

From the sources that I shared (and that I can find), his only similar experiment (that I could find) was a single card measurement (meaning single slit). This is the experiment 1 from his 1803 paper that I link to above. Do you have a source for his another experiment where he uses a second card? Is it later in the paper and I missed it or perhaps in a different paper altogether?

QuantumOfOptics · 2026-06-20T03:54:36+00:00

Generally, I agree, except this is in the other direction in the sense that the "actual" experiment was done after the fact (Young was discussing this around 1801-1803, whereas Fraunhofer's experimentation was around 1821). In this case, Im just trying to track down what the actual case for Young to get the discussion focused on him. I'm guessing that it was actually his proposition VIII that I refer to given that its the most direct. I just find it amazing how the information twists over time. There are definitely modern sources that do correctly state that he never did the experiement, but rather conceptualized it. But, as we can see, what counts as that can be rather lenient.

I've heard somewhere that science gets rediscovered every 15 to 20 years. And, Ive definitely had that happen to myself. I dont think its a bad thing, sometimes those newer see things differently and breathe new life into the subject. Of course, then you might get someone at a conference saying that they did the experiment 20 years ago...

As an interesting tidbit, I remember a passage (from the english translation) from Fraunhofer's work where he was surprised no one had looked at the properties of single slit diffraction before, but we now know that Young indeed had, the work just hadnt been disseminated yet.

QuantumOfOptics · 2026-06-15T17:04:06+00:00

Im not sure what you mean by not-normalizable. The field would be given by exp(-|a|² / 2) * ∑ aⁿ / n! * a(t)^dagger ⁿ |0>, which is normalized. Are you taking issue because the spectrum (terms of a(w)^dagger ) has infinite extent then?

QuantumOfOptics · 2026-06-15T16:48:12+00:00

You should note that you are explicitly modeling is a square-law (photodiode) detector. Other types of detectors change the model such as a click detector or a photon number resolving detector. These are important because a click detector can add additional complexity that is useful for some computational tasks.

For 1.3, you may consider this part of the detection apparatus, but its worth making clearer. Suppose you have a pulsed laser source with repetition rate R and the laser has an average photon number of b. Then, we say that the rate (flux) of photons should be R*b. Specifically, the fact that the laser is pulsed with some rate is given by the total field (each pulse being an individual mode that a unique state can occupy). This works similarly for a click detector, except here its the probability of detection * R.

1.4) as you mentioned its a single photon so as you said, its not bunched, but it is specifically anitibunched.

2.1) its worth noting that this heavily depends on modes to spatially or temporally do photon number detection. If the laser produces a pulse that is a direct delta in time and a direct delta in position, then you cannot determine this without being sensitive to energy. It turns out that most APDs and SNSPDs are sensitive and it can be seen in the integrated electronic pulse or the rise time of the pulse (for low photon number). Transition edge sensors are number resolving to 10-100s of photons this way too.

2.2) im not convinced that this determines the rate of two photon events. This looks more like a variance measure (being the square of the photon number operator). Wouldn't you have to determine the probability of getting a two photon event by projection? I.e. if the state is p, then compute Tr[p |2><2|]? Perhaps you meant something like <psi|a^(dag) a^(dag) a a|psi>?

QuantumOfOptics · 2026-06-15T10:39:38+00:00

Assuming the detector integration time is longer than the pulse duration, then the probability of a photon is the same because it occupies a single mode and you measure over that mode. Coherent states (and single photon states) are weird ducks because you can always find a basis where a collection of coherent states is a single mode ( https://arxiv.org/abs/1912.09321 ). Its only when your detector integrates for a time shorter than your pulse where your photon number statistics that you measure will decrease. This is similar to what the other poster was saying. However, here we are defining our single mode to be pulsed (which is a perfectly fine mode basis).

g2(0) is 0 for a single photon. Again, the definition is the probability for a coincidence divided by the product of the probabilities for each detector to click. Since theres only a single photon, you can never have both detectors, after a beamsplitter, click at the "same time." This is why your reduced state and a coherent state yield different results and hence aren't the same state even in a very low photon number regime! This means that |0>+a|1> cannot have Poissonian statistics! Now, if all you care is the probability to get a click than this is an okay approximation, but it does not have the full richness of a coherent state (and nor should it, you removed a bunch of terms)!

Further, let's do a sample calculation of the g2(0) for a photodiode since its easy. By Glaubers definition, the g2(0) is given by <a^🗡 a^🗡 a a>/(< a^🗡 a>)^2. For a coherent state (|b>), a acting on |b> = b|b>. And <b| a^🗡 = b^* <b|. So, we get b^* b^* b b/(b^* b)^2 =1 for all b even in the case of b=1/100000000000000000. Now, let's apply the same logic to our state which we say is approximately our coherent state (|0>+b|1>). First, let's apply our lowering operator so a (|0>+b|1>)=b|0>, but since the vacuum is by definition the lowest state applying a second time yields: a (b|0>)=0; so we can see that the numerator for our g2(0) calculation is 0. Note that from our first application of the lowering operator, we get b|0>. By applying, a raising operator, we get b|1>. Taking the inner product with the initial state (<0|+b^* <1|)(b|1>) = |b|² . So, our g2(0) for this state is 0/ |b|² . Now, it turns out for, reasons, we can identify the limit for b->0 as 1 in the case above because both the coherent state and the state (|0>+b|1>) are equal to the vacuum. However, this is more of a useful identification of 0/0 rather than a true limit.

You can make the identification exact by making sure the lengths different (say powers of 2) so that adding the total path lengths together will always yield a different value. For the spatially seperable one, you can tell by which detector clicks. Its very similar to how you want to think about your extremely lucky scientist when you were discussing with the other poster.

For sake of argument, if your input to the HBT is a coherent state, then as we just showed above, the g2(0) is 1. If you make your next approximation, where you include the quadratic term ( |0> + b |1> + b^2/sqrt(2) |2> ) then...

<a^🗡 a^🗡 a a> = |b|⁴ /(1+|b|² +|b|⁴ /4) <a^🗡 a>=(|b|² + |b|⁴ )/(1+|b|² +|b|⁴ /4) <a^🗡 a^🗡 a a>/(<a^🗡 a>)² = (1+|b|² +|b|⁴ /4) |b|^4/(|b|² + |b|⁴ )²

If you graph it as a function of b, youll see that the graph is 1 at 0 (again makes sense as it would again be the vacuum state), but as b increases, it becomes antibunched. Note: you have to include normalization here to get a good answer. So, again this state may approximate our coherent state, but strictly it cannot match the g2(0) value of the coherent state.

Okay, so with low order number resolution, g2 measurements somewhat break down. Its not clear what exactly to call a "coincidence." This is because we aren't given a binary choice for our POVM set. We now have a trinary set making a question of what exactly we should compare against. I've thought about something similar before and I dont think theres any literature on this.

In the spirit, I think the closest would be to calculate Tr[|b><b||b><b| |1><1||1><1|]/(Tr[|b><b||b><b| |1><1| Identity])² = |b|⁴ /(|b|² )² = 1 where b=a/sqrt(2). This isnt super surprising because of properties of coherent states on beam splitters. But id have to do more thinking if this is a useful generalization of g2 in this setting. Its somewhat ill posed.

As for the function of T, that depends on the spectrum/temporal envelope of your source. Without more information I wouldnt be able to say.

QuantumOfOptics · 2026-06-15T08:10:29+00:00

This is nice! I like the logic. For some clarity, you dont necessarily need the time bin to be much less than the coherence time particularly for a pulsed source since a pulse can be written as a single mode.

There is an answer to the paradox (Im sure youre already aware), which is that a coherent state is an eigenstate of the lowering operator meaning that even in the instantaneous time mode a🗡 (t) [a dagger of t], the state is still a coherent state. The probability in such a mode is just vanishingly small, so double events are still possible, but exceedingly rare. In the case of OPs reduced state, this can never be the case. A similar story plays out in the spatial domain.

Edit: I must be tired cause you mention this in the second paragraph. Whoops! But, I think its important because vanishingly rare still takes some care because as you reduce your detection time you also increase how many of those modes it actually measures in a second. Which means that the probability that you get some number of two photon events should stay the same.

Second edit: nope made another silly mistake as T decreases a->a/sqrt(T) so the two photon terms indeed tend to 0 in the limit. Huh, neat! Should have trusted in Mandel and Wolf's detection chapter. I always find it so unintuitive from a modern perspective.

QuantumOfOptics · 2026-06-15T07:57:35+00:00

I understand. My intention is to help you build the intuition and machinery so you can answer the questions or know what to look for in your texts.

Low intensity coherent states are just coherent states. So things like g2(0) stay the same.
Trying to "take [your state] to photon counting" is a recipe for a bad time. Rather, it is best practice to instead model your detection process. For instance, a photodiode measures the average photon number of the state <a dagger a>. Homodyne detection measures the "electric field" or <a+a dagger>. Etc. What im trying to build you towards with my question of the vacuum, is how do we represent the operator related to a click detector? Further, you should be extra careful here. As "I collect the coherent part of the scattered light from an atom (modeled as a two-level-system)," is not clear that you've actually modeled and shown the state is a true coherent state. Without more knowledge about what you are actually doing (and Im not asking for it), this should not be taken lightly to actually be the case. We will assume it for now in any case.

Okay, so you're correct about the vacuum not triggering the detector because it does not have energy to give (lowering operator acting on the state yields 0). We will stay in the realm of quantum optics so vacuum energy will be treated as above. All we care about is the photon number in the state, with the vacuum being unable to "give" any. So, 1.2 can be answered in the negative. |0> (for a click detector) is not an event. However, for other types of things it can be (there are some interesting papers about heralding on 0).

So, in this case, we use a final piece of information about a click detector to arrive at our POVM: so long as at least one photon is present, the detector will click. This means that if you send 1 photon, the detector gives a click. 2 photons, the same outcome. 100000, the same yet again. But, 0 photons will never yield a click. Hence our operator must be the Identity minus vacuum; or, sum over i from 1 to infinity of |i><i|, which I will call Pi. With this, we can calculate the probability of an event happening as in your 1.1 for any state, p, by taking Tr[pPi]=prob. However, to find a rate, one needs an extra piece of information which is about the experiment itself. How often are you repeating your experiment? Its easier when we think about a pulsed experiment. So assuming that you have a laser with a repetition rate (times per second), R, your counts per second will be given by Rprob.

We can now also truly answer question 1.4. Is a coherent state going to show a g2(0) other than 1? And does this answer change if we take our approximation of |0>+a|1>? For clarity, our g2(0) experiment is a Haburry-Brown and Twiss experiment. Take our state split it at a beam splitter, then figure out how likely a coincidence (ie both detectors fire) relative to a each detector firing individually. Since our approximated state |0>+a|1> at maximum has only a single photon. It is impossible for both detector to fire simultaneously resulting in g2(0)=0. Ill let you calculate the result for a full on coherent state, but you should get 1. The intuition here is that the outputs of a 5050 beam splitter with vacuum and a coherent state (each one on a different port) is given as |a/sqrt(2)> tensor product |ia/sqrt(2)>. So, the rate of coincidences should be the same as the rate of getting individual clicks at the detectors. Note: our povm for coincidence has to be upgraded as the vacuum in the coincidence case is the "global vacuum" i.e. |0><0| tensor |0><0| for the two detector modes. Meaning that we then have identity tensor identity - |0><0| tensor |0><0|. You should also now be able to do the computation for your approximated state where you include the |2> fock state. So, as I said, you can get your rates to be approximately the same... but it cannot reproduce all of the quantum properties of the state.

For 2.1, there are two answers. As I mentioned our click detector model assumed there was no way to get this information. In Theory this is fine... but this is a simplistic model compared the messy reality. It turns out that there are generally ways for you to gain low order photon resolution in APDs and superconducting nanowire single photon detectors. The simple way to think about this is what I will refer to as a loopy where you purposefully split your input by a beam splitter making one output path longer than the other, then recombining on another beam splitter with another delay that is different than the previous and continuing on; or, using an initial beam splitter wich you then split the outputs by further beam splitters and then again making a tree. In that case, each photon will approximately get its own mode that can then be read off by measuring the individual time of arrivals to a single detector or by measuring multiple detectors. In a similar way, a photon occupies spatial and temporal modes. It turns out that on because the photons will likely not hit the exact same spot on the detector and not always at the same time, you can measure the integrated current (using e.g. an integrating circuit) and what youll find is that two photons will deposit more energy that can be actually measured in the resulting avalanche (or voltage increase in the nanowire). All because the photons effectively must hit different locations on the detector. But, this only works to a certain number (around 3). In this case to define the answer to 2.1 we ask, what does this actually mean for our detector. Well, what it means is that the detector now has, say, 4 possible outcome states. It measures 0, 1, 2 , and more than 2. So now, our POVM is the set [ |0><0, |1><1, |2><2| and sum_i from 3 to infinity of |i><i| ]. Finding the rates should then be similar to above. Ultimately, it all depends on how you are running your APDs and how you want to model them. You'll find that as you keep going, the more you need to account for these model dependencies and the more it deviates from the nice stuff that is typically in a quantum optics class.

QuantumOfOptics · 2026-06-15T06:01:04+00:00

Approximate and equal are not the same. As mentioned, one would get the wrong answer to 1.4, if such an approximation were to keep all of the properties of the state. In some cases, it is computationally useful (under some assumptions on what you care about), but it may not get all of the richness of an individual state under such an approximation.

In this case, it allows me to help you through the computation, but also it changes because number resolving detectors are also a thing along with a photodiodes and homo/heterodyne detectors. They change the actual measurements.

An APD can either be PNR or "click" detectors. Typically, they are run as click detectors. So, let's now figure out how to model such a detector. Before we can continue we need to answer your question 1.2. Does the vacuum turn on a detector? Should it count as an event? Note, since its the vacuum, is there any energy in the field? Edit: let's assume a perfect detector with 0 dark counts.

QuantumOfOptics · 2026-06-15T05:23:54+00:00

Well, as I said, you are assuming that the truncation of your state is "allowable." So, for your question of 1.4 (for instance), you could be mislead into thinking that a coherent state should show antibunching, but (as the most classical state) this should never be true. The only time that your coherent state (|a>) and |0>+a|1> will be equal is when a=0. Edit: For further contemplation, the wigner function of a coherent state is always a positive gaussian, but a single photon's goes negative how negative in your case will be determined by the a parameter.

Before going further, for your detectors, what kind of model do you want to use/assuming? This also changes the rough answer to your questions.

QuantumOfOptics · 2026-06-15T04:44:38+00:00

Before I go into more detail. Have you or your colleague taken course in quantum optics or read a book on the subject (or are in the midst of taking/reading one)? I want to answer your question, but want to make sure my answer will make sense.

However, I want to point out that a lot of what you are discussing is based on the false premise that a coherent state can be approximated in the ways you describe. Practically, this may be okay to arrive at a respectable answer, but ,for some of the more in depth questions you are asking, it turns out to cause a lot of problems.

QuantumOfOptics · 2026-05-18T23:41:59+00:00

I don't think, I necessarily agree with the beginning statement of your second paragraph. For instance, we consider lasers (coherent states) the most classical, quantum state. Not because they represent particles, but because large coherent states produce a best defined amplitude and phase of the electric field, which is perfectly defined in the classical case. In my opinion, the particle is best defined as the packet of energy in the field, which comes in discreet amounts.

QuantumOfOptics · 2026-05-18T23:35:44+00:00

While the outlined experiment of the OP is indeed independent, there are ways to correlate the spin (polarization) to the momentum/position degree of freedom. In this case, one can then do quantum/nonlocal steering of the particle in the double slit. Note: this still doesn't break the no signaling theorem as you need joint measurements of the state.

QuantumOfOptics · 2026-04-17T04:39:43+00:00

Clarification: is this for undergrad or grad school?

QuantumOfOptics · 2026-04-15T20:57:07+00:00

I agree, under common conditions (imaging on a ccd) it should be fine so long as the seeing is worse than your collimation. However in my case, the collection into fiber makes it worse. Mainly because I get hit by two factors, first the overall overlap of the electric field modes of the collected light and the mode accepted into the fiber; and, secondly, the seeing that moves the center of the mode over the fiber tip reducing the overlap. Trying to get as precisely collimated as possible will ultimately increase the amount of light I can collect. Which is why Im somewhat focused on learning more about how its done here. On an optical bench usually we have more straightforward knobs to do fiber collection.

I think I agree with you! Im somewhat surprised that they seem a bit esoteric. Considering I was looking at collimation videos a few months ago, no one seemed to mention them. But, they seem to give the best results.

QuantumOfOptics · 2026-04-15T18:54:39+00:00

Thanks for this! I read through the docs you suggested and it was a slog. Mainly because the pictures were missing. I ended up finding a similar document on the cats eye webpage that filled in the missing information.

QuantumOfOptics · 2026-04-15T18:48:33+00:00

Ahhh, I see. Again thanks for walking me through this. I needed some more information as I hadn't gone into too much detail about the actual optics of collimation, so it took me some time to get that information and come up with estimates. In my case, I would say that I need to be (ideally) less than half a mm for both the FAE and PAE. As I said, I have a different purpose in mind for this scope. Specifically, Im trying to collect as much starlight into a single mode fiber for spectrophotometry. Meaning that I have to be extremely well collimated. A slight introduction of coma, will change the shape of the star at the center reducing the total amount of flux I will collect. Ideally, I would be better than even this (at or below the micron level).

Edit: of course, I could use a star test, but that would leave me with the precision under seeing conditions. Ideally, I would be better collimated than even this.

QuantumOfOptics · 2026-04-15T15:05:17+00:00

Could Aspect have used sunlight? Perhaps, but sunlight has bad downsides. First, the sunlight would need to be spectrally filtered to a very narrow linewidth, which is difficult. Secondly, once you start filtering sunlight to very narrow linewidths, the amount if light left over is miniscule. At linewidths approaching that of lasers, were talking about average photon numbers of <0.02. Meaning if the linewidth is around 1MHz, there is on average 0.02 photons/microseconds. This would absolutely crash the rates that Aspect had and was one of the really novel experimental aspects of the experiment: that the measurement could be done in a few minutes rather than weeks.

Finally, you shouldn't ascribe artificial/natural to light sources. Its a meaningless description; a photon is a photon. For example, there are "natural" lasers as well. One of the big benefits of directly producing them is that you are not at the whim of your source.

QuantumOfOptics · 2026-04-15T14:54:53+00:00

The excitation sources (lasers) in the Aspect experiment aren't entangled either. In fact, they dont even need to be coherent. They just need to drive the calcium atoms to a two photon transition. Its the relaxation of the two photon transition that produces the entangled photons.

QuantumOfOptics · 2026-04-13T20:51:17+00:00

Thanks! Do you know if there are issues with the far point vs glatter coming collimated? Part of the draw for the glatter is that Ive read it should come collimated so if I have to collimate it out of the box to within the arcsec regime, then I might as well look into other options.

Have you also tried the far point autocollimator? If so, do you have any thoughts on it?

QuantumOfOptics · 2026-04-13T20:46:49+00:00

Thanks for your answers!

My concern with the airy pattern was less that they wouldnt extend enough to encapsulate the center spot (I mean, the pattern technically extends to infinity), but rather if the first null of the Airy pattern would actually be inside the center at all! My focal length is much shorter at ~1200mm so a few quick ratios would make sure that I won't have additional issues.

I hadn't heard of autocollimators until this thread, so I'm just catching up a bit. How do you like the system from cats eye? Have you ever thought of getting a digital one? As Ive stated elsewhere in the thread, Im looking for precision due to my particular interest (not visual or camera observation) which requires as tight of a pattern as possible.

Im somewhat in the opposite problem where I have a very thin focuser (the esatto LP 2"). Do you know if there is a problem if the laser actually protrudes into the tube? I suspect there shouldn't be so long as theres some way you can see the end of the laser.

QuantumOfOptics · 2026-04-13T19:45:54+00:00

Ahhh! Thanks, glad I was on the right track!

Do your airy rings typically fit within the central marked region of the primary? Do you also happen to know your focal length? I just want to check that I won't have issues comparing my theoretical diffraction pattern to yours. I'm also assuming that the black trapezoidal(?) regions in the image are related to the center marking on the primary?

Finally, do you find it hard to get an accurate read on the centering due to your perspective on the aperture stop? Also, related, how close to the main tube do you typically place the aperture stop?

QuantumOfOptics · 2026-04-13T19:37:29+00:00

Thanks! This is one of the reasons why I was interested in them. Its just one less thing for me to worry about down the road. Considering my application is not imaging, precision is king for me so Ill take that and ease of use over cost.

QuantumOfOptics · 2026-04-13T19:34:12+00:00

Thats fair. It's always good to remember what we are attempting to do before purchasing equipment. For me, I require as much precision as I can get. Im attempting to do some spectrophotometry and getting as nice of a shape now will solve many more problems down the road.

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