# How does the light source fire a single photon in the double-slit experiment

All the youtube videos I have seen on the double slit experiment broadly fall into one of the following three categories:

• Documentaries and fan made videos heavy on animation which 'admire' the wave-particle duality of light rather than 'explaining' it.
• University lectures where the professors draw diagrams on the board and go into the mathematics (e.g. how to calculate the distance between two bright bands) but never explain how the actual apparatus is built.
• Double slit DIY videos on how to do the experiment with sunlight or laser but they seem to leave out the most critical part, how do you make the light source so faint that only one photon is emitted at a time?

As far as I understand, the classical wave theory of light fails to explain the interference only when you do the experiment with one photon at a time. My question here is, how do you make the light source fire a single particle of light at a time when you are trying to determine whether or not light is made of particles in the first place? I am more interested in the engineering of the apparatus rather than the mathematical explanation.

A follow up question might be, how can you be sure that indeed only one photon is coming out of the source at a time? (a single photon is hitting the screen at a time is not the same as a single photon is coming out of the source at a time, especially given the weird wave-particle duality).

• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Aug 1 '16 at 20:32
• It's not exactly the double slit experiment, but here is a paper that describes an actual class-room experiment for a single-photon Mach-Zehnder interferometer: Dimitrova T. L. and Weis A., The wave–particle duality of light: a demonstration experiment, Am. J. Phys., 76 (2008). – Emil Aug 2 '16 at 9:14
• @Emil: The sad part about these description is that they teach 100% false physics. Photons in the optical part of the spectrum don't interact. It doesn't matter how many there are in the apparatus. It would take conditions far beyond those of a nuclear explosion to observe photon-photon interaction. Someone needs to tell me why people teaching these things don't understand this basic fact. – CuriousOne Aug 2 '16 at 11:42
• @CuriousOne: The experiment sets out to disprove a false hypotheses: "Photons need other photons to interfere." I would say it successfully shows that this is not the case. From there, a good physics instructor can explain how this is not even the case when there are many photons involved. – Emil Aug 2 '16 at 12:17
• @Emil: The experiment can't tell you anything about photons. It works with water waves just as well. It's a perfectly classical interference experiment. That photons don't interact with other photons can only be proven by spectroscopy on crossed light beams. That is what gamma-gamma colliders do... or better, will do, once the first one has been built. A good instructor will stay away from presenting physics experiments in a way that confuses students like this one. – CuriousOne Aug 2 '16 at 19:15

I think there's a bit of confusion here. The double-slit experiment was not performed with "single photons" - it's very hard to even consider what that would mean. At its heart, it is a thought experiment, and it's not really possible to make a real-life device that tests it.

The first low-intensity experiment (Taylor 1909) was challenging the EM field interpretation of photons - the idea was that if photons were localised concentrations of the EM field, as you lowered the intensity, there would be no photons to interfere with each other, and the diffraction pattern would disappear. When the experiment was low-intensity enough that Taylor couldn't distinguish between photons emitted and photons absorbed, he noted that the diffraction pattern still exists, so the photons couldn't just be localised concentrations of the EM field. Dirac had a different explanation - he considered that each individual photon was capable of interacting with itself.

Later, the experiments were repeated not with light, but rather, electrons. Electrons are a lot more convenient than photons, since they obey the Pauli principle: it makes a lot more sense to say "an electron here, an electron there". And you can emit individual electrons, which was first tested in 1974, and it was found that individual electrons do in fact display the same interference pattern. Later, it was found that the same pattern also appears for atoms and complex molecules (the current "world record" has the experiment done with a molecule with more than 800 atoms, at ~10 000 atomic weights; the experiment gets much harder with bigger "particles", since it requires much more precision). But we'll stick to electrons, since they're quite convenient.

Emission of individual electrons is still quite tricky, but they have a few important properties. They carry a charge, and they have mass. Both of these can be measured, and while this does disturb the electron (change its trajectory), it doesn't absorb it. Photons, on he other hand, will be absorbed by any measurement, which makes them tricky to deal with.

So you can measure with precision to individual electrons how many electrons were emitted from your emitor, and how many were absorbed on the detection surface. More importantly, you can try experimenting with what happens when you measure the electrons on the way between the emitor and the detector - and that's when real quantum phenomena come in.

Which path does the electron take through the double-slit apparatus? If you're already in the modern quantum physics mindset, the question doesn't really make any sense (and by modern, I still mean the first half of the 20th century :). But as we've seen, electrons can be measured on the way, so what if we put detectors on the two slits? Well, as was of little surprise to the people involved, the pattern disappears completely (this was predicted by Feynmann and others long before the actual experiment was done in 1961). What's more interesting (and still predicted in advance), putting the detectors behind the two slits also destroys the pattern.

But this is really a matter for another question entirely. At the core, the answer to your question is "the full experiment cannot be done with light, and was never attempted with light". The best that was attempted was to reduce the intensity of the incident light so low that the energy corresponded to one photon of a given energy - by adding distance and barriers in the light's path. But we know of no way to emit individual photons, or to detect them with certainty, or to measure their paths - and it may very well be outright impossible, not just infeasible. Now, this in itself is quite enough to discredit the dual theory of light (unless you go with Dirac's assertion), but the real experiment was done with single electrons, and continued with bigger and bigger quantum particles, up to the 800+-atom monster I mentioned earlier.

• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Aug 2 '16 at 11:36
• Single photon experiments are a waste of time. Optical photons don't interact. It takes a gamma-gamma collider to perform that trick. There are thousands of people worldwide who would love to build such a facility, they are just about a billion dollars short. :-) – CuriousOne Aug 2 '16 at 11:40
• Some references to go with those dates would be handy. – Eric Towers Aug 2 '16 at 16:22
• @Luaan what does photon-photon interaction, which CuriousOne spends days going on about to everyone's confusion, have to do with questions about double slit experiments? – hobbs Aug 2 '16 at 16:44
• @hobbs: Light interferes because Maxwell's equations are linear. Linear equations obey the distributive law of multiplication a(b+c)=ab+ac, which is the only thing that happens when one dims the light. If there was a term that didn't obey the linearity, then one would expect two photons to mix to a different frequency, i.e. intense red light would go in and faint blue light would have to come out. That is what multi-photon physics actually looks like. – CuriousOne Aug 2 '16 at 19:05

A light beam is not like a swarm of photons flying through space - the relationship between light beams and photons is rather more complicated than that. This is discussed in What is the relation between electromagnetic wave and photon? though a proper discussion requires going deeper into the subject of quantum optics that many of us consider sane.

Anyhow, the photon is basically the unit of energy exchange with the light beam i.e. if the power of the light beam is $W$ joules per second then the number of photons emitted per second is this power divided by the energy per photon:

$$N = \frac{W}{h\nu}$$

The travel time of the light is $\tau = \ell/c$, where $\ell$ is the length of the light beam in your experiment, so to have only one photon at a time you need the number of photons per second to be of order $1/\tau$.

$$N = \frac{W}{h\nu} = \frac{c}{\ell}$$

And solving for the power gives:

$$W = \frac{h\nu c}{\ell}$$

So that's all you need to do. Make your light beam power equal to or less than $W$ calculated as above and you'll have only one photon present at a time.

• I've deleted an unconstructive comment discussion. As John suggested, discussions can be held in Physics Chat. – David Z Aug 1 '16 at 20:31
• @Luaan I understand duality no longer being satisfactory if you abandon the wave but what proof is there to abandon the photon? Your saying photons can take every path and they can be measured but they don't exist??? You can't have it both ways. On the other hand you can't measure a wave or even describe one but still you have unwavering faith in the idea. – Bill Alsept Aug 1 '16 at 21:50
• I don't think that's all the story. For example, consider a thermal light source (any kind of light source except laser and a single atom). It's considerably probable to see photons arriving together which is called "photon bunching". That's why you can't obtain a single-photon generator just by reducing the power (if we neglect various sources of noise). – N.S. Aug 5 '16 at 18:44
• Also, according to the paper I am linking below, by adding 'some kind of filters', and 'increasing the order of the filters', the photons can be further separated from one another to achieve evidence for single photon interference. So, technology to carry out single photon double slit experiment exists, it seems. optics.rochester.edu/workgroups/lukishova/QuantumOpticsLab/2013/… – Prem kumar Nov 28 '16 at 7:17

You can use a single photon source, such that there is (provably) only one photon in the system at any one time. These are quantum mechanically single-photon emitters, as distinct from just turning down the intensity of a source emitting random photons.

One convenient example is the nitrogen-vacancy centre in diamond. A single NV centre only has one excitation and so can only emit a single photon each time it's excited.

Because the experiment is so inefficient (very few photons actually make it through the slits) finding a nice paper that directly shows it is tricky. However there are plenty of similar experiments using these sources (example) for duality experiments.

You keep reducing the energy of the source until the number of photons impacting the screen are small enough to count one at a time. The easiest way for me to make a slit experiment is with guitar strings and a laser. It's really easy if you direct a laser at one string. That is the way Young did it the first time when he used a single human hair. The light will wrap around the edges of the string and make a fringe pattern on the distant wall.

• Thank you for your answer. "You keep reducing the energy of the source until the number of photons impacting the screen are small enough to count one at a time". Are you aware of any kind of light source that is delicate enough to do that? As for the rest of your answer, there are several ways to do the experiment and find the patterns. The challenge is, how do you make the source shoot one single particle at a time? – user58110 Aug 1 '16 at 9:45
• @Rahat, the attenuation doesn't have to happen at the source. You can introduce increasing numbers of filters until the power past the final filter is as desired. – BowlOfRed Aug 1 '16 at 23:53
• @BowlOfRed yes that's probably better than trying to weaken the source – Bill Alsept Aug 2 '16 at 0:58