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Double Slit Experiment: How do scientists ensure that there's only one photon?

+ 8 like - 3 dislike
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Many documentaries regarding the double slit experiment state that they only send a single photon through the slit. How is that achieved and can it really be ensured that it is a single photon?

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user Felicitus
asked Sep 3, 2013 in Experimental Physics by Felicitus (25 points) [ no revision ]
Most voted comments show all comments
Related: physics.stackexchange.com/q/70855/2451

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user Qmechanic
@Qmechanic I think this is a different question. This question is more about experimental design as far as I can tell as opposed to some conceptual difficulty with what's happening to a single photon that is fired.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user joshphysics
Tag added in keeping with @joshphysics' take on the question.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user dmckee
@joshphysics That's why Qmechanic said "related" and not "possible duplicated"

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user Tobias Kienzler
@TobiasKienzler I could have sworn it was originally a "possible duplicate" comment, but I could also be going insane.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user joshphysics
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... the title of the question looks also slightly popular, which may also explain the immense number of upvotes...

It should be zeroed out, this is a silly question. I don't know what it was imported, except maybe to save it from deletion.

6 Answers

+ 9 like - 2 dislike

quantum dots. nanoscale semiconductor materials that can confine photons in 3 dimensions and release them a measurable time after. based on material used the decay time is known empirically. frequency is also known. the latter is sufficient to calculate the energy of one photon. the former is then sufficient to calculate the rate of photon re emission from the QD. if the peaks at the detector are further apart than the decay time and each peak is measurable to one photon's worth of energy then you know you have a beam of single photons.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user gregsan
answered Sep 4, 2013 by gregsan (70 points) [ no revision ]
I'm sure this will work, but I doubt that anyone bothers.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user dmckee
The double slit experiment was performed over a century ago, there were was no nanotechnology back then. So how do you think people made sure that just one single photon was involved in the experiment, other than deducing the existence of light as quanta by the observation of experiments?

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user user1800
@user1800 The double slit experiment with single photons was first performed in ... 1986, in a pioneer experiment by Aspect, Grangier and Roger, please see dx.doi.org/10.1209/0295-5075/1/4/004 especially Fig.4, which shows the appearance of the fringe for large number of photon counts. The gregsan answer is a modern version of the photon cascade effect that Aspect and Grangier used in the 80's for photons on demand. Studying quantum dots semi-conductor is still an active field of research for unique-photons-on-demand, too.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user FraSchelle
@gregsan To comment further about the above cited Aspect/Grangier/Roger experiment alongside the question Felicitus asked: they are sure they have only one photon because they have two detectors (avalanche photo-diode) which never clic at the same time. Say differently, when they clic together at the same time, you can discard the counting controllably. Thanks you for this interesting answer.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user FraSchelle
Indeed this has been our method of choice. We've been working particularly with CdS QDs. The great advantage is the ability to grow the QDs to any arbitrary size and therefore tune the photon frequency as desired.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user John Cuna

This doesn't help with double slit anything. You don't "trap photons in quantum dots" usually, you excite electrons. At this point, there are measurement artifacts, decoherence from phonons or whatever in the quantum dot, and the experiment is not as clean as photons through air, which are coherent for a long time. If you are just looking for a dilute photon source, look at x-ray diffraction. X-rays are arbitrarily dilute, often photon by photon.

+ 7 like - 0 dislike

The practical answer (which I also wrote in a comment on the linked question) is that you turn the intensity of the light source down until the expectation value for the number of photons on the optical path is low enough to suit you.

If $\bar{n} = 0.1$ then very few of the events that are recorded on the screen will come from events where more than one photon was present on the optical path and the data will be dominated by single photon event.

Not good enough for you? Turn it down until $\bar{n} = 0.01$. Or $0.001$ or whatever suits you.

At some point the exercise becomes silly.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user dmckee
answered Sep 4, 2013 by dmckee (420 points) [ no revision ]
Unfortunately, this answer is incorrect. Turning the intensity off do not provide you're working with photon eigenstate of the light field. Coherent state with an average photon number is not a Fock state of photon. To go to single photon effect, you need matter-light interaction, as the experiments by Aspect/Grangier and Haroche/Raymond/Brune clearly shown. The first team using photon cascade effect, the second one using flux-number commutation relation on quasi-classical field (coherent) and matter (Rydberg).

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user FraSchelle
@Oaoa That is not material that I'm familiar with, but I found what looks like the slides from a graduate summer school presentation on the matter by Haroche, and he seem to say (lecture 1, slide 11) that for a coherent state, the distribution of fock states is Poisson, in which case very low mean occupation number does imply that the data is dominated by the single-photon fock state, which is what I said. Sure, there is some contamination, but you simply turn down the intensity until you are happy with the results.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user dmckee
Thanks a lot for the material you shared. I'm also always puzzled with this issue. Let me try to me a little bit more clear. For me, a single photon is a wave packet. Reducing (using filter of whatever) the intensity of coherent (laser, say) field will never produce wave packet: instead you will always find a continuous in time field, at rather low intensity. Maybe this is too much naive point of view. I never heard about an experiment with low intensity which results in discrete photon state. The Haroche argument is perfectly right: low occupation means 0 or 1 photon dominant (...)

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user FraSchelle
(...) but 0 photon state is a coherent state (perhaps the easiest to do coherent state), and the 1 photon state is already in the queue of the Poisson distribution. Really puzzling issue. Of course the detection scheme you're using is of primary importance, too: you must detect one-photon energy per unit time for instance. My understanding of the Haroche/Brune/Raimond experiments is that you need an atom to make a Fock state. You start from a thermal field, and the atom-light interaction select the Fock mode. The atom clearly acts as an extremely precise clock, mandatory for the experiment.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user FraSchelle
@Oaoa I think the difference is that I am not worrying about being able to show that any particular count came from a single photon event, just being sure that the pattern I see is dominated by them. That kind of thing won't do for quantum computing or entanglement measurements. Or maybe I just don't get the argument. As I said, I'm a bit out of my depth with that stuff.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user dmckee
+ 7 like - 0 dislike

In the double slit experiment, if you decrease the amplitude of the output light gradually, you will see a transition from continuous bright and dark fringe on the screen to a single dots at a time. If you can measure the dots very accurately, you always see there is one and only one dots there. It is the proof of the existence of the smallest unit of each measurement which is called single photon: You either get a single bright dot, or not.

So, probably you may ask why it is not a single photon composite of two "sub-photon", each of them passing through the slit separately and then interference with "itself" at the screen so that we only get one dot. However, the same thing occurs for three slits, four slits, etc... but the final results is still a single dot. It means that the photon must be able to split into infinitely many "sub-photon". If you get to this point, then congratulation, you basically discover the path-integral formalism of quantum mechanics.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user hwlau
answered Sep 4, 2013 by hwlau (85 points) [ no revision ]
I chose this answer because it explains it better to me than the highest-rated one; but thank you all for answering. I'm not a physicist, so a simple explanation like this is more clear to me than formulas. Thank you!

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user Felicitus
+ 4 like - 0 dislike

I'd like to add to gregsan's answer about using quantum dots by also raising diamond nanowire one photon sources or similar devices made on CVD-grown diamond waveguides in contemporary or future experiments - of course this this method was not used in the historical one-photon experiments!. See:

Mark P. Hiscocks, Kumaravelu Ganesan, Brant C. Gibson, Shane T. Huntington, François Ladouceur, and Steven Prawer,"Diamond waveguides fabricated by reactive ion etching", Optics Express, 16, Issue 24, pp. 19512-19519 (2008) http://dx.doi.org/10.1364/OE.16.019512

This one can be freely downloaded, and I should declare my close contact with several of the authors, although I was not part of this work.

Thomas M. Babinec, Birgit J. M. Hausmann, Mughees Khan, Yinan Zhang, Jeronimo R. Maze, Philip R. Hemmer & Marko Lončar, "A diamond nanowire single-photon source", Nature Nanotechnology 5, 195 - 199 (2010)

This one is paywalled, but independent from me.

These devices might find experimental uses in more sophisticated one photon experiments both now and in the future because one can trigger them to emit lone photons almost on demand, whereas dimmed light sources simply emit photons randomly following a Poisson process and cannot be triggered. I have heard several times the opinion of several very bright (much brighter than I am) experimentalists that "no-one would bother", but I just can't shake the feeling that triggering might be useful in yet-to-be thought of experiments - I'd call myself a lousy experimentalist but once upon a time I could drive an analogue oscilloscope pretty well and triggering is surely what makes a useful piece of kit ten times more useful!

Here's how the diamond waveguide one-photon source works. One lays down a diamond waveguide by chemical vapour deposition and shapes the surrounding environment by reactive ion etching. Then a highly controlled amount of nitrogen is included into the diamond lattice. A nitrogen atom normally only makes three covalent bonds with its neighbours, whereas carbon normally makes four, so one is left with carbon atoms with one covalent bond forming electron "dangling" in the lattice wherever there is an included nitrogen. This "dangling electron" is then a fluorophore. One controls the concentration of nitrogen carefully and dices the waveguides so that there is exactly one fluorophore centre in each waveguide device: you can do this by building many devices at once then testing each and throwing out any with none or more than one fluorophores. The waveguide device can then be coupled readily to a single mode optical fibre which links the device to the outside world. The extremely high diamond to air refractive index difference (diamond $n=2.4$) and the ability to build waveguides down to below wavelength dimensions means that we can make "nanowires", which are single moded but also their strong waveguiding properties means that there is near unity probability for the photon fluoresced from the dangling electron into the single mode waveguide. So now we have something which is very readily connected to other experimental kit and doesn't need alignment.

When the time comes to use the device, a trigger signal causes an optical pump laser in the device to pulse the device with an intense beam of light, so that the lone fluorophore in the device is almost certainly raised to its metastable state. At the same time, a very high performance optical switch gates the device's optical output so no light at all escapes to the outside world. After the pumping pulse has ebbed, the output optical gate opens and the fluorophore relaxes a short time later - the fluorescence lifetime is a few nanoseconds. Although this fluorescence is governed by an exponential time-till-emission probability distribution so that, strictly speaking, it is stochastic just like the traditional dimmed light sources, the gating process means that one can control the time of photon emission to within a few nanoseconds.

Moreover, the near unity photon coupling probability into the single mode waveguide system means that there is an extremely low coupling probability for outside photons incoherent with the waveguide mode to couple in. This system is therefore extremely immune to stray light even though one photon experiments are being done: the exposure of the waveguide system to normal room light levels means that almost no photons get in - and that's before one blackens the system or puts it in a box. There is zero contamination in this device in the hands of even the most ham fisted experimentalist.

The original and main motivation for these devices is not one-photon double slit experiments, but rather as sources for quantum cryptography protocols such as the Bennett and Brassard (BB84) quantum key distribution protocol.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance
answered Sep 5, 2013 by WetSavannaAnimal (480 points) [ no revision ]
I suspect that triggered single photon sources will interest people. Perhaps interest them a lot. And if they become common, off-the shelf turn-key devices they will be used for basic interference demonstrations, but as yet such sources are a long way from that easy to use and people will only get them going for things that matter.

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user dmckee
@dmckee Alas you are right that they are a long way off, owing mainly to the difficult quest for funding!

This post imported from StackExchange Physics at 2014-03-30 15:43 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance
+ 3 like - 0 dislike

How do scientists ensure that there's only one photon ?

The simplest answer is: there are many processes in nature that emit only a single photon. Among many, let me give you a very common example from nuclear physics. Unlike chemistry, \(A+B\rightarrow A+B+C\),  where A and B interact to give A, B and C, is a common process in nuclear physics. An example of such process is deeply virtual Compton scattering (DVCS) where an electron of momentum about 10 GeV hits a stationary proton  and an electron, a proton and  only a single photon are produced. But it does not mean that every time when an electron hits a proton there will always be  an electron, a proton and a photon. To ensure that there is only one photon, it is enough to ensure that the scattering process is DVCS. To ensure that the process is DVCS conservation of four momenta before and after the scattering are checked; if conserved then the process is DVCS. This is how it is done in general in the experimental nuclear physics labs. 

answered Jun 24, 2014 by Nottherealwigner (135 points) [ no revision ]

Nice example and explanation :-)

+ 2 like - 0 dislike

No experimental apparatus generates precisely a single photon state or generates precisely an entangled state of two photons, etc.

If you look at experimental data, it's always at least a little messy, both because of slightly less than ideal state preparation and because of slightly less than ideal measurement, but, we can look for ways to prepare light and to measure light that give experimental results that are pretty close to what we would expect for a given single photon state and measurement device (or for whatever entangled state we want to generate and characterize). It's a fact that experimentalists have been able to find light sources, transformations, and measurement devices that do come close, and over decades experimentalists have managed to construct steadily better approximations to many quantum theoretically ideal states and measurements that people have thought they would like to make (to a classical imagination, of course, it is something of a surprise this can be done). Each decade of work by thousands of physicists and engineers provides a significant amount of ingenuity for each small advance of precision: it's no small thing to figure out what novel material and surface preparation will give a 10% improvement on some measure or another and to make it happen reliably, so there's no guarantee that this process of refinement will continue forever.

Some of the other answers here touch on what might now almost be called a crude distinction, between coherent states and single-photon states. Others mention quantum dots, which are more recent and significantly closer to a quantum single-photon ideal, but still are not perfect. Each generation has its own better preparations and/or better measurements, closer to the ideals of the moment.

answered Jun 25, 2014 by Peter Morgan (1,075 points) [ no revision ]

Dear Peter

May be you are answering this question at a level much deeper than I am aware of. Let's consider an exclusive electro-production of neutral pion (an electron scatters off a nucleon and the nucleon and a neutral pion is produced) the neutral pion than decays to two photons which are entangled. In a multiparticle detector it can checked very precisely. Cannot this whole detector set-up be considered a source that generate precisely two entangled photons? Thanks 

I'm happy to say that "Precisely two entangled photons" is approximated pretty well by your setup, but experimental data is never precisely clean. Also, to verify that you really have a given state, you have to gather statistics, typically for multiple incompatible observables, not just one pair of observed events.

Dear Peter

Of course "we have to gather statistics" for any kind of experiment at the quantum level.  And also the fact that every detector has a finite resolution. As an experimental physicist what I can assure is that in the particle detector I am working at, it is possible to get \(1000\pm1\) two entangled photons one pair at a time from a neutral pion decay. Thanks

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