# How does one emit a single photon?

1. Sep 10, 2008

### Greylorn

The double-slit experiment produces its mysterious interference pattern only when photons (or electrons) can be sent through the slit pair one at a time. I'd like to find an explanation of the experimental apparatus which is guaranteed to emit only single photons, at time intervals which are significantly greater than c (such a millisecond apart).

The question is relevant because if two or more photons are emitted simultaneously, or nearly so, the double-slit experiment ceases to be interesting.

2. Sep 10, 2008

### atyy

3. Sep 10, 2008

### Cthugha

Usually one uses some transition of a single emitter, which is blocked for a while after the emission of a photon. As an easy picture imagine an atom in an excited state, which emits a photon. After the emission it will take some time for the atom to get into the excited state and decay again, so there is a certain dead time.

For a closer look, you might look here:http://www.sciencemag.org/cgi/content/abstract/290/5500/2282
or here:http://www.sciencemag.org/cgi/content/abstract/303/5666/1992

4. Sep 10, 2008

### cesiumfrog

In practice, for the single slit experiment, you would just turn down the intensity (and add neutral density filters to further attenuate the beam) until statistically there is almost never two photons at the same time.

5. Sep 10, 2008

### f95toli

No, modern single photon sources ARE really single photon sources; they are not just "sources with low intensity". I.e. in theory they fire ONE photon everytime they are triggered and at a well definied time.
In reality they "misfire" occasionally (i.e. emitt a photon even when there is no trigger) and occasionally they will fire more than one photon; so they are not perfect (yet) but since e.g. quantum crypthography relies on such sources there is a lot of money and effort being put into their development.

When the photons are generated using parametric downconversion one can also use a scheme where the experiment is gated; due to conservation of momentum the two photons travel in different directions meaning you can use one of them to gate the other; i.e. you let the "extra" photon hit a detector which in turn generates a signal that can then be used to to tell when a photon reached the experimental apparatus; then you discard any "extra" photons. This is how "slit experiments" (single,double, tripple...) are usually done nowadays.

it is possible to test single photon sources by using e.g. a Hanbury-Brown and Twiss interferometer and measuring the second-order correlation function (known as g(2) ); low intensity light would have a different signature than a true single photon source (since the latter is in a number state).

See also the review article by M Oxborrow, AG Sinclair - Contemporary Physics, 2005

6. Sep 10, 2008

### cesiumfrog

f95, I'm not disagreeing, just pointing out that there is also a much cheaper and more practical way to accomplish this simple experiment.

7. Sep 10, 2008

### Cthugha

Well, one can do so, but I think this completely misses the interesting point of using single photons for a double slit experiment. By attenuating a beam you just shift the average value of the Poisson or Bose-Einstein distributed photon number, but everything else remains the same: You have several emitters, so you also have several fields, which are superposed. So - from a classical point of view - you have contributions to the intensity, which arise from the field of one emitter alone and you have contributions, which arise from the combination of the fields of two different emitters. Now the interesting point in using real single photon sources for a double slit experiment is, that those mixed contributions are not present anymore, which makes the double slit experiment even more interesting in my opinion.

8. Sep 10, 2008

### f95toli

No, because it is not the same experiment. An attenuated laser is NOT the same thing as a single photon source.
Specifically because the light from a laser is always Poissonian(coherent) and if you measure the second order temporal correlation function for a laser you will ALWAYS find
$g^{(2)}=$1 for all times (regardless of the intensity) wheras a true single photon source has $g^{(2)}(\tau=0)=0$
Hence, there is a fundamental difference between two and and they are not -in general- equivalent from an experimental point of view.

Also, note that even the imperfect sources I mentioned above are much "better" than a laser beam; some of the sources that are now becoming available exhibit $g^{(2)}(\tau=0)$ very close to 0.

9. Sep 10, 2008

### cesiumfrog

In what way is the double slit experiment "more interesting" if the photons (which you count one at a time reaching the screen) were created by an ideal single-photon source rather than a mundane lamp (with attenuation and zeroeth slit for coherence)? How would I tell the patterns apart? (It seemed as though the OP's aim was just to eliminate the possibility that all interference is due to interaction of multiple separate photons..)

Last edited: Sep 10, 2008
10. Sep 10, 2008

### Cthugha

The surprising point of the double slit experiment with single photons is, that one sees wave-like behaviour contrary to the almost ballistic naive picture of the photon the layman usually has. A layman expects a particle, which travels from some source towards the screen. The emission of a single photon source is very close to this naive view.
Coherent light is already a different topic. You now usually have an ensemble of emitters. Due to coherence, emitted photons inside a coherence volume are indistinguishable anyway. You can never know, which one of the emitters was the source for a detected photon. Even worse, you do not for sure, whether your single photon is a "product" of a single emitter or already the "product" of interference between several emitters.

Usually people do not ask, what a double slit pattern looks like, if photons are detected one at a time, but what a double slit pattern looks like, if photons are sent through the slits one at a time, which is just a small difference, but opens up a few more loopholes and makes the situation a bit more complicated, if you have several emitters and interferences. Using real single photon sources simplifies the situation in my opinion.

11. Sep 10, 2008

### cesiumfrog

Right. But provided we measure only one photon per minute reaching the screen detector (so that we can be confident there was never two different photons going through both slits simultaneously) is there any reason for the layman to care whether the source of the photon was a true single-photon source (triggered once per minute) or even just very well-attenuated sunlight (with another slit preceding the double slit as is normally always the case)?

To the layman, who obviously lacks the QM to fully scrutinise the former "black box" source, wouldn't the latter source actually be preferred?

12. Sep 10, 2008

### Cthugha

Well, this does not change the problem. The attenuation process is still a superposition of statistically independent random absorption/transmission processes, so you will just achieve an average time span of 60 seconds between two photons, but there will be events, when there are 2 minutes in between and there will be moments, when two photons arrive simultaneously, although those will not occur very often.

Well, I agree, that when a layman really wants to build a double slit at home attenuated light is the easier and better way for a do-it-yourself experiment. But if you just want to give a theoretical explanation, I don't see why the usage of a strongly attenuated light source should be easier to understand than a device, which fires a single photon every once in a while.

I just do not see, why a wrong explanation should be used when a correct explanation is not more complicated. Quantum optics is such a huge and alive field nowadays, that in my opinion one should not ignore its results.

13. Sep 10, 2008

### cesiumfrog

To be precise, we can make those coincidences arbitrarily rare; it is not merely a simplification if we ignore that fraction of detections completely.

I do not see why you think either explanation (of the source of particles in a classic "one particle through the apparatus at a time" double slit experiment) is actually "wrong". I'm familiar with experiments which do require ideal single photon (on-demand) sources, but this seems not to be one of them. Please correct me. I was even under the impression that true single photon sources did not exist yet at the time in history when the results of these classic experiments convinced physicists that each single particle must follow a superposition of multiple paths?

Last edited: Sep 10, 2008
14. Sep 11, 2008

### Cthugha

Yes, you can, but in practice this will also mean, that most pulses will be empty. As your detectors need to be switched on during every pulse, this really spoils your signal to noise ratio. If pseudo single photon sources are used, the mean photon number per pulse is almost never below 0.1 due to this.

Well, whether this description is wrong or not, depends a bit on what you are actually trying to say. If you just want to show, that there is an interference pattern, when photons are detected one at a time, it works. But if you want to show, that there is an interference pattern when photons are present one at a time, it gets more complicated. As an example now you already have to assume perfect detectors to rule out, that there are other photons present, which are just not detected. The number of loopholes just increases.

Good question. I always thought this was shown with electrons first as it is rather easy to get single fermions, but I must admit, that I don't know.

15. Sep 11, 2008

### lightarrow

So, now the question turns into: how does one know to have fired a single photon (regardless of detection)?

16. Sep 11, 2008

### f95toli

Measure $g^{(2)}$ using a HB&T interferometer, a perfect single photon source will exhibit perfect anti-bunching, i.e. $g^{(2)}(\tau=0)=0$.

17. Sep 11, 2008

### edguy99

Could you view this experiment as testing the intensity of light needed to set off the detector? If so, it seems there could be a lot of lower intensity photons going through that are not being detected.

18. Sep 12, 2008

### lightarrow

Sorry, but this means that you can't say to have fired a single photon before detection.

19. Sep 12, 2008

### Cthugha

No, but the second order correlation function usually depends just on the source. Once you have shown, that a photon source emits nonclassical light and shows perfect antibunching, you would assume that it will still do so, when you put it in front of a double slit. The good thing in $$g^{(2)}$$ measurements is, that a non-ideal detector efficiency is no problem anymore as you can clearly distinguish between Poissonian and sub-Poissonian statistics.

20. Sep 12, 2008

### DrChinese

There is no such thing as "lower intensity photons" that are not being detected. Certainly no detector can be claimed as being of perfect efficiency, but the scenario you imagine has been ruled out experimentally. Entangled beams are used by detector manufacturers to rate their detector efficiency, which is extremely high these days. So for every detection at Alice, there is essentially always one at Bob.

The conclusion is that "lower intensity photons" are no photons at all. Otherwise, we would often have detections at Alice without a matching detection at Bob (and vice versa), but that doesn't happen.