# Double slit with time measurement

1. Dec 31, 2013

### DParlevliet

Propose this measurement:

It is a well know interference measurement, but now asymmetrical.
Suppose the red path is 1 m and the blue path 10 m. If you measure the time between detector 1 and 2 I expect you will find two times: 3.3 ns and 33 ns. Then you know which path the photon took. But between BBO source and detector 2 the photons are not disturbed, so according Feynman rules there will be interference.
Who know what is wrong here, because QM predicts different.

2. Dec 31, 2013

### M Quack

In order for your light pulses to have a defined length in time, they have to have a certain spread in wave length. Basically this is the uncertainty principle, delta E x delta T > hbar/2.

I have not done the calculation, but I strongly suspect that if delta t is small enough to distinguish the path, then delta E will be so large that you don't get an interference pattern anymore.

3. Dec 31, 2013

### Cthugha

No. What you have built is a version of a Mach-Zehnder interferometer. Even for classical light, unless the coherence time of the light is longer than 33 ns, you will not get interference. For single photons created by PDC, coherence time is typically short - usually below 1 ps. If you use pulses instead of cw light, you need good overlap of the two pulses at your detector to get interference. See the huge amount of literature on the Hong-Ou-Mandel effect for typical timescales.

4. Dec 31, 2013

### DParlevliet

But if you calculate it with the Feynman paths, what is the result? Feynman does not take coherence time into account and also works with paths of different length.

If coherence time is less then 1 ps there will be no interference if the path difference is less more then 0.3 mm.

5. Dec 31, 2013

### Cthugha

What do you mean by that? Photons are only indistinguishable within one coherence volume. For longer timescales/spatial scales they are distinguishable particles. Probability amplitudes just add for indistinguishable pathways leading to the same result.

edit: Just to make it a bit more clear on the technical side: Assuming a cw light beam, phase coherence gets lost on a timescale corresponding to coherence time. Calculating the path integral over two terms with random relative phase, you will not get an interference pattern.

Last edited: Dec 31, 2013
6. Dec 31, 2013

### DParlevliet

Then I am not sure what you mean with coherence time. According the results of the double-slit (and Feynman rules), as soon as a single photon is generated there are spherical (probability) waves with intensity 1/r2 everywhere in space. So if a photon following the red path arrives at the detector, then its wave through both red and blue path are already present and interfere. Because of the 1/r2 the wave through the blue path will be 100 times lower, so only an interference of 1%.

Or strictly following Feynman: there are two possible (probability) paths (histories), each with its own phase rotation depending on path length. At the detector those add up and result in interference.

7. Dec 31, 2013

### DrChinese

Obviously there is no interference (under any regime) if you know which way particle 2 went. That will always occur when the path lengths are different. On the other hand, if the path lengths are the same: you have the Mach-Zehnder interferometer, as Cthugha said.

Please be aware also that depending on your exact setup, there are issues with coherence generally for entangled photons. For example, particle 2 will not exhibit interference in a double slit setup (neither will particle 1). Not sure if the same applies to an MZI, but I would guess it would. (I am not sure if there is something that can be done to a particle after entanglement ends to make it coherent.)

8. Dec 31, 2013

### DParlevliet

And Feynman? all his paths have different lengths.
An interference pattern exist when both waves (along both paths) has phase differences, caused by path length differences (again see Feynman). So interference patterns are based on (and calculated) path length differences.

9. Dec 31, 2013

### Cthugha

You did not state whether you are assuming cw excitation or pulsed single photons. As pulsed single photons are completely pointless in that scenario I assumed cw pumping of the BBO.

There is no double slit in your setup. Anyway, you get the kind of spherical wave for a point source, but not for a laser or parametric downconversion. These have a very different distribution of emission angles.

This is comparing oranges and apples. You do not have a point source. There is no wave already present. There WAS a wave present some nanoseconds ago. Anyway, back to your scenario in the reply to the next part:

Oh, no. This is not what Feynman implies. You are neglecting any time dependences. Feynman tells us that you need to sum over all paths/possible histories leading to the same final detection event. A detection event now and a detection event 33 ns later are not the same detection event. You have two possible paths, but why should they interfere? They are perfectly distinguishable as there is no moment in time when both probability amplitudes connected with the two possible paths are nonzero. At the earlier possible detection time the probability amplitude for the photon having taken the second path (later detection time) is zero and vice versa. Only if both of these amplitudes are non-zero at some instant, you will get interference terms. The presence of both of these terms is what roughly translates to having indistinguishable pathways.

edit:
Yes, but with a path difference smaller than one coherence length/time such that the relative phase is still well defined, but simultaneously the uncertainty in photon emission time is larger than the time difference introduced by the two different paths. The more you can nail down the exact time (in principle) when photon emission happened, the smaller the path difference may be, because otherwise the two waves just "miss" each other. This critical timescale is just the duration of the wavetrain for pulsed light and it is the coherence time for continuous wave light.

Last edited: Dec 31, 2013
10. Dec 31, 2013

### DrChinese

Very small ones, true enough. But they must be indistinguishable to the detection event. You keep ignoring this critical point by insisting that there is interference when there is a significant timing difference. If you can, in principle, determine the path taken, there is NEVER inference between the paths.

There is nothing special about the Feynman perspective that makes it different in any way from other ones. They all yield the same answers eventually.

11. Dec 31, 2013

### DParlevliet

How small? Where does this value come from?

In Feynman: can you evaluate this situation with his paths, and how they show that there is a limit where it does not interfere anymore?

With the BBO I used the setup of Kim e.a. It is a crystal which, when radiated by laser, now and then emits two photons at the same time each with half intensity of the laser photons. So I am talking about single photons

Last edited: Dec 31, 2013
12. Dec 31, 2013

### DrChinese

You already have this answer, courtesy Cthugha: "with a path difference smaller than one coherence length/time such that the relative phase is still well defined, but simultaneously the uncertainty in photon emission time is larger than the time difference introduced by the two different paths. The more you can nail down the exact time (in principle) when photon emission happened, the smaller the path difference may be, because otherwise the two waves just "miss" each other. This critical timescale is just the duration of the wavetrain for pulsed light and it is the coherence time for continuous wave light."

I think my point is that there are details of the process that you are glossing over. Before you say that there is an inconsistency, you need to understand one of the methods in more detail. Then check the other one in detail. Cthugha has already given you several choice nuggets to look at. It should be obvious to you by now that a 30 ns difference is far too long to have interference in this case. And just as obviously, if there is going to be interference: it must be shorter, but we can assume there is some point at which it is short enough (since experiments of this type have been run previously). And consequently, since the path is short enough now that the path is unknown: there is no contradiction as you imply in the OP.

13. Jan 1, 2014

### DParlevliet

Yes, but he did not told where this time was based on, how it is calculated, which formula. I have discussed this item on several forums and this is the first time someone mentioned this about single photons and I have never seen it in double-path discussions.
As mentioned before, 1 ps is about 0.3 mm. So with Feynman all paths which differ more larger then 0.3 mm with the main path should be disregarded. That is certainly not the case.

I have seen enough other measurements and a lot jump too easily to the QM statement without actually analyzing what is happening with the wave. Therefore I start with the simplest one. It could be wrong, but then needs a clear description why.

Last edited: Jan 1, 2014
14. Jan 1, 2014

### Staff: Mentor

... Disregarded when calculating single-photon interference effects, not in general.

15. Jan 1, 2014

### Cthugha

Ok, there is much to it, but I try to keep it understandable.

To keep in Feynman's picture, he states that you need to take the sum over all possible paths from the same initial state to the same final state, which is in this case a detector detecting a photon at some position and time. Detections at different times are different events and the paths leading to detection events at different times do NOT add up. Now that seems to be at odds with adding up paths of different length as the time a photon would need to travel these paths differs. However, it is important to consider the correct initial state. The initial state is not "a photon gets emitted at time 0". The initial state is rather "Some emitter, say a single heavy atom, is in an excited state". The act of photon emission itself matters a lot. Just like in classical em, we need a dipole moment to get some dipole radiation. The excited state itself does not have a dipole moment. What is needed to create it, is some external perturbation, which puts the atom into a superposition of the excited state and the ground state. This superposition state can give rise to spatial oscillations of the electron probability density and therefore to a dipole moment. This superposition of course also involves the electromagnetic field as it has to carry away the energy difference between the excited state and the ground state. However, this is not a steplike process, but rather a damped oscillation. This also means that there is no exactly defined photon emission time. It is as uncertain as the duration of this damped oscillation is. As a handwaving visualization one might say that we have a superposition of the atom in the excited state and the atom in the ground state and 1 photon in the light field.

Now going back to Feynman's paths, we need to remember that the initial state was just an atom in the excited state. Now we need to take all possible emission times and all possible paths to the position of the detector into account. Earlier emission times now correspond to slightly longer paths. Later emission times correspond to shorter paths. But all paths which correspond to emission times where the atom is not in superposition - it is either for sure in the excited state or the photon has for sure already been emitted - are ruled out.

Please note, that this is just simple spontaneous emission. You can also further increase this emission time uncertainty by using stimulated emission or putting the emitter inside a resonator, but this is rarely the case. The exact duration of this emission uncertainty time interval depends strongly on your light source and geometry. Technically speaking, this coherence time is the decay constant of the Fourier transform of the power spectral density of your light field, but I doubt that this definition helps you right now. As a rule of thumb it is long for stimulated emission and lasers (ns and upwards), can be long or short for single atoms and is pretty short when the emission is governed by geometry. This is the case in a BBO crystal. You have this crystal and need to fulfill phase matching conditions. These crystals are pretty thin and the emission time uncertainty is roughly similar to the time the initial light beam needs to pass through the BBO crystal - typically about 1 ps.

You intended to write half the energy, I assume. That is correct. For single photons from BBOs coherence time is short and only few "paths" contribute and it is hard to create a good double slit experiment. This is why you do not really see a true single photon double slit experiment very often. You usually see setups which use a low mean photon number and have a small probability of having more than one photon in the setup at any repetition of the experiment. For this kind of experiment you can just use an attenuated laser which has huge coherence time. In that case a lot of paths contribute.

16. Jan 1, 2014

### DrChinese

Your OP spec was about a 30 ns difference. That is pretty large, no interference to consider in any picture. So no conflict.

Now suppose it is small as you are now saying. There would be interference effects to consider in any picture. So no conflict.

And as previously mentioned, entangled photons generally lack the coherence needed for double slit interference. I am not sure if a MZI setup has that specific problem or not. Still no conflict between the views as you had imagined in your OP.

17. Jan 2, 2014

### naima

Would seeing fringes violate some Heisenberg inequality (Energy/time or phase/number)?

18. Jan 2, 2014

### DParlevliet

This is quite a different explanation then I read everywhere. I am not yet sure if I understand what you mean, so let me first ask about:

1 When a single photon travels from the emitter, at what moment does the above superposition ends, and is the photon on its own?
2 If the photon is on its own, how does its (propability) wave looks like?
3 If the photon is not on its own, what happens when the superposition time is over?
4 Take a double-slit with 1 m between a BBO emitter and the detector (double slit in between). A photon reaches the detector after 3 ns. If you add up photon positions, will it form an interefernce pattern?

19. Jan 2, 2014

### Cthugha

True. In the optics regime such experiments are usually introduced for very coherent light sources, so one does not have to worry about that very much. The details get quite messy if you want the general result for any arbitrary light field.

That depends as there are several factors here. Assuming an otherwise unperturbed system the time uncertainty depends strongly on the actual system you have. Things like the energy difference between ground and excited state and the different charge distributions in the ground state and the excited state have to be taken into account. You can get the transition rate using Fermi's golden rule.

Another way to end the superposition is of course any interaction which leaves a "mark" that emission happened. This involves of course detection of the photon, but may also include recoil of the atom. Usually the recoil drowns in uncertainty, but if you have a very light emitter, the recoil upon photon emission may be enough to push the atom into a different state, thus ending superposition.

That depends on the emitter. For a point like emitter, it will be a spherical wave. For an emitter in a resonator it depends on the orientation of the resonator. For parametric processes like in a BBO you get an emission cone at some angle to the incident beam which is given by phase matching conditions. Please note that parametric processes do not work by spontaneous emission.

Hmm, I do not get that question. If you can know that the photon is "on its own", the superposition is necessarily over already.

This is complicated as it depends on spatial coherence, too. What a double slit measures is spatial coherence. This roughly translates to the angular size of your light source as seen by the slits. As a rough visualization: Imagine you have a point source and a large light source illuminating the same double slit. Now you can measure the path from the point source to the first slit and the path from the point source to the second slit. The path difference will correspond to some phase difference which introduces an overall shift to your interference pattern, but still gives good visibility. Now you can do the same for the large light source. As the light source is spatially extended, you can calculate a path length difference and relative phase for each point on the surface of the emitter. The phase difference will vary depending on the position on the emitter you choose. As a consequence you average over several different interference patterns which results in an interference pattern that is smeared out and has smaller fringe visibility.

Now there is one easy way to change the size of the source as seen by the slits. You can put it closer to the slits or further away. Close to the slit, you get huge emission angle differences, while far away from it, you just have a narrow range of angles reaching the slits and therefore also small relative phase variation. This just increased the spatial coherence of your light beam. So the question is: How far away from the double slit do you place the BBO?

This question has been investigated in detail and the interesting conclusion is that if you see an interference pattern with good visibility, you cannot violate Bell's inequality in the same experimental setup. In other words: Entanglement and creation of a simple interference pattern are mutually exclusive. See for example Phys. Rev. A 63, 063803 (2001) (ArXiv version: http://arxiv.org/abs/quant-ph/0112065) for details.

20. Jan 2, 2014

### naima

Decoherence is not easy to be explained with the language of path integrals.
were ther attempts?
It is easier with density matrix.