Double slit with time measurement

[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.

 

[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.

 

[Cthugha]

Propose this measurement:
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.

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.

 

[DParlevliet]

No.

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.

 

[Cthugha]

Feynman does not take coherence time into account and also works with paths of different length.

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.

 

[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.

 

[DrChinese]

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.

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.)

 

[DParlevliet]

That will always occur when the path lengths are different.

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.

 

[Cthugha]

Then I am not sure what you mean with coherence time.

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.

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.

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.

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%.

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:

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.

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:

And Feynman? all his paths have different lengths.

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.

 

[DrChinese]

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.

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.

 

[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

 

[DrChinese]

How small?

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.

 

[DParlevliet]

You already have this answer, courtesy Cthugha: "with a path difference smaller than one coherence length/time...

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.

 

[Nugatory]

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

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

 

[Cthugha]

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?

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.

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

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.

 

[DrChinese]

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.

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.

 

[naima]

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

 

[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:

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.
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.

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?

 

[Cthugha]

This is quite a different explanation then I read everywhere.

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.

1 When a single photon travels from the emitter, at what moment does the above superposition ends, and is the photon on its own?

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.

2 If the photon is on its own, how does its (propability) wave looks like?

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.

3 If the photon is not on its own, what happens when the superposition time is over?

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

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?

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.

 

[naima]

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

 

[DParlevliet]

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.

So in my measurement 1 ps after emission in the BBO a single photon is on its own with a wave cone as you mention above?

 

[Cthugha]

So in my measurement 1 ps after emission in the BBO a single photon is on its own with a wave cone as you mention above?

I would not phrase it that way. It sounds as if there was some well defined emission time which is not the case. You know that there is some time window with the length of 1 ps during which the photon could have been emitted. The emission will lie on a narrow cone as given by phase matching. If you measure the direction of emission of the second photon you can in principle get some information about which direction on the cone is realized in a single run of the experiment, but one does not have to do that.

Generally speaking, what you will get is a nonzero detection probability for your photon inside a volume which is given by the cone in x- and y-direction and the uncertainty of the emission time times the speed of light in z-direction.

If you forget about single photons for a moment and just think about classical optics, this volume corresponds to a simple wave train emitted by some light source. For interference in classical optics, you just need to create two different wave trains from the initial one (using a double slit or a beam splitter for example) and have these overlap with well defined phase somewhere. If the wave trains miss each other and do not overlap, there will be no interference. Going back to the quantum optics picture, you do not have these wave trains with some mean intensity at every point inside some volume, but some finite detection probability for the single photon inside each little part of the volume. Still the prerequisite for interference is the same: you need to split the volume inside which the detection probability is non-zero into two and then need to have a part of these volumes overlap somewhere with well defined relative phase.

In the limit of a large number of repetitions, the results of any ideal linear measurement are the same, no matter whether you do it once with many photons or many times with single photons which is why it is rarely really necessary to take the quantum nature of photons into account. You can get very far using classical optics.

 

[DParlevliet]

But it is my intention to look to single photons, because it is often explained that with single photons you will get the same interference pattern (after adding a lot of single measurements)

Let me make it simpler. Suppose I have a point source with a "superposition" time of 1 ps. Then after 10 ps, how does the wave of the photon look like. Spherical I suppose?

 

[Cthugha]

Let me make it simpler. Suppose I have a point source with a "superposition" time of 1 ps. Then after 10 ps, how does the wave of the photon look like. Spherical I suppose?

Ok, draw a sphere with radius of 10 ps times the speed of light. Draw a second sphere with radius of 9 ps times the speed of light.Subtract the second sphere from the first one. Inside the remaining spherical shell you have a non-zero probability to detect the photon.

I tend to avoid terminology like "wave of the photon" because strictly speaking you cannot apply this concept in the general case. There is no fully satisfactory definition of wave functions for photons.

But it is my intention to look to single photons, because it is often explained that with single photons you will get the same interference pattern (after adding a lot of single measurements)

You get the same pattern if all other parameters are the same (experimental setting, coherence properties of the beam and so on).

 

[DParlevliet]

Ok, draw a sphere with radius of 10 ps times the speed of light. Draw a second sphere with radius of 9 ps times the speed of light.Subtract the second sphere from the first one. Inside the remaining spherical shell you have a non-zero probability to detect the photon.

Indeed I always mean wave-property of the photon, not a real wave.
But how does the wave look like outside this spherical shell? According Feynman that is a spherical, sinus-shaped, with amplitude 1/r². In quadrate that represents the probability to be absorbed at the place of absorption

 

[DrChinese]

But it is my intention to look to single photons, because it is often explained that with single photons you will get the same interference pattern (after adding a lot of single measurements)

I think we have wandered quite a distance from your original example. That's fine, but it is very unclear what you are asking about at this point. A photon created from a BBo crystal and a photon from a laser act quite differently as to self-interference. Your example used BBo entangled photons, so that is a special case.

It is also unclear whether you are still trying to argue some particular point of view, or whether you are trying to understand how photons behave in certain situations.

 

[DParlevliet]

I try to argue that the start measurement does show an interference pattern. Cthugha disagrees because of a time argument I have never seen elsewhere. Step by step I am trying to close the gap between his opinion and mine. I am not yet sure how it will end: one of us is right or we misunderstand each other. Because my knowledge is minor to his, my official goal is to work from his opinion to mine, to understand what he means.

 

[audioloop]

But it is my intention to look to single photons, because it is often explained that with single photons you will get the same interference pattern (after adding a lot of single measurements)

you need heralded photons.



.

 

[DrChinese]

you need heralded photons.
.

Maybe. And the entangled photon setup in the OP does provide that. But the general case is that heralded particles are not necessary for interference effects to be demonstrated.

 

[audioloop]

Maybe. And the entangled photon setup in the OP does provide that. But the general case is that heralded particles are not necessary for interference effects to be demonstrated.

yes, but he wish to

[STRIKE]But it is my intention to look to single photons, because it is often explained that with single photons [/STRIKE]
you will get the same interference pattern (after adding a lot of single measurements)

highest symmetric heralding efficiency..

 

[DParlevliet]

For single photons created by PDC, coherence time is typically short - usually below 1 ps.

Back to the original argument. Are there references or explanation for that? According the formula in Wikipedia the coherence length of a single photon should be large because the spectral width is official zero. Of course this is not useful, because interference shows only with multiple photons, but then the spectral wide seems to me the same as the laser beam, so also the coherence length.

 

[Cthugha]

Back to the original argument. Are there references or explanation for that? According the formula in Wikipedia the coherence length of a single photon should be large because the spectral width is official zero.

Official zero? The spectral width of PDC is huge. See for example Optics Express, Vol. 20, Issue 7, pp. 7507-7515 (2012) (ArXiv version: http://arxiv.org/pdf/1201.0367.pdf) where the spectral width versus gain is measured. In the high gain regime you get 80-100 nm spectral width. If you go to the very low gain regime you might be able to reduce that by 15 nm or so, but still the spectral width in common down conversion is really large.

 

[DParlevliet]

Official zero? The spectral width of PDC is huge.

I did mean the single photon. This has one exact wavelength, so official no spectral width. But from the PDC I expected that the two photons would be exactly half the energy of the laser photon, and that seems not to be the case. However it is not that bad as in your article, because these are high-gain PDC. I found a reference there http://www.google.nl/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CD0QFjAB&url=http%3A%2F%2Fwww.ornl.gov%2Fsci%2Fqis%2Fpublications%2FOptics%2520Letters%25202005%2520Kim.pdf&ei=ezXVUpraKamJ0AWPlYCYAQ&usg=AFQjCNHTprZXfLq9G3NLEi17ff0aa7HpvQ&bvm=bv.59378465,d.d2k&cad=rja of the normally used PDC, which is much better (0,8%). But I agree, that is still not good enough for my measurement.
So practical the measurement as mentioned above cannot be done. But in theory, as a thought experiment, what principle objections can be made? Suppose a much better laser (which can reach very high coherence length) and ideal PDC? Or do you think that the beam width of the PDC fundamentally disturbs the time measurement?

 

[Cthugha]

I did mean the single photon. This has one exact wavelength, so official no spectral width.

Single photons are Fock states and they can have any spectral width. You can do a Fourier decomposition into the spectral components, but this is an outdated usage of the term photon and completely unrelated to what is happening in experiments. Single photons in the lab are never monochromatic.

But from the PDC I expected that the two photons would be exactly half the energy of the laser photon, and that seems not to be the case.

No. The sum of the photon energies needs to match the initial photon energy. There is no need for a 50/50 split. You get a random distribution around that value.

However it is not that bad as in your article, because these are high-gain PDC. I found a reference there http://www.google.nl/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CD0QFjAB&url=http%3A%2F%2Fwww.ornl.gov%2Fsci%2Fqis%2Fpublications%2FOptics%2520Letters%25202005%2520Kim.pdf&ei=ezXVUpraKamJ0AWPlYCYAQ&usg=AFQjCNHTprZXfLq9G3NLEi17ff0aa7HpvQ&bvm=bv.59378465,d.d2k&cad=rja of the normally used PDC, which is much better (0,8%). But I agree, that is still not good enough for my measurement.

The exact width depends on a lot of stuff: The mode of the pump laser. The way you cut the crystal. The thickness of the crystal. The duration of the pump laser and so on and so forth.

So practical the measurement as mentioned above cannot be done. But in theory, as a thought experiment, what principle objections can be made? Suppose a much better laser (which can reach very high coherence length) and ideal PDC? Or do you think that the beam width of the PDC fundamentally disturbs the time measurement?

In that case you do not have entanglement anymore or rather you cannot beat Bell inequalities. However, you still cannot beat the uncertainty relation which gives a minimal spectral width depending on the pump laser duration and the crystal thickness. The coherence length of the excitation laser rarely matters. The ones used are highly coherent.

 

[DParlevliet]

In that case you do not have entanglement anymore or rather you cannot beat Bell inequalities. However, you still cannot beat the uncertainty relation which gives a minimal spectral width depending on the pump laser duration and the crystal thickness. The coherence length of the excitation laser rarely matters. The ones used are highly coherent.

According Wikipedia lasers can have coherence lengths of more the 100 m, so that cannot be a principle problem. If you would have an ideal PDC, what would be the minimal spectral width caused by the uncertainty relation? Is that Planck constant compared to the photons energy?