Can a single photon of a single frequency pass through a FP cavity?

[Daniel Petka]

TL;DR Summary

Photons should pass through a cavity if they resonate (wavelength matches 2 times cavity length), as they can interfere with themselves. According to a video, this doesn't happen

I was watching this amazing video about FP cavities and the creator mentioned that a single photon shouldn't pass through that cavity. Only multiple photons can. This doesn't make sense to me. The cavity works thanks to interference and single photons still interfere with themselves, so shouldn't every photon you send pass strsight through the cavity if it has a well defined frequency? I asked this in a comment but got no answer unfortunately

 

[hutchphd]

Exactly where in the video is this statement made? Please quote it exactly so I don't have to listen to this guy. Paraphrase his reason if you can.
Otherwise you are on your own with this foolishness.

 

[sophiecentaur]

TL;DR Summary: Photons should pass through a cavity if they resonate (wavelength matches 2 times cavity length), as they can interfere with themselves. According to a video, this doesn't happen

I was watching this amazing video about FP cavities and the creator mentioned that a single photon shouldn't pass through that cavity.

I have tried two different browsers and the link opens in a blank screen

 

[hutchphd]

I had no problems with google chrome (Win10 machine) but soon wished it hadn't. That's a trifle overstated but I do resent the generic natue of the question. The OP should explain what he thinks the video says and where it says it, so we are not required to listen to the so-so lecture on Fabry-Perot all the way through, waiting for the incorrect bits.

 

[Daniel Petka]

Exactly where in the video is this statement made? Please quote it exactly so I don't have to listen to this guy. Paraphrase his reason if you can.
Otherwise you are on your own with this foolishness.

27:47, sorry my bad!

 

[Daniel Petka]

I have tried two different browsers and the link opens in a blank screen

Try typing "Fabry Perot Week 6B", the statement is at 27:47

 

[hutchphd]

Fabry-Perot guy says no such thing. Explain your interpretation of the physics please. This is a singularly bad way to attempt to do science, and I will henceforth not comment upon "what he said" questions.

 

[hutchphd]

TL;DR Summary: Photons should pass through a cavity if they resonate (wavelength matches 2 times cavity length), as they can interfere with themselves. According to a video, this doesn't happen

so shouldn't every photon you send pass strsight through the cavity if it has a well defined frequency?

No. Each will have a probability that depends upon the interfernce curve as shown

 

[Daniel Petka]

Fabry-Perot guy says no such thing. Explain your interpretation of the physics please. This is a singularly bad way to attempt to do science, and I will henceforth not comment upon "what he said" questions.

Alright, I'm gonna quote: "So how does a photon know to go through that first mirror? The answer is that it doesn't. You send a 100 photons through into the mirror, 99 bounce off and 1 goes through. "
Let me know if I misunderstood something

 

[Daniel Petka]

No. Each will have a probability that depends upon the interfernce curve as shown

The point is that the interference curve has a peak of 1 (in the ideal case) if the wavelength matches 2 times the cavity length. So a photon that has that wavelength (created by attenuated laser light) will always pass

 

[hutchphd]

Look at your original question please. Are you asking whether there are very narrow band optical thin film notch filters that transmit nearly 100% of the energy? The answer to that is emphatically yes. Thay cost $100 at Edmund Scientificand are typically multilayered interference filters.

 

[Daniel Petka]

Look at your original question please. Are you asking whether there are very narrow band optical notch filters that transmit nearly 100% of the energy? The answer to that is emphatically yes. Thay cost $100 at Edmund Scientificand are typically multilayered interference filters.

I think the title and the post say exactly what I'm asking, here it is again: if a single frequency beam can pass through a cavity, then a single photon of that beam has to pass as well. According to the quote, the author of that video doesn't agree with this. The reason why I asked this on the forum is that the guy is a physics professor and I'm not, so maybe he knows something that I don't.

 

[hutchphd]

One cannot analyze photons as BB's. The probability for the energy to be transmitted through the filter at resonance approaches 100 %. I don't honestly care what he says. His description of photons is incorrect in a host of ways tyat have been discussed m here at Physics Forum., I shall not add to that. The concept of the geometric sum of phased reflections producing a resonant denominator is a useful one in many circumstances and corresponds to a sum over possible paths for each detection event. (I did a pretty good dissertation on it once upon a time for Helium atomic beams.) I will not argue with some unknown guy on a video, by proxy.

 

[Daniel Petka]

One cannot analyze photons as BB's. The probability for the energy to be transmitted through the filter at resonance approaches 100 %. I don't honestly care what he says. His description of photons is incorrect in a host of ways tyat have been discussed m here at Physics Forum., I shall not add to that. The concept of the geometric sum of phased reflections producing a resonant denominator is a useful one in many circumstances and corresponds to a sum over possible paths for each detection event. (I did a pretty good dissertation on it once upon a time for Helium atomic beams.) I will not argue with some unknown guy on a video, by proxy.

Gotcha, I agree. He didn't really provide an explanation why they shouldn't pass. Thanks for the help!

 

[hutchphd]

You might want to look also at Fiber Bragg Gratings which are similar and perhaps easier to analyze.

 

[Daniel Petka]

You might want to look also at Fiber Bragg Gratings which are similar and perhaps easier to analyze.

I think a fabry perot is as easy as it gets. There are just 2 interfaces. For bragg gratings, there are many

 

[sophiecentaur]

One cannot analyze photons as BB's. The probability for the energy to be transmitted through the filter at resonance approaches 100 %. I don't honestly care what he says. His description of photons is incorrect in a host of ways tyat have been discussed m here at Physics Forum., I shall not add to that. The concept of the geometric sum of phased reflections producing a resonant denominator is a useful one in many circumstances and corresponds to a sum over possible paths for each detection event. (I did a pretty good dissertation on it once upon a time for Helium atomic beams.) I will not argue with some unknown guy on a video, by proxy.

This is a very difficult concept for me. You have a temporal filter with a response time of many cycles of the wave passing through. You can calculate its passband characteristic in terms of CW waves. Not hard.

Then you have a 'single photon', which has a finite duration (?) so it's like a burst of waves with a non zero bandwidth. How do you define that - if all photons of a particular energy are supposed to be indistinguishable? If the source has a wide bandwidth then the frequency of the one photon that gets through the selector can lie anywhere within the source bandwidth. Then it comes to a very narrow bandwidth of the etalon so it may not get through.

There must be a quantum explanation of this because this doesn't make senses as it stands.

 

[Daniel Petka]

This is a very difficult concept for me. You have a temporal filter with a response time of many cycles of the wave passing through. You can calculate its passband characteristic in terms of CW waves. Not hard.

Then you have a 'single photon', which has a finite duration (?) so it's like a burst of waves with a non zero bandwidth. How do you define that - if all photons of a particular energy are supposed to be indistinguishable? If the source has a wide bandwidth then the frequency of the one photon that gets through the selector can lie anywhere within the source bandwidth. Then it comes to a very narrow bandwidth of the etalon so it may not get through.

There must be a quantum explanation of this because this doesn't make senses as it stands.

If laser light is attenuated to single photon intensities, the frequency doesn't change.
I think the FP is a perfect example why photons aren't bullets. How does the photon know that there is a second mirror? It doesn't and yet, it can still pass through both mirrors. That's kind of amazing. Quantum mechanics will never give you an intuitive answer, it just describes the photon as a probability wave function that somehow collapses into a single point and gets absorbed by a single electron without an explanation

 

[sophiecentaur]

the frequency doesn't change.

Correct. But any source has a bandwidth; a distribution of photon frequencies about a mean value. As has been established, the source is likely to have a wide bandwidth (wider than an FP)

How does the photon know that there is a second mirror? It doesn't and yet, it can still pass through both mirrors

But not all photons. If they are off frequency then they will not pass; they will all need to 'know' as you say. This implies that the 'equivalent transit time' for the photon must elapse before additive or subtractive interference can take place. So the photon must have a great extent (undefined but dependent on the situation)

That's kind of amazing

Certainly head scratching stuff.

 

[Daniel Petka]

Correct. But any source has a bandwidth; a distribution of photon frequencies about a mean value. As has been established, the source is likely to have a wide bandwidth (wider than an FP)

But not all photons. If they are off frequency then they will not pass; they will all need to 'know' as you say. This implies that the 'equivalent transit time' for the photon must elapse before additive or subtractive interference can take place. So the photon must have a great extent (undefined but dependent on the situation)

Certainly head scratching stuff.

Well you are talking about a FP with a very high finesse (so very high mirror reflectivity) Even for such an FP, you can build a laser source with a lower bandwidth, usually frequency locked (in the MHz or even kHz range or even less) If such laser light resonates and goes through completely, then it does so even if you decrease the intensity to single photon levels.
Also, let's say that the FP has a relatively high bandwidth, so I don't know.. a mirror reflectivity of just 90%. In that case, if the photons are off frequency, they still pass, all of them pass.

I don't think the extent of the photon depends on the situation. It's called coherence length and is directly related to its bandwidth. Here, I mentioned a near zero bandwidth laser, so the extent is near infinity. Today, you can have lasers with kilometres of coherence length if you have the $€

 

[Daniel Petka]

Correct. But any source has a bandwidth; a distribution of photon frequencies about a mean value. As has been established, the source is likely to have a wide bandwidth (wider than an FP)

But not all photons. If they are off frequency then they will not pass; they will all need to 'know' as you say. This implies that the 'equivalent transit time' for the photon must elapse before additive or subtractive interference can take place. So the photon must have a great extent (undefined but dependent on the situation)

Certainly head scratching stuff.

Well you are talking about a FP with a very high finesse (so very high mirror reflectivity) Even for such an FP, you can build a laser source that is frequency locked and because that it has a ridiculously low bandwidth (in the MHz or even kHz range or even less) If such laser light resonates and goes through completely, then it does so even if you decrease the intensity to single photon levels.
Also, let's say that the FP has a relatively high bandwidth, so I don't know.. a mirror reflectivity of just 90%. In that case, if the photons are off frequency, they still pass, all of them pass.

I don't think the extent of the photon depends on the situation. It's called coherence length and is directly related to its bandwidth. Here, I mentioned a near zero bandwidth laser, so the extent is near infinity. Today, you can have lasers with kilometres of coherence length if you have the $€

 

[Charles Link]

See https://journals.aps.org/pra/abstract/10.1103/PhysRevA.58.4348
The subject is something that I don't have much expertise with, but it looks like there has been some fairly recent research in this area.

 

[hutchphd]

You might want to look also at Fiber Bragg Gratings which are similar and perhaps easier to analyze.

Then you have a 'single photon', which has a finite duration (?) so it's like a burst of waves with a non zero bandwidth

I think the fundamental disconnect is thinking of the photon as a wavepacket in space. I know it took some effort to disabuse me of this notion (mostly by folks at PF). It is an eigenexcitation of the (free) field. For (solid state) me the analog to phonons in solids was useful. So a single photon does not imply a spatially constrained object, although interactions with charged matter can be localized and treated the usual Wigner way.
I was certainly sold a bill of goods in my early exposure to this long ago.

 

[sophiecentaur]

So the photon must have a great extent (undefined but dependent on the situation)

I guess the 'extent' is determined by the loss / Q factor of the resonator. It will only be when the resonance is there that the photon will pass or be reflected / absorbed.

I can 'see' a classical picture of this process by referring to what you get with a Time Domain Reflectometer and a transmission line resonator; the displayed time domain image relates to the frequency response of the resonator. But then there's the other consideration of the fact that all photons (same frequency) are identical yet whilst they are interacting with matter, they modify to fit the situation.

There must be a way to reconcile these classical views and the QFT model.

 

[sophiecentaur]

I don't think the extent of the photon depends on the situation.

This is difficult if we recognise that 'all photons' are alike when travelling through the space between.

When you talk about coherence then I can see how it's relevant for a wave of long duration (CW from a laser or a radio transmitter) but the coherence of a photon would need to be the same from the same source that started with many photons. I suppose the idea is that (and this is a problem in all circumstances) you can't talk about the spatial nature of the photon until it's interacting with matter.

 

[Daniel Petka]

This is difficult if we recognise that 'all photons' are alike when travelling through the space between.

When you talk about coherence then I can see how it's relevant for a wave of long duration (CW from a laser or a radio transmitter) but the coherence of a photon would need to be the same from the same source that started with many photons. I suppose the idea is that (and this is a problem in all circumstances) you can't talk about the spatial nature of the photon until it's interacting with matter.

Hm I'm not sure I understand, the coherence of a single photon is the same as the coherence of the whole beam. That being said, I don't think it makes sense to talk about "photons" before the wave collapses.
Btw I think you might like this this video by Huygens Optics, although it's definetly not 100% correct

 

[kered rettop]

I think the title and the post say exactly what I'm asking, here it is again: if a single frequency beam can pass through a cavity, then a single photon of that beam has to pass as well.

It's possible for two photons to interfere. The most popular example is the Hong-Ou-Mandel effect which is explained in this Wikipedia article. Yes, it came a shock to me too.
I have no idea whether that's what the guy is talking about, but it should caution you against taking the "beams act exactly the same as single photons" mantra at face value.

 

[f95toli]

There must be a way to reconcile these classical views and the QFT model.

yes, there are. There are "recipes for how to create "quantum versions" of classical equations for things like waveguides and filters. These are frequently used by people who design quantum circuits.
That said, most of the time you don't really need to worry about this. When designing linear circuits meant for single photons Maxwell's equations still works, and if something is a filter or a resonator in the "classical" case then it will be so for single photons as well.
Note that for non-linear circuits (components such as amplifiers, mixers etc) things are not so straightforward.

 

[sophiecentaur]

if something is a filter or a resonator in the "classical" case then it will be so for single photons as well.

So that would imply that, because you can't have less than a quantum of the wave getting through, a photon of a frequency that's not on the peak of the FP cavity response there is a probability that it just won't get through - depending on the place on the resonance curve. This makes sense and doesn't involve any contradictions.

 

[kered rettop]

It's possible for two photons to interfere. The most popular example is the Hong-Ou-Mandel effect which is explained in this Wikipedia article. Yes, it came a shock to me too.
I have no idea whether that's what the guy is talking about, but it should caution you against taking the "beams act exactly the same as single photons" mantra at face value.

Second thoughts, forget it, I'm pretty sure it has no relevance to the matter here,