Detecting single photons - conceptual question

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Imagine I have a detector which is "single photon sensitive". What that means, I gather, is that if I shine a pulsed laser at the detector with just the right amount of attenuation, I should get to the regime where there is less than 1 photon arriving each time the laser fires a pulse, so the detector shows a sporadic and random blip. (I have never seen this happen, but my research advisor claims that this is what we would observe if the detector is really that good.)

My question is about whether this "photon" is a monochromatic wave or a wavepacket. It is a basic definition that a photon is a quantization of an electromagnetic field of frequency v, so that each photon has energy hv. If I shine my (say) 50 fs pulse on the detector, and attenuate to just less than hv, I must have a single photon arriving at my detector. And by the textbook definition, the photon has a well defined, discrete energy. But by some sort of Fourier argument, my laser light is no longer in a 50 fs pulse - in fact, for the light to be monochromatic it has to extend infinitely in time. Physically, it doesn't make any sense for the pulse to broaden any more than it would due to normal dispersion through the optic. But then I have the exact same bandwidth that I had before.

So how many photons are reaching my detector? Is it one photon, defined by the fact that the detector only registers an interaction for a small fraction of the incident pulses? In that case the photon could only be viewed as a wavepacket, which seems contrary to how it is always presented in the (admittedly) introductory courses that I have taken.
 

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  • #2
sophiecentaur
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Why do you need a laser for this? Any source would do. You would need to know its spectrum (the bandwidth of the transmitted signal) and you would also need to know the bandwidth of the receiving equipment.
The probability of registering a 'hit' and detecting that single photon would relate to the overlap of these two curves and the total intensity of e.m. being transmitted. But we are only dealing with one photon at a time.
The emission of the photon will take a finite (very short) time, whilst the emitting atom is changing energy state. so there is no need to think in terms of a 'single' frequency for it; it is only a 'burst' of energy with an appropriate spectrum (that of a pulse modulated cw wave). The energy, hf, will relate to the mean f of the photon.
This photon will interact with its target (by chance) if there is a corresponding resonance / energy gap there which coincides (closely enough) with the transmitting atom.

The introduction of a laser is a bit of a red herring, I think, because the long coherence of laser light relates to the co-phase nature of all the stimulated photons - effectively producing a 50fs pulse with a long chain of successive photons. That is not relevant if you have whittled it down to just one photon.
It is very difficult to have a foot in both classical and quantum camps whilst discussing something like this because these sorts of perceived paradoxes keep coming up. You could almost make an arm waving analog of the whole thing, using RF pulses and a simple tuned receiver.
 
  • #3
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We would commonly test the gain of photomultiplier tubes (PMT) to single incident photon response by putting very short (~ 10 ns) pulses of light into the PMT cathode at 60 Hz using special stimulated-emission lead telluride light sources (like laser diodes, except broad spectrum), and putting the signal from the laser diode pulser into coincidence with the PMT anode output. We would insert neutral density filters in front of the light source, until the PMT output was about 3 or 5 pulses per second. Using Poission distribution counting statistics, most of the PMT output pulses were due to single photons. Using a pulse height analyzer (PHA) on the PMT showed the separate amplitude peaks due to single-photon and two-photon inputs. Typical PMT gains were 106 or 107, depending on the PMT.

Bob S
 
  • #4
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The emission of the photon will take a finite (very short) time, whilst the emitting atom is changing energy state. so there is no need to think in terms of a 'single' frequency for it; it is only a 'burst' of energy with an appropriate spectrum (that of a pulse modulated cw wave). The energy, hf, will relate to the mean f of the photon.
The introduction of the laser was not critical, I just wanted to give some background into why I'm asking the question. The real question I was trying to ask relates to this section of your response. You said that the single photon has a frequency bandwidth (ie it is a wavepacket). But the definition of the photon is a quantization of the electromagnetic field with energy hv. You can define this energy in terms of the mean frequency of the transition, fine. But I don't understand how it can be quantized with that particular energy if the transition can have an arbitrary bandwidth (within reasonable physical limits). Suppose I compare a narrow transition to a broad one, both centered at the same mean frequency. How do both of these "single photons" have the same energy? That seems to imply some strange, intrinsic normalization of the energy over the bandwidth of the transition.
 
  • #5
Cthugha
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So how many photons are reaching my detector? Is it one photon, defined by the fact that the detector only registers an interaction for a small fraction of the incident pulses? In that case the photon could only be viewed as a wavepacket, which seems contrary to how it is always presented in the (admittedly) introductory courses that I have taken.
Attenuating a laser pulse will never give you single photons. A single photon pulse is given by a Fock state or a linear combination of Fock states belonging to different modes. Assuming you have a detector of 100% efficiency that means you will get one photon all the time and never a different result. Attenuating a laser pulse means you will get a Poisson distribution of photon numbers with small mean photon number. Depending on how small you choose this mean photon number, you will mostly detect no photons, quite often 1 photon and on very rare occasions more than one photon. Real single photon states are therefore identified by their nonclassical photon number variance properties.

johng23 said:
But the definition of the photon is a quantization of the electromagnetic field with energy hv. You can define this energy in terms of the mean frequency of the transition, fine. But I don't understand how it can be quantized with that particular energy if the transition can have an arbitrary bandwidth (within reasonable physical limits).
You are mixing up two things. What you mean by definition of a photon you mean is the elementary excitation of a single mode of the em field. The (probability) amplitude belonging to this field mode is finite, leading to an occupation number of 1 in this mode and zero in all others. The energy is well defined, but the "pulse train" has infinite duration.

The single photon pulse is a state with a fixed total photon number of 1, but here the amplitudes belonging to different modes of the field are nonzero. However they will sum up such, that the corresponding expected photon number is exactly one. However, if you repeat this process several times, the energy of the detected photons will give a distribution according to the squares of the amplitudes connected with each mode of the field. Here you will get some energy uncertainty, but a finite pulse train duration.
 
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You are mixing up two things. What you mean by definition of a photon you mean is the elementary excitation of a single mode of the em field. The (probability) amplitude belonging to this field mode is finite, leading to an occupation number of 1 in this mode and zero in all others. The energy is well defined, but the "pulse train" has infinite duration.

The single photon pulse is a state with a fixed total photon number of 1, but here the amplitudes belonging to different modes of the field are nonzero. However they will sum up such, that the corresponding expected photon number is exactly one. However, if you repeat this process several times, the energy of the detected photons will give a distribution according to the squares of the amplitudes connected with each mode of the field. Here you will get some energy uncertainty, but a finite pulse train duration.
Ahh, thanks, makes much more sense.
 
  • #7
sophiecentaur
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Attenuating a laser pulse will never give you single photons. A single photon pulse is given by a Fock state or a linear combination of Fock states belonging to different modes. Assuming you have a detector of 100% efficiency that means you will get one photon all the time and never a different result. Attenuating a laser pulse means you will get a Poisson distribution of photon numbers with small mean photon number. Depending on how small you choose this mean photon number, you will mostly detect no photons, quite often 1 photon and on very rare occasions more than one photon. Real single photon states are therefore identified by their nonclassical photon number variance properties.
I find this hard to square, 'intuitively'. You seem to be implying that the photons from a laser, however the beam may be attenuated or spread out, are, in some way, different from atoms produced randomly. That would imply that would be expected to interact differently with the target atoms they interact with. Can you explain this? Does it relate to coherence?
I appreciate that you may think you've explained it already but be patient! ;-)
 
  • #8
Cthugha
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I find this hard to square, 'intuitively'. You seem to be implying that the photons from a laser, however the beam may be attenuated or spread out, are, in some way, different from atoms produced randomly.
At this point I am not sure I get the question right, so let me ask what you mean exactly. Are you asking whether a beam of photons is different from a beam of atoms or are you asking whether the photons themselves are different?

Does it relate to coherence?
If I get your question right, then in some sense yes. But please rephrase your question a bit. I am very good in misunderstanding even simple questions. ;)
 
  • #9
sophiecentaur
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Sorry - for "atoms" substitute "photons". A freudian slip, I think - I usually proofread my replies.
 
  • #10
Cthugha
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I find this hard to square, 'intuitively'. You seem to be implying that the photons from a laser, however the beam may be attenuated or spread out, are, in some way, different from photons produced randomly.
Ok, I requoted it the way it was supposed to be.
I will try to give a detailed explanation. Please tell me, if it gets too complicated.

Although that may sound strange photons emitted from a laser are as random as it gets. You can distinguish different photon sources in terms of the statistical dependence or independence of the photons emitted. If you detect one photon out of an ensemble of independent photons, you should not get any new information about when the other photons will arrive. That sounds rather abstract, but can be quantified rather easily. If you have a look at the probability to detect a photon pair, it will just be the product of the single photon detection rates for statistically independent photons.
If you have some detector allowing you to take single pictures and count the photons inside it, you would for example expect to see a photon pair every 100 pictures if you see a single photon every 10 pictures.

In other words, the ratio of the classical pair count rate [tex]\langle n^2 \rangle[/tex] and the product of the mean count rates [tex]\langle n \rangle^2[/tex] is equal to one for independent photons (for example a fixed number of them).
If it is larger than one, this means photons will have a preference to stick together in groups. If it is smaller than 1, they have a tendency to avoid each other. However, this case is classically impossible.

However, going to quantum mechanics, you have to account for the change of the light field you induce when you detect a photon: You destroy that photon. So the above ratio changes to:
[tex]\frac{\langle n (n-1) \rangle}{\langle n\rangle^2}[/tex]
to account for that destructed photon.

In this case, a fixed photon number state will not give you a ratio of 1 all the time. This is easy to picture for a single photon. You will never detect a photon pair if there is just a single photon present and a detection destroys one photon. To get the ratio back to one, you need to add some photon number fluctuations. It turns out that the fluctuations exactly get that ratio back to one if the noise corresponds to a Poisson distributed photon number distribution. That is the case for stimulated emission and lasers in general. This gives the remarkable result that you do not change the state of the light field when one photon is detected, which makes coherent states very stable and immune to loss.

As you asked about the connection to coherence: There are basically two kinds of coherence involved here: first and second order coherence. Second order coherence is what I just explained. If the photons are statistically independent and the ratio above equals 1, the light source is second order coherent. First order coherence is what is measured in a Michelson interferometer. If the ratio is different from one - so you have the tendency for photons to arrive in groups or to avoid each other - this tendency will be strongest for photons detected simultaneously, but it will vanish for long waiting times between photons. The timescale on which this tendency vanishes is also the coherence time of the light.

I hope this explains it somehow. I can illustrate it better, but that would need some simple math.
 
  • #11
sophiecentaur
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Go on then. Hit me with some maths as long as it isn't too hard. I think my prob is with my too-classical picture of things.
 
  • #12
Cthugha
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Sorry for the late answer. I missed your reply.

I will try to discuss the ratio I mentioned above in easy terms. It is called the second-order intensity correlation [tex]g^{(2)}[/tex].

So I introduced
[tex]g^{(2)}=\frac{\langle n (n-1) \rangle}{\langle n \rangle ^2}[/tex].

The averages are either time-averages or ensemble averages, depending on whether you look at a continuous stream of photons or pulses. Now you can deconstruct the non-averaged photon number [tex]n[/tex] into [tex]\langle n \rangle + \delta[/tex], the mean value and and the actual deviation from the mean at one instant. So you get:
[tex]g^{(2)}=\frac{\langle (\langle n\rangle +\delta) (\langle n\rangle +\delta-1) \rangle}{\langle n\rangle ^2}[/tex]
or
[tex]g^{(2)}=\frac{\langle (\langle n\rangle^2 +2 \delta \langle n \rangle + \delta^2 - \delta - \langle n \rangle \rangle}{\langle n\rangle ^2}[/tex].

Of course all terms of the kind [tex]\langle \delta \rangle[/tex] vanish as they describe the expectation value for the deviation from the mean. So simplifying, the following terms are left:

[tex]g^{(2)}=1-\frac{1}{\langle n \rangle}+\frac{\langle \delta^2 \rangle}{\langle n \rangle^2}[/tex].

Now one can clearly identify several "kinds" of light. To have statistically independent photons, this term needs to be unity. This is approximately true for large photon numbers as the photon number dependent terms decay towards 0.

It is not true for Fock states with well defined photon numbers. Here [tex]\langle \delta^2 \rangle[/tex] should be zero as there are no photon number variations and you get

[tex]g^{(2)}=1-\frac{1}{\langle n \rangle}[/tex], which is for example equal to 0 for a single-photon state. In fact this is the quantity used in experiments to identify single photon sources.

However, it is also possible to have exactly statistically independent photons. In this case we must get
[tex]\langle \delta^2 \rangle= \langle n \rangle[/tex] so that the extra terms cancel out. In other words: the variance of the photon number must be equal to the mean to get independent photons. A probability distribution where this is fulfilled is the Poisson distribution, which is the distribution found for the photon number of a coherent laser pulse. So the intrinsic photon number fluctuations cancel the change of the light field induced by detecting a photon such that the overall state of the field remains unaltered.

There are of course also other photon number distributions for photons, most well known is the Bose-Einstein distribution of thermal light. Here the variance is given by [tex]\delta^2 =\langle n \rangle^2 + n[/tex]. Inserting this into the equation, you get [tex]g^{(2)}=2[/tex]. This means that the photons will have a tendency to arrive in pairs or groups. In easy terms this is a consequence of gaining information about the light field at some instant by detecting a photon. Having such a large variance will mean that there are times with huge photon numbers and times with almost no photons present. The detection of a photon now means that it is very probable that the light field is now in a high intensity instant. Therefore it will be very probable to detect another one as the photon number is now very high. In this way you got some information about the light field because of the absence of statistical independence of photons.

This concept of classifying the light field (or light source and emission process) is indeed usually independent on how you spread the beam out and on how you attenuate classical light (obviously you can change this ratio by attenuating photon number states). Having such a correlation function equal to unity is a part of the modern definition of coherence (according to Glauber's treeatment of coherence). The other part is the classical coherence time concept.
 

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