What's the meaning of mean photon number?

[Tspirit]

I am studying Quantum Optics. A single photon state will give the mean photon number of 1, as shown the following equation:
$$<\hat{n}>=<1|\hat{n}|1>=1.$$
For a two-photon number state, the similar calculation will be
$$<\hat{n}>=<2|\hat{n}|2>=2.$$
And for a coherent state, it is
$$<\hat{n}>=<\alpha|\hat{n}|\alpha>=|\alpha|^{2}$$.
However, for the real detection system, the mean photon number is related to the integration time: the more time, the more mean photon number. Then I have several questions:
(1) Given a light state, is the mean photon number related to the integration time of the detector?
(2) For a light of 1 Watt in the single photon state, or coherent state, how to calculate their mean photon numbers?

 

[Mentz114]

I am studying Quantum Optics. A single photon state will give the mean photon number of 1, as shown the following equation:
$$<\hat{n}>=<1|\hat{n}|1>=1.$$
For a two-photon number state, the similar calculation will be
$$<\hat{n}>=<2|\hat{n}|2>=2.$$
And for a coherent state, it is
$$<\hat{n}>=<\alpha|\hat{n}|\alpha>=|\alpha|^{2}$$.
However, for the real detection system, the mean photon number is related to the integration time: the more time, the more mean photon number. Then I have several questions:
(1) Given a light state, is the mean photon number related to the integration time of the detector?
(2) For a light of 1 Watt in the single photon state, or coherent state, how to calculate their mean photon numbers?

The mean photon number per unit time is given by the Poisson distribution. So the mean count is related to the integration time. The time between events is the negative exponential of the mean $e^{-\mu}$ if I remember correctly.

You'll need the power spectrum of the light source tro make an estimate of the mean photon count for different modes based on the energy of a single photon.

 

[Cthugha]

I am studying Quantum Optics. A single photon state will give the mean photon number of 1, as shown the following equation:
$$<\hat{n}>=<1|\hat{n}|1>=1.$$
For a two-photon number state, the similar calculation will be
$$<\hat{n}>=<2|\hat{n}|2>=2.$$
And for a coherent state, it is
$$<\hat{n}>=<\alpha|\hat{n}|\alpha>=|\alpha|^{2}$$.
However, for the real detection system, the mean photon number is related to the integration time: the more time, the more mean photon number. Then I have several questions:
(1) Given a light state, is the mean photon number related to the integration time of the detector?
(2) For a light of 1 Watt in the single photon state, or coherent state, how to calculate their mean photon numbers?

First, you are confusing several things here. What a detector does is not equivalent to application of the photon number operator. Applying the photon number operator leaves the state of the light field unchanged, while a photodetector destroys a photon upon detection and thus changes the light field.

Typically, introductory texts to quantum optics do not consider free space cw fields, but light fields inside a perfectly reflecting cavity. There the meaning of the mean photon number is quite clear as no additional energy enters the system and no light leaves the cavity. The mean photon number is just given by the light field inside the cavity. As you incorporate losses to this cavity mode, you will usually create a balance of pump and decay. Some additional light enters the cavity and some leaves the cavity. A steady state will form and the mean photon number of the state inside the cavity is still well defined. The light field outside the cavity, however, is defined by the leakage of photons through the cavity walls. This is given by a leakage rate (some number of photons per second). For the light field inside the cavity, the notion of a mean photon number makes sense as you can measure the total number of photons inside the cavity and this does not depend on integration time. For the light field outside the cavity, such a notion does not make sense as you will always measure a photon flux: The number of photons that pass through a certain detector diameter in a certain amount of time. To get to a total detected photon number you will integrate the photon flux over the detection time. Based on that, you can calculate a mean photon flux or a mean photon number for a certain given integration time. So in response to your questions:

(1) You can calculate a kind of mean photon number, but it is not necessarily related to the concept of a mean photon number that one would use for the state of the field inside the cavity. The difference is most obvious for a single photon state. There, the maximum photon number you can detect is 1 and it will not increase with time.

(2) For a single photon state of 1 W, you would need a REALLY high-energy single photon. However, the typical meaning of a single photon state in real devices and experiments is somewhat different. Usually you take a single photon emitter such as a single atom or a quantum dot and place it inside a cavity. You excite it using a pulsed laser and after some time, it will emit a single photon. The cavity has a finite lifetime (say, 1 ns), so the single photon will leave the cavity within this time range and you will have a single photon state outside the cavity. Then you simply excite the system again using the next pulse. A typical time between consecutive pulses often found in experiments is 13 ns. So 13 ns after the first photon another single photon will leave the cavity within a time range of 1 ns. This way, you get a consecutive train of single photons. So here, the power in the light beam is mainly governed by the pulse repetition rate.
Accordingly, your question about "mean photon numbers" is meaningless. You can define a mean photon number per arbitrarily chosen integration time (which is a mean count rate) or (more common) for pulsed light fields you can define a mean photon number per pulse, but for free space light fields the notion of a mean photon number does not make much sense without reference to some fixed time range and volume.

The mean photon number per unit time is given by the Poisson distribution.

Only for coherent light fields or classical light fields with a coherence time shorter than the integration time (or of course horrible detector efficiency).

 

[Tspirit]

Thank you very much, Cthugha. Your reply is of great help. If the mean photon number of a single photon state in the free space is meaningless, how can we understand the operator equation of <n>=<1|n|1>=1? Can I think that there is only one photon in unlimited-short integration time? Or for two-photon, only two photons in an unlimited-short integration time?

 

[Cthugha]

Again, it is important to emphasize that in standard formulation the photon number operator gives the total photon number present within the field. This is not what you measure using a detector. If you have a look at standard textbooks in quantum optics (e.g. Mandel/Wolf), you will find that they spend a good amount of time on detector theory, which is the link on how to get from a well defined initial state to detection probability amplitudes of detectors.

The operator equation you give just means that your field contains one photon on average. It may be a laser or a single photon state or a thermal state. You need to develop your light fields into a hierarchy of correlation functions to have more information about this. If this is a real single photon state it means that there is one photon in the field and nothing more. This operator is not associated with any actual acts of measurement. If you perform a measurement on this single photon state, you will find that you can detect exactly one photon, no matter how long you integrate. By detecting this photon you have changed the light field and the state afterwards will be the vacuum. As it is not really possible to detect photons without perturbing the light field, this is a problem that always arises.

Accordingly, the answer to your question is: <n> gives the photon number in the total field or in some well defined spatial volume and time span. For example it is quite common to give the number of photons inside one coherence volume. What depends on the integration time is the photon detection rate. Here you can assume an effective volume: (detector area*speed of light*integration time) and consider the mean photon number inside this volume. If your detector is faithful (linear, reasonable quantum efficiency and so on) the detector will give average detected photon numbers that are linearly proportional to the mean photon number inside this volume.

 

[deepalakshmi]

The mean photon number per unit time is given by the Poisson distribution. So the mean count is related to the integration time. The time between events is the negative exponential of the mean $e^{-\mu}$ if I remember correctly.

You'll need the power spectrum of the light source tro make an estimate of the mean photon count for different modes based on the energy of a single photon.

Then what is the mean photon number for even coherent state( Schrodinger cat state)?

 

[Cthugha]

The mean photon number for cat states is quite a lengthy expression.
Check, e.g., this paper for details:
Statistical properties of cat states

The original publication was in Physics Letters A 199, 123 (1995).

 

[vanhees71]

This brings me to a general question, I have for quite some time. Is "photon number" a proper observable?

One should note that the annihilation and creation operators refer first to the (fully gauge-fixed quantized free-field) vector potential, but their corresponding commutation relations do not fulifill the microcausality conditions, because of the transversality constraint.

Thus to build proper local observables you have to use rather the field operators $\vec{E}$ and $\vec{B}$ which obey commutation relations fulfilling the microcausality conditions, and indeed as it turns out when considering what's measured with a photo detector, the "intensity" is defined in close analogy to the classical theory, i.e., as the gauge-invariant energy density $u=\frac{1}{2} (:\vec{E}^2+\vec{B}^2:)$.

Isn't it then more consistent to define energy spectra as measuring the intensity than "photon number" spectra? Isn't then the "average photon number" rather to be defined for a single-frequency photon state the average total em.-field energy divided by $\hbar \omega$?

In my own field of research (relativistic heavy-ion physics) what's indeed measured are "invariant photon-pT spectra", derived from $E \mathrm{d} N/\mathrm{d}^3 p$, where $E=|\vec{p}|c$. This implies that at least here one uses the energy spectrum to define "photon number distributions".

 

[Cthugha]

Maybe I do not get your point, but usually the Hamiltonian is constructed such that you get something akin to $H=\frac{1}{2}\int (\varepsilon_0 E^2 +\mu_0 H^2)dr$
which is then equivalent to $H=\sum_k \hbar \omega_k (a^\dagger_k a_k + \frac{1}{2})$.

The standard approach towards a theory of photodetection then usually involves trying to map the incoming light field to what happens inside the detector. Most of these mappings rely on the photoelectric effect and one of the standard mappings is to consider an ensemble of atoms that may be photoionized, where the number of photoelectrons is the measured signal. Here, it is most convenient to relate the signal to the photon number as the vacuum contribution to the energy will not result in any photoionization events.

However, depending on what measurement one has in mind, other approaches may make more sense. For example, if I use homodyning to measure the Wigner or Husimi Q function of a light field, it makes more sense to use a description in terms of fields.

I am not a theorist, but from what I hear the description given the quantum optics book by Vogel and Welsch (chapters 2 and 6) is a pretty thorough take at a solid theory of photodetection.

 

[vanhees71]

Exactly. That's the Hamiltonian, which is gauge invariant and the corresponding Hamiltonian density fulfills the microcausality condition for all other local observables (which are also necessarily gauge invariant expressions in terms of the fields rather than the potentials).

Indeed, what you describe as the standard quantum theory of photo detection also underlines this view. So what's measured is the gauge-invariant energy density of the em. field (or equivalently the "time integrated" energy-flow density/Poynting vector) rather than the "photon-number density", which is not definable in an gauge-invariant way.

I think that's also pretty much the summary of what's written on photodetection in the Quantum Optics textbooks I know (mainly Scully+Zubairy, Garrison+Chiao). I'll check, whether I can get a look at the book by Vogel and Welsch.

 

[Cthugha]

I see. I am not really familiar with the intricate details of this topic in optics. I know that the problem of gauge has been discussed in terms of how to formulate light-matter interaction in a consistent way and it has recently become important again in non-perturbative optics in the ultrastrong coupling regime where one cannot simply ignore the counterrotating terms which are usually neglected in the rotating wave approximation.

The former topic has been discussed by Loudon already in the 1990s:
Journal of Modern Optics 37, 685 (1990)

The latter topic is still discussed currently:
Nature Physics 15, 803 (2019)

Maybe these papers are related to what you are interested in.