Number of photons in an arbitrary EM field

[Dale]

TL;DR Summary

Number of photons in an arbitrary EM field

Correct me if I am wrong. I understand that if $\vec E$ and $\vec B$ are solutions to Maxwell’s equations then $\Psi= \vec E + i \vec B$ is a solution to Schrodinger’s equation.

Is there an easy way to calculate the statistical distribution of the number of photons, or at least the expected number of photons for such a wavefunction?

 

[Irishdoug]

Someone with more experience will likely be able to give a better answer. But I fail to see how one could answer this question without first knowing what type of light you have i.e. coherent, thermal, squeezed etc.

 

[Vanadium 50]

Is there an easy way to calculate the statistical distribution of the number of photons, or at least the expected number of photons for such a wavefunction?

You wanty the number of virtual photons in a static field? If so, you can't get it - virtual particles cannot be counted.

 

[PeterDonis]

I understand that if $\vec E$ and $\vec B$ are solutions to Maxwell’s equations then $\Psi= \vec E + i \vec B$ is a solution to Schrodinger’s equation.

This can't be correct as you state it, because $\vec{E}$ and $\vec{B}$ are 3-vectors and $\Psi$ is a wave function. They don't even "live" in the same mathematical spaces.

Also, in QM, the fields $\vec{E}$ and $\vec{B}$ are observables, not wave functions--i.e., they are operators.

The closest QM equivalent to a classical EM field would be a coherent state. The probability distribution for a photon number measurement on a coherent state is a Poisson distribution. See, for example, section 7.3 here:

https://courses.cit.cornell.edu/ece531/Lectures/handout7.pdf

 

[Dale]

Ok, so what would be the correct way? I want to start with a solution to Maxwell’s equations and calculate the expected number of photons or (preferably) the distribution.

 

[HomogenousCow]

You could construct the equivalent coherent state in QFT and then calculate its inner product with photon fock states. There's a problem in Peskin & Schroeder where you calculate the number distribution of a scalar field in response to a classical source, maybe look at that first.

 

[Dale]

You wanty the number of virtual photons in a static field? If so, you can't get it - virtual particles cannot be counted.

Ideally that fact would drop out of the math and you could get a definite count (or distribution) for a non-static field and something undefined or impossible to compute for a static field.

 

[Vanadium 50]

Virtual particles still can't be counted.

If you were talking about light, we could talk about a mean number of photons, but those photons are real.

 

[Dale]

I am talking about all EM fields. Including light. How would you calculate it for a dipole antenna?

 

[Vanadium 50]

For EM radiation, the easiest way is to calculate the power radiated, divide by the energy per photgon of that wavelength and getg the mean number of photons per second. It won't be exact, because you are not in an eigenstate of photon number, but you probably didn't know the power to ten decimal places either.

For EM fields in general, where you have virtual photons, there is no answer because virtual photons cannot be counted in any way.

 

[Dale]

So is there no way to calculate the distribution of the photon number? Is the expectation the best we can do?

 

[Vanadium 50]

I am still not clear if you are talking about radiation or not.

 

[Dale]

I am talking about all EM fields, including radiation. If the calculation must be limited to specific classes of fields then I want to know how to get it for whatever fields it applies.

I was hoping that there was some operator that could simply be applied to get this for an arbitrary field, similarly to how you might calculate the momentum of any wavefunction using $-i \hbar \frac{\partial}{\partial x}$. It seems it is not that simple for photon number.

 

[Vanadium 50]

Well, for radiation, the answer in #10 holds. There is something called a Fock state, which is a state of definite photon number, but no light source you are likely to run across is one.

For fields, again, #10 applies. There is no answer because virtual photons cannot be counted in any way.

 

[PeterDonis]

I was hoping that there was some operator that could simply be applied to get this for an arbitrary field, similarly to how you might calculate the momentum of any wavefunction using $-i \hbar \frac{\partial}{\partial x}$. It seems it is not that simple for photon number.

Applying the momentum operator to a wave function is not the same as "calulating the momentum" of that wave function. If the wave function is not a momentum eigenstate, it does not have a definite momentum at all.

Similarly, there is a photon number operator, but applying it to a wave function is not the same as "calculating the number of photons" for that wave function. If the wave function is not an eigenstate of photon number (i.e., not a Fock state), it does not have a definite "number of photons" at all.

So it is actually "not that simple" for any observable, not just photon number.

 

[PeterDonis]

is there no way to calculate the distribution of the photon number?

Yes, just expand the state in the photon number basis. As noted already, when you do this for a coherent state, you get a Poisson distribution. But you can do it for any state.

 

[PeterDonis]

For EM radiation, the easiest way is to calculate the power radiated, divide by the energy per photgon of that wavelength and getg the mean number of photons per second. It won't be exact, because you are not in an eigenstate of photon number, but you probably didn't know the power to ten decimal places either.

For EM fields in general, where you have virtual photons, there is no answer because virtual photons cannot be counted in any way.

I think the relevant distinction here might be between source-free fields ("radiation") and fields in the presence of sources.

 

[vanhees71]

The only correct way known is QED, and there is no photon wave function in the usual sense, because a photon has no position operator to begin with. Also photons are easily destroyed and created in interactions. So photon number is not conserved. Another point to consider is gauge invariance.

The most pragamatic approach is to start with classical Maxwell electrodynamics, introducing the potentials and fix the gauge completely by, e.g., choosing the Coulomb gauge.

To define "photons" you consider the free em-field case first and canonically quantize it. The Coulomb gauge in this case leads to the radiation gauge, where the classical (!) potentials are subject to the constraints
$$\vec{\nabla} \cdot \vec{A}=0, \quad \Phi \equiv A^0=0.$$
Now you quantize the fields canonically with the qualification that the canonical equal-time commutation relations must be made consistent with the transversality condition.

Finally this leads to the mode decomposition of the quantized field $\vec{A}$ with only two physical field-degrees of freedom. The most convenient choice of a single-photon basis is the momentum-helicity eigenbasis with the corresponding annihilation and creation operators $\hat{a}(\vec{k},\lambda)$ and $\hat{a}^{\dagger}(\vec{k},\lambda)$ with $\vec{k} \in \mathbb{R}^3$ and $\lambda \in \{-1,1\}$. These operators follow the usual commutation rules (for bosons, which is necessary to have a local (microcausal) QFT with the Hamiltonian bounded from below),
$$[\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=(2 \pi)^3 \delta^{(3)}(\vec{k}-\vec{k}') \delta_{\lambda \lambda'}.$$
A complete basis is the Fock basis, which is the (generalized) eigenbasis of the number operators $\hat{N}(vec{k},\lambda)=\hat{a}^{\dagger}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda)$.

To calculate a photon-number distribution you need to specify the state. For thermal photons, described by the canonical ensemble, the state is given by
$$\hat{\rho}=\frac{1}{Z} \exp \left (-\frac{\hat{H}}{T} \right), \quad Z=\mathrm{Tr} \exp \left (-\frac{\hat{H}}{T} \right).$$
The Hamiltonian is given by
$$\hat{H}=\sum_{\lambda \in \{-1,1\}} \int_{\mathbb{R}^3} \mathrm{d}^3 k |\vec{k}| \hat{N}(\vec{k},\lambda).$$
The photon-number distribution then turns out to be the Planck distribution (as it should be), i.e.,
$$f=\frac{2}{\exp \left (\frac{|\vec{k}|}{T} \right)-1}.$$
The $2$ comes from the two helicities.

Another example is a coherent state, which you get when assuming the source being (approximately) described by a classical charge-current distribution $(\rho,\vec{j})$ (the socalled "hemiclassical approximation", to be distinguished from the "semiclassical approximations, where the charges are quantized but the em. field kept classical). That's also how you describe quantum-field theoretically the emission of an antenna (e.g., the usual Hertzian dipole approximation treated in the standard classical-electrodynamics lecture). Considering harmonic time dependence you get a Poisson distribution for the photon number.

Last but not least there are also Fock states of definite photon numbers, which you can prepare by simply using an excited atom and wait till it deexcites and emits precisely one photon in a dipole transition or, more efficiently, by parametric downconversion using a laser (emitting never photon Fock states but coherent states) and a BBO crystal to produce entangled two-photon Fock states.

 

[Dale]

If the wave function is not a momentum eigenstate, it does not have a definite momentum at all.

I know that I asked multiple times for a way to calculate the distribution. That is what I would prefer over the expectation.

Yes, just expand the state in the photon number basis.

Ok, that is what I am looking for. How do you do that?

 

[Dale]

The most pragamatic approach is to start with classical Maxwell electrodynamics, introducing the potentials and fix the gauge completely by, e.g., choosing the Coulomb gauge.

Thanks @vanhees71 so I cannot start with the classical E and B fields. I must start with the potentials?

 

[vanhees71]

That's a good question. I guess for the free fields you could get along with the fields, but I've not seen a solution to formulate interacting QED with the fields alone.

I think there is some attempt to introduce a "photon wave function" using the Riemann-Silberstein-field formalism you mentioned in #1. Indeed $\vec{\psi}=\vec{E}+\mathrm{i} \vec{B}$ (in Gauß or Heaviside-Lorentz units) transforms under Lorentz transformations in the representation of the Lorentz group as $\mathrm{SO}(3,\mathbb{C})$, and you get some equations which are analogous to the Dirac equation for spin-1/2 particles (which is the representation $(1/2,0) \oplus (0,1/2)$ of the Lorentz group including spatial reflections).

For a nice review, see this talk (including a discussion of the free-photon treatment in terms of the Riemann-Silberstein vector):

https://inis.iaea.org/collection/NCLCollectionStore/_Public/37/036/37036055.pdf

 

[Vanadium 50]

Potentials will not help with fields in the presence of sources. These are not composed of photons one can count in any sense.

As far as radiation, if you take a Gaussian approximation (and this situation, this approximation is far, far better than in other cases where we use it), a light bulb might emit a mean number of 10²⁰ photons per second with a standard deviation of 10¹⁰. Is this good enough? If not, why not?

Does it matter that no physical light bulb is stable to one part in 10 billion?

Yes, there are quantum mechanical reasons why the number of photons per unit time varies, but there are classical reasons why tehy vary much more.

 

[vanhees71]

Standard QED is formulated in terms of the potentials. The potentials are the way to formulate interacting QED as a local, microcausal field theory. You can of course count only asymptotic free photons, because only for them you can define Fock states of definite photon number.

 

[jim mcnamara]

FWIW - @Dale

I think that you may have come up with "counting photons" as a way to solve something else. What are you really trying to do? The answer to my question probably is not "counting photons".

 

[Dale]

FWIW - @Dale

I think that you may have come up with "counting photons" as a way to solve something else. What are you really trying to do? The answer to my question probably is not "counting photons".

This actually is about counting photons. I would like to show, e.g. as you take a classical dipole wave and reduce the amplitude to the point that you have a small expected number of photons that it still is a dipole field. I should wind up with a wavefunction that matches the classical dipole fields and gives a very small probability of detecting a photon anywhere, but with no gaps.

 

[Dale]

light bulb might emit a mean number of 10^20 photons per second with a standard deviation of 10^10. Is this good enough?

That is good enough. How did you obtain the mean and standard deviation?

 

[Dale]

That's a good question. I guess for the free fields you could get along with the fields, but I've not seen a solution to formulate interacting QED with the fields alone.

I think there is some attempt to introduce a "photon wave function" using the Riemann-Silberstein-field formalism you mentioned in #1. Indeed $\vec{\psi}=\vec{E}+\mathrm{i} \vec{B}$ (in Gauß or Heaviside-Lorentz units) transforms under Lorentz transformations in the representation of the Lorentz group as $\mathrm{SO}(3,\mathbb{C})$, and you get some equations which are analogous to the Dirac equation for spin-1/2 particles (which is the representation $(1/2,0) \oplus (0,1/2)$ of the Lorentz group including spatial reflections).

For a nice review, see this talk (including a discussion of the free-photon treatment in terms of the Riemann-Silberstein vector):

https://inis.iaea.org/collection/NCLCollectionStore/_Public/37/036/37036055.pdf

That seems good. On slide 9 it looks like he takes the classical complexified fields and expands them in a basis that seems similar to a filtered Fourier transform. Then on 10 and 11 he makes them quantized. On slide 12 he then identifies the classical expansion coefficients with the photons.

That basis, seems similar to the Fourier transform, but not the same. Do you have any insight on that?

I don’t see him clearly identify the complexified field with the wavefunction. In your opinion, is that implied? I have seen it somewhere, but I don’t remember the source and therefore I could also be misremembering the statement that you could write $\vec \Psi = \vec E + i \vec B$.

 

[Vanadium 50]

hat is good enough. How did you obtain the mean and standard deviation?

Mean is total energy / energy per photon, which is 2 eV or so for visible light.
Standard deviation is the square root of the mean.

 

[Dale]

Standard deviation is the square root of the mean.

That is true for a coherent state. Is it also true for a light bulb?

That would certainly make things easier if it were general for most classical radiation.

 

[Vanadium 50]

Is it also true for a light bulb?

It is an extremely good approximation. (Probably a million or more times better than the approximation a 100W bulb puts out a constant amount of light.)

 

[Dale]

It is an extremely good approximation. (Probably a million or more times better than the approximation a 100W bulb puts out a constant amount of light.)

Thanks, that is really helpful. Do you think the same approximation would be valid for a dipole antenna?

 

[Vanadium 50]

Sure. Probably better since the power is higher (usually) and the energy per photon lower.

 

[vanhees71]

I don’t see him clearly identify the complexified field with the wavefunction. In your opinion, is that implied? I have seen it somewhere, but I don’t remember the source and therefore I could also be misremembering the statement that you could write $\vec \Psi = \vec E + i \vec B$.

No, not in these slides. There's a nice review here:

https://iopscience.iop.org/article/10.1088/1751-8113/46/5/053001

This actually is about counting photons. I would like to show, e.g. as you take a classical dipole wave and reduce the amplitude to the point that you have a small expected number of photons that it still is a dipole field. I should wind up with a wavefunction that matches the classical dipole fields and gives a very small probability of detecting a photon anywhere, but with no gaps.

There is no wave function, but of course a corresponding state. You get it by calculating the dipole solution for the field operators assuming a classical charge-current distribution. You can just copy the calculation from the classical theory because it's solving linear equations and there's not much formal difference between operator and "c-number"-valued fields. The state is then given by applying the positive-frequency part of the field-operators to the vacuum. The result is a coherent state, and the photon-number probabilities are given by a coherent state. The calculation for this "hemiclassical approximation" is discussed in detail in

C. Itzykson and J.-B. Zuber, Quantum Field Theory,
McGraw-Hill Book Company, New York (1980).

Of course a coherent state can have as low a "intensity" as you like. You can even make the average photon number as small as you want. It's just that in such cases the largest contribution in the decomposition in terms of Fock states is the vacuum contribution. Thus you find with the probabilities of a Poisson Distribution,
$$P(N)=\frac{\lambda}{N!} \exp(-\lambda N)$$
$N$ photons.

 

[Dale]

The result is a coherent state, and the photon-number probabilities are given by a coherent state.

Interesting, that explains why @Vanadium 50’s approach above works for many sources.

Many thanks everyone! I found this helpful.

 

[Paul Colby]

I might point out that $E+iB$ is a self dual field combination. Your field combination only describes one of the two possible polarizations. The other equally valid combination is $E-iB$.

Also, it's my understanding that $E$ and $B$ as operators in QM still obey Maxwell's equations. The natural thing to do would be to view them as such. Then the photon number question is as others have said, an operator expectation based on the given field state which you must first supply.