I'm wondering whether the number of photons of a system is a Lorentz invariant. Google returns a paper that seems to indicate that yes it's invariant at least when the system is a superconducting walls rectangular cavity.

However I was told in the hbar chatroom that it's not an invariant and it's proportional to the 1st term of the 4-momentum which is related to the Hamiltonian of the "free field theory".

Today I've talked to a friend who studies some GR (no QFT yet) and he couldn't believe that this number isn't Lorentz invariant.

So all in all I'm left confused. Is it a Lorentz invariant for some systems and not others? If so, what are the conditions that a system has to fulfil in order for the number of photons to be invariant?

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    $\begingroup$ A counted number is a counted number, no matter which coordinate system you write it down in (you don't turn into triplets by sending your twin on a space mission in a really fast rocket) . The first term in the four momentum would be an energy and it does, of course, transform under a Lorentz transformation. That's the Doppler effect. I don't think this is quite as trivial for the case of thermal photons, which do not have a fixed number, to begin with. Only the average number of photons of a thermal state is meaningful. $\endgroup$ – CuriousOne Mar 5 '16 at 0:49
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    $\begingroup$ I would not take that one too seriously. It's not a simple question, by the way. The photon number is definitely NOT an invariant in accelerated coordinate systems. I don't think it's an invariant in the case of thermal photons which have to be in thermodynamic equilibrium with a non-Lorentz invariant thermal bath. If you throw seven atoms into a fixed volume, and you have them emit seven photons, those seven non-thermal photons will always stay seven in any coordinate system, though... so it's a yes, but... $\endgroup$ – CuriousOne Mar 5 '16 at 2:17
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    $\begingroup$ Seven detected photons would be the same for any observer; there's no way to know about photons witout detection. $\endgroup$ – Peter Diehr Mar 5 '16 at 2:28
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    $\begingroup$ @J-T: I will let a theoretician take this one. For my taste that's too much handwaving on an important question. I am sure one can give a much better answer than that. If my intuitive one agrees with the correct theoretical one I'll have a drink on the house, but I don't want to claim to have sufficient expertise in a area where I can only wing it. $\endgroup$ – CuriousOne Mar 6 '16 at 0:20
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    $\begingroup$ @Rococo For instance, if we have two non-equivalent representations of a free system together with dynamics, at time $t$ in a given frame in a manifold, the number operator may achieve different values in each system, but they're not even comparable, because they live in different Hilbert spaces. $\endgroup$ – user40276 Mar 6 '16 at 17:48

Alice prepares an electromagnetic field in a state with a sharp number of photons $\hat{N}|n\rangle=n|n\rangle$ where $\hat{N}$ is the number operator. Alice is boosted with respect to Bob. In Bob's reference frame the field is in state $\hat{U}(\Lambda)|n\rangle$. The question asks if a measurement of the number of photons for Bob's state gives the sharp answer $n$. In other words, is it true that $\hat{N}\hat{U}(\Lambda)|n\rangle=n\hat{U}(\Lambda)|n\rangle$? Bob will get the sharp result $n$ if the boost operator commutes with the number operator. We just need to show that the commutator $[\hat{U}(\Lambda),\hat{N}]_{-}=0$.

The number operator for photons of helicity $\lambda$ is, \begin{equation} \hat{N_{\lambda}}=\int \frac{d^{3}p}{2\omega}\hat{\eta}_{p\lambda}\hat{\eta}^{\dagger}_{p\lambda} \end{equation} where $\hat{\eta}_{p\lambda},\hat{\eta}^{\dagger}_{p\lambda}$ are emission and absorption operators respectively for a photon of momentum $p$ and helicity $\lambda$ (the notation for emission and absorption operators is from Dirac's monograph "Lectures on Quantum Field Theory"). We also have $\omega = p^{0}$ in the Lorentz invariant measure.

Single photon states transform as, \begin{equation} \hat{U}(\Lambda)|p,\lambda\rangle=e^{-i\theta(p,\Lambda)}|\Lambda p,\lambda\rangle \end{equation} where $\theta(p,\Lambda)$ is the Wigner angle. Creating a single particle state from the vacuum $|S\rangle$ by $|p,\lambda\rangle=\hat{\eta}_{p\lambda}|S\rangle$ implies that the emission operators transform like states, \begin{equation} \hat{U}(\Lambda)\hat{\eta}_{p\lambda}=e^{-i\theta(p,\Lambda)}\hat{\eta}_{\Lambda p\lambda} \ . \end{equation} Taking the Hermitian conjugate, using unitarity, and replacing $\Lambda$ by $\Lambda^{-1}$, \begin{eqnarray} \hat{\eta}^{\dagger}_{p\lambda}\hat{U}^{\dagger}(\Lambda)&=& e^{i\theta(p,\Lambda)}\hat{\eta}^{\dagger}_{\Lambda p\lambda}\\ \hat{\eta}^{\dagger}_{p\lambda}\hat{U}(\Lambda^{-1})&=& e^{i\theta(p,\Lambda)}\hat{\eta}^{\dagger}_{\Lambda p\lambda}\\ \hat{\eta}^{\dagger}_{p\lambda}\hat{U}(\Lambda)&=& e^{i\theta(p,\Lambda^{-1})}\hat{\eta}^{\dagger}_{\Lambda^{-1} p\lambda} \ . \end{eqnarray} Now evaluate the commutator, \begin{eqnarray} [\hat{U}(\Lambda),\hat{N}_{\lambda}]_{-}&=& \int \frac{d^{3}p}{2\omega}\hat{U}(\Lambda)\hat{\eta}_{p\lambda}\hat{\eta}^{\dagger}_{p\lambda}- \int \frac{d^{3}p}{2\omega}\hat{\eta}_{p\lambda}\hat{\eta}^{\dagger}_{p\lambda}\hat{U}(\Lambda)\\ &=&\int \frac{d^{3}p}{2\omega}e^{-i\theta(p,\Lambda)}\hat{\eta}_{\Lambda p\lambda}\hat{\eta}^{\dagger}_{p\lambda}- \int \frac{d^{3}p}{2\omega}\hat{\eta}_{p\lambda}e^{i\theta(p,\Lambda^{-1})}\hat{\eta}^{\dagger}_{\Lambda^{-1} p\lambda} \ . \end{eqnarray} Make a change of variable in the second integral, $p'=\Lambda^{-1}p$. \begin{equation} [\hat{U}(\Lambda),\hat{N}_{\lambda}]_{-}= \int \frac{d^{3}p}{2\omega}e^{-i\theta(p,\Lambda)}\hat{\eta}_{\Lambda p\lambda}\hat{\eta}^{\dagger}_{p\lambda}- \int \frac{d^{3}p'}{2\omega'}\hat{\eta}_{\Lambda p'\lambda}e^{i\theta(\Lambda p',\Lambda^{-1})}\hat{\eta}^{\dagger}_{p'\lambda} \end{equation} The Wigner angle $\theta(p,\Lambda)$ corresponds to a rotation matrix $R(p,\Lambda)=H^{-1}_{\Lambda p}\Lambda H_{p}$ where $H_{p}$ is the standard boost. Now, \begin{equation} R(\Lambda p,\Lambda^{-1})=H^{-1}_{\Lambda^{-1}\Lambda p}\Lambda^{-1}H_{\Lambda p}=H^{-1}_{p}\Lambda^{-1}H_{\Lambda p}= (H^{-1}_{\Lambda p}\Lambda H_{p})^{-1}=(R(p,\Lambda))^{-1} \end{equation} so that the Wigner angle $\theta(\Lambda p,\Lambda^{-1})$ is $-\theta(p,\Lambda)$. Upon putting this result into the last integral the commutator vanishes $[\hat{U}(\Lambda),\hat{N}_{\lambda}]_{-}=0$ and so Bob's electromagnetic field also has the same sharp number $n$ of photons as Alice's field.

Edit: Explanation of why the invariant measure appears in the number operator

The method of induced representations, which is used to get the response of the single particle states to a Lorentz boost (second equation in main text), is simplest if one chooses a Lorentz invariant measure so that the resolution of unity for the single particle states is, \begin{equation} \sum_{\lambda=\pm 1}\int \frac{d^{3}p}{2\omega}|p,\lambda\rangle\langle p,\lambda|=1 \ . \end{equation} This choice implies that the commutator for the emission and absorption operators is, \begin{equation} [\hat{\eta}^{\dagger}_{p\lambda},\hat{\eta}_{p'\lambda'}]_{-}= \langle p,\lambda|p',\lambda'\rangle= 2\omega\delta_{\lambda,\lambda'}\delta^{3}(p-p') \ . \end{equation} In turn, this implies that the normal-ordered Hamiltonian for the free electromagnetic field is, \begin{equation} \hat{H}=\frac{1}{2}\int d^{3}p(\hat{\eta}_{p\lambda=-1}\hat{\eta}^{\dagger}_{p\lambda=-1}+\hat{\eta}_{-p\lambda=+1}\hat{\eta}^{\dagger}_{-p\lambda=+1}) \ . \end{equation} Now create $n$ photons from the vacuum with a state, \begin{equation} |\Psi\rangle=(\hat{\eta}_{p\lambda})^{n}|S\rangle \end{equation} and demand that the number operator $\hat{N}_{\lambda}$ measures the sharp result $n$ on this state. This implies that the Lorentz invariant measure must be used in the definition of the number operator (first equation in main text). So, one sees that there are no assumptions here, just a choice of the invariant measure (instead of a quasi-invariant measure) to make the method of induced representations used to get the irreps of the Poincare group for massless particles as simple as possible.

  • $\begingroup$ The big assumption here is in how the Fock space states and creation operators transform wrt the Lorentz group (i.e. \begin{equation} \hat{U}(\Lambda)|p,\lambda\rangle=e^{-i\theta(p,\Lambda)}|\Lambda p,\lambda\rangle \end{equation}). Once those assumptions are made, the result is straightforward. Are those assumptions valid? Since the wave equation is Lorentz covariant, and its solutions bring creation operators, I'd say those assumptions are valid in the free case and consistent. In the interacting case it's less clear. In an accelerating frame, the Unruh effect clearly shows this doesnt work $\endgroup$ – Saleh Hamdan Mar 6 '16 at 7:14
  • $\begingroup$ As far as any talk about the zeroth component of the 4-momentum, this is irrelevant. No one would expect the Hamiltonian to be Lorentz covariant. Clearly there would be red/blue shift, but that says nothing of the number of particles being red/blue shifted, which should be Lorentz invariant if we make the above assumptions on how the Fock space transforms. $\endgroup$ – Saleh Hamdan Mar 6 '16 at 7:20
  • $\begingroup$ @Saleh Hamdan : The response of the single particle states to a Lorentz boost (second equation in my answer) is not an assumption. It is a consequence of finding the irreducible representations of the Poincare group using the method of induced representations (see "Induced Representations of Groups and Quantum Mechanics" by George W. Mackey, W.A. Benjamin, 1968) for the massless case. These representations are only derived for free particles and the Poincare group only works between non-accelerating reference frames as you point out. $\endgroup$ – Stephen Blake Mar 6 '16 at 10:59
  • $\begingroup$ The "Explanation of why the invariant measure appears in the number operator" killed any doubt. Thanks $\endgroup$ – Nogueira Mar 8 '16 at 13:37
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    $\begingroup$ How does the answer reconciles with the accepted answer of physics.stackexchange.com/questions/21830/…? $\endgroup$ – thermomagnetic condensed boson Feb 8 '18 at 20:17

An experimentalist's answer.

There are innumerable experiments measuring two gamma events. Lorentz invariance is a basic assumption for all measured interactions. Each interaction is in a different Lorenz frame depending on the energies and momenta involved. When we make the distributions of crossections and angles, we depend on this invariance of the number of particles in the interaction under observation. As the standard model manages to fit all these to a very good approximation this assumption holds.

Now each individual photon is coming from a Lorenz invariant interaction by construction of electromagnetic interactions, even though nothing is recording it, so the numbers should stay constant.

For the numbers to change if the Lorenz frame changes for an ensemble of already created photons, it means that the imposed Lorenz frame interacts somehow with the photons under observation. If energy is exchanged, more photons may appear which will look like non conservation of numbers, but should not be considered so.

  • $\begingroup$ You talk about experiments with gamma events, but they might just be not too sensitive to e.g. some microwave photons appearing in another frames. $\endgroup$ – Ruslan Mar 5 '16 at 7:55
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    $\begingroup$ @Ruslan If the production of photons , the number of photons, were not Lorenz invarian the calculated crossections which come from the hypothesis of Lorenz invariance would not fit. The data for pi0 decay for example cover an enormous range of energies of the pi0, each is a different lorenz frame with respect to the other decays. My point is that to generate a photon you need an interaction, they are not generated out of the vacuum. $\endgroup$ – anna v Mar 5 '16 at 10:37

I am not sure who is deleting my comments, arguments and critics challenging Mr. Blake's derivation. Just because we don't have enough reasoning to argue against an idea should never mean to erase that voice. As far as I know, science is not meant to be monopolized as the same is true with wealth.

In counting the number of photons in a system, we are limited by detection which in turn it is limited by Heisenberg Uncertainty Principle something that can never be overcome in Nature. Theoretically, a photon can have any non-zero energy over an infinitely wide spectrum. So, there is a non-zero probability of some photon having any particular energy. Let's not forget that photons are only one ingredient of the fundamental particles in Nature. This means that they can interact with at least every other charged fundamental particle of Standard Model including their self-interactions. (The interaction channels would exponentially increase as we go beyond Standard Model theories such as Extra Dimensions and Supersymmetry). All this being said, it is perfectly fine that photons get lost and/or created (through their interactions) and never be compensated in any given finite volume of our Universe (to be called system) during any finite period of time (within the finite age of Universe).Hence, being only one subset of all fundamental particles and having an infinitely long lifetime during which they have many interaction channels, their number will never be conserved in any finite volume of our Universe. However, it would be more challenging to generalize the question into, "whether the total number of all fundamental particles inside Visible Universe is conserved." To answer this one though, we need to know the Theory of Everything in which ALL fundamental particles are known including their lifetimes and masses and interaction channels along with the knowledge of Dark Matter and Dark Energy (accounting for 96% of Universe budget of matter and energy which are invisible as of today). So, I am certain that the answer to your question is, "No. Photons number in a finite volume known as system will never be conserved." And I would like to leave the more challenging one to future generations of physicists.

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    $\begingroup$ I didn't downvote. But photons are not charged so don't say "every other charged" particle. And conserved is different than frame invariant so your answer has completely misunderstood the question. And the universe doesn't even have to be in a state of definite particle number and even if it momentarily were we know we can change the number of particles, we do experiments and observations that involve this all the time. So it isn't conserved, and the question was about frame invariance anyway which is totally different. $\endgroup$ – Timaeus Mar 6 '16 at 22:24
  • $\begingroup$ I am not sure who is deleting my comments, arguments and critics challenging Mr. Blake's derivation. Just because we don't have enough reasoning to argue against an idea should never mean to erase that voice. As far as I know, science is not meant to be monopolized as the same is true with wealth. $\endgroup$ – Benjamin Mar 7 '16 at 2:57
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    $\begingroup$ I haven't deleted your comments to Stephen Blake, but this is the second comment of yours I've seen that seems to be placed as a response to my comment. Are you sure you are commenting on Blake's answer? And comments aren't for discussions anyway, they are focused on improving answers. They are also second class in that there is no expectation that they survive. You can vote up or vote down and that stays. A comment is supposed to encourage modification of an answer and that's really what it's for. But you can't force a change to an answer. $\endgroup$ – Timaeus Mar 7 '16 at 3:07
  • $\begingroup$ Hi Timaeus, thanks for letting me know. Yes, I am sure I am commenting on Blake's answer. I really don't know what is going on here. I was waiting for him to respond me but it seems my question is being constantly deleted. I am going to be more patient though until the problem resolves. Sincerely, $\endgroup$ – Benjamin Mar 7 '16 at 3:09
  • $\begingroup$ You are free to craft a better answer yourself. $\endgroup$ – Timaeus Mar 7 '16 at 3:09

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