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I changed again the mind experiment to avoid any explanation involving doppler effect or nonlinear interaction

I expressed the same problem under a different angle but it will probably make the situation more clear in: Creating light pulse via: turning on and off coherent state VS putting photons in many mode, different in the quantum regime? I hope it helps.

There is something that always confused me to link the laws of physics and signal processing.

Consider you have a monochromatic laser of frequency $\omega_0$, for example, made with hydrogen transition. Consider the beam has a given width of $w$.

At $t=-\infty$ you start your laser. You never turn it off.

You have at your disposal a device that measures the intensity of the electric field. I call A the point representing this device.

For $t<0$ the device is outside of the beam. I start to move it so that it crosses in an orthogonal direction the light beam. At $t=0$ it is in the beam and it stays like this until $t=T$, then it is outside of the beam.

If you analyse the electric field in A you will have something like (I work with complex signal for simplicity but replace my $exp$ by $cos$ or $sin$ if you wish):

$$ E(t<t_0)=0 $$ $$ E(t_0<t<t_0+T)=e^{j \omega_0 t} $$ $$ E(t_0+T<t)=0 $$

If I do the spectral analysis of such signal I won't have a single mode $\omega_0$.

But from a physical point of view the only frequency of light I used in all this is $\omega_0$, I didn't excite any other mode than this one.

What does that physically mean?

In the classical regime there is no paradox for me, it is just that I can represent my wave from two points of view: I turn on a monochromatic source for a finite amount of time, or I send many modes on different frequencies with Fourier amplitude corresponding to the Fourier transform of $E(t)$. Both approaches are mathematically and physically totally equivalent here.

The paradox comes in the quantum regime.

I used a device that produces photons at a single frequency of $\omega_0$. But if I study the classical signal on A, I find many modes occupied. It is a kind of paradox because I never created other photons than the ones in $\omega_0$ initially.

Then two possible answers: either photons have been created in the other modes afterward. Either there is indeed the only photon in $\omega_0$ in all the experiments.

I am not convinced by the first possibility, indeed if I reason in the frame associated with A, it will see the signal I described having many frequencies. And it is just a change of frame I cannot create photons in another mode because of this.

It cannot be explained by the Doppler effect as well as in my mind experiment I move at a constant speed in an orthogonal direction to the light beam. Furthermore, the doppler effect is a shift. Here I don't have a shift in frequency, but I have a "creation" of many other frequencies. It is different.

In addition, the laser is based on hydrogen transition which emits photons of a very specific frequency: it couldn't know that a device will cross the beam later on. The emission frequency is independent of what I will do in the lab.

As I proposed in an answer, I think it means that there are different way to describe the signal, but only one of them corresponds to how it was physically produced ?

In practice, in this example, the interpretation is that photons only occupy the mode $\omega_0$, no other photons exist in all this experiment.

The thing is that in A I will see those photons only for $0<t<T$ and not after. By doing a Fourier transform I can describe the light I see as if other light mode were excited. Which is not the case.

Would you agree with this and my proposed answer? People proposed another answer but as I explained in the comment I am not convinced by those answers. I justified my point. If I made a mistake in my answer where is it? I think I am convinced by it but I would like some external point of view.

A different way to ask the question:

The laser being monomode and always turned on, the quantum state inside of its cavity or of the light it radiates can be described using coherent states of frequency $\omega_0$. If photons exist at other frequency could you write down the interaction that created them.

(I don't think those other frequencies photons exist but if they do I would like to see in the answer precisely written the mechanism that created them).

Note: A kinda similar question has been asked here Fourier transform paradox(?) of a wave packet but I am not very convinced by the answers. Also, to avoid explanation like nonlinearity induced by some shutter I would put on the laser path, I choose to take an example with a laser rotation.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Chris Jan 11 at 21:45
  • $\begingroup$ I substantially edited my answer to more directly address your paradox. $\endgroup$ – knzhou Jan 12 at 2:10
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Reviewing the mode expansion

The laser being monomode and always turned on, the quantum state inside of its cavity or of the light it radiates can be described using coherent states of frequency $\omega_0$. If photons exist at other frequencies could you write down the interaction that created them?

The first issue here is that you're assuming that modes must have definite frequencies, and hence that photons in those modes have definite frequencies. This is an idealization, which doesn't hold in the real world. Let's review where it comes from in the textbooks:

  1. Consider the classical electromagnetic field, either in complete vacuum, or in a perfectly closed system with time-independent non-dissipative boundary conditions, and without any matter present to absorb energy.
  2. Under these unrealistic assumptions, the field has solutions that oscillate periodically in time forever, with absolutely no decay, which we'll call modes.
  3. Upon quantizing the system, we find that the quantum state of the field is described by a quantum harmonic oscillator for each mode, and we call the excitations within each mode photons.

Thus, modes of the classical field only have definite frequencies under idealized assumptions. In the real world, modes don't need to have definite frequencies, and neither do photons. In fact, in the real world, there are often cases where there are multiple valid sets of modes to use, which correspond to multiple valid definitions of photons; this will resolve your paradox below.

A toy example

Here's a toy model to illustrate subtleties with the mode expansion. (It actually won't be relevant to the final answer, but it might help build intuition.)

In free space, we can describe the evolution of a single degree of freedom of a field by a quantum harmonic oscillator. So more generally, consider a degree of freedom evolving under the Hamiltonian $$H(t) = \frac{p^2}{2m} + \frac12 m \, \omega(t)^2 x^2.$$ The time-dependent of $\omega(t)$ could represent, e.g. the effect of fluctuations of the cavity walls. The classical solutions to the equations of motion are not sinusoids, and hence don't have a definite frequency.

The same remains true when we quantize. At every time, we can define instantaneous raising and lowering operators in the usual way, along with an instantaneous vacuum, corresponding to an instantaneous mode which oscillates sinusoidally at the instantaneous frequency. Similarly, at every time, we can define a ladder of instantaneous energy eigenstates, $$|n(t) \rangle = \frac{(a^\dagger(t))^n}{\sqrt{n!}} |0(t) \rangle$$ In the case where $\omega(t)$ changes slowly, the adiabatic theorem applies, so $|n(t) \rangle$ at time $t$ evolves into the state $|n(t') \rangle$ at a later time $t'$. Similarly, you can define instantaneous coherent states, $$|z(t) \rangle \propto e^{z a^\dagger(t)} |0(t) \rangle$$ which in the adiabatic limit evolve into other instantaneous coherent states.

The adiabatic limit demonstrates that coherent states do not necessarily have definite frequency. Recall that for the electromagnetic field, the "position" variable is the vector potential $\mathbf{A}$, and the conjugate momentum is $\mathbf{E}$. A reasonable physical definition of "definite frequency" is that the observed electric field is sinusoidal, i.e. $\langle p(t) \rangle$ is sinusoidal for this coherent state. But it isn't, because Ehrenfest's theorem tells us that $$\frac{d \langle p(t) \rangle}{dt} = - m \, \omega(t)^2 \langle x(t) \rangle$$ or, differentiating again, $$\frac{d^2 \langle p(t) \rangle}{dt} = - \omega(t)^2 \, \langle p(t) \rangle $$ which does not have sinusoidal solutions when $\omega(t)$ varies. (This isn't actually related to your paradox, but it illustrates how you can get frequency spread inside a cavity even if only "one mode" is excited.)

In the non-adiabatic case, we can get even weirder behavior. For example, suppose that $\omega(t)$ suddenly changes at $t = 0$, $$\omega(t) = \begin{cases} \omega_< & t < 0, \\ \omega_> & t > 0. \end{cases}$$ We can define two sets of ladder operators before and after $t = 0$ corresponding to frequencies $\omega_<$ and $\omega_>$, and thereby define two independent sets of states, $|n_< \rangle$ and $|n_>\rangle$. In particular, if you start in the state $|0_< \rangle$, you won't end up in $|0_> \rangle$. Instead, you end up with some "$t > 0$" photons, not because there was an explicit source term, but because the natural definition of photons changed at $t = 0$.

Addressing the paradox

Let me boil down your paradox to the following:

  1. Start with a monochromatic plane wave in free space, containing only photons of frequency $\omega$.
  2. Couple a detector to this plane wave for a finite time $T$.
  3. The detector "sees" photons of frequency $\omega'$ in a width $\sim \hbar/T$ about $\omega$. In other words, to the detector, it's as if the laser pulse were only a time $T$ long, even though it really is infinite.

There's really no problem here, you just have to be careful with what it means for a detector to "see photons". In your situation, the state of the electromagnetic field is perfectly well-defined. Your detector can't perfectly capture that state, but no detector can see literally everything, nor should we expect any to.

For example, if I were color blind, a red photon and a green photon would look the same to me. That doesn't mean that my eyes are converting red photons to green, or a mixture of red and green, it just means they can't tell the difference. If your detector just measures the electric field for a short time, it's effectively color blind, so that's it.

Refining the paradox

This might not be satisfying, so let's consider an alternative detector which explicitly measures photons, following the question you linked. Suppose the detector works as follows: at a prescribed time, two perfectly conducting metal plates suddenly sweep down. The plates are separated by a distance $L = c T$, so they effectively "cut out" a time $T$ of the pulse. Then, the detector just counts up the photons inside it, along with their frequencies. The paradox is that the detector sees photons of frequency $\omega'$ in a width $\sim \hbar/T$ about $\omega$.

You can probably now see the trick, given the first section. The detector plates have changed the boundary conditions of the electromagnetic field. That means the photons the detector measures correspond to a different set of modes than the free space photons. The free space modes look like $e^{ik x}$ with no boundary conditions, while the detector modes look like $\sin(k' x)$ with the $k'$ defined by hard wall boundary conditions.

Upon quantizing each set of modes separately, we find that a state of the electromagnetic field corresponding to only photons in one free space mode also generally corresponds to photons in multiple detector modes. The standard mathematical tool used to swap between the equivalent mode descriptions is the Bogoliubov transformation.

This appeared in a simple form in the previous section, where $|0_< \rangle \neq |0_> \rangle$. It is also the reason behind the Unruh effect, the fact that an accelerating detector sees a thermal bath of photons, even in vacuum: this is due to the mismatch between detector-defined photons, and the plane wave photons defined in inertial frames in free space. Hawking radiation also runs on the same principle.

So in some sense, the resolution to your paradox is quite "exotic". But really, this ambiguity of modes was always there into the formalism of quantum field theory. Most textbooks ignore it only because there is a unique set of modes if you stay in inertial frames in free space, but this breaks down quickly.

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  • $\begingroup$ Thank you for your answer. It is interesting. I however have questions. First, it is important for me that we don't consider any practical limitations in all this. For example I would like to avoid the fact the laser is not monomode in practice. Conceptually we can make its emission spectrum as narrow as possible. So I would like to stick to this consideration to really understand the underlined physics, which you did in your second part. $\endgroup$ – StarBucK Jan 12 at 15:10
  • $\begingroup$ In the end, would you agree with me if in the end what you say is that the notion of photon is highly dependent on the frame you reason in (which I know from doppler effect), but it is deeper than that in the sense that conceptually you could have a given spectrum in one frame and a very different spectrum (not deduced from a simple shift) in another one. It is what is happening here with the detector I put along the path. $\endgroup$ – StarBucK Jan 12 at 15:11
  • $\begingroup$ Thus here it means that the "particle" interpretation of photons is not possible in this experiment because it would mean that in one frame some photons would exist but not in another frame. Which is not possible with a particle vision (if a particle exist it must exist in all frame, maybe it will look like different but it will exist) $\endgroup$ – StarBucK Jan 12 at 15:12
  • $\begingroup$ @StarBucK Exactly, the only well-defined thing is the state of the quantum field, which need not have a single interpretation in terms of photons. You can read a lot more about this in textbooks on QFT in curved spacetime, where these considerations really matter. $\endgroup$ – knzhou Jan 12 at 18:01
  • $\begingroup$ Allright, to be really convinced there is another thing I would like to clarify. Consider a radiating laser. You can compute from model inside of the cavity of the laser its emission spectrum. It will correspond to what you would have measured for an infinite time: see physics.stackexchange.com/questions/524571/… all this really makes sense for me (writing the next) $\endgroup$ – StarBucK Jan 12 at 18:09
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The apparent paradox is analogous to the problem of blind men describing an elephant https://en.wikipedia.org/wiki/Blind_men_and_an_elephant (it's like a rope, a tree, a tent, a snake---). A Fourier transform is only one example of a way to represent a wave form. The same wave form can be represented as a sum of delta functions, Gabor wavepackets, and even square waves. They all describe the same thing, and not one of them is quite “correct”. Each is a blind man's description of something that can't be perfectly described from any one perspective.

A “monochromatic” beam is one whose wave peaks are perfectly in step for all time and for an infinite distance. To describe a laser beam as monochromatic is, of course not fully meaningful because we can never know whether it has been and will continue shining for all time. At best, it can only be “effectively monochromatic”: monochromatic enough for whatever the practical purpose may be.

Internal to the laser, the emission events do not take an infinite time to occur, so in one view the laser beam is composed of a lot of superimposed non-monochromatic pulses whose fundamental frequency components are in phase but whose higher frequency components are randomly out of phase. Add all those up, (for a very long time!) and the beam is effectively monochromatic.

So a monochromatic beam can be described in two very different ways (as one long thing or a superposition of short things), but still be the same thing.

Even the idea of “beam” has similar problems. If you have an infinitely wide plane wave, it will propagate as a perfectly collimated beam. But what is a “beam” if it's infinitely wide? If you block the infinite wave so that it has a finite diameter, it will no longer propagate as a collimated beam; it will spread at an angle inversely related to its diameter and directly related to its wavelength. Huygens showed that a plane wave can be represented both as a simple propagating plane wavefront, and as a superposition of an infinite number of spherical wavefronts diverging from points on the wavefront. Neither description is “correct”, but each is useful in different situations.

There is a direct correspondence between these two representations of wave propagation and the two representations of monochromaticity. In each case, both representations are equally valid; and neither is the “correct” representation. We use whichever representation is most useful for analyzing any given beam scenario.

You are seeking an intuitive understanding of the fact that the frequency spectrum of a light beam is altered by shuttering the beam to form a long pulse. Perhaps the easiest way to understand it is the second representation of a “monochromatic beam”: as the superposition of a lot of short pulses whose central frequencies are all in phase. It doesn't matter whether you consider the short pulses to be femtoseconds long or microseconds long; the math works out the same. When the beam is shuttered, it limits the number of such short pulses that can be summed to represent the resulting long pulse, and thus prevents full cancellation of the portions of the pulses that are out of phase (which correspond to the off-center frequencies of the pulses).

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  • $\begingroup$ I think I agree with your explanation and it is related to my answer as well. In the end, mathematically we have infinite ways of describing the signal, two examples here are monochromatic wave that is turned on for a finite time. The other is a combination of monochromatic waves of different frequencies that leads to the same wave. However if we think about what physically produced the wave, there is only one answer. Indeed, a laser emit because of the physics photons at a given frequency. So the second physical interpretation cannot hold here. $\endgroup$ – StarBucK Jan 9 at 23:19
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    $\begingroup$ The idea that a single photon, produced by a single emission event (e.g., from a single atom in a gas laser) is single-frequency would be very difficult to support with experimental data. And, it conflicts with the fact that single-frequency requires infinite extent in time and length, both forward and back in time and space. $\endgroup$ – S. McGrew Jan 10 at 0:53
  • $\begingroup$ I like this answer and I don't think it conflicts with the one I wrote, but I'm afraid I lack the skill to reconcile them in a comment. $\endgroup$ – rob Jan 10 at 2:45
  • $\begingroup$ There are two separate things. First, problematic of measuring the frequency of a signal. To know if a signal is monomode you need to measure on an infinite time. But it is a problem of how you measure it it is different. Next I agree that indeed the photon will not be fully monochromatic. But it is more a practical problem. Consider it slightly non monochromatic if you wish. In the end its frequency is independent from the fact I rotated my laser. How can the physical process know this ? $\endgroup$ – StarBucK Jan 10 at 9:39
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    $\begingroup$ You are attributing a fixed inherent frequency to the light, independent of measurement. The important point is that the narrowness of the frequency spectrum that is measured is inversely related to the time taken to do the measurement. It is not physically possible to define frequency in a measurement-independent way. $\endgroup$ – S. McGrew Jan 11 at 13:40
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The nonlinearity that you can't find in your monochromatic laser is here (or was in v1 of the question):

Then you let the light pass until a time $t_0+T$ when you cut again the light using the plate again.

A metal plate is a charge distribution where the lattice of positive ions has a different spatial arrangement and response function that the Fermi gas of conduction electrons. In a conductor we usually talk about the "skin depth" of electromagnetic radiation with a given frequency, and we can show that the intensity of the radiation falls off exponentially over a few skin depths --- a nonlinear interaction. (It's possible to make a metal plate that's thin enough to transmit some of your signal, but that's mechanically challenging for the frequencies of visible light.) Moving the plate introduces nonlinear components to the electromagnetic field in time as well, because the plate isn't rigid: the information that the plate has moved propagates from its points of support to its free end at approximately the speed of sound, by deformations of that positive-ion lattice.

How is it that you've designed this apparatus to move the plate at $t_0$? Maybe you've built some actuator arm that's attached to a clock --- a clock which is, from a signals perspective, an electro-mechanical oscillator with a complicated frequency spectrum. Or maybe you just decide when it's time and move the plate with your hand. In the astounding future where we live, it's possible to see the electro-chemical and electro-mechanical parts of that system as well, using tools like EEGs or fMRIs.

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    $\begingroup$ This answer bothers me, because it suggests that the composition of the shutter is somehow encoded into the spectrum of the "pulse" generated by turning the shutter on and off. I see the question as seeking an intuitive explanation of why a finite pulse of single-frequency light contains a broadened range of frequencies. $\endgroup$ – S. McGrew Jan 9 at 13:53
  • $\begingroup$ I might counter by saying that the difference between a low- and high-quality shutter is the extent to which information about the shutter leaks into the signal. For example, imagine a shutter that sticks halfway and changes the dominant laser cavity mode briefly from TEM00 to TEM01. That we can construct systems where such effects are negligible does not mean that those effects vanish entirely --- that was my understanding of the question. But I'd be interested to see a more pedestrian explanation. $\endgroup$ – rob Jan 9 at 14:39
  • $\begingroup$ OK, I've risen to the "pedestrian" challenge and posted an answer. $\endgroup$ – S. McGrew Jan 9 at 16:11
  • $\begingroup$ @rob I don't think I agree with the answer. You could imagine a different experiment that would lead to the same interpretation. Consider you have an infinitely small width laser (a "line"). You do not enlight a target, then you rotate the laser to enlight it for a time $T$, then you rotate again the laser to not enlight it. If you look at your signal on the target you will see the function I described. However it wouldnt make sense that some process changed the frequency of the photons just because I rotated the laser. $\endgroup$ – StarBucK Jan 9 at 23:25
  • $\begingroup$ I don't think that moving the laser instead of moving a shutter substantially changes the answer: everything that I wrote about the motion of the shutter being an electro-mechanical process applies to the motion of the entire laser cavity. For instance, suppose that instead of out-and-back, you rotate the laser cavity by spinning it on a turntable with frequency $ω_\text{slow}$. Then your intermittent illumination of the target will happen with fundamental frequency $ω_\text{slow}$, though the briefer the illumination the more harmonics of $ω_\text{slow}$ will appear in a Fourier transform. $\endgroup$ – rob Jan 10 at 2:43
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This is the answer that I misunderstood the question. If you want to find an answer using Lorentz transformation and quantum fields, then skip this and see below.

The interaction you want to have is simply the interaction between electron bound to an atom of blocking material and electromagnetic waves. The monochromatic laser in your experiment drive dipole oscillation of the atom of blocking material and the dipole radiation from the atom of the blocking material generates wave with completely out-of-phase wave because the blocking material will not transmit any photon after the blocking material. Then by removing your blocking material, you create new modes, simply by moving it away, because the dipole radiation from the blocking material is turning off, as you start to move the blocking material away from the beam path. Main factors in the dipole radiation intensity can be written as $$ I(t) \propto \frac{d}{dt}\int d\omega|A(\omega)|^2|M(\omega)|^2\Theta(t,\omega-\omega_0), $$ (I will drop the spatial variables in the notation for simplicity in writing... spatial variables are assumed to be there if necessary) where $I(t)$ is the time-dependent intensity of the dipole radiation from the blocking material, $A$ is the vector potential of the incident wave (in this case the monochromatic wave), $M$ is the cross-section of the dipole transition, which is related to the reflectance or transmittance of the blocking material, and $\Theta$ is the Fourier transform of the temporal window function for the dipole radiation, which is just a Fourier transformation of unit rectangular pulse, but in our case let us assume that we blocked the monochromatic laser for a long time and $\Theta(t,\omega-\omega_0)\rightarrow\pi t\delta(\omega - \omega_0)$. Then

$$ I(t) \propto |A(\omega_0)|^2|M(\omega_0)|^2 $$

right before you put away the blocking material from the beam path of your experiment, which is constant because the monochromatic laser field has constant power (or intensity). This dipole radiation $I(t)$ exactly cancels the monochromatic wave after the blocking material after the blocking material as the dipole radiation generates a wave that destructively interferes with the monochromatic wave after the blocking material.

The cross-section $M$ of the dipole radiation is decreasing over time as we put away the blocking material from the beam path of the monochromatic laser. This means that $M$ now has time dependence, which has the time scale of moving the blocking material away. If we put the blocking material away fast, then the time scale the $M$ has will also be fast, and the same for the other case (slow case). Then we have

$$ I(t)\propto|A(\omega_0)|^2|M(\omega_0,t)|^2 $$

where the time dependence of $M$ can be anything you want, for example, Gaussian, rectangular pulse, etc.. Then, as the intensity of the laser field is proportional to the number of photons,

$$ I(t) = \int d\omega I(\omega)\hbar\omega \hat{a}^{\dagger}(\omega)\hat{a}(\omega)e^{-i\omega t} $$

where $\omega$ is the angular frequency distribution of the dipole radiation from the blocking material and $\hat{a}$ is photon creation operator. Thus

$$ \int d\omega I(\omega)\hbar\omega \hat{a}^{\dagger}(\omega)\hat{a}(\omega)e^{-i\omega t} \propto |A(\omega_0)|^2|M(\omega_0,t)|^2. $$

If I rewrite this by taking Fourier transformation of the right hands, then I get

$$ \int d\omega I(\omega)\hbar\omega \hat{a}^{\dagger}(\omega)\hat{a}(\omega)e^{-i\omega t} \propto \int d\omega |M(\omega_0,w)|^2e^{-i\omega t}, $$

where M(\omega_0,w) is the Fourier transformation of $M(\omega_0,t)$, essentially the Fourier transformation of the blocking time function (Gaussian, rectangular pulse, etc..). Thus we get finally

$$ I(\omega)\propto|M(\omega_0,w)|^2. $$

Because our total electromagnetic field can be written as

$$ A_{total}(t) = A_{Laser}(t) + A_{Blocking Dipole}(t), $$

and as $A_{Blocking Dipole} = A_{Laser}(t)(G(t)-1)$ behind the blocking material, where $G(t)$ is the temporal gate function that you open and close the blocking, the transmitted wave (pulse) can be written as

$$ A_{transmitted} = G(t)\times A_{Laser}(t). $$

Therefore, the dipole radiation from the blocking material has a mode that does not exist in the monochromatic laser field in this way. This means that the transmitted monochromatic wave after putting away the blocking material will have modes that do not exist in the monochromatic laser field in this way.

If you need full interaction Hamiltonian and expression for the dipole transition cross-section, I will provide them for you and please leave a comment.

This is the answer using Lorentz transformation and quantum fields.

Alright. I misunderstood the real question you are asking for. Let me define the monochromatic laser beam $A$ you have, first.

$$ A(t,\vec{r}) = \int dk^3 U(\vec{k})\hat{a}^{\dagger}(\omega_0,\vec{k})e^{-i\omega_0 t + i\vec{k}\cdot\vec{r}}, $$

where $U$ is the Fourier transform of the spatial distribution of the laser beam (If you have a plane wave, then you cannot imagine such a mind experiment because you will observe the laser field all the time at any place), $\hat{a}^{\dagger}(\omega_0,\vec{k})$ is a photon creation operator applying to vacuum state, $\omega_0$ is the temporal angular frequency of the monochromatic laser beam and $\vec{k}$ is the spatial wave vector of the laser beam.

Assume that you are moving toward the transverse direction of the laser propagation axis. Let us say z-axis is the laser propagation axis and x-axis is the axis that your moving frame moves with the speed of $v$. Since you are in the moving frame, the laser you observe $A'(t',r')$ is then

$$ A'(t',\vec{r}') = A\left(\gamma(t'+vx'/c^2),\gamma(x'+vt'),y',z'\right), $$

where $\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$ and

$$ A'(t',\vec{r}') = \int dk^3 U(\vec{k})\hat{a}^{\dagger}(\omega_0,\vec{k})e^{-i\omega_0 \gamma(t'+vx/c^2) + ik_x\gamma(x'+vt') + ik_yy+ik_zz}. $$

That is, the integration over $k$ now affects $\omega$ because you have to integrate $-ik_xvt$ term where $k_xv$ works as a new mode which is not necessarily $\omega_0$. Therefore,

$$ A'(t',\vec{r}') = \int dk^3 U(\vec{k})\hat{a}^{\dagger}(\omega_0,\vec{k})e^{-i\gamma(\omega_0 - k_x v)t' + i\gamma(k_x-\omega_0v/c^2)x' + ik_yy+ik_zz}. $$

Up to this point, you observe that the $\gamma k_xv$ works modification of $\omega_0$ which can "create" modes other than $\omega_0$. There is no interaction at all. You simply transformed the frame into this and that. Let us show you further, assuming $U$ is identically zero along the z direction other than $\omega_0/c$ and a Gaussian distribution in space that

$$ U(k_x,k_y,k_z) = exp(-Q(k_x^2+k_y^2))\delta(k_z-\omega_0/c) $$

and then

$$ A'(t',\vec{r}') = e^{iz\omega_0/c}\int dk_xdk_y exp(-Q(k_x^2+k_y^2))dk^3 U(\vec{k})\hat{a}^{\dagger}(\omega_0,\vec{k})e^{-i\omega_0 \gamma(t'+vx'/c^2) + ik_x\gamma(x'+vt') + ik_yy}. $$

After integrating Gaussian functions, we finally get (ignoring creation operator for a moment to see the envelope)

$$ A'(t',\vec{r}') \sim \frac{\pi}{Q}e^{\gamma^2(x'+vt')^2/Q}. $$

Thus, the beam observed in your moving frame has an envelope with a finite temporal duration of $\frac{\sqrt{Q}}{\gamma v}$. Therefore, simply moving the frame, you observe or "create" new modes other than $\omega_0$. There is no interaction, but this is just a matter of boosting an inertial frame. Fourier transformation is not "Classical" but "Mathematical".

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Jan 12 at 1:11
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Consider you have a monochromatic laser of frequency ω0. Consider it is infinitely small in width (see it as a line).

An infinitely small line width source has to be stationary from negative infinity to positive infinity. Yours isn't (you have it change at t=0), so it must contain additional frequencies.

But this doesn't make sense. Indeed, I only rotated my laser. There were absolutely no interaction occuring between my photon and anything else (I assume everything is in vacuum).

Rotation implies motion, so you are adding new frequencies through Doppler shifting and probably many other mechanisms.

Let's consider photons. Here if we interpret each mode as occupied by photon,

Individual frequencies do not map to individual photons, so this is not a valid interpretation.

The real answer here is that your source was never monochromatic, so when you measure it's spectrum you correctly record that it has more than one frequency.

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  • $\begingroup$ Hello, thank you for your answer. About your first point I agree but as I said for me it is only a mathematical description, it doesn't mean photons at those frequencies are actually excited. For your second point: doppler effect is a shift in frequency, it could eventually change the frequency $\omega_0$ but it would be a shiffting only. Furthermore one can imagine that laser is always turned on and the measurement device moves at constant speed in orthogonal direction to the beam. Doing so he won't see doppler effect. $\endgroup$ – StarBucK Jan 11 at 14:25
  • $\begingroup$ And for the final point, if the laser has been created with for example hydrogen transition, it cannot by itself create photons in other modes than this one without involving some non linear effect after the emission which are by construction of this mind experiment not present. (I assume for simplicity that the hydrogen laser is perfect so that it only emits at a single frequency) $\endgroup$ – StarBucK Jan 11 at 14:26
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    $\begingroup$ @StarBucK you have some misconceptions about how lasers work. The photons emitted by transitions in hydrogen for example span many GHz due to a variety of effects. Look up Doppler broadening. In addition, moving your laser really will broaden it slightly. $\endgroup$ – user1850479 Jan 11 at 15:16
  • $\begingroup$ But in quantum optic experiment, the light emitted by a laser during a finite time $T$ (thus a "monochromatic pulse") is modelled by a coherent state having $|\alpha|=\sqrt{\frac{P*T}{\hbar \omega_0}}$. This coherent state is single mode. Also, forget about moving the laser but reason in a frame moving in orthogonal direction to the laser beam (I edited my message to switch to this experiment). In this case there would not have doppler effect. And doppler effect is a shift in frequency, it cannot from a single frequency create a continuum (from my understanding) $\endgroup$ – StarBucK Jan 11 at 15:19
  • $\begingroup$ Light emitted during a finite interval has a continuum of frequencies (see Fourier uncertainty). In addition, molecules have continuously distributed velocities, so the Doppler shifts are continuously distributed, and thus the spectrum of a hydrogen atom is continuous. Spectra of gas molecules is a complex topic, ask a new question if you want to know more. $\endgroup$ – user1850479 Jan 11 at 15:25
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Consider you have a monochromatic laser of frequency $ω_0$, for example made with hydrogen transition. Consider the beam has a given width $w$.

The contradiction begins right here. Even if your device emits light from $t=-\infty$ to $t=\infty$, a finite beam width precludes monochromaticity. Especially so if we consider that your measured field abruptly changes from $E\equiv0$ to $E=\mathrm{var}$ as you move through the beam.

So your additional harmonics are present in the radiation field of the laser itself, and the laser isn't monochromatic.

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When the field detector is not in the beampath it is not sensitive to the field of the beam in the beampath. It is sensitive to the field in the mode next to the beam path which is zero in this experiment.

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