# About quantization of EM energy for amplitude-modulated wave

Consider an electromagnetic monochromatic plane wave at frequency $f$ and E-field amplitude $E_0$ in vacuum. From the quantum electrodynamics point of view, we could say that the energy that EM wave is carrying is transmitted in form of discrete packets (photons), each one with energy $E = hf$. The time-average power of the EM wave for unit of surface is $S = 0.5E_{0}^2/\eta_0$ (with $\eta_0$ the vacuum impedance) and must agree with the time-average amount of photons crossing an unity surface (we could average in a long time to neglect quantum noise). If we, for example, decrease by a half the amplitude $E_0$, the time-average amount of photons crossing an unity surface will decrease by $1/4$, but they will still be photons of frequency $f$.

Now consider that for example, $f$ is in the optical spectrum, and we modulate the amplitude of such EM wave with a sinusoidal signal at a very low frequency $f_\ell$ (let's say 1 Hz). My question is, what is actually happening?:

1) Are only $f$ photons crossing an unity surface and the time-average amount of them is changing sinusoidally at 1 Hz rate?,

or

2) Are $f$ and $f \pm f_\ell$ photons crossing all the time?

I believe the answer must be 2), because the Fourier analysis of this AM modulation lead to additional frequency components at $f \pm f_\ell$. However the fact that 'dimming' light at a frequency $f_\ell$ produces additional photons with different energies, is counter-intuitive for me.

I am puzzled that you think that the answer is 2). Whence will the new photons come? There exist optical modulators:

. Often the easiest way to obtain modulation of intensity of a light beam, is to modulate the current driving the light source, e.g. a laser diode.

The laser will still have its characteristic frequency, at less intensity and no source of the modulation frequency with new photons.

Electromagnetic waves can be modulated when one has access to the electric (or magnetic) field that is building up the wave, and the modulation is imposed in coherence with known phases, between the plane wave and the modulating one. In the case of a laser the plane wave is built up by the coherent de- excitation (stimulated emission) of the atoms making up the laser . There is no way to modulate coherently this by modulating the input current.

More complex modulation than the amplitude one may introduce new photon frequencies but it will need a specific example to know if it does or not.

• Yes, that is why I have a physical interpretation problem. But after the modulation takes place, the resulting signal has frequency components at $f \pm f_\ell$ according to Fourier transform. – Gabriel Santamaria Dec 9 '15 at 11:46
• I think there is not discrepancy as this EM signal can be produced by an antenna as you say, placed in vacuum. That is why I go with answer 2), although is hard for me to physically understand why photons with different energy arise. In the case of the modulated laser light, the change of amplitude, in some way must generate different energy photons, although this is not intuitive. – Gabriel Santamaria Dec 9 '15 at 13:48
• with a laser, instantaneously plane waves are leaving the laser. The photons are not generated by a changing electric field but by a fixed quantum transition that cannot be modulated as in the function of the link. – anna v Dec 9 '15 at 13:58
• Yes, but thinking only in the signal, we can consider the EM field in the far zone of the antenna which is locally plane. This EM field in the far zone must carry $hf$, and $hf \pm hf_\ell$ energy photons. And I think this is valid for a modulated laser too. In general, in my opinion, any 'monochromatic' carrier modulated in amplitude must have these 3 different energy photons. – Gabriel Santamaria Dec 9 '15 at 14:05
• On second thoughts I think it has to do with coherence. In the radio antenna the phases are coherent when the modulation is introduced. In the laser case there can be no way that the laser plane wave could be made coherent with the modulation of the power to the laser. The laser wave arises because of quantum mechanical coherent transitions of individual atoms, and these cannot be accessed by the modulation of the input currents. – anna v Dec 9 '15 at 15:14

According to classical theory of radiation, if you modulate EM emission so that its amplitude is varied with separate frequency, the resulting EM field will indeed have Fourier spectrum with at least three separate peaks and in practice also some low intensity background in between. Also if you modulate only average intensity of the radiation (proportional to $E^2$), the resulting EM field will acquire different spectrum where the frequency of this variation will make its mark by satellite peaks.

Emission of masers/lasers can be described with classical theory extremely accurately, so this applies to pocket laser pointer whose light gets blocked periodically by some obstacle, too.

If now you want to describe this situation with quantum theory of light and talk about photons, you may try to use the standard procedure - describe the multi-harmonic classical EM field with Fock states. These states will generally be coherent states with no definite occupation numbers for the relevant modes of mentioned frequencies ( = "number of photons").

• This was helpful. I am not an expert in QED, but only in classical electromagnetism, so I will take a look to Fock states. However, could we conclude that in the case of AM-modulation of a monochromatic source (even periodically blocking a laser beam with an obstacle) different energy photons are present? This is counter-intuitive, specially in the periodic blocking scheme. – Gabriel Santamaria Dec 10 '15 at 10:39

I'm not sure to understand your statement, but you seem to be mixing up the frequency $f$ of the plane wave (which is related to the energy $E = h \nu$ of the photons) and the rate at which photons cross some control surface. I don't see how they would be related. Here is how I understand the situation : if you try to collect photons from a very low amplitude plane wave, for instance with a CCD captor, you will get one photon every once in a while, the interval of time between two detected photons being random from one event to the other. If you wait for a time $\tau$ until you get a reasonably high number of photons, say $N=1000$, you will get the following identity :

\begin{equation} P \tau \simeq N h \nu , \end{equation}

$P$ being the power oh your light source. This equality will get more and more accurate as you wait for a longer time and collect a higher number of photons ; as you will even out the randomness of photon detections.

• They are related because (for the same E-field amplitude) the greater the frequency, the lower the rate at which photons cross the control surface (due to the fact that every photon has more energy, and less of them are needed to produce the same time-average power). I agree with your following analysis, but my question is that if you dim light at a frequency $f_\ell$ (AM-modulation) is the same as having three components of light at $f$ and $f \pm f_\ell$ (according to Fourier transform). So, dimming light, apparently produces different frequency photons which is counter-intuitive. – Gabriel Santamaria Dec 9 '15 at 16:57