1
$\begingroup$

In the following, we assume that the polarization is aligned such that the scalar treatment of the electric field is justified. Furthermore, we limit the discussion to a fixed coordinate $x=0$ to drop the wave vector dependency.

Classical description

Let's consider a classical electric field

$$ E(t)=A(t)\cos\omega_c t \tag{1} $$

with angular carrier frequency $\omega_c$ where we modulate the field amplitude with

$$ A(t)=A_0\cos\omega_m t \tag{2} $$

where $\omega_m$ is the angular modulation frequency. For illustration, we assume $\omega_m\ll\omega_c$, for example, $\omega_c$ could be in the optical whereas $\omega_m$ could be in the low-frequency domain.

In this picture, we can think of the electric wave propagating with frequency $\omega_c$ while it's amplitude slowly oscillates with $\omega_m$.

Nevertheless, we can also combine eq. (1) with eq. (2) and write

$$ E(t) = A_0\cos\omega_m t\cos\omega_c t = \frac{1}{2}A_0\bigl(\cos\omega_+t+\cos\omega_-t\bigr) \tag{3} $$

where we define $\omega_\pm=\omega_c\pm\omega_m$.

In this picture, we actually have two waves, one oscillating rapidly with $\omega_+$ and one oscillating slowly with $\omega_-$.

So far, so good. Although, eq. (1) and eq. (3) suggest a different point of view both are equivalent and should give the same (classical) predictions.

Quantum description

Now, we turn to the quantum description where we define the electrical field operator to be

$$ \hat{E} = i\sum_i\omega_i\left\{\hat{a}_ie^{-i\omega_it}-\text{c.c}\right\} \tag{4} $$

with $\text{c.c.}$ referring to the complex conjugate term, and where $\hat{a}_i$ is the annihilation operator of mode $i$ that satisfies the canonical commutation relation

$$ \left[\hat{a}_i,\hat{a}_j^\dagger\right]=\delta_{ij} \tag{5}. $$

We assume a two-mode coherent state $\vert\alpha_1,\alpha_2\rangle$ with $\alpha_i\in\mathbb{C}$ and calculate the expectation value of the electric field operator for that state

$$ \begin{aligned} \langle\alpha_1,\alpha_2\vert\hat{E}\vert\alpha_1,\alpha_2\rangle &= i\omega_1\left(\alpha_1e^{-i\omega_1t}-\text{c.c.}\right)+ i\omega_2\left(\alpha_2e^{-i\omega_2t}-\text{c.c.}\right)\\ &= -2\omega_1\operatorname{Im}\left\{\alpha_1e^{-i\omega_1t}\right\}+ -2\omega_2\operatorname{Im}\left\{\alpha_2e^{-i\omega_2t}\right\} \end{aligned}\tag{6}. $$

With the choice $\omega_1=\omega_+$ and $\omega_2=\omega_-$ as well as $\alpha_i=-iA_0/(4\omega_i)$ (upto some unit-preserving factors) we can recover our classical result given in eq. (4).

On the other hand, we should also be able to reproduce eq. (1) by asserting a single-mode coherent state $\vert\alpha(t)\rangle$ with $\alpha(t)\propto A_0\cos(\omega_mt)/\omega$.

However, this time, we could think of an experiment that distinguishes between these two states!

Recap the photoelectric effect describes electron emission by a photon hitting a metal. The photon energy $\hbar\omega$ needs to be greater than the work potential $W$ which binds the electron to the metal. Suprisingly, the photoelectric effect is (neglecting multi-photon absorption) independent of the intensity.

Let's assume that we are in possession of a metamaterial where the work potential is tailored to $\hbar\omega_c$. In this case, we could distinguish between the single- and two-mode coherent state because the amplitude $\vert\alpha\vert$ of the coherent state $\vert\alpha\rangle$ fixes the mean of the (Poissonian) photon statistics but the energy, which determines if electron emission occurs, is given by the mode frequency.

If we remind ourselves on how we derived eq. (3) from modes in a confined cavity this also makes sense.

Is it correct to conclude that amplitude modulation of an optical laser signal, shifts the energy of the participating photons?

$\endgroup$
1
$\begingroup$

This is an answer to your classical description of an amplitude modulation of EM wave.

Let us start with a few basic observations:

  • Most of EM radiation are without modulation. Examples are the radiation from thermic sources (light bulb, sun, radiant heater).
  • EM radiation is caused by the relaxation of electrons (and other particles and quasi-particles ) and the emission of photons this time. Another source of EM radiation does not exist. Therefor every EM radiation consists of photons. EM radiation is a stream of photons.
  • Every photon has an electric and a magnetic field component and these field components are oscillation during the movement of the photon through space.

Only with two conditions you will be able to measure wave properties on a stream of photons:

  • the photons are polarized (their electric and magnetic field components are polarized)
  • the radiation is modulated (the best example is the antenna of a communication device)

Your equation (1) describes a modulated EM wave. To be precise, it is a radio wave, with its periodically changing number of emitted photons. Remember, that a wave generator pushes the electrons in the antenna rod fourth and back and the accelerated electrons emit the photons. The electrons get accelerated at any moment all in the same direction and this makes the polarization of the emitted photons.

Equation (2) describes the amplitude modulation of a communication. Clearly the number of photons changes.

One last thing. The photons emitted by the accelerated electrons on the rod have a different wavelength than the carrier frequency. In dependency from the power of the wave generator, the length of the rod, the material of the rod, ... the photons are in the range from infrared to X-rays. It is not for nothing that you should never stand in front of a military radar. However, equation (3) describes the number of photons (in the equivalent of the intensity, named unfortunately amplitude) under the influence of a carrier frequency and a modulation of this frequency.

$\endgroup$
1
  • $\begingroup$ I think it a bit problematic to talk about "photons" in a classical picture. You are correct that in the semiclassical picture, the number of photons corresponds to the amplitude. But then again, according to the quantization procedure of the electric field, clearly the photon frequency changes while the number of photons splits evenly across the two frequency modes? $\endgroup$
    – bodokaiser
    Nov 15 '20 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.