# Does a modulated electromagnetic wave contain photons of different energy?

1. Aug 23, 2011

### sriecewit

Hi Guys,

I have a doubt. When we think of a modulated wave, consider frequency modulation for example, does the wave contain photos of gradually changing energy (or frequency)?

2. Aug 23, 2011

### Staff: Mentor

Are you referring to something like FM radio? If so, I believe that is correct.

3. Aug 23, 2011

### sriecewit

Yep. I'm reffering to something like an FM radio. But, trying to extend the context to something like binary modulation, as in BPSK, how can one explain what kind of photos then wave has?

BPSK signal has photons of constant energy in between until the carrier wave encounters a phase shift. When it hits a phase shift of 180 degree, what kind of photons will it contain. I mean, it can be explained by wave theory well rather than quantum mechanics I believe.

I'm stuck at this right now :-(

4. Aug 23, 2011

### Delta Kilo

Well, E=hv, so you just look at the BPSK spectrum and that would be the distribution of your photon energies.

5. Aug 23, 2011

### Staff: Mentor

I'm not familiar enough with modulation and brodcasting techniques to really say for sure. I believe the photons are all of the same frequency still, just the phase of the carrier wave changes.

6. Aug 23, 2011

### Delta Kilo

No, this is incorrect. Any kind of modulation causes frequency spread. Here is a typical BPSK spectrum plot

7. Aug 23, 2011

### Staff: Mentor

So a phase shift requires a frequency shift to accomplish this?

8. Aug 23, 2011

### Delta Kilo

It is not so much as frequency shift but frequency spread as in continuous range of frequencies. At the moment when phase shift occurs you waveform is not a sine wave anymore but something different. Therefore its Fourier transform (that is frequency spectrum) is not a delta-function at single frequency but a distrinbution of frequencies.

Example : consider amplitude modulation with carrier freq Fc, modulation freq Fm, modulation depth k:
$A(t) = [ 1 + k \sin (2\pi F_m t)] \sin (2\pi F_c t)$
which can be re-written using trig identities as
$A(t) = \sin (2 \pi F_c t) + \frac{k}{2} \sin (2\pi (F_c-F_m) t + \frac{\pi}{2}) +\frac{k}{2} \sin (2\pi (F_c+F_m) t - \frac{\pi}{2})$
So it s exactly the same as unmodulated sine wave with frequency Fc plus two sidebands at Fc+Fm and Fc-Fm. The information is actually carried in those sidebands, not in the center frequency band. In fact the center frequency and one of the sidebands can be discarded altogether (SSB modulation).

9. Aug 24, 2011

### f95toli

One of the primary uses of phase modulators is to add sidebands to a carrier*. E.g. if you send a carrier with frequency fc through a phase modulator driven at fm; the spectrum of the outgoing signal will contain fc, fc+fm and fc-fm (as well as fc+2fm etc, but the amplitude of those tend to be much smaller).

This will of course be true for all forms of modulation schemes that use phase modulators, including bpsk etc.

And yes, the answer is that the signal will contain photons with those frequencies.

*If is often better to use PM than FM to add sidebands.

10. Aug 24, 2011

### haael

I think the main problem here is the wave-particle duality. You consider some wave-like phenomenon and ask what happens to the particles at the same time.

11. Aug 24, 2011

### G01

It's not really a problem. As mentioned above, your photon energy distribution is just given by the spectrum (Fourier transform) of your signal scaled by $\hbar$.

12. Aug 25, 2011

### Zarqon

There is usually no problem scaling down the ideas to the single quantum level, as long as you look at the probability distribution instead. So if the light spectrum at large intensities had a particular energy/frequency distribution, then the corresponding situation for a single photon is a probability distribution for that photon to be measured with a particular energy.