# Monochromatic waves

1. Jan 26, 2012

### Niles

Hi

As the word suggests, a monochromatic wave has a single frequency. However, usually it also said in various books that it has a constant amplitude and phase as well, and thus it is coherent. I don't quite see why a monochromatic wave *has* to have a constant phase and amplitude? I can easily imagine an infinite wavetrain varying in both amplitude and phase (not necessarily deterministically) and still being monochromatic.

Best,
Niles.

Last edited: Jan 26, 2012
2. Jan 26, 2012

### torquil

The frequency content of a signal is defined by its Fourier transform. If the amplitude or phase is dependent on time, the Fourier transform will be nonzero at several frequencies, hence the signal is not monochromatic.

3. Jan 26, 2012

### sophiecentaur

Monochromatic means only a single frequency and this implies that the phase does not change (or the frequency would be changing about the nominal value). If you change the phase, you are introducing a change in frequency (frequency is dϕ/dt) so ot's no longer monochromatic.
If the amplitude changes you are Amplitude Modulating it. This means that you will be splitting some of the unmodulated wave power into 'sidebands'.

You can take this to an extreme and say that a truly monochromatic wave cannot exist because it would have to extend over all time (turning it on and off, constitutes modulation which means sidebands.)

So when you modulate either the phase or amplitude, some of the wave

4. Jan 26, 2012

### Niles

Thanks for the good replies. That was kind of you.

Best,
Niles.

5. Mar 3, 2012

### Niles

I have been thinking about that statement for a little while now. I assume you mean that ϕ = ωt + θ is the instantaneous phase, and dϕ/dt is the instantaneous frequency. However, when I have an EM-wave, is the frequency of it given by ω or dϕ/dt? I have to admit that I am not 100% sure about what "instantaneous frequency" physically is.

Best,
Niles.

Last edited: Mar 3, 2012
6. Mar 3, 2012

### sophiecentaur

ω is the same as dΦ/dt. It's the rate of change of phase or angular velocity.

What I was getting at is, if the value of θ is varying (there is a non zero dθ/dt), this is effectively the same as having a different value of ω in that expression
ϕ = ωt + θ
becomes
ϕ = (ω + dθ/dt)t + θ
which involves a different frequency.
If you added a steadily increasing phase in the form of a 'ramp', this would be the equivalent to a permanent frequency offset.

7. Mar 3, 2012

### Niles

So to recap, whenever I have a monochromatic wave given by e.g. Acos(ωt + θ(t)) with constant amplitude, then the angular frequency (= rate of change of phase) of the monochromatic wave is given by ω + dθ/dt, which enables the permanent frequency offset by a steadily increasing phase.

Thanks.

8. Mar 3, 2012

### sophiecentaur

Yes. Look at the Doppler effect. Frequency shift by adding or subtracting phase constantly, every cycle.

9. Mar 3, 2012

### Niles

So the instantaneous frequency ωt + θ(t) is what we "see". If I look at a plane EM-wave going from one medium n1 to another n2, then there is also a change in the phase due to the $kz$-factor.

In principle this should also change the frequency. This contradicts what I have been taught, namely that only the wavelength changes when an EM-wave propagates in a medium, and not the frequency (?).

10. Mar 3, 2012

### sophiecentaur

Why should it change the frequency? It's just a change in phase. The frequency before and after the interface will be identical. If you accept that there is phase continuity throughout, there can be no frequency change to a stationary observer. Nothing I have said implies it.
btw, why are you using the term "instantaneous frequency" in this context? It doesn't have any more meaning than plain 'frequency'. Do you really mean something like 'resultant frequency'?

11. Mar 3, 2012

### Niles

I think it is generally termed the instantaneous frequency: http://en.wikipedia.org/wiki/Instantaneous_phase#Instantaneous_frequency. I don't know why it is called that, it is a good question.

The instantaneous frequency is now $d(\omega t + kz + \theta(t))/dt$, where k = k(t) since we have an interface. It is not a constant contribution since dk/dt is a delta function, but it still changes the (instantaneous) frequency.

Best regards,
Niles.

12. Mar 3, 2012

### sophiecentaur

But k is a function of z at an interface, not t. If you take a material with a variable phase constant then you could, indeed, produce fm but that's not what you're talking about, is it?

13. Mar 3, 2012

### DrDu

In fact this whole question becomes much more complicated once you take the quantum nature of light into account. So you can have e.g. incoherent and coherent monochromatic light. Any state with a fixed number of photonsof equal frequency has a completely undefined phase while states with a well defined phase are coherent superpositions of states with different number of photons.

14. Mar 3, 2012

### Niles

You are right, k=k(z), not t. No, that wasn't what I had in mind.

Thanks to all for helping me.

Best wishes,
Niles.

15. Mar 3, 2012

### sophiecentaur

So where does a frequency change come in?

16. Mar 3, 2012

### DrDu

I don't know whether you were refering to my comment but it makes me think about how we measure frequency in quantum mechanics. I took it for granted that the energy of a photon is e=h f, but in fact this seems to be only a classical correspondence.
So even a beam of photons of equal energy - which is usually called monochromatic- will have a frequency spectrum due to quantum fluctuations.

17. Mar 3, 2012

### sophiecentaur

Sorry, I should have quoted Nile's last post because that was what I was referring to. But, to respond to your post, you give a good connection between classical and quantum when you imply that it is only when all photons are in the same state that there is truly single frequency light. My EM knowledge comes mainly from Radio, in which one assumes a monochromatic source, perturbed by frequency or amplitude noise (sort of the other way round from the quantum approach). I have a feeling that a lot of people who post about photons do not see it as clearly as you have put it. There's nothing magical about laser light - the single quantum state is largely a way of saying there is not the 'jitter' that normal light sources have - hence the annoying and disturbing grainy patterns (just like TV transmitter service areas).

18. Mar 3, 2012

### Niles

My original reasoning was that if we say that at time t=0 the wave is in medium n1, then at some other time t=t' the wave is in medium n2. Hence the wavevector of the EM-wave changes during the interval 0 to t', so dk/dt is a delta function. This I thought could introduce a frequency shift only at the interface, but I somehow think my reasoning is wrong.

What do you say?

19. Mar 3, 2012

### sophiecentaur

You've got it. There is a phase change with z not t. I think you were looking at the wave from the frame of the moving wave. But even then, the velocity changes to eliminate the effect you suggest.

20. Mar 3, 2012

### DrDu

I think the important difference between quantum mechanics and classical mechanics is the meaning of monochromatic. It's meaning is "single colour". From the theory of vision this means quanta of defined energy, but does not really say that the power spectrum consists of a delta peak at a single frequency. However, I am not sure whether this is really the convention in quantum optics.
There is an uncertainty relation between e.g. photon number and phase. In a laser the product of these uncertainties is minimal, although there are other non-classical forms of light where this product is minimal, like squeezed light.

Edit: I just looked it up. For photons of exactly one energy, the coherence length is infinite for both chaotic and coherent state radiation. So the radiation has well defined frequency in both cases. Hence in this respect, quantum radiation is not different from classical one.

Last edited: Mar 3, 2012