# I have a very basic question about wave interference and more

1. Dec 11, 2011

### amir11

I have a very basic question about wave interference and more specifically electromagnetic wave interference.
Suppose two beams of light each with amplitude A interfere with the same phase. the amplitude of the resulting beam is 2A. the two beams at the start have powers p= $A^{2}$ each. The resulting beam, on the other hand, has a power of 4$A^{2}$!
Which part of this conclusion is wrong?

2. Dec 11, 2011

### DrGreg

Re: interfernce

Two interfering beams of light cannot be perfectly parallel to each other. Therefore the waves cannot constructively interfere everywhere. In some places they are in phase and interfere constructively, with amplitude 2A and power 4A2. In other places they are out of phase and interfere destructively, with amplitude 0 and power 0. When you average out the power everywhere, it comes to 2A2.

3. Dec 12, 2011

### amir11

Re: interfernce

Thx for your reply. But This is not a good answer. In theory you can assume that the two beams are sumed perfectly and have the same phase everywhere.

4. Dec 12, 2011

### Cthugha

Re: interfernce

Whether or not you like the answer given, it is the correct one. By the way the answer to the more frequently asked question "Isn't it a violation of conservation of energy when two fields of same frequency and amplitude completely cancel each other?" is the same.
There is no way - not even theoretically - to create such a superposition state of two waves being completely in (or out of ) phase with each other over some distance unless they are completely in (or out of) phase everywhere. So if they are in phase everywhere, you directly create a beam with amplitude 2A and thus conserve energy. Any attempt to merge two beams of amplitude A (via beam splitters or any other geometry) automatically creates regions of constructive and destructive interference such that energy is conserved. This also works the other way round. Conservation of energy is also the reason why it is impossible to create a beam splitter where the light fields at both exit ports have the same amplitude and phase. Any symmetric beam splitter must necessarily introduce a phase shift of Pi between the fields exiting via the different output ports.

5. Dec 12, 2011

### Staff: Mentor

Re: interfernce

The plane waves that we often talk about are only idealizations. Real electromagnetic waves are produced by finite-sized sources, and the waves always have some curved configuration in three-dimensional space. The only way you could get two waves to interfere completely constructively or destructively everywhere would be to have two identical sources (oscillating either in or out of phase with each other) in exactly the same location.

6. Dec 15, 2011

### amir11

Re: interfernce

you are simply trying to persuade me that 2+2 is not 4!!! This is the most basic question to ask in theory and the fact that you give such a silly answer to it, you are simply saying that physics has no answer it, is really unacceptable, I urge you no to answer question you don't know the answer to!!!

7. Dec 15, 2011

### Cthugha

Re: interfernce

1) I did not say that physics has no answer to it. I said that the situation you mention is not realizable, not even in theory and therefore unphysical. The correct answer has been given to you several times in this thread: there is no way to create constructive interference without having destructive interference elsewhere, not even in theory. Period. This is a basic accepted result you can find in many books. See e.g. the treatment of the 50/50 beam splitter in the Mandel/Wolf.

2) Insulting people who take the time to address your question and explain the situation to you in detail and throwing around with exclamation marks not a good practice. I doubt you will get a lot of answers this way.

8. Dec 15, 2011

### amir11

Re: interfernce

Ooops. this happened to me once before in this forum too. I aksed a question about CI in quantum dots. A guy popped out of nowhere saying that this is not possible and it is very so and so and I have no understanding of the theory of CI and wasted my time for a while. But later another guy came and sent me some articles about the issue and it turned out that CI was a somehow common method!!!

Regarding impossibility of the perfect interference. Can you tell me what do you expect to find in waveguide number 2 of the below image if I lunch two waves of the same amplitude but 180 out of phase in waveguides 1 and 3?

waveguide1 ========
waveguide2 ===========
waveguide3 ========

assume that the overlap distanse is so that the 1 and 3 compeletly couple to central waveguide!

9. Dec 15, 2011

### Cthugha

Re: interfernce

Funny, some time ago I wrote a post in these forums in answer to a guy who was looking for configuration interaction methods in QDs and linked some papers from our theoreticians.

That guy was told before that configuration interaction is not a suitable method for QDs which was, however, only based on the misunderstanding that it was assumed that CI should be used for the whole story, while in QDs you start with the bands and consider the pssible electron/hole configurations that can occur using CI.

To be honest, I am not firm enough concerning waveguides that I could ad hoc write down the phase relationships and calculate the situation. Most of the cases where interference occurs can be traced back or projected on some kind of beam splitter problem. Is that possible here, too?

There are lots of papers on ideal beam splitters and it is fantastic how nontrivial such a simple element is. Some literature on that is e.g.
U. Leonhardt, "Quantum statistics of a lossless beam splitter: SU(2) symmetry in phase space", Phys. Rev. A 48, 3265–3277 (1993) or
A. Zeilinger, "General properties of lossless beam splitters in interferometry", American Journal of Physics (1981) Volume: 49, Issue: 9, Publisher: AAPT, Pages: 882-883.

There is also a good didactic paper on these issues, but I am out of office and cannot find the citation right now.

Again, I am not sure how to interpret your scenario. What exactly do you mean by "they completely couple to the central waveguide". Does it act as if 1 and 3 are not having guide character on the inside (100% transmittivity) or do you assume 50% transmission or something completely else?

10. Dec 15, 2011

### sophiecentaur

Re: interfernce

"In theory" you cannot say that at all (there is no theory to support that idea). Two "beams" will have edges to them and because of the spatial extent of them, there will be diffraction. The vector addition at the edges of the beams will give a resulting beam that is narrower than either of the individual beams. Total Energy is conserved so the beam centre may have four times the power but there is less power at the periphery.

Even if you want to modify your original model and say that you have two coincident point sources then you still can't be correct for two reasons. Any coherent em radiator will not be isotropic and they will still interfere with each other to satisfy energy conservation. Also, to use an RF transmitting approach to prove the point, if you were to try to generate these two outward flowing powers from two coincident radiators, the impedance that each would 'see' would constitute the impedance of the free space around it in parallel with the impedance of the other. To maintain the volts at its drive point, it would need to supply twice as much current - resulting in four times as much power as just one isolated source but only twice as much power as two sources under those conditions. Still no paradox.

11. Dec 15, 2011

### amir11

Re: interfernce

OK so sorry for my previous response but that is due to a previous experience. the scenario I am putting forward is some how different form a simple beam splitter, in that a beam splitter is essential a four port element, two input two out but and it is obvious if you write down the governing eqs that if beam interfere positively in one of the ports they will interferometer negatively in the other port. The element I am proposing is have three ports, that is two waveguides that evancently couple to a central wave guide(waveguide2) this device is then three port device. that is two waveguides 1 and 3 compeletly couple their light to the central one.
or you can also think of the image below( machzender interferometer). what happens if the two arms are 180 out of phase?

Actually this issue has taken me a lot of time I know the answer with the fact that if you have constructive interference here then you get destructive somewhere else but it realy dose not apply here dose it?

:D Actually I found that post, That was you. here is it.

Last edited by a moderator: May 8, 2017
12. Dec 15, 2011

### amir11

Re: interfernce

In no sense do I mean to say that conservation law is wrong, That is definitely right. In very simple words what I mean is that superposition principle for electromagnetic fields is NOT consistent with energy conservation law! That is if you double the amplitude of a field you get for times the power. but you can superpose to fields and get 2 times higher amplitude but definitly two times the power not four times!

13. Dec 16, 2011

### Cthugha

Re: interfernce

Oh, I see. I am no expert in terms of waveguides, but here is the stuff I know. Any real implementation of such waveguide-based MZ-interferometers uses four-port directional couplers instead of three-port ones. The group of Jeremy O'Brien over at Bristol does that excessively for example (see e.g. "Manipulation of multiphoton entanglement in waveguide quantum circuits" by J. Matthews et al., Nature Photonics 3, 346 - 350 (2009)).

Now the question of course is why three-port devices are never used. Wikipedia is not really a good resource, but has the following to say about the topic: "These suffer from very poor isolation between the output ports – a large part of the power reflected back from port 2 finds it way into port 3. It can be shown that it is not theoretically possible to simultaneously match all three ports of a passive, lossless three-port and poor isolation is unavoidable." The following book is cited, but I do not have access to it and cannot say whether it includes an in-depth study of the problem: Antti V. Räisänen, Arto Lehto, Radio engineering for wireless communication and sensor applications, Artech House, 2003.

My guess is really that the problem is similar to the problem that creating an ideal three-port beamsplitter is not possible and the crosstalk between the output ports will automatically create a situation in which energy is conserved. However, I am not aware of any experimental or theoretical study of crosstalk properties of three-port directional couplers, but I am sure there must be some out there.

Indeed, the superposition principle holds only for linear systems. Two independent waves that are superposed somewhere form in good approximation a linear system. Two waves that are indeed completely identical throughout all space and time are indeed not a linear system at all.

14. Dec 16, 2011

### sophiecentaur

Re: interfernce

Of course superposition only works with a linear system but that isn't the issue I think.

I think the key to the apparent paradox in the waveguide splitting scenario is that half the energy goes into each guide but for this to happen, the resultant impedances of the two branches must be identical to that of the input guide. I'd rather treat this as a transmission line problem because it's easier yet a totally equivalent situation. For transmission lines, at the split, the voltages would be the same but the currents would need to be half - no violation of energy conservation here because the impedances HAVE to be double the input line impedance for this to work. Same thing applies where the lines rejoin.

If you don't get the impedances matched then you will get standing waves and the situation is not comparable with the free space 'two beam' model. Still, the power flow will be such that half of the transmitter power goes down each leg. The existence of the standing wave can be regarded as being actually due to necessary energy conservation at the interface (the power that doesn't get into the branches goes back into the transmitter and is dissipated).

15. Dec 16, 2011

### sophiecentaur

Re: interfernce

I have a problem with that statement. You can regard any wave as the superposition of two waves (or any number of waves, for that matter). It's just that you need to split / resolve both E and H fields appropriately so that each has equal power - not just 'half the E field'.

Linearity seems to be a 'convenient' red herring (avoiding actual details). Free space is linear, is it not? I can't see how total coherence would constitute the one and only special case of when it's not. I think that the essence of the explanation must involve considering Impedance (as in the waveguide scenario). It may be easier to think in terms of Admittance, actually. If you split the wave into two, then each is 'sharing' the space and each one will have half the total admittance. So they will each have identical E fields but half of the H field. OK, I can hear you saying "Why choose to keep the E fields the same and reduce the H fields?" I haven't an answer to that yet! If my argument involved free space Impedance then the H fields would be equal and the E fields would be halved.

16. Dec 16, 2011

### sophiecentaur

Re: interfernce

Further to the impedance idea. Treat the two waves as having been produced by two dipoles and consider the waves in a plane in a direction normal to the two dipoles (at a radiation pattern max), to make it easier.
Imagine you had a dipole, being powered at its feed point and radiating em waves. If it is a half wave dipole, its input impedance will be about 72Ω. Bring another dipole near it. The two dipoles will have a Mutual Impedance. There is a voltage at the feedpoint of one due to the current in the other times this mutual impedance. The act of bringing the other driven dipole in is to add this mutual impedance in series with the existing self impedance. The value of mutual impedance is about 72 +j45Ω. This will reduce the current in the dipole - and, by symmetry, will also reduce the current in the other dipole too. The two dipoles will have the original drive voltage on them and half the original current, producing two waves (at infinity, away from the near field) with half the power each. This is in the limit as the spacing between dipoles approaches zero. The waves will be co-phased.

p.s. I'm a bit fuzzy about the imaginary term in the mutual impedance but I think it will come out in the wash if you do it formally. Any opinions about that, anyone?

17. Dec 17, 2011

### Cthugha

Re: interfernce

Ok, let me expand on that. My point was rather that infinitely long waves popping up somewhere are an approximation anyway as that would break conservation of energy. When considering simple interference effects you throw the sources, usually driven dipoles or oscillating circuits, away. That is ok, as they do not matter much in such experiments and you assume that everyone knows how these work. However, you can only do so as long as the source of wave A and wave B are far from each other and wave A never sees source B and vice versa. If both waves fully overlap everywhere the situation is different. Imagine the simple situation where source A and B are placed along the direction the waves of same frequency and phase travel, but one wavelength apart. Now you have to consider the driven dipole sources and the effect of wave A driving dipole B and wave B driving dipole A which spoils the possibility of simple superposition.

So, yes. I would agree that free space is linear, but that is not the problem here. By intuition I would say you get a solution similar to the problem of having two coupled oscillators undergoing forced motion.

18. Dec 17, 2011

### sophiecentaur

Re: interfernce

I would agree that the inverse square law will always apply, meaning that power will always be attenuated at 6dB each time the distance is doubled - but how is this relevant? You are suggesting a discontinuity for the special case of two concurrent waves. But I have already pointed out that one field can always be resolved into two equal components. Are you suggesting that a difference of 1 second of arc, in angle of arrival, of these two waves would have to alter things so much?

There is no spoiling of superposition of the waves from two dipoles - all that happens is that the impedances change as they get closer together. So each source will launch a different amount of power. That's nothing to do with linearity - it's just to do with impedance and reflected power. I don't see how you can think that vectors won't always behave themselves properly. The problem with your model is in the assumption that the waves would be 'equal' to what they would have been if they had existed, individually, on their own. My point is that two such waves could not be generated or even re-directed by reflection or refraction without affecting and modifying each other.

19. Dec 17, 2011

### Cthugha

Re: interfernce

Huh? Wait, I never said that. I never said that they would be equal to the "individual" situation. I said you have to replace the "bare" waves by the solution of the full problem of two coupled driven dipoles and the interaction of the waves with the other dipole into account which is imho not different from the conclusion you arrive at. I thought it would be trivial that two coupled driven dipoles affect each other and give more complicated outputs.

Last edited: Dec 17, 2011
20. Dec 17, 2011

### sophiecentaur

Re: interfernce

OK, fair enough. There has been a bit of misunderstand here -(not for the first time on these fora haha) and I'm not even sure with whom I am taking any particular issue. But there seems to be a feeling that there's a paradox somewhere and there just isn't. Taking one region where two waves are interfering and producing the 2 times / 4 times paradox is to ignore what is going on elsewhere. When the sources are not coincident, all you need to say is that there is somewhere else in space where you get destructive interference. (That's the easy one to deal with)
In any other case (waveguides / dipoles / whatever) there will either be standing waves which imply that power is going in a direction other than 'forward' OR the impedances will have been modified so that the actual forward power of each wave is still half of the resultant.

Trying to visualise arrangements where this is not the case is a bit like devising perpetual motion machines - they look almost believable but there's always something the designer forgot, other than just the general principle that it can't be done. Analysing the situation correctly will always give a good reason why there's no violation of any principle.