Hi Derek,

The above answers are all correct for the optical version of the Mach-Zehnder, and the principle that is articulated is absolutely correct, namely that the energy needs to be distributed into other degrees of freedom, typically the other output port (although there are many instances where it is implied that the beams need to be at right angles, and this is not really the case).

The issue is that sometimes in the electrical case, a Y-junction is seen as being the same as an optical beamsplitter, it is not the same and has a different transfer matrix (the Y junction is a 3 port device, whilst the spatial beamsplitter is always a 4 port device, but where one port is often closed), and I wonder if your statement "where two transmission lines join" is referring to this.

If you create something that is like a Mach-Zehnder interferometer out of two Y junctions, instead of two beamsplitters, then the energy is typically reflected when destructive interference occurs. This is like the case for an Aharanov-Bohm loop. In some instances, the interference can go into higher order modes of the transmitted loop (this is more likely the case for optical Y junctions), but also, one can consider overlap of states and indeed you could conceivably have a case where the extra energy is radiated.

Andy

The final beam splitter in a Mach-Zehnder has two outputs directed at right angles to each other. When the relative path difference in the arms is such that none of the light passes out of one of the outputs, then all of the light passes out of the other. Absorption is not involved, and so no energy is lost and no heat is generated.

What about when an MZ is formed using optical waveguide?

In optical waveguides the destructive interference for the guided mode coincides with energy transferring into cladding modes. These (higher order) modes are generally lossy so that the light dissappears on the chip away from the waveguide.

Would the profile outside the waveguide be exponentially decaying? If so, that doesn't seem to fit well with dumping lots of optical power.

Yes! it is.

I think I disagree with the previous statement although it depends a bit on the interpretation that was meant. When you solve for modes in waveguides you obtain a number of possible solution that still contain some freedom in the parameters. One of those freedoms is the sign of the (complex) propagation constant k (or beta). At that point you generally apply the condition that physically possible modes must go to zero at infinity so that you chose the solutions that have exponentially decaying profiles. These modes can still contain some loss (imaginairy part of k) but they always go to zero at infinity. The other solutions are however perfectly good descriptions of modes as well, it is just that they describe modes that run off to infinity so that the stable solution would contain an infinite amount of energy (an exponentially growing amount) at infinity. When you try to figure out what happens when you have a boundary condition given by a particular mode profile at the entrance (e.g. given by a particular interference pattern at that point), you have to match the condition at that point (phase and amplitude) to a combination guided and non-guided modes. You can subsequently generate the particular solution for your situation by calculating the propagation of the set of modes that match the boundary condition. For the non-guided modes this means that they start radiating outward from that point so that their amplitude decreases exponentially over the propagation direction whereas the guided modes will propagate in the waveguide without considerable attenuation. How fast the modes "sheds" the non-guided modes depends on the exact solutions. When the non-guided modes are far above the cutt-off they will decay faster, when they are close to the cutt-off they can actually propagate quite far and might still show up at the end of teh waveguide with a measurable amplitude. It is therefor not so obvious how much energy is dumped over what length of the chip.

The general problem of the profile of leaky modes is that they are unbounded solutions and therefor cannot be normalized in the regular way but you can use them locally to match their overlap at the waveguide and calculate form there. For intuition you can think of them as modes that have angles under the angle for total internal reflection such as the light that leaves a fiber that is bent too tightly. Often the loss can be approximated as some fixed number per unit length so that the pattern in the plane of the waveguide is something that decays exponentially along the waveguide and radiates outwards. The decay and angle of the radiation depends on the precise refractive index numbers. In the end this energy will decay somehow (for example by also radiating up and down) so that it will decay exponentially at some point but close to the waveguide I would say that it is pattern that radiates away (like an outward propagating beam).

The key of all these energy transfers and conservations is the Poynting's theorem.

This holds true and even explains the effect of two identical waves shifted by half wavelength phase "absolutely" canceling each other.

We are considering the system to be perfectly aligned, I suppose. A single fringe occupying the whole field of the interferometer. On the other hand, is hard to believe that energy goes to a lateral displacement when a longitudinal displacement of the mirror is done (the same for the Michelson interferometer). But I would like to know if somebody measured a perfect zero energy fringe in one system and a maximum fringe at the complementary, and conversely, I tried but was not sure of a perfect matching. One problem is we have reflections in two sides of a beamsplitter.

Dear Jose, I think you are comparing the wrong things. In a classic interferometer where the beams are at 90 degrees to each other and meet at a beamsplitter, you have no problem understanding that a shift of half a wavelength in one beam would cause destructive interference in one arm and make for constructive interference in the other arm. When you think of an (on-chip) interferometer, the beam-splitter seems to be "in-line" so that it is not clear which port is the "reflection arm"and which one is the "transmission arm". However, these two ports exist in much the same way as in the bulk-optics version. When you use a 2x2 splitter, the beams from each input port couple (through evansecent coupling) to the other channel and the energy "bounces" back and forth betwen the two channels. For a 50% spliter, the bouncing ends at a mid-bounce so that half of the energy goes down each arm. If light enters in both arms, both arms do the same thing (superposition) and the addition of the two beams is constructive or destructive depending on the relative phase. The extreme cases are those where the light in one arm is 90 degrees out of phase with the other arm so that transmitted part (no phase change) + bounced part (with phase change) add up or destruct.

If you use a y-junction to merge the two beams, it may look like there is no other output channel but this is just because your 2x2 splitter that ends in mid-bounce has a change of geometry at that point, whereby the energy in the middle is captured in a waveguide in the middle. In this picture it is now easy to envisage what happens for the situation that corresponds to all the energy in one arm (for the 2x2): It would put all the energy at the entrance of one of the output waveguides but since the geometry is changed to a single waveguide in the middle, half the energy is lost due to mismatch. It shows the interesting diference betwen a y-splitter and a 2x2 which is that for maximum energy in the through chanel you need both arms to be in-phase so that they both half couple and those half couple bits interfere constructively whereas both crosscoupled bits interfere destructively with the beam from the other input channel.

This diference is also obvious from the time-reversed picture where light that is split in ay-junction results in two beam that are in-phase in both channels whereas light at the output of a 2x2 splitter (when only one input is used) is out-of-phase.

I hope this helps to explain the apparant lateral movement as a result of a longitudinal movement.

It is an electromagnetic wave, the electric fields cancel. In the case of a optical waveguide (modulator) the light in the output guide is either in the 00 (supported) mode or in the 01 (unsupported) mode. Light in the unsupported mode is not guided and scatters into the substrate.

Thank you, Mr. Herman. Mine was answer to Sarka and only meant an ordinary interferometer, not a reduced size one with more elements. By the way, what is a 2x2 splitter?

OK, got it, sorry. A 2x2 splitter is a fiber device with two input ports and two output ports that acts like beamsplitter. They can be made by evanescently coupling two fibers or by coupling to a waveguide with a coupling section. They can be made (like beamsplittters) with 50/50 splitting or almost any other ratio.

See attached paper

I cannot understand the lengthy discussion after a satisying answer to the initial question was given by David. A Mach-Zehnder, like the Michelson and other interferometers has two outputs with complementary intensities, which add up to the incident intensity (when no absorption of scattering loss occurs).

@Vittorio. Where do I find the attachment?

See: V. Degiorgio

"The phase shift between the transmitted and the reflected optical fields of a semireflecting lossless mirror is /2"

American Journal of Physics 48, 81-82 (1980)

@Antoine. The reason for the lengthy discussion is there are two cases: the beamsplitter MZ and the waveguide MZ. The way the energy is redistributed appears to be different in those cases.

See attached documents

http://www.sljol.info/index.php/SLJP/article/viewFile/201/210

Addendum to V. Degiorgio's answer: the general case is discussed using an elegant well known formalism by A. Zeilinger, see attachment below.

Attachment: Zeilinger's paper in Am. J. Phys. 1981

Answer is in losses energy at mergering beams in an interferometer. You can splitter beam without losses of energy, but you can't to merge because in the second case the entropy is decreased.

:)

The answer is simple: to the other port.

I like Am. J. Phys. 1981

I'm looking to dig deeper than the beam spiltter case. What if we make an electrical analog with transmission lines? When destructive interference occurs, where two transmission lines join, what happens then? Is all the power merely reflected back or are there conditions under which the transmission line also radiates?

Hi Derek,

The above answers are all correct for the optical version of the Mach-Zehnder, and the principle that is articulated is absolutely correct, namely that the energy needs to be distributed into other degrees of freedom, typically the other output port (although there are many instances where it is implied that the beams need to be at right angles, and this is not really the case).

The issue is that sometimes in the electrical case, a Y-junction is seen as being the same as an optical beamsplitter, it is not the same and has a different transfer matrix (the Y junction is a 3 port device, whilst the spatial beamsplitter is always a 4 port device, but where one port is often closed), and I wonder if your statement "where two transmission lines join" is referring to this.

If you create something that is like a Mach-Zehnder interferometer out of two Y junctions, instead of two beamsplitters, then the energy is typically reflected when destructive interference occurs. This is like the case for an Aharanov-Bohm loop. In some instances, the interference can go into higher order modes of the transmitted loop (this is more likely the case for optical Y junctions), but also, one can consider overlap of states and indeed you could conceivably have a case where the extra energy is radiated.

Andy

@Andy. I like that answer. But I would add that you can have a Y junction optically too, if you use optical waveguides rather than beam-splitters.

Hi Derek,

Yes, absolutely correct. Typically in the optical Y junction you will get both reflection and coupling into higher order modes of the output waveguide (modes which are generally not applicable in the conventional free-space Mach-Zehnder).

There are two beams exiting the last beam splitter of the Mach Zender. When one of them is canceled by destructive interference the other is is increased an equal amount by constructive Interference.

To see how this happens, imagine the last beam splitter slanting across the page and a down pointing beam incident from the top and a right pointing beam incident from the left. There will be a phase shift on reflection less than pi, assume it to be -pi/2. Also assume that the right pointing beam leads the down beam by pi/2. Then the two components exiting to the right are out of phase by pi/2 + pi/2 or pi and interfere destructively to give zero output. The reflected component of the right pointing beam, now pointing down, has phase pi/2 initially plus a reflective phase shift -pi/2 or phase 0 which is in phase with the transmitted part of the down pointing beam and the two interfere constructively giving twice the amplitude. Although we don't know the true phase shifts, whatever they are, they must must total to Pi for the one beam and zero for the other.

Jerome, and now if the MZ is made of optical waveguide with only one output port?

Well Dr. Derek if I had know the question was about integrated Optics wave guides I would have had to punt because that stumped me when I ran into it. But I see you did get some good answers.

Hi,

Interference only stands for monochromatic waves, otherwise one speaks of wave beating and the story is much different.

When a "monochromatic" wave enters an interferometer, whatever the interferometer, Michelson, Fabry-Perot, Mach-Zender etc, the interferometer acts as a variable miror that reflect between 0% to 100% of the impinging light depending on the phase difference between the two arms of the interferometer. So, with 0 ≤ p ≤ 1, if p of the total intensity of light passes through the interferometer, 1-p is reflected and the total energy is conserved.

In the case of optical waveguides - technically speaking and putting aside a theory - energy goes into substrate modes of cladding modes.

So while you are building some optical-waveguide M-Z device, you have to remember to keep the output waveguide long, to get rid of this "optical noise", guided by cladding or substrate.

Yes, a fiber interferometer is a lossy miror, but still when the exit is blocked (minus the losses, absorption, diffraction, diffusion) the remaining energy goes back to the source.