# Mach-Zehnder Interferometer: two output interference pattern question

I have drawn diagram so not to confuse.

So far, I've heard that in Mach-Zehnder interferometer, two output should have one constructive interference, and one destructive interference for other.

But, what I had calculated above for the phase shift, doesn't fit with the argument above.

What am I missing? Where is the constructive interference?

• I hate optics... thinking about the polarization always throws me off! Can you walk an old man trough the beam polarizations? – CuriousOne Sep 19 '15 at 5:04
• What is the "Phase 180° addition chamber? I mean, is there such a device, and if so, what is the make and model? I don't know of one. (will/am reading Princeton article in answer). I'm presuming the splitters are all 'classical phase shifting' type. Anyway, @user65452 this isn't an Ideal venue for you and I to continue this discussion from a previous video post, it is not a forum. In addition the video in question from MIT used a Michelson interferometer with extra legs, an entirely different animal. My apologies to the moderators – Cyberchipz Sep 19 '15 at 18:30
• Well, in my Optics textbook, there is a chamber that controls the phase of the lower part EM wave, so it's just my notation for that the chamber that lags 180 degree of phase respect to upper part EM wave. Also, you SHOULD read the Princeton article. – user65452 Sep 19 '15 at 19:14
• So far as I understood, some of the beamsplitters are different with the one in Wikipedia. I'm trying to figure out the work with metal coated splitter. – user65452 Sep 19 '15 at 19:16

When you use a real beamsplitter, it has a finite thickness. When such a splitter is placed at the second position in a particular way, the green light beam going to screen 1 will be suffering Zero phase shift because it suffers a reflection coming from the denser medium and going back into the denser medium. The red beam going to screen 2 will suffer a phase shift of $\Pi$ radians because it is getting reflected coming from lighter medium and going back into lighter medium. This will be reversed if you change the position of the splitter but essentially, only one screen will have a constructive interference.

• denser to denser? does that really make sense? not denser to lighter? – user65452 Sep 20 '15 at 21:44
• Yes. Please refer to this [question]{physics.stackexchange.com/questions/32122/…}. – sarat.kant Sep 21 '15 at 6:57
• Ah.. I see what you meant – user65452 Sep 21 '15 at 17:20

Use actual beamsplitter, from wikipedia

https://en.wikipedia.org/wiki/Beam_splitter

This is the solution!!!!

https://www.cs.princeton.edu/courses/archive/fall06/cos576/papers/zetie_et_al_mach_zehnder00.pdf

Forget about most of the instrument, just think about the final beam recombiner. You have two mutually coherent beams of some arbitrary phase $e^{\pm i\,\phi}$ at the two inputs to the final beamspliter (without loss of generalness, subtract out the common mode phase, so we can represent the two beams as $a_{\pm}\,e^{\pm i\,\phi}$ where $a_\pm$ are the real-valued amplitudes).

Now if the beamsplitter reflects negligible light back whence it came (which is a pretty good approximation with appropriately coated optics), and it is lossless then the sum of the powers of the two outputs has to be constant and the two variations of output power from each output as functions of the input phase difference $\phi$ must be two antiphase functions of $\phi$ so that their sum is the total input power, simply by conservation of energy.

To explore this idea in more detail, the two outputs $(y_1,\,y_2)^T$ written as a column vector are related, in the linear system case, to the two inputs $(a_+\,e^{i\,\phi},\,a_-\,e^{-i\,\phi})^T$ by a homogeneous, linear, unitary relationship:

$$\left(\begin{array}{c}y_1\\y_2\end{array}\right) = e^{i\,\chi}\left(\begin{array}{cc}\alpha&\beta\\-\beta^*&\alpha^*\end{array}\right)\left(\begin{array}{c}a_+\,e^{i\,\phi}\\a_-\,e^{-i\,\phi}\end{array}\right)$$

where $\chi\in\mathbb{R}$ and $\alpha,\,\beta$ are any two complex number fulfilling $|\alpha|^2+|\beta|^2=1$. Any $2\times 2$ unitary relationship can be written in the above form. Try working out the powers $|y_1|^2,\,|y_2|^2$ as functions of all the parameters $\chi,\,\alpha,\,\beta$: set it up in Mathematica or Maple or the like. You'll find that no matter what you put for these parameters, the powers $|y_1|^2,\,|y_2|^2$ will be antiphase, sinusoidal functions of the relative input phase $\phi$.