# Fully destructive interference

I'm aware that this is somewhat of a frequently asked question (not only here), and i've dug through respective answer for a few hours before i decided to ask here. None of those answers helped me predict/explain my case though, so here's my problem:

Suppose we have this setup:
A laser source on the left, two semi-transparent mirrors (grey) and two normal mirrors (black). The red lines designate the laser path. The bottom arm introduces a phase difference of half the wavelength to the respective beam (integer multiples of $\lambda/2$ could be added to the vertical distance without affecting the setup).

From what i understand, the dotted lines should both exhibit destructive interference (assuming the semi-transparent mirrors both reflect and let pass exactly 50% of the light).

To the question(s): I'm aware that energy conservation should still be at work, but i simply cannot figure out where the energy ends up. Assuming that full interference actually happens, it would either suggest that no work is being done in the laser source (how/why?) or that the energy is going somewhere else (where?). And if the beams don't actually cancel each other out, i'd like an explanation as to why.

• What you understand is incorrect. No matter how you set up the mirrors, one of the dotted lines will pass $1-p$ fraction of the light, and the other will pass $p$ percent of the light for some $p$. I think you're neglecting what the semi-transparent mirrors do to the phase of the light. Nov 30, 2013 at 1:34
• @PeterShor: I'm not sure i understand. Don't Michelson interferometers use a very similar technique? Nov 30, 2013 at 1:43
• ...but I'm not going to bet against anyone named Peter Shor :) Nov 30, 2013 at 1:53
• @MaxLanghof Great question. This video may be helpful: video.mit.edu/watch/…
– Joe
Nov 30, 2013 at 3:38
• You seem to think that there's a way to set up the two beams so that there is destructive interference on both dashed lines. My guess is that you think this because you're just looking at the path lengths. What actually happens is that the beam splitters add a phase that combines with the phase due to the path lengths, and which ensures that the two paths are complementary … if the probability increases for the photon to go down one path, the probability decreases that it goes down the other. Nov 30, 2013 at 5:19

I'd like to add to Peter Shor's comments. An ideal partly silvered mirror comes very close to conserving energy. This actually implies quite a lot about the phase shift imparted by both mirrors, particularly if the mirrors are symmetric in their action. The total power of the two outputs must sum to a constant and so, as the relative phase of the interferometer's arms changes, the two output powers both vary sinusoidally with the relative phase and these two sinusoids are exactly out of phase. Sometimes in practice one finds that they are not exactly out of phase and this means that energy is "leaking" somewhere: something is absorbing or (if we think of an optical fibre interferometer as below) the optical couplers can be wasting some light by coupling it out of the fibres to the radiation field.

However, it might help to derive the defining equations for this system in the case of symmetric mirrors. A good way to understand your problem is to imagine one optical mode. This kind of working is realized in the single mode optical fibre Mach-Zehnder interferometer. The single-modedness means we don't have the complicated beam structure as shown in Anupam's answer. This structure can certainly be present in multi-moded beans in interferometers, so for now simpler things to one optical mode, as happens in a single mode fibre Mach-Zehnder interferometer (the mirror's are replaced by $2\times2$ optical fibre couplers in that case). If the mirror is working in its linear regime we can then characterise the incident and reflected fields on both sides of such a mirror by a $2\times2$ scattering matrix:

$$\left(\begin{array}{c}b_1\\b_2\end{array}\right) = \left(\begin{array}{cc}s_{11}&s_{12}\\s_{21}&s_{22}\end{array}\right) \left(\begin{array}{c}a_1\\a_2\end{array}\right)$$

or, as simple matrices and column vectors:

$$b = S a$$

where we define the waves by single, scalar complex numbers (recall we assume everything is working one-modedly) as in the following diagram:

Energy is conserved, thus $b^\dagger b = a^\dagger a$ or, otherwise written, $a^\dagger (S^\dagger S - I) a = 0;\;\forall a\in \mathbb{C}^2$. By choosing arbitrary vectors in $\mathbb{C}^2$, one can show that this statement is the same as $S^\dagger S = I = S S^\dagger$ i.e. the scattering matrix is unitary.

Now let's assume that a mirror is symmetric in its action, i.e. its action is the same if the beams number one and two were to swap roles (or if it is flipped over), i.e:

$$\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\left(\begin{array}{cc}s_{11}&s_{12}\\s_{21}&s_{22}\end{array}\right)\left(\begin{array}{cc}0&1\\1&0\end{array}\right) = \left(\begin{array}{cc}s_{11}&s_{12}\\s_{21}&s_{22}\end{array}\right) = \left(\begin{array}{cc}s_{22}&s_{21}\\s_{12}&s_{11}\end{array}\right)$$

so that the most general form the scattering matrix can have is:

$$\left(\begin{array}{cc}\alpha e^{i\gamma}&\beta e^{i\delta}\\\beta e^{i\delta}&\alpha e^{i\gamma}\end{array}\right);\quad \alpha,\beta,\gamma,\delta\in\mathbb{R}$$

with the unitaryhood condition expressed by $\alpha^2+\beta^2 = 1$ and $\cos(\gamma-\delta)=0$. This means that the most general form the scattering matrix can have is:

$$S = e^{i\,\theta}\,\left(\begin{array}{cc}\cos\frac{\phi}{2}&i\,\sin\frac{\phi}{2}\\i\,\sin\frac{\phi}{2}&\cos\frac{\phi}{2}\end{array}\right);\quad \phi,\theta\in\mathbb{R}$$

which, if the mirrors are exactly half silvered ones (so that they split powers equally so that $\cos(\phi/2)^2 = \sin(\phi/2)^2 = 1/2$:

$$S = \frac{e^{i\,\theta}}{\sqrt{2}}\,\left(\begin{array}{cc}1&\pm i\\\pm i&1\end{array}\right)$$

The phase relationships between the matrix elements are what enforce the conservation of energy.

The action of the interferometer's two arms is described by the simple phase delays:

$$D = e^{i\xi}\left(\begin{array}{cc}e^{i\,\zeta/2}&0\\0&e^{-i\,\zeta/2}\end{array}\right)$$

where we can, without loss of generality, we can suck the phase $\xi$ common to both arms as a scalar forefactor. Therefore, the interferometer's two output amplitudes are:

$$S D S \left(\begin{array}{c}1\\0\end{array}\right) = e^{i(\xi+\theta)}\left(\begin{array}{c}\sin\frac{\zeta}{2}\\i\,\cos\frac{\zeta}{2}\end{array}\right)$$

Here we define the system "input" to be the column vector $\left(\begin{array}{c}1\\0\end{array}\right)$; note that the system input is such a vector because we could also drive the interferometer system with a beam at right angles to the input beam you have drawn from your source at the left of your diagram. In this case, one of the column vector's entries is nought because this input is not driven in your diagram.

We see that the output square magnitudes $\sin( \zeta/2)^2$ and $\cos( \zeta/2)^2$clearly vary sinusoidally exactly out of phase so that the total power sums up to one unit.

With general, unitary scattering matrices $S$ standing for general asymmetric mirrors, the equations are more complicated, but the results are the same. Unitaryhood enforces phase conditions at the mirrors and always leads to sinusoidally varying out-of-phase outputs. In the more general case, the "visibility" of the fringes is less as the sinsusoids in general have offsets that mean that neither output, notwithstanding the sinusoidal variation with phase, can ever be "nulled" i.e. can never output exactly zero power. Only the special, symmetric case considered above leads to full fringe visibility.

Although WetSavannaAnimal has worked out the answer to your problem in full detail, it might be nice to have a concise summary. You are forgetting the $180^\circ$ phase flip of one of the fields at the beamsplitter. This means that if the interference is perfect, one of the output ports is dark and the other is bright. They cannot both be dark or both bright. One of these cases would imply the destruction of energy, and the other would imply the creation of energy.

The standard convention is that the field reflected off of the front face picks up $180^\circ$ of phase (a minus sign), but the requirement is actually that the phase picked up by all of the fields sum to $180^\circ$. This is mandated by energy conservation at the beamsplitter, and forgetting about it will end up with you violating energy conservation.

Incidentally, energy conservation also implies that the reflectivity of the beamsplitter be the same for the field entering from the bottom as the field entering from the left ( a fact which is fairly intuitive), and that the transmissivity plus the reflectivity sum to one, $R+T=1$, in a perfect beamsplitter with no absorption.

• Could you just comment on 180$^\circ$ phase? I count $0\times$ flip for straight beam and $4\times$ flip for its side interfering branch. And $1\times$ flip for upward beam and $3 \times$ flip for the side interfering branch. They both come in the same phase, where I do mistake in my counting from 1 to 4? Apr 18, 2017 at 11:50