Since the very day in 1905 when Einstein presented his hypothesis that light is made of quanta, "lichtQuanten" a question immediately arose: does a single quantum, a photon, interfere with itself? This is an intriguing puzzle. On the one hand, physicists have been convinced since the time of Young and Fresnel in the early 1800s that phenomena of interference and diffraction can be described consistently only within a wave model of light. This model implies that light follows many paths simultaneously, and the observed variations in the light intensity result from the difference in the optical lengths of the various paths. But on the other hand, if there is only one photon, which is never detected simultaneously in two different paths, how could it follow several paths simultaneously? You know I'm sure that the answer to that question can be given only in the framework of quantum mechanics: a quantum particle, such as an electron or a neutron, also behaves as a wave, a matter wave, a de Broglie wave: it is the famous wave-particle duality. But a photon is not a particle similar to an electron or a neutron: it has no rest mass, and the notion of a particle of light is not a primary concept when one develops a consistent model of optics. It is thus interesting to revisit the question of wave-particle duality in the context of optics, where the primary concept is a wave. You are going now to find how the question can be addressed and answered clearly in quantum optics considering one photon states of light sent into a Mach-Zehnder interferometer. The Mach-Zehnder interferometer is the beloved interferometer of theorists, since the scheme is especially clear. It is also used in experiments, but building a stable Mach-Zehnder interferometer with standard optics components demands advanced technology, since the various elements are separated from each other, and the stability must be ensured at the level of a fraction of a micrometer. In contrast, it is widely used in integrated optics, for instance, with optical wave guides engraved in a piece of glass, or optical fibers. Let us describe the scheme of a Mach-Zender interferometer. The input is a beam splitter on which the input beam in mode 1 is divided into beams 3 and 4. Two mirrors, M3 and M4, then reflect these two beams, which are recombined on a second beam splitter. Two detectors, D5 and D6, are placed in the output beams 5 and 6, and counters register the number of counts N5 and N6 for the duration of the experiment. You know that the crucial parameter is the difference delta L between the optical lengths of path 4 and path 3. This path difference can be varied thanks to a piezo transducer on which one of the mirrors is mounted, here mirror M3. A PZT is a remarkable device whose length can be controlled by a voltage with an accuracy much better than a micrometer. I could cite cases where one can control its displacement at the sub-nanometer level, believe it or not. The path difference can then be varied with an exquisite accuracy, definitely better than one optical wavelength. You know that interference is revealed by the fact that the photodetection signals N5 and N6 vary periodically with the path difference delta L, while the total number of counts N5 plus N6 is constant. The classical optics calculation of this dependence is standard, but let us remind it, since part of it will be useful in the quantum optics calculation. We use the notations already introduced in this lesson. Ignoring the polarization perpendicular to the figure for all waves, we write again the relations between the input complex amplitude E1+ at the input beam splitter and the complex amplitudes E3+ and E4+ just after the input beam splitter at point O. All frequencies are the same, so we do not write explicitly the exponential of i omega t factors which factorize out. After propagation to the second beam splitter, the amplitudes at point O' have taken propagation factors that depend on the optical path length L3 and L4. Now, what happens at the second beam splitter? The amplitudes E3 and E4 at O' are recombined, and we write E5 and E6 as a function of E3 and E4 at O' using the relation that we have seen earlier in this lesson. Are they really the same relations? You can pause the video to check. Haha. There is a difference. There is a minus sign in the reflection term of E5+ expressed as a function of E3+, in contrast to the similar term in E3+, expressed as a function of E1+ at point O. Can you guess why? Observe carefully the figure. The second beam splitter at O' is reversed compared to the first one at O. In one case, the reflection happens out of the dielectric plate of the beam splitter, while in the second case, a reflection happens inside the dielectric. Illuminating calculations done for the first time by my hero, Augustin Fresnel, show that the signs should be opposite for these two cases. Note, however, that the transformation matrix is still unitary, as you can check using the sufficient condition I wrote explicitly earlier. So it is still a good description of an ideal lossless beam splitter. As an experimentalist in optics, I can give you a good reason to reverse the second beam splitter. If you ever have to implement a Mach-Zehnder interferometer, the scheme with reverse beam splitters is symmetric. I mean, each path 3 or 4 passes once only in the dielectric material of a beam splitter. If the second beam splitter was reversed, beam 4 would pass twice, and beam 3 would not pass at all. A symmetric scheme is always preferable, it is crucial if you have non-monochromatic light because of the dispersion of the dielectric. But let us return to our classical interference calculation. It is now a simple exercise to complete the calculation. We express E5+ as a function of E3+ and E4+ at O' and then E3+ and E4+ at O as a function of E1+, and we take the squared modulus to have the rate of single detections rate at D5. We replace the squared amplitude reflection and transmission coefficients by capital R and capital T and develop. Replacing the path difference L4 minus L3 by delta L, we obtain a cosine modulation with period 2 pi over k, that is to say, the optical wavelength. We can just write the number of counts at detector 5 during a measuring interval as proportional to [1 - V cos(k delta L)] where V is the visibility of the fringes. You can calculate the number of counts at 6 by a similar development. The result is obtained by changing the minus of N5 into a plus, which should be obvious looking into E6 compared to E5 at O'. But it does not hurt to do the calculation. In the expressions of N5 and N6, N total is the total number of counts N5 plus N6 which does not depend on delta L. Note that for so-called balanced beam splitters with equal reflection and transmission coefficients, one has visibility of 1.