Let us now come back to quantum optics and learn how to calculate what happens in a Mach-Zehnder interferometer with quantized radiation. More specifically, we want to know what happens if we have a single photon at input 1 and the vacuum at input 2. Will we have the count rates of detectors 5 and 6 depending on delta L? this would be the signature of an interference for a single photon. Let us embark into the quantum optics calculation. You will learn that, in fact, it is not much more complicated than the classical calculation. Can you guess why? Do you remember what you have learned about beam splitters in quantum optics? That in fact, one can transform the field operators exactly as a classical amplitudes replacing E+ by E+_hat. Let's do it at each beam splitter. Note that when writing the relation of the first beam splitter O, we have explicitly involved the input 2, although there is vacuum. You must remember that in quantum optics, vacuum plays a role and we must take it into account. For this calculation, it will turn out not to play a role, but we will find other cases where it is crucial not to forget the input with vacuum so better keep it and find out explicitly if it plays a role or not. We must now take into account propagation from O to O' in each channel. In fact, the same recipe as the one for beam splitters does apply. You can express E3+ and E4+ at O' as a function of their value at O by introducing the classical propagation factors. The expressions of the operators E5+ and E6+ as a function of E1+ and E2+, are thus readily obtained. I have written here, E5+. According to the recipe, in order to calculate the rate of single detections at detector 5, we need to apply E5+ expressed as a function of E1+ and E2+, to the input state |psi_12> with 1 photon in input 1 and 0 photon in input 2. Remember, that the operators E+ are proportional to the annihilation operators a. So the terms E2+ applied to 0 photon in mode 2, gives 0. As I told you in that calculation, the vacuum in the empty input does not play a role but we see where exactly it happens in the calculation. When applying E5+ to |psi_12>, we are thus left with an algebraic expression times E1+ applied to 1 photon in the mode 1. That is to say a prefactor times the vacuum state. That prefactor has a part that resembles the one of the classical calculation and when we take the squared modulus, we obtain an expression analogous to the semi-classical result for a classical field. The number of counts for a certain period of time then, has the same form, that is to say, its value is modulated as a function of cosine of k times delta L. The exactly complementary result would be obtained for the output 6. So according to those quantum optics calculations, if we send single photon wave packets in a Mach-Zenhnder interferometer, the probability to detect that photon in one or the other output depends on the path difference, exactly as in classical optics. Actually, the same result would be obtained for any interferometer and more generally for any optical setup. You know the reason. It is because the expressions of the output field operators as a function of the input operators are exactly the same as the relations for the classical field amplitudes. When applying these expressions to one photon at the input, one gets the vacuum state, which factorizes. The expression in factor of the vacuum state is then exactly the same as in the classical description, and when you take the squared modulus, you get the same interference pattern as with a classical calculation. If this is not yet fully clear to you, I suggest that you look again into the calculation we have just done with this in mind.