Chapter 2: Experiments with Photons

Chapter 2: Experiments with photons Contents 1 Mach-Zehder Interferometer 1 2 Elitzur-Vaidman Bombs 6 1 Mach-Zehder Interferometer We have discussed before the Mach-Zehnder interferometer, which we show again in Figure 1. It contains two beam-splitters BS1 and BS2 and two mirrors. Inside the interferometer we have two beams, one going over the upper branch and one going over the lower branch. This extends beyond BS2: the upper branch continues to D0 while the lower branch continues to D1. Figure 1: The Mach-Zehnder Interferometer Vertical cuts in the above figure intersect the two beams and we can ask what is the probability to find a photon in each of the two beams at that cut. For this we need two probability amplitudes, or two complex numbers, whose norm-squared would give probabilities. We can encode this information in a two component vector as   α . (1.1) β Here α is the probability amplitude to be in the upper beam and β the probability amplitude to be in the lower beam. Therefore, |α|2 would be the probability to find the photon in the upper beam and |β|2 the probability to find the photon in the lower beam. Since the photon must be found in either one of the beams we must have 2 2 α + β = 1. (1.2) 1 Following this notation, we would have for the cases when the photon is definitely in one or the other beam:     1 0 photon on upper beam: , photon on bottom beam : . (1.3) 0 1 We can view the state (1.1) as a superposition of these two simpler states using the rules of vector addition and multiplication:           α α 0 1 0 = + = α +β . (1.4) β 0 β 0 1 In the interferometer shown in Figure 1 we included in the lower branch a ‘phase shifter’, a piece of Figure 2: A phase shifter of phase factor eiδ . The amplitude gets multiplied by the phase. material whose only effect is to multiply the probability amplitude by a fixed phase eiδ with δ ∈ R. As shown in Figure 2, the probability amplitude α to the left of the device becomes eiδ α to the right of the device. Since the norm of a phase is one, the phase-shifter does not change the probability to find the photon. When the phase δ is equal to π the effect of the phase shifter is to change the sign of the wavefunction since eiπ = −1. Let us now consider the effect of beam splitters in detail. If the incoming photon hits a beam  1 splitter from the top, we consider this photon to belong to the upper branch and represent it by . 0 If the incoming photon hits the beam-splitter from the bottom, we consider this photon to belong to   0 the lower branch, and represent it by . We show the two cases in Figure 3. The effect of the beam 1 splitter is to give an output wavefunction for each of the two cases:         1 s 0 u → , Right BS: → . (1.5) Left BS: 0 t 1 v As you can see from the diagram, for the photon hitting from above, s may be thought as a reflection amplitude and t as a transmission coefficient. Similarly, for the photon hitting from below, v may be thought as a reflection amplitude and u as a transmission coefficient. The four numbers s, t, u, v, by linearity, characterize completely the beam splitter. They can be used to predict the output given any incident photon, which   may have amplitudes to hit both from above and from below. Indeed, an α incident photon state would give β                α 1 0 s u αs + βu s u α =α +β → α +β = = . (1.6) β 0 1 t v αt + βv t v β In summary, we see that the BS produces the following effect      α s u α → . β t v β 2 (1.7) Figure 3: Left: A photon incident from the top; s and t are the reflected and transmitted amplitudes, respectively. Right: A photon incident from the bottom; v and u are the reflected and transmitted amplitudes, respectively. We can represent the action of the beam splitter as matrix multiplication on the incoming wavefunction, with the two-by-two matrix   s u . (1.8) t v We must now figure out the constraints on s, t, u, v. Because probabilities must add up to one, equation (1.5) implies that 2 2 2 2 s + t u + v = 1, (1.9) = 1. (1.10) The kind of beam splitters we use are called balanced, which means that reflection and transmission probabilities are the same. So all four constants must have equal norm-squared: s 2 = t 2 = u 2 = v 2 = 1 2 . (1.11) Let’s try a guess for the values. Could we have √1 2 √1 2   s u = t v √1 2 √1 2 ! ? (1.12) This fails if acting on normalized wavefunctions (or column vectors) does not yield normalized wavefunctions. So we try with a couple of wavefunctions !  ! ! 1 !   √1 √1 √1 √1 √1 √ 1 1 2 2 2 2 2 2 = , = . (1.13) 1 1 1 1 √1 √1 √ √ √ √ 0 1 2 2 2 2 2 2 While the first example works out, the second does not, as |1|2 + |1|2 = 2 6= 1. An easy fix is achieved by changing the sign of v: !     √1 √1 1 1 s u 1 2 2 √ = = . (1.14) √1 √1 t v 1 −1 − 2 2 2 3   α Let’s check that this matrix works in general. Thus acting on a state with |α|2 + |β |2 = 1 we find β 1 √ 2      1 α+β 1 1 α =√ . 1 −1 β 2 α−β (1.15) Indeed the resulting state is well normalized. The total probability is what we expect 1 2 α+β 2 + 1 2 α−β 2 2 = 21 ( α + β 2 =α + β 2 2 2 + αβ ∗ + α∗ β) + 12 ( α + β 2 − αβ ∗ − α∗ β) (1.16) = 1. The minus sign in the bottom right entry of (1.14) means that a photon incident from below, as it is reflected, will have its amplitude changed by a sign or equivalently a phase shift by π (check this!). This effect, of course, is realized in practice. A typical beam splitter consists of a glass plate with a reflective dielectric coating on one side. The refractive index of the coating is chosen to be intermediate between that of glass and that of air. A reflection causes a phase shift only when light encounters a material of higher refractive index. This is the case in the transition of air to coating, but not in the transition from glass to coating. Thus the beam splitter represented by (1.14) would have its coating on the bottom side. Transmitted waves have no phase shift. Another possibility for a beam splitter matrix is   1 −1 1 √ , (1.17) 1 1 2 which would be realized by a dielectric coating on the top side. You can quickly check that, like the previous matrix, its action also conserves probability. We will call the left beam-splitter BS1 and the right beam splitter BS2 and their respective matrices will be     1 −1 1 1 1 1 . (1.18) BS1 : √ , BS2 : √ 1 1 2 1 −1 2 The two beam splitters are com  to form the interferometer shown in Figure 4. If we now assume bined α an input photon wavefunction from the left, the output wavefunction that goes into the detectors β is obtained by acting first with the BS1 matrix and then with the BS2 matrix:             1 1 α 1 1 −1 1 α 0 2 α β 1 √ input : output: √ =2 = . (1.19) 1 1 β −2 0 β −α β 2 1 −1 2 With the help of this result, for any input photon state we can write immediately the output photon state that goes into the detectors.    0 1 If the input photon beam is , the output from the interferometer is , and therefore a 1 0 photon will be detected at D0. This is shown in Figure 5. We can make a very simple table with the possible outcomes and their respective probabilities P : Outcome P photon at DO 1 photon at D1 0 4 (1.20) Figure 4: The Mach-Zehnder interferometer with input and output wavefunctions indicated. Figure 5: Incident photon from below will go into D0. Now, block the lower path, as indicated in Figure 6. What happens then? It is best to track down things systematically. The input beam, acted by BS1 gives      1 −1 1 0 1 1 √ √ = 2 . (1.21) 1 1 1 1 2 This is indicated in the figure, to the right of BS1. Then the lower branch is stopped, while the upper ! branch continues. The upper branch reaches BS2, and here the input is coming from the lower branch. We therefore get an output   √1 ! 1 1 1 2 √ = 0 2 1 −1 1 2 1 2 √1 2 0 , because nothing is ! . (1.22) In this experiment there are three possible outcomes: the photon can be absorbed by the block, or 5 Figure 6: The probability to detect the photon at D1 can be changed by blocking one of the paths. can go into any of the two detectors. As we see in the diagram, the probabilities are: Outcome P photon at block 12 1 photon at D0 4 1 photon at D1 4 (1.23) It is noteworthy that before blocking the lower path we could not get a photon to D1. The probability to reach D1 is now 1/4 and was increased by blocking a path. 2 Elitzur-Vaidman Bombs To see that allowing the photon to reach D1 by blocking a path is very strange, we consider an imaginary situation proposed by physicists Avshalom Elitzur and Lev Vaidman, from Tel-Aviv University, in Israel. They imagined bombs with a special type of trigger: a photon detector. A narrow tube goes across each bomb and in the middle of the tube there is a photon detector. To detonate the bomb one sends a photon into the tube. The photon is then detected by the photon detector and the bomb explodes. If the photon detector is defective, however, the photon is not detected at all. It propagates freely through the tube and comes out of the bomb. The bomb does not explode. Here is the situation we want to address. Suppose we have a number of Elitzur-Vaidman (EV) bombs, but we know that some of them have become defective. How could we tell if a bomb is operational without detonating it? Assume, for the sake of the problem, that we are unable to examine the detector without destroying the bomb. We seem to be facing an impossible situation. If we send a photon into the detector tube and nothing happens we know the bomb is defective, but if the bomb is operational it would simply explode. It seems impossible to confirm that the photon detector in the bomb is working without testing it. Indeed, it is impossible in classical physics. It is not impossible in quantum mechanics, however. As we will see, we can perform what can be called an interaction-free measurement! We now place an EV bomb on the lower path of the interferometer, with the detector tube properly aligned. Suppose we send in a photon as pictured. If the bomb is defective it is as if there is no detector, the lower branch of the interferometer is free and all the photons that we send in will end up in D0, 6 Figure 7: A Mach-Zehnder interferometer and an Elitzur-Vaidman bomb inserted on the lower branch, with the detector tube properly aligned. If the bomb is faulty all incident photons will end up at D0. If a photon ends up at D1 we know that the bomb is operational, even though the photon never went into the bomb detector! just as they did in Figure 5. Outcome P photon at D0 1 no explosion photon at D1 0 no explosion bomb explodes 0 (2.24) If the bomb is working, on the other hand, we have the situation we had in Figure 6, where we placed a block in the lower branch of the interferometer: Outcome P bomb explodes 12 1 photon at D0 4 no explosion 1 photon at D1 4 no explosion (2.25) Assume the bomb is working. Then 50% of the times the photon will hit it and it will explode, 25% of the time the photon will end in D0 and we can’t tell if it is defective or not. But 25% of the time the photon will end in D1, and since this was impossible for a faulty bomb, we have learned that the bomb is operational! We have learned that even though the photon never made it through the bomb; it ended on D1. If you think about this you will surely realize it is extremely surprising and counterintuitive. But it is true, and experiments (without using bombs!) have confirmed that this kind of interaction-free measurement is indeed possible. 7