Rule A, you will recall, applies “if the intermediate measurements are made (or if it is possible to find out what their outcomes would have been if they had been made),” and Rule B applies “if the intermediate measurements are not made (and if it is impossible to find out what their outcomes would have been if they had been made).” The two-slit experiment we will discuss in this section demonstrates the rationale behind the cryptic clauses in parentheses.
In this experiment[1,2] atoms are used instead of electrons. All atoms are of the same type — Cesium-133, say — and all start out in the same excited state. Placed in front of the slits are two initially separate microwave resonance cavities, each tuned to the energy difference ΔE between this excited state and the atoms’ ground state and thus capable of “holding” photons “carrying” the energy ΔE. The design of each cavity moreover ensures that the probability with which an atom is found to emerge from it in the ground state equals unity — provided, of course, that the appropriate measurement is made.
In what sense can such a cavity hold a photon? A reasonably safe interpretation is that if a (100% efficient) photodetector were inserted into the cavity, it would detect (and absorb) a photon. And in what sense does a photon carry the energy ΔE? A reasonably safe interpretation is that the photodetector would absorb (and gain) this energy.
The two resonance cavities are separated from each other by a pair of electro-optical shutters,which remain closed for now. Atoms are launched, one at a time, with nothing to predict the particular cavity through which any given atom will pass. (Before an atom is launched, the photon left behind by the previous atom is absorbed and thus “removed” from the cavity.) Each atom leaves a mark on the screen. How will the marks be distributed?
Focus on a single atom, after it has hit the screen but before the photon is removed. This is a situation in which it is possible to find out what the outcome of an intermediate measurement would have been if it had been made. The intermediate measurement, had it been made, would have determined the slit taken by the atom. The reason why we can find out what its outcome would have been is the following strict correlation between the outcome of this measurement and the cavity containing the photon: if the atom were found to emerge from the left slit, the probability of absorbing the photon in the left cavity would be 1, and if the atom were found to emerge from the right slit, the probability of absorbing the photon in the right cavity would be 1. Hence if the photon is detected in the left cavity, a measurement of the slit taken by the atom would have indicated the left slit, and if the photon is detected in the right cavity, a measurement of the slit taken by the atom would have indicated the right slit. Thus Rule A applies.
Let us color the marks: those made by atoms that left a photon in the left cavity green, and those made by atoms that left a photon in the right cavity red. The dotted curve in Fig. 1.8.2 gives the distribution of the green marks, the dashed curve that of the red marks. The solid curve is the sum of the two distributions. The green marks are distributed as we expect from atoms that went through the left slit, and the red marks are distributed as we expect from atoms that went through the right slit. (Compare Fig. 1.8.2 with Fig. 1.4.2.)
Between the shutters there is a (100% efficient) photosensor. If the shutters are opened before the photon is absorbed, quantum mechanics predicts that the sensor will absorb the energy ΔE with probability 1⁄2. Since we now have a single cavity instead of two, information about the slit taken by the photon is no longer available. (It has become customary to say that the information has been “erased.” What has actually been erased, however, is merely the possibility of obtaining the information.) Does this mean that Rule B now applies? If this experiment is done with sufficiently many atoms, will the overall distribution of marks exhibit interference fringes?
The answer has to be negative, for the measurement involving the photon is made after the atom has hit the screen. The decision about which measurement to perform — to determine the cavity that held the photon or to determine the behavior of the photosensor upon opening the shutters — comes too late to affect the overall distribution of marks.
But we now have another way of coloring the marks: yellow if the photosensor responds, blue if it fails to respond. Quantum mechanics predicts that the yellow marks will exhibit the same interference pattern as electrons in a two-slit experiment under the conditions stipulated by Rule B (Fig. 1.4.3). Because the overall distribution of marks is the same in both versions of the experiment, the blue marks will exhibit the complementary interference pattern, having maxima where the other has minima, and vice versa.
All “yellow” atoms and all “blue” atoms have something in common, but as their respective behaviors lack classical counterparts, we have no ready name for it. We may say that the “yellow” atoms went through the slits in phase} while the “blue” atoms went through the slits out of phase. We use these phrases for the following reason. If there is to be a maximum at the center of the screen, the phases of the amplitudes associated with the alternatives “through L” and “through R” must differ by an even multiple of 180° — the alternatives must be “in phase.” And if there is to be a minimum instead, the amplitudes must differ by an odd multiple of 180° — the alternatives must be “out of phase.” (As you will remember, the phase of the amplitude associated with a particle’s propagation from point A to point B is proportional to the distance between A and B.)
The “green” atoms, we noted, behave like atoms that went through L, while the “red” atoms behave like atoms that went through R. Likewise, the “yellow” atoms behave like atoms that went through the slits in phase (inasmuch as they display the corresponding interference pattern, while the “blue” atoms behave like atoms that went through the slits out of phase. Cannot we conclude from this that the “green” atoms actually went through L, that the “red” atoms actually went through R, that the “yellow” atoms actually went through the slits in phase, and that the “blue” atoms actually went through the slits out of phase? After all, if it looks like a duck, swims like a duck, and quacks like a duck, then it probably is a duck. The problem with this conclusion is that it seems to imply the possibility of influencing the past!
If the experimenters determine the cavity that held the photon, they learn through which slit the corresponding atom went. If they open the shutters and observe whether or not the sensor responds, they learn how the atom went through the slits — in phase or out of phase. They cannot make the atom go through L or through R, yet by doing the former experiment, they can make sure that it went through a single slit — either L or R. Nor can they make an atom go through the slits in phase or out of phase, yet by doing the latter experiment they can make sure that it when through both slits. (Further discussion of the meaning of “both” can be found here.) And since they can choose between the two experiments after the atom has made its mark on the screen, they can, by their choice, contribute to determine the atom’s past behavior.
1. [↑] Englert, B.G., Scully, M.O., and Walther, H. (1994). The duality in matter and light, Scientific American 271 (6), 56–61.
2. [↑] Scully, M.O., Englert, B.G., and Walther, H. (1991). Quantum optical tests of complementarity, Nature 351 (6322), 111–116.