According to a Physics World poll conducted in 2002, the most beautiful experiment in physics is the two-slit experiment with electrons. According to Richard Feynman, this classic gedanken experiment “has in it the heart of quantum mechanics” and “is impossible, absolutely impossible, to explain in any classical way.” The setup consists of an electron gun G, a plate with two slits L and R equidistant from G, and a screen at which electrons are detected.
To apply the rules of the game, we need to identify the initial measurement (on the basis of whose outcome probabilities are assigned), the final measurement (to the possible outcomes of which probabilities are assigned), and a set of alternatives. The initial measurement indicates that an electron has been launched at G. The final measurement indicates the position — along an axis across the backdrop — at which the electron is detected. (If we assume that G is the only source of free electrons, then the detection of an electron behind the slit plate also indicates the electron’s launch at G.) A single intermediate measurement, if made, indicates the slit through which the electron went. Thus there are two alternatives:
- The electron went through the left slit (L).
- The electron went through the right slit (R).
The corresponding amplitudes are AL and AR. The event whose probability we wish to calculate is the detection of the electron by a detector D situated somewhere at the backdrop. (As we will use the letter G for both the gun and its position, so we will use the letter D for both the detector and its position.) Here is what we need to know in order to be able to perform this calculation (→ proofs):
- AL is the product of two complex numbers, which we will refer to as propagators, and for which we will use the symbols <D|L> and <L|G>. Thus AL = <D|L> <L|G>. Likewise AR = <D|R> <R|G>.
- The magnitude of the propagator <B|A> — that is, of the amplitude associated with a particle’s propagation from A to B — is inverse proportional to the distance between A and B.
- The phase of <B|A> is proportional to this distance.
According to Rule A, the probability with which an electron launched at G is found by a detector situated at D thus is
pA(D|G) = |<D|L><L|G>|2 + |<D|R><R|G>|2.
Here is how pA(D|G) depends on the horizontal position of the detector at the screen:
Calculated according to Rule B, the probability with which an electron launched at G is found by a detector situated at D is
pB(D|G) = |<D|L><L|G> + <D|R><R|G>|2.
Here is how this depends on the horizontal position of the detector at the screen:
Strictly speaking these graphs are plots of probability densities (in one dimension) — that is, they plot a probability per unit length. One has to pick an interval on the horizontal axis and draw (or imagine) two vertical lines at the interval’s end points to obtain the probability with which an electron is detected within the interval. This probability is given by the area enclosed by the plot, the horizontal axis, and the two vertical lines. (What then is the area between the entire horizontal axis and the entire plot? Since the probability of detecting the electron anywhere along the axis equal 1, this area also equals 1.)
Two remarkable facts are worth noting. The first is that near the (local) minima of Fig. 1.4.3 the probability of detection is less if both slits are open than if either of the slits is shut, as a comparison with Fig. 1.4.2 reveals. The second is that the central maximum of Fig. 1.4.3 is twice as high as the maximum of Fig. 1.4.2. The first fact is generally attributed to “destructive interference,” the second to “constructive interference.” This terminology is likely to cause much confusion. Except in the imaginary world of classical physics, interference is not a physical process. Whenever we speak of constructive interference, all we mean is that a probability calculated according to Rule B is greater than the corresponding probability calculated according to Rule A. And whenever we speak of destructive interference, all we mean is that a probability calculated according to Rule B is less than the corresponding probability calculated according to Rule A.
Here is how the interference pattern plotted in Fig. 1.4.3 builds up over time:
Now suppose that the conditions stipulated by Rule B are met: there is nothing — no event, no state of affairs, anywhere, anytime — from which the slits taken by the electrons can be inferred. Further suppose that, nevertheless,
- each electron goes through a particular slit (either L or R),
- the behavior of electrons that go through a given slit does not depend on whether the other slit is open or shut.
If assumption (1) is true, then the distribution of hits across the backdrop, when both slits are open, is given by
n(x) = nL(x) + nR(x),
where nL(x) and nR(x) are the respective distributions of hits from electrons that went through L and electrons that went through R. If assumption (2) is true, then we can observe nL(x) by keeping the right slit shut, and we can observe nR(x) by keeping the left slit shut. What we observe when the right slit is shut is the left dotted hump in Fig. 1.4.2, and what we observe when the left slit is shut is the right dotted hump. If both assumptions are true, we thus expect to observe the sum of these two humps. But this is what we observe under the conditions stipulated by Rule A. At least one of the two assumptions is therefore false.
1. [↑] Crease, R.P. (2002). The most beautiful experiment, Physics World, September, 19–20.
2. [↑] Feynman, R.P., Leighton, R.B., and Sands, M. (1965). The Feynman Lectures in Physics Volume 3, Section 1–1, Addison–Wesley.
3. [↑] Tonomura, A., Endo, J., Matsuda, T., and Kawasaki, T. (1989) Demonstration of single-electron buildup of an interference pattern, American Journal of Physics 57 (2), 117–120.