This Quantum World/Feynman route/two slits

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An experiment with two slits
In this experiment, the final measurement (to the possible outcomes of which probabilities are assigned) is the detection of an electron at the backdrop, by a detector situated at D (D being a particular value of x). The initial measurement outcome, on the basis of which probabilities are assigned, is the launch of an electron by an electron gun G. (Since we assume that G is the only source of free electrons, the detection of an electron behind the slit plate also indicates the launch of an electron in front of the slit plate.) The alternatives or possible intermediate outcomes are The corresponding amplitudes are $$A_L$$ and $$A_R.$$
 * the electron went through the left slit (L),
 * the electron went through the right slit (R).

Here is what we need to know in order to calculate them:
 * $$A_L$$ is the product of two complex numbers, for which we shall use the symbols $$\langle D|L\rangle$$ and $$\langle L|G\rangle.$$
 * By the same token, $$A_R = \langle D|R\rangle\,\langle R|G\rangle.$$
 * The absolute value of $$\langle B|A\rangle$$ is inverse proportional to the distance $$d(BA)$$ between A and B.
 * The phase of $$\langle B|A\rangle$$ is proportional to $$d(BA).$$

For obvious reasons $$\langle B|A\rangle$$ is known as a propagator.

Why product?
Recall the fuzziness ("uncertainty") relation, which implies that $$\Delta p\rightarrow\infty$$ as $$\Delta x\rightarrow0.$$ In this limit the particle's momentum is completely indefinite or, what comes to the same, has no value at all. As a consequence, the probability of finding a particle at B, given that it was last "seen" at A, depends on the initial position A but not on any initial momentum, inasmuch as there is none. Hence whatever the particle does after its detection at A is independent of what it did before then. In probability-theoretic terms this means that the particle's propagation from G to L and its propagation from L to D are independent events. So the probability of propagation from G to D via L is the product of the corresponding probabilities, and so the amplitude of propagation from G to D via L is the product $$\langle D|L\,\rangle\langle L|G\rangle$$ of the corresponding amplitudes.

Why is the absolute value inverse proportional to the distance?
Imagine (i) a sphere of radius $$r$$ whose center is A and (ii) a detector monitoring a unit area of the surface of this sphere. Since the total surface area is proportional to $$r^2,$$ and since for a free particle the probability of detection per unit area is constant over the entire surface (explain why!), the probability of detection per unit area is inverse proportional to $$r^2.$$ The absolute value of the amplitude of detection per unit area, being the square root of the probability, is therefore inverse proportional to $$r.$$

Why is the phase proportional to the distance?
The multiplicativity of successive propagators implies the additivity of their phases. Together with the fact that, in the case of a free particle, the propagator $$\langle B|A\rangle$$ (and hence its phase) can only depend on the distance between A and B, it implies the proportionality of the phase of $$\langle B|A\rangle$$ to $$d(BA).$$

Calculating the interference pattern
According to Rule A, the probability of detecting at G an electron launched at D is

p_A(D) = |\langle D|L\rangle\,\langle L|G\rangle|^2 + |\langle D|R\rangle\,\langle R|G\rangle|^2. $$ If the slits are equidistant from G, then $$\langle L|G\rangle$$ and $$\langle R|G\rangle$$ are equal and $$p_A(D)$$ is proportional to

$$ Here is the resulting plot of $$p_A$$ against the position $$x$$ of the detector:
 * \langle D|L\rangle|^2+|\langle D|R\rangle|^2 = 1/d^2(DL) + 1/d^2(DR).

$$p_A(x)$$ (solid line) is the sum of two distributions (dotted lines), one for the electrons that went through L and one for the electrons that went through R.

According to Rule B, the probability $$p_B(D)$$ of detecting at D an electron launched at G is proportional to

$$ where $$\Delta$$ is the difference $$d(DR)-d(DL)$$ and $$k=p/\hbar$$ is the wavenumber, which is sufficiently sharp to be approximated by a number. (And it goes without saying that you should check this result.)
 * \langle D|L\rangle + \langle D|R\rangle|^2 = 1/d^2(DL) + 1/d^2(DR) + 2 \cos(k\Delta)/[d(DL)\,d(DR)],

Here is the plot of $$p_B$$ against $$x$$ for a particular set of values for the wavenumber, the distance between the slits, and the distance between the slit plate and the backdrop:

Observe that near the minima the probability of detection is less if both slits are open than it is if one slit is shut. It is customary to say that destructive interference occurs at the minima and that constructive interference occurs at the maxima, but do not think of this as the description of a physical process. All we mean by "constructive interference" is that a probability calculated according to Rule B is greater than the same probability calculated according to Rule A, and all we mean by "destructive interference" is that a probability calculated according to Rule B is less than the same probability calculated according to Rule A.

Here is how an interference pattern builds up over time :



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