The setup

To be able to apply Rule A or Rule B to this setup, we need to identify the initial measurement M₁ (on the basis of whose outcome probabilities are assigned), the final measurement M₂ (to the possible outcomes of which probabilities are assigned), and a set of alternatives.

The initial outcome is the launch of an electron by the electron gun G. (If G is the only source of free electrons, then the detection of an electron behind the slit plate also indicates the launch of an electron in front of the slit plate.)

The final outcome is the detection of an electron at the backdrop, by a detector D at x. The alternatives (that is, the possible intermediate outcomes) are

The corresponding amplitudes are A_L and A_R. To be able to calculate them, all we need to know is that

Thus according to Rule A the probability of detecting at D an electron launched at G is

p_A(D) = |<D|L><L|G>|² + |<D|R><R|G>|².

If the slits are equidistant from G, then <L|G> = <R|G>, and p_A(D) is proportional to

|<D|L>|²+|<D|R>|² = 1/d(DL)² + 1/d(DR)².

Here is the resulting plot of p_A against the position x of the detector:

p(x) according to Rule A

Integrate p_A(x) over any interval I=[a,b] of the x axis, and obtain the probability p_A(I) with which the electron is found by a detector monitoring this interval:

p_A(I) = ∫_I p_A(x) dx.

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

According to Rule B the probability of detecting at D an electron launched at G is proportional to

|<D|L> + <D|R>|² = 1/d(DL)² + 1/d(DR)² + 2 cos(kΔ)/[d(DL)·d(DR)],

where Δ stands for the difference d(DR)–d(DL). The last term, which is missing from p_A(D), is often referred to as  interference term. The proportionality constant k in the argument of the cosine depends on the momentum with which the electrons are launched. Here is the plot of p_B against x (for a particular choice of values of k, of the distance between the slits, and of the distance between the slit plate and the backdrop):

p(x) according to Rule B

Observe that near the minima the probability of detection is less if both slits are open than it is if only one slit is open.

Here is how such an "interference pattern" builds up over time:

Number of electrons detected: b: 100, c: 3000, d: 20000, e: 70000

And here is our next million-dollar question: Suppose that the conditions stipulated by Rule B are met; there is nothing — no event, no state of affairs, anywhere, anytime — from which the slit taken by an electron can be inferred. Can it be true, in this case,

The answer, once again, is a resounding NO.

To keep the language simple, we will say that an electron leaves a mark where it is detected at the backdrop. If each electron goes through a single slit, then the observed distribution of marks when both slits are open is the sum of two distributions, one from electrons that went through L and one from electrons that went through R:

p_B(x) = p_L(x) + p_R(x) .

If in addition the behavior of an electron that goes through one slit does not depend on whether the other slit is open or shut, then we can observe p_L(x) by keeping R shut and we can observe p_R(x) by keeping L shut. What we observe if R is shut is the left dashed hump, and what we observed if L is shut is the right dashed hump:

p(x) according to Rule A

Hence if these two conditions (as well as those stipulated by Rule B) are satisfied, we will see the sum of these two humps. In reality we see this:

p(x) according to Rule B

It follows that those those two conditions and the conditions of Rule B cannot be simultaneously satisfied. If Rule B applies, at least one of those two assumptions is wrong.