Feynman's non-rigid double slit experiment

Let's say you measured the momentum of the screen to be $p_m$, with uncertainty $Delta p$. You want to know if $p_m$ is consistent with $p_1$ (the momentum imparted on the screen if the electron goes through hole 1) or if it is consistent with $p_2$ (the momentum imparted on the screen if the electron goes through hole 2). Note that I have changed notation slightly from Feynman in that I am using $p_1$ instead of $delta p_1$ for the momentum of the screen after the electron goes through hole 1, to avoid needing to distinguish $Delta p$ from $delta p$.

Let's think about the different possibilities.

1. If $p_1$ lies in the interval $[p_m-Delta p/2,p_m+Delta p/2]$, and $p_2$ does not lie in this interval, then the results of your experiment are consistent with the electron going through hole 1, and not with the electron going through hole 2. You conclude the electron went through hole 1.
2. Similarly, if $p_2$ lies in the interval $[p_m-Delta p/2,p_m+Delta p/2]$, and $p_1$ does not lie in this interval, then you can conclude from your experiment that the electron went through hole 2.
3. If both $p_1$ and $p_2$ lie in the interval $[p_m-Delta p/2,p_m+Delta p/2]$, then you cannot conclude which hole the electron went through.
4. If neither $p_1$ nor $p_2$ lie in the interval $[p_m-Delta p/2,p_m+Delta p/2]$, then something has gone wrong with your setup; we won't consider this case further.

The key question is whether the experiment lies in cases 1 or 2, or in case 3. If you do some thinking, you can see (assuming we are not in case 4) that we will be in case 3 if

begin{equation} Delta p > |p_1 - p_2| end{equation}

Similarly, we will be in either case 1 or case 2 if $Delta p < |p_1-p_2|$.

If we are in cases 1 or 2 ($Delta p < |p_1-p_2|$), then we can tell which slit the electron has gone through.

If we are in case 3 ($Delta p > |p_1-p_2|$), then we can't tell which slit the electron has gone through.

An important point is that how to interpret the measurement uncertainty. A simple (but common) model is to imagine that the measuring device produces a random outcome drawn from a Gaussian distribution, with a mean equal to the true value of the momentum, and a variance given by $(Delta p)^2$. A key point for this experiment is that the measuring device will report a continuous value (or if there is some discretization in the values it is reporting, that discretization is so small we don't care). Rather than a meter stick, think of a stop watch with a lot of digits, or a digital voltmeter.

Let's suppose the true value of the momentum was $p_1$, then the measuring device will produce an output $p_m$ given by

begin{equation} p_m = p_1 + p_{rm err} end{equation} where $p_{rm err}$ (the error due to the fact that the measurement device is not perfect) is a random variable drawn from a Gaussian distribution. The probability distribution for $p_{rm err}$ is explicitly begin{equation} P(p_{rm err}) = frac{1}{sqrt{2pi (Delta p)^2}} e^{-frac{(p_{rm err}-p_1)^2}{2 (Delta p)^2}} end{equation} If $Delta p$ is small, then the probability that $|p_{rm err}|$ will be large enough that $p_m$ will be closer to $p_2$ than $p_1$ will also be very small -- this means we can very cleanly distinguish $p_1$ and $p_2$ (we are in case 1 above). If $Delta p$ is large, then there is a reasonably large probability that $|p_{rm err}|$ will be large enough that $p_m$ will be closer to $p_2$ than $p_1$, so we won't really be able to tell with much confidence whether the momentum is $p_1$ or $p_2$.

Note that the origin of the measurement uncertainty $Delta p$ is not quantum mechanical, it is just that any real measurement is not perfect. (Just as an example, maybe there are thermal fluctuations that cause the screen to jiggle up and down randomly and $Delta p$ is a typical value of the momentum due to these thermal fluctuations).