I am currently reading Quantum Mechanics and Path Integrals by Feynman and Hibbs, and in the very first chapter, Feynman qualitatively tries to motivate the Uncertainty Principle. He started out with his modified double-slit experiment with detectors placed behind the slits and argued that you can't detect which slit the electron went through if you want to preserve the interference pattern.

In the second experiment, he uses a non-rigid slit screen, and says that we can use the measurement of momentum imparted to the screen as a proxy for detecting which slit the electron went through. His argument is as follows-

"An electron passing through slit 1 is deflected by a different amount than the electron passing through slit 2. The change in momentum of the screen can be measured to detect the slit through which the electron went through. Call $\delta p$ the difference in momentum change between electrons passing through slit 1 and slit 2. Then an unambiguous determination of the slit requires a momentum determination of the screen to an accuracy better than $\delta p$." Now, I was puzzled with two questions about this experiment-

Question 1 Let's say an electron goes through slit 1, and suffers a momentum change $\delta \vec{P_1}$. Similarly, for an electron going through slit 2, let the momentum change be $\delta \vec{P_2}$. How can then $\delta p=|\delta \vec{P_1}-\delta \vec{P_2}|$ possibly be a measure of the accuracy required for momentum determination?

enter image description here

To frame my confusion better, let's say that $\delta \vec{P_1}=-1.3 \hat x$ (a downward impulse) and $\delta \vec{P_2}= 1.1 \hat x $ (an upward impulse). This means that $\delta p = |1.3+1.1| = 2.4$. Now in no way will an accuracy around $2.4$ be useful because it will measure impulses like $\delta \vec{P_1}$ and $\delta \vec{P_2}$ to be $0$. So, Feynman's argument for defining $\delta p$ in that fashion makes no sense to me. I'll link the excerpt from the book at the end.

Question 2 A further question this raises is as follows. Let's say this definition of $\delta p$ does make sense, then Feynman argues using interference arguments that $d= \frac{h}{\delta p}$ where $d$ is the distance between 2 maxima of the interference pattern. Now according to experimental observation, the interference pattern is killed, which must mean that $\delta x$ due to the uncertainty in the position of slits was $\delta x > \frac{d}{2}$ (which averages out the pattern). Thus-

$$\frac{d}{2}=\frac{h}{2 \delta p} \implies \delta x >\frac{h}{2 \delta p} \implies \delta x \delta p > \frac{h}{2}$$ which agrees with the statement of the uncertainity principle.

But unlike the previous experiment where we used scattering of light and the wavelength of light was the limiting factor, I don't see any such lower bounds for momentum measurements here. If I send electrons with lower and lower energy, they will cause lesser jerkiness in the slit-screen, and eventually at some point, I will recover the interference pattern. All I will need to do is improve my momentum accuracy and I don't see any lower bound on that. Of course, we are only using classical arguments here, but that is how Feynman concluded all these things and I think I am missing some point here.

Any help is deeply appreciated.

From Feynman and Hibbs, page 10 enter image description here

  • $\begingroup$ The uncertainty principle has absolutely nothing to do with quantum mechanics. It applies to all wave phenomena including classical waves. Technically it's not even a "physics thing". It's a more or less trivial lemma about operators in functional analysis. I would also warn you not to take textbooks in which quanta have paths too seriously. Feynman was a very smart guy, but some of the things in his textbooks are outrageously false. $\endgroup$ Commented Oct 3, 2022 at 14:06

1 Answer 1


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{2\pi (\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).

  • $\begingroup$ sure, that argument would have made sense if momentum were just a scalar quantity. but since the momentum also has a direction attached to it, even if the magnitude of momenta $p_1$ and $p_2$ both lie in the same interval, they still correspond to different momenta, one of them being imparted upwards and other downwards. That's precisely where I am confused. If this was a measurement like energy, this would have made complete sense, but the same magnitude of momentum $p$ can mean both going through the upper, or going through the lower slit. $\endgroup$
    – Tachyon209
    Commented Dec 28, 2020 at 22:09
  • $\begingroup$ @Tachyon209 Hm, I interpret Feynman's and Hibbs's setup to involve a screen that can only move along one direction, in which case the momentum imparted on the screen is a (signed) scalar quantity. Then $|p_1-p_2|$ is a meaningful metric for the difference in the two momenta. For example, perhaps $p_1=+p$, for some positive constant $p$, representing an "upward" momentum, and $p_2=-p$. Then $|p_1-p_2|=2p$, which is a sensible result. Do you disagree? $\endgroup$
    – Andrew
    Commented Dec 28, 2020 at 22:13
  • $\begingroup$ I don't really understand what you mean by a signed scalar quantity. If you do have a sign attached to the momentum, that should mean that the screen moves in both directions. As for your example, if $|p_1-p_2|=2p$ is our accuracy, that is what creates all the problems. Say an electron imparts a momentum $p_m=+0.9p$ to the screen. Your setup considers all the momenta between $+p$ and $-p$ to be 0, and thus such a measurement would imply that no momentum was imparted to the slit-screen. And that is what confuses me because this asks for such low accuracy. $\endgroup$
    – Tachyon209
    Commented Dec 28, 2020 at 22:44
  • $\begingroup$ Also, this might be a dumb question but doesn't the accuracy of $\Delta p$ mean that the range $[p_m-\Delta p,p_m+\Delta p]$ would be interpreted as $p_m$ instead of the range $[p_m-\Delta p/2,p_m+\Delta p/2]$ since a measurement written with an error is always written as $p_m \pm \Delta p$? $\endgroup$
    – Tachyon209
    Commented Dec 28, 2020 at 22:48
  • 1
    $\begingroup$ @Tachyon209 OK I have added a bit more, hopefully this helps! $\endgroup$
    – Andrew
    Commented Dec 29, 2020 at 13:45

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