Why doesn't the double-slit experiment violate the uncertainty principle?

In the double-slit experiment, when an electron reaches the detector after passing through the holes, it has a certain momentum which we can measure with arbitrary accuracy.

From this data, we can calculate what momentum the electron had when it was passing through the hole.

So we can know the electron's position (as it was inside the hole) with arbitrary accuracy and also simultaneously know its momentum.

Doesn't this violate the uncertainty principle? Where have I gone wrong?

• "From this data, we can calculate what momentum the electron had when it was passing through the hole." How do you intend to make the measurement? A beam of particles passing through a double slit would give a interference pattern on the screen. Suppose you place a detector at an angle from the particle beam, from the probability distribution, the probability of you detecting a particle there is non - zero, but that would violate the conservation of momentum, so how are you going to calculate the momentum of the electron at the hole? Jun 22 '21 at 11:05
• It says, "simultaneously and with arbitrary precision." Jun 22 '21 at 11:21
• What is calculated retrospectively isn't what was measured at the time… Jun 23 '21 at 0:09
• @YoungKindaichi Yeah arbitrarily. Didn't I include the word in the question...Anyway the momentum can be measured with arbitrary accuracy. Also we can make the hole as small as we want and can have arbitrary precision in momentum. Jun 23 '21 at 8:16
• I have difficulties with the statement “So we can know the electron's position (as it was inside the hole)…”. Isn’t the whole point of the double-slit experiment that there are two of them, so even the phrase “the hole” isn’t fitting, not to speak of claming to know a position? Jun 23 '21 at 14:36

The narrower the slits ($$dx$$), the broader the expected (measured and/or calculated) distribution of momentum ($$dp$$) of the photons passing said slits, so the product ($$dx.dp$$) cannot get arbitrarily small, therefore the Heisenberg principle ($$dx dp \geq \hslash$$) is respected in the double slit experiment.

• In fact, as the slit get narrower, the interference pattern becomes wider. Jun 22 '21 at 14:19
• But that didn't address my question. My question is that we can calculate the position and momentum with arbitrary accuracy. How does it not violate the uncertainty principle. Please read my question again Jun 23 '21 at 8:19

Where have I gone wrong?

Here, the noted with italics.

when an electron reaches the detector after passing through the holes, it has a certain momentum which we can measure with arbitrary accuracy.

Any measurement means new quantum mechanical interactions, i.e. new wavefunctions and boundary conditions on them. In any case momentum is measured also within the Heisenberg uncertainty bounds.

Also here:

So we can know the electron's position (as it was inside the hole)

The width of the hole is the accuracy we can experimentally have, there is no way it can be better.

• I think the question's idea is that one can, in principle, make the hole as small as you want, and given any fixed hole you can, in principle, measure the momentum as accurately as you want (the uncertainty principle does not rely on practical experimental constraints). I believe neither of the claims you pinpointed are incorrect in this perspective, but rather the claim that we can infer the electron's momentum at the time it passed through the hole from its momentum at the time of measurement. Jun 22 '21 at 18:21
• @jawheele "the uncertainty principle does not rely on practical experimental constraint" it is not experimental uncertainty I am talking about but the quantum mechanical uncertainty of the dimension of atoms and molecules that define the slit. They are constrained by the HUP. The HUP is the result of the properties of the commutators in the quantum field theory which describes the world of atoms and molecules. Even theoretically one cannot make a slit that will defy the HUP of the particular interaction. Also momentum is a conserved quantity unless there is an interaction. Jun 22 '21 at 18:29
• I maintain that's a practical constraint. One can model this problem as an electron in a potential that is infinite on a thin plane with a small opening and zero everywhere else. The constraint you discuss is being able to construct a system with this potential, but the model system obeys the HUP regardless of this. Also, if the inference were as simple as momentum conservation, it seems to me we could only ever measure expectation values (the quantity actually conserved). Jun 22 '21 at 19:00
• The momentum conservation comment goes on the observation of measuring momentum at one point. The momentum of the particle is known after that until there is another interaction. The double slit is an experiment in the real world, not in the mathematics. Jun 23 '21 at 3:56
• OP is not trying to infer the momentum after-- they measure it at the screen, and want to deduce its value before, when the particle was passing through the hole. Yes, the double slit is a real world experiment, but the OP is clearly idealizing and asking a question about that idealization. I highly doubt OP actually believed they could construct an apparatus to measure anything to arbitrary accuracy. Citing that as the heart of the issue feels akin to responding to a question on the speed of a gravitational circular orbit with "circular orbits don't exist, so don't worry about it". Jun 23 '21 at 5:41

In the double-slit experiment, when an electron reaches the detector after passing through the holes, it has a certain momentum which we can measure with arbitrary accuracy.

The weird thing about quantum mechanics is that the electron actually does not have a definite momentum. It would only have a definite momentum if it were in an eigenstate of the momentum operator (which is not physical, due to not being normalisable). When you measure the momentum, the electron transitions to (an approximation of) an eigenstate of the momentum operator, corresponding to the measured (eigen)value; i.e. measurement changes the state of the electron.

So we can know the electron's position (as it was inside the hole) with arbitrary accuracy and also simultaneously know its momentum.

Note the tense here. The Heisenberg uncertainty principle places a limit on how precisely the position and momentum can be defined (not just known) for a single state. However, your measurements of the position and momentum were at different times, and since each measurement changes the states of the electron, the measured values describe different states. Thus, the potential arbitrary precision in the measurements does not violate the Heisenberg uncertainty principle.

From this data, we can calculate what momentum the electron had when it was passing through the hole.

We can't do this. It is true that in principle, we can measure all three components of momentum $$p_x,p_y,p_z$$ with arbitrary precision at single time, but in general, we do not know how to extrapolate this to past to find "particle's components of momentum" at some known previous time or assign momentum to some region of space ("hole in the wall").

We can do naive extrapolation by assuming that momentum did not change at all, so the particle always had the momentum we measured. But this is unphysical, because we know the particle had to interact with the wall, and with other bodies before that. So this naive extrapolation is good only for limited span of time/region of space where interaction is believed to be negligible.

So we can know the electron's position (as it was inside the hole) with arbitrary accuracy and also simultaneously know its momentum.

No, because we can't extrapolate the momentum measurement back to time when the particle was "inside the hole", and also we do not know exactly where the particle is when we find its momentum, so we can't just draw a line from "position of particle when its momentum was measured"; there is no such single point.

• Thanks, I didn't thought about the fact that the extrapolation will not work because of the "quantum weirdness" . This helped me.... Jun 23 '21 at 8:28
• This answer is not correct. Real momentum measurements never measure the particle's momentum at the original location. Rather, the particle flies for a bit and then through extrapolation the average momentum over its flight time is measured. Take for example recoil ion momentum spectroscopy, or measurements at the LHC. This is Jun 23 '21 at 9:24
• @doublefelix I fail to see how this answer is meaningfully different from yours, which also emphasizes the distinction between the times of the measurement. This answer's point is that one should not conflate the momentum (and hence its uncertainty) at the time of screen measurement with the same quantity at the time of passing through the slit, particularly given the particle's interaction with the wall-- this seems like exactly the same point your answer makes. Ján just used words instead of symbols like $\psi_\text{after}$ and $\psi_\text{before}$. Jun 23 '21 at 22:32
• @doublefelix My answer is in the context of standard quantum theory, where it is assumed that physical quantities like momentum components can be measured at single point of time, even if that is practically impossible or very hard. The point is that even if we could do that, knowledge of momentum at some point between the slits and the screen still does not allow us to extrapolate momentum to the slits in the wall. Jun 23 '21 at 22:38

I don't believe the accepted answer is correct. Here is the resolution: The uncertainty principle $$\sigma_x \sigma_p \geq \hbar/2$$

is derived for the state at a particular time $$\psi(x,t)$$. However in your case, there are two different states involved: the position measurement$$^*$$ is done on the wavefunction before interaction with the screen, and the momentum measurement is done after the interaction with the screen and wavefunction.

So your question compares $$\sigma_x$$ for $$\psi_{before}$$ against $$\sigma_p$$ for $$\psi_{after}$$, and the uncertainty relation does not apply for these two unrelated quantities. The interaction with the screen causes $$\psi_{before} \neq \psi_{after}$$.

*Note: Since it is not done at a fixed and chosen time, the double slit screen is not a "textbook" position measurement either, so the usual derivation for the uncertainty relation does not apply. This is a separate discussion.

In order to measure through which hole the electron goes you need a device with at least two, orthogonal, states, say L and R. The setup must be such that L and R are coupled to the left and right part of the wave function, respectively, so that $$\psi = |L {\cal L} \rangle + |R {\cal R} \rangle \,.$$ This means that the two parts of the wave function are orthogonal so that no interference occurs

S, yes, you can in principle measure through which slit the particle went but this eliminates the interference pattern.