My question is in regards to the stance that the 'wave function collapse' is not an actual physical occurrence. That is, you are not, by observation, changing the particles position from a wave to a particle, it is merely that the wave function is the probability of finding the particle at a particular point. In other words, the wave function is a figment of our mind, and merely represents our in-ability to know its position without measurement.

Here I am going to make an important assumption: This argument implies that the electron does not actually travel as a wave (1.0).


Assuming this argument is true. Consider electrons being fired one, by one, through the slits. How can an electron know that there are two slits open if it is not travelling as a wave (1.1)? A point particle cannot know if the slit above it, or below it, is open, unless it is at the two slits at the same instance (1.2). Therefore, at the point of the slits, it must be propagating as a wave (1.3). But, through observation of the slits (specifically behind the slits), we can force the electron to act as a particle (because it is observed to only go through one slit(1.4)) and create two simple bands on the plate (1.5). As such, the electron has gone from being a wave, before the slit, and at the slit, to a particle, through the act of observation(1.6).

I have included (1.i) to represent my argument in parts. Obviously I have made some error/(s) somewhere since many bright minds have this stance. I would appreciate it if my error could be pointed in terms of its position with (1.i)

A further note on (1.4) - is it even possible to observe which slit the electron has gone through, or does the uncertainty principal rule that out?

  • $\begingroup$ On your final comment: The uncertainty principle says nothing about measuring (only) the position of a particle. It only rules out knowing both the particle's position and its momentum (or any other observable corresponding to an operator that does not commute with the position operator) at the same time to arbitrary accuracy. $\endgroup$
    – Danu
    Jul 1, 2014 at 23:19

1 Answer 1


You have fallen prey to the same confusion that many people have with regards to the wave/particle duality:

The quantum objects that constitute our world are neither waves nor classical particles, and it is an error to believe that electrons/photons/whatever can "propagate as a wave" in one moment and "behave like a particle" in the next. The quantum objects do not change their behaviour, they are always states in a Hilbert space [1], which have some properties we would classically associate with particles (they possess position and momenta in a certain sense) and some properties we would classically associate with waves (they can show behaviour often referred to as superposition or interference).

Your whole argument rests on the premise that quantum objects somehow switch between being waves and particles. They don't. They have always been, are, and will always be states in a Hilbert space, which are, through observation/interaction (no consciousness required, ere someone gets the wrong idea here), sometimes projected onto a certain eigenbasis of an observable, and whose time evolution is governed by the Schrödinger equation [2]. How this measurement/interaction process is to be understood is the subject of many debates. Though I tend to favor a kind of Many Worlds Interpretation in my own thoughts, the fact of the matter is that is doesn't matter. Quantum mechanics, the incredible machinery of Hilbert spaces and $C^*$-algebras, of operators and eigenbases, works no matter how you interpret it. And its predictions are not influenced by our "interpretation" of its workings. This is what Feynman's (alleged) "Shut up and calculate!" is intended to convey.

To answer the double-slit experiment in question: Yes, it is possible to observe which slit the electrons go through by shooting photons at them, and indeed, the interference pattern vanishes. This is because the quantum state that represented $|\mathrm{left}\rangle + |\mathrm{right}\rangle$, i.e the electron going through both slits, was forced to interact with something (the photon) that forced it to take definite value in the position base, either $|\mathrm{left}\rangle$ or $|\mathrm{right}\rangle$, thus "collapsing" one of the two states (or, in another diction, entangling the state of the photon/observer with the state that they interacted with). Thus the state arriving at the screen where interference took place is now no superposition of the different ways the electron "could have taken" anymore, but a state which has a definite slit it has come through. At no point needs the quantum mechanical explanation to make any reference to "waves" or "particles". These concepts are relics from a classical age, and we should never forget that.

May the disagreements with what I have written begin ;)

[1] I would say Fock space, but I was reminded of the fact that that is not wholly universally true. It doesn't matter for arguments like these.

[2] QM is not the end of the story, though. QFT takes the divorce between the "intuitive" classical world of waves and particles and the "real" world even farther.

  • $\begingroup$ I am glad to have identified an error in my thinking of quantum mechanics before I started studying this subject in more detail. I have one more, slightly more innocent question to ask. You said by shooting photos at them you force the electron to take either left or right. Why doesn't this happen when doing the simple Young's Double Slit Experiment? When you can see the slits in person, there must be photons hitting the slits, why don't these interact with the photons of the laser and have the same effect of forcing the photons to a definite position value, causing two narrow bands? $\endgroup$
    – Blake
    Jul 2, 2014 at 0:07
  • $\begingroup$ Good point. It's not as simple as I wrote, you have to ensure that you somehow "catch" the photon afterwards, i.e actually interact with the entangled photon/electron state before the electron state hits the screen [I admit this sounds a bit weird to me myself, but I will believe it's true until someone convinces me otherwise] in a way that depends on the slit the electron has taken. If no interaction actually dependent on that takes place, the "collapse" doesn't happen, and photons which are merely there by accident do not constitute no such dependency. $\endgroup$
    – ACuriousMind
    Jul 2, 2014 at 0:47
  • $\begingroup$ Great post! "Shut up and calculate" is almost always the best approach. When first learning QM, I had some animosity towards that view and I wanted to know the "correct" interpretation. But as I began to learn more, I realized that "interpreting quantum mechanics" does a huge injustice to a beautiful theory. Humans think classically. That's just a fact. Looking for an intuitive explanation of a quantum effect is just trying to squeeze non-classical objects into a classical framework. My point is you can choose any interpretation you want, as long as it works. $\endgroup$ Jul 2, 2014 at 3:32

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