How exactly does which-slit information destroy interference?

I've seen in so many places online claiming:

"which-slit information destroys interference pattern"..."the act of observing which-slit causes the object to behave as a particle"... "by erasing the which-slit information, the object returns to behaving as a wave"

As if particles magically know to behave differently when somebody is watching. There is so much false information and I want to know if this is really correct. I feel like there must be a disturbance caused by measurement, or some other explanation as to why phenomena changes depending on how we measure it.

I'm OK with the double slit single-particle experiments. A single particle still produces an interference pattern, meaning it interferes with itself, the thing goes through both slits at once, is in superposition of states until collapsing according to wavefunction probability. OK. It's a quantum object. Fine. But the which-slit detectors trouble me.

I've looked at which-way interferometer experiments, polarization analyzer experiments, Stern-Gerlach experiments, quantum eraser experiments. I can't say I understand them completely. I'm not expecting a full answer, but hoping somebody can enlighten me, or point me in the right direction to get started. I've realised that it's probably impossible to understand this stuff without grasp some of the other quantum concepts, which I probably lack.

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    $\begingroup$ I think the answers and comments here might be helpful: physics.stackexchange.com/questions/343121/… $\endgroup$
    – JPattarini
    Commented Apr 8, 2021 at 18:59
  • $\begingroup$ Don't you think, that there must be a disturbance - or rather, some unspecified difference - caused by measurement, or some other explanation as to why phenomena change depending on how we measure it is the Answer, not the Question? $\endgroup$ Commented Mar 22 at 23:19

2 Answers 2


You can think of the double slit pattern as caused by the interference of two branches of the wave function, say $|L\rangle$ and $|R\rangle$. Now assume that there is which way information because of the presence of some form of detection. The detection apparatus can also be in two states $|\cal L\rangle$ and $|\cal R\rangle$. If the detector is perfect, without false positives or negatives, then these two states should be orthogonal. Also the slit wave function branches and the detector states should be entangled such that the combined states are $|L\cal L\rangle$ and $|R\cal R\rangle$. These states are orthogonal so there is no interference.

  • $\begingroup$ +1 for making Dr. Everett's ghost happy ;) $\endgroup$
    – user87745
    Commented Apr 9, 2021 at 0:09
  • $\begingroup$ +1 from Niels... $\endgroup$ Commented Apr 9, 2021 at 4:20
  • $\begingroup$ I read up "orthogonal states" and am not sure if I understand. Would I be correct to put this in layman's terms that basically the detector collapses the probability cloud at the point of detection, as if the particle had hit a "curtain", therefore its position is revealed and it cannot be superimposed? This seems like a disturbance to me! Can you confirm if detection measurement imposes a disturbance that destroys interference? If there were truly no disturbance, particles would have to be psychic! $\endgroup$
    – Tian
    Commented Apr 9, 2021 at 9:57

The answer is basically decoherence, which can be understood easily if you're familiar with some basic quantum information theory. Let's simplify away the physics of a double-slit experiment to simply a qubit in a superposition $\phi = (|0\rangle + |1\rangle)/\sqrt{2}$. Naively you might think this is a "qubit that's $|0\rangle$ or $|1\rangle$ but I just don't know which", but actually if you measure it in the $a = (|0\rangle + |1\rangle)/\sqrt{2}, b = (|0\rangle - |1\rangle)/\sqrt{2}$ basis you will get the $a$ outcome 100% of the time (assuming no experimental error), where as if you had $|0\rangle$ or $|1\rangle$ with 50% probability each you'd get $a$ or $b$ with 50% probability each. This is basically the same type of thing that goes on with the double-slit experiment in terms of interference.

The point is, outside of special conditions, we can't just think of $\phi$ by itself -- it interacts with the rest of the universe. Suppose it interacts with a qubit in a state $|0\rangle$ so that they as a system evolve to $(|00\rangle + |11\rangle)/\sqrt{2}$ (this is just an example meant to abstract away complicated physics of particles interacting -- things essentially like this can and do happen). You could still detect an interference by performing a measurement on both particles (e.g. measure in the Bell basis). But what if this other qubit was just a stray particle bouncing around and you have no idea where it is now? If you only have access to the first qubit, the information available to you is given by the local density matrix, which we obtain from $(1/2)(|00\rangle \langle00| + |00\rangle \langle11| + |11\rangle \langle00| + |11\rangle \langle11|)$ by tracing out the second qubit, which gets us $(1/2)(|0\rangle \langle0| + |1\rangle \langle1|)$. This is now the mixed state which represents a 50% chance of $|0\rangle$ and a 50% chance of $|1\rangle$ -- we've lost the interference.

The only way to observe the interference is to perform a joint measure on both qubits. Even in the case of a single particle or something like that, it might be impossible because you've lost track of it (it could be receding away from you at the speed of light and now good luck catching it) -- but more to the point, instead of a single qubit it could be an immensely complicated system of zillions of qubits, like a macroscopic which-way device.

If you push this logic an $\epsilon > 0$ further you end up arriving at the many-worlds interpretation, which lots of people don't like, but it's what the math says. For more, see e.g. here https://arxiv.org/abs/1908.03920


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