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Can we say that a measurement on a system of $k$ entangled particles is the cause of the collapse of the wave function into $k$th states simultaneously?

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    $\begingroup$ Just a small remark regarding entanglement: It isn't particles that are entangled, but rather quantum modes that they excite. One could very well have a photon entangled with the vacuum. It may seem like a small remark but it makes a world of difference in understanding that the excitations themselves aren't the carriers of entanglement. (Cf. for example, teleportation schemes with a single photon.) $\endgroup$
    – Tfovid
    Commented Oct 7, 2022 at 11:05
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    $\begingroup$ @FlatterMann Of course there is such a thing: the concept comes directly from experiment. That's physics. You can come up with an elaborate rationalization, but that's not physics: it's mathematics. $\endgroup$
    – John Doty
    Commented Oct 7, 2022 at 12:32
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    $\begingroup$ @FlatterMann That's your mathematical interpretation. What experiment can you perform to demonstrate that collapse doesn't happen? $\endgroup$
    – John Doty
    Commented Oct 7, 2022 at 12:49
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    $\begingroup$ @FlatterMann You assert that wavefunction collapse, a concept with a long and productive history, doesn't happen, and then you accuse me of arguing for argument's sake? $\endgroup$
    – John Doty
    Commented Oct 7, 2022 at 13:15
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    $\begingroup$ @FlatterMann I understand what you're saying but my point is that simply refuting something that is 'generally accepted' (I put generally accepted in quotes on purpose) is confusing for someone reading this. Are the textbooks wrong? Is this random comment on the internet wrong? It would be better to put this in context, for example: many introductory textbooks say wavefunction collapse is X, however I think this is not the case because of Y. $\endgroup$
    – AccidentalTaylorExpansion
    Commented Oct 7, 2022 at 14:30

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The following state is the GHZ entangled state for 3 particles: $$|\mathrm{GHZ}\rangle=\frac{1}{\sqrt2}(|000\rangle+|111\rangle)$$ where each ket $|q_1q_2q_3\rangle$ represents the state of the first, second and third particle in that order. In this case, the particles that can only be in two states $0$ or $1$.

If you measure (or "collapse") one of the particles in the GHZ state (in the basis given above), you will know the state of the other two. For example, if you measure the first particle to be in state $0$, you know already that particles #2 and #3 are also in state $0$.

Now if you have a W entangled state, such as: $$|\mathrm{W}\rangle=\frac{1}{\sqrt3}(|001\rangle+|010\rangle+|100\rangle),$$ measuring the state of a single particle will not necessarily disentangle the other particles. Example: Let us say that you collapse the first particle and you measure $0$, then the other two particles remain maximally entangled in the following Bell state: $$|\Psi^+\rangle=\frac{1}{\sqrt2}(|01\rangle+|10\rangle).$$

So no, collapsing the state of a single particle does not necessarily disentangle (or "collapse") the rest (in this basis).

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    $\begingroup$ A minor comment that might be interesting to the OP: In the GHZ state as well, if you measure one of the particles in the $\vert \pm\rangle$ basis, the other two particles will be left in the maximally entangled Bell state $\vert{00}\rangle\pm\vert{11}\rangle$. $\endgroup$
    – FreeAssange
    Commented Oct 10, 2022 at 13:31