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As a supplement to this question, as to whether particles can be observers, let us suppose that the answer is yes. One could suppose a setup where particle $A$ is observing particle $B$, but what is to stop us switching viewpoints around here and supposing particle $B$ is observing particle $A$?

I find this is an intriguing possibility considering the importance of symmetry in Physics.

Question: Is there a symmetry of observed & observer in QM?

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  • $\begingroup$ Observers are irrelevant to quantum mechanics. They are only relevant in interpretations that involve wave function collapse, which itself is unphysical. $\endgroup$
    – my2cts
    Commented Apr 12, 2021 at 22:28
  • $\begingroup$ @my2cts: Observers are not irrelevant to QM. It's wrong to think that interpretations of QM are not QM; they manifestly are. This is why Bohr, Heisenberg and Einstein discussed the ontology of QM. It's also wrong to say that collapse is unphysical - it's only in Many Worlds and similar interpretations that the collapse postulate is said to be unphysical; there are many others where it is taken to be physical. $\endgroup$ Commented Apr 12, 2021 at 22:58
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    $\begingroup$ I couldn't disagree more. $\endgroup$
    – my2cts
    Commented Apr 12, 2021 at 23:36
  • $\begingroup$ Same here. Anything beyond the minimal interpretation is philosophy, not science. It is fine for scientists to have philosophy, but that doesn’t make it science. $\endgroup$
    – Dale
    Commented Apr 12, 2021 at 23:52
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    $\begingroup$ @Mozibur Ullah I am not philosophically unaware, I am anti-philosophy in the context of science. Science has progressed substantially since the days of Rutherford, et al. Philosophy, not so much. $\endgroup$
    – Dale
    Commented Apr 13, 2021 at 1:40

3 Answers 3

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Particles don't observe each other, they simply interact according to the Hamiltonian (if they didn't quantum mechanics wouldn't be able to explain the hydrogen molecule, for instance).

If you have two macroscopic observers, and they try to observe each other they would both collapse individually (though their collective wave function might remain the same, depending on interpretation).
However, this effect is absolutely unnoticeable because most states are experimentally indistinguishable from thermal equilibrium (look into quantum equilibration to understand more on this).

So: yes, symmetry is preserved.

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I guess this sort of question depends heavily on the interpretation of QM you are "believing" in. However, IMO, measurement or observing is a undirected relation. Consider a two-dim. system with orthogonal states $| 0\rangle$ and $| 1\rangle$. The von Neumann "pre-measurement" will establish entanglement between you, the experimentalist observing the system, and the system itself: $(|0\rangle|\text{you measure 0}\rangle+|1\rangle|\text{you measure 1}\rangle)/\sqrt{2}$.

The Copenhagen interpretation would suggest that the state collapses into one of the two terms. In relative state interpretations, each term would correspond to a distinct reality or "world". But regardless what interpretation you choose, the measurement process is symmetric and it's meaningless to ask whether you measure the system or the system measures you. Note however, that I'm not aware to all interpretations of QM.

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Oops. Read only the title, so answered the question, if an observer can observe themselves. The actual question is meaningless because if one person observes the other, then the one who is observed is not the observer.

Answer to the question I (wrongly) inferred from the title:

Short answer: no.

Full answer: in a system that properly includes the observer, there are states that the observer cannot distinguish in principle. Thomas Breuer called it "subjective decoherence".

So, the observer can try to observe a system where he is properly included, but he will fail determining its quantum state. In other words, the quantum state of such system does not exist. The wavefunction is not well defined.

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