# What is the Philosophy Behind Relational Quantum Mechanics?

I watched this video on Relational Quantum Mechanics yesterday and my brain has been trying to comprehend it since.

The interpretation I currently have is this: If an observer O measures the state S of a system, in doing so they themselves become entangled with that system.

To another observer O', the state of both S and O are still completely probabilistic until he measures them himself. However the state of S to O is now deterministic as they have become perfectly correlated with one another (through the measurement/entaglement). This is why to observer O it seems the wavefunction of S has collapsed as there is no longer any randomness in S relative to O (but this is not to say there is no randomness in the system (S+O) to an outside observer).

Further if I were not measuring a system (my friend let's say) they could be in one of a very large number of states with each having a certain probability. If I were to now look at them and 'measure' them, in effect one of these states is being sampled and I have become entangled/perfectly correlated with it.

My question is, have I completely missed the point or is this the general idea behind "Relational Quantum Mechanics"?

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Assuming that I understand it (I have an opinion that makes sense, but I don't guarantee it's the same as what they're trying to communicate), yes, that is the general idea. Wave function collapse is not a property of isolated quantum systems, or a real effect of observation- it's a perceptual effect on the observer. Once you interact with something to observe it, you are now entangled with it. You yourself are also in a superposition, but any state of you can no longer observe all the other superposed states of the system that you're part of. Other observers outside the system still can. –  Logan R. Kearsley Aug 13 '14 at 18:01
I would argue that relational quantum mechanics replaces one difficulty (what's a measurement?) with another one (how can we separate the quantum system from the observer in a universe that has only one wave function?). As such it is a refinement of the original axiomatic definition of QM, which states that there are quantum systems (for which it can give examples from inside the theory), and then there are classical observers (which are external, since the theory can't model them itself). If I understand it correctly, the hope of RQM would be to define the observer internally. –  CuriousOne Aug 13 '14 at 19:08
@CuriousOne I thought the observer was being defined as a quantum system also (albeit a very complex one). Then what we think of as classical observers are in fact manifestations of very complex quantum systems. This is why the observer themselves become entangled with whatever they are measuring and is why the wave function of what they are measuring appears to collapse. –  rwolst Aug 13 '14 at 21:28
@CuriousOne one wavefunction for the entire Universe means that the Universe has entropy 0 which is impossible. –  Bubble Oct 21 '14 at 11:43
@CuriousOne I was thinking in terms of von Neumann entropy and the fact that pure states have entropy 0. –  Bubble Oct 22 '14 at 10:05

You used the letter $S$ ambiguously to refer to both a system and the result of a measurement made on that system. It's important to distinguish them.

So the situation is that two observers $O_1$ and $O_2$ independently make a measurement $M$ of a system $S$ and agree that the result of that measurement is $X$.

It is tempting to conclude that the reason they agree is that $S$ is in fact in a state that corresponds to $X$, and that $X$ is a reflection of that underlying reality.

But that is not the case. The math tells us that $O_1+O_2+S$ is a macroscopic system of mutually entangled particles.

If you trace over $S$ (i.e. remove the degrees of freedom associated with $S$ from the description of the system) what you are left with is a description of a macroscopic system of mutually entangled particles in classical correlation. That is the reason $O_1$ and $O_2$ agree.

It has nothing to do with the state that $S$ is "actually" in. (I put "actually" in scare quotes because S is not actually in any particular state.)

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Hi user1136471, you should space you answer out so it's more easily readable. Also consider using MathJax. meta.math.stackexchange.com/questions/5020/… –  Brandon Enright Aug 14 '14 at 5:32