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Observables correspond to hermitian operators on the quantum state.

But in the Everett interpretation, the wave function doesn’t collapse since we consider the entire universe as a single quantum state at time $t$, so observation only happens “within that quantum state” (or something like that).

This confuses me. Given that observations are not treated differently than other events in the Everett interpretation, how does this relate to observations being hermitian operators?

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Even in traditional interpretations of quantum mechanics it is easier to define what is an observable than what is an observation: it is a property of a quantum system or subsystem that we in principle have access to through observation (whatever the latter means).

If we say that this is a decomposition in orthogonal subspaces of the state space, which are distinguished by a real numerical value assigned to it, called the outcome of the measurement, as does the measurement postulate, we don't need to know anymore what we mean by an observation.

In Everett's interpretation an observation is described by unitary evolution just like everything else. A decomposition of the state space in orthogonal subspaces however remains meaningful, just like in traditional interpretations. Let's still call the associate number an outcome. Disregarding subtleties when working in infinite dimensional state spaces, orthogonal decompositions in subspaces indexed by some real numbers are exactly equivalent to hermitian operators (just pick your orthonormal basis $|\lambda\rangle$ with corresponding outcome $\lambda$. The corresponding hermitian operator is $\sum\lambda|\lambda\rangle\langle\lambda|$. This happens to be one of the occasions where I really like the Dirac notation). It could be argued that the former, the decomposition with outcomes, actually is the more fundamental definition of an observable.

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    $\begingroup$ This post is a bit unclear. Just because you can decompose the Hilbert space into orthogonal subspaces doesn't mean you have explained which outcome or outcomes we (and our many worlds counterparts) actually experience. Furthermore, the decomposition of the Hilbert space is not unique so the question can be asked why it appears to us, after measurement, that we have measured in a particular basis when the underlying structure prefers no particular basis. $\endgroup$ – jgerber Sep 25 '18 at 2:32
  • $\begingroup$ @jgerber Decoherence does have the ability to pick out a preferred basis. Typical decoherence processes such as particle scattering have been extensively studied starting in 1970s. They show that due to the local nature of QED interactions, the position basis is typically preferred - a process call localization. This is what justifies describing a particle using a wave packet. $\endgroup$ – Bruce Greetham Sep 25 '18 at 3:25
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    $\begingroup$ @jgerber thanks. What I describe is what observables are and why they are encoded in hermitian operators, independent of whether we can actually make an observation in that basis, and independent of what an observer would experience. An observable encodes a basis, but doesn't prefer one. An experimental setup tries to mould the world to (temporarily and locally) "prefer" one decomposition over another, e.g. in a Stern-Gerlach experiment two orthogonal spins are preferred, in an idealized CCD array you have one subspace for each pixel and a large component for whatever misses the array etc. $\endgroup$ – doetoe Sep 25 '18 at 6:13
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While many worlds interpretation provides a complete and thorough description of the state of quantum systems (including observers as quantum system), it doesn't provide a prescription for how the physical state of the system is related to the experience of observers.

In many worlds there is no significance to any operators on the Hilbert space unless those operators happen to be part of the Hamiltonian (which is Hermitian) which is used to determine the time evolution of the universal wavefunction.

Otherwise there is no special significance to Hermitian operators. One might say that many worlds doesn't give a prescription for how observations happen so there are no such thing as observations in the many worlds interpretation and thus there is no need for observables.

To answer your question directly

Given that observations are not treated differently than other events in the Everett interpretation, how does this relate to observations being hermitian operators?

Since observations are treated no differently than other events (unitary evolution through the Schrodinger equation for all events) there are no observation so there is no relationship between observations and hermitian operators.

Because MWI doesn't give a prescription relating our experiences to the physical states I see it as being a fundamentally incomplete and scientifically insufficient interpretation of quantum mechanics. I would think that anyone who says otherwise is implicitly or explicitly adding additional postulates to what is usually understood as the many worlds interpretation, that is pure wave mechanics.

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    $\begingroup$ I think this is wrong. you say: "Since observations are treated no differently than other events (unitary evolution through the Schrodinger equation for all events) there are no observation so there is no relationship between observations and hermitian operators." But that's really not true I think. observation still happens, but it is done by one part of the quantum system in relation to another part of the quantum system. We can simply treat the whole observing apparatus as part of the quantum system (and it will be in a superposition with the thing it's observing) $\endgroup$ – user56834 Sep 25 '18 at 7:34
  • $\begingroup$ I agree that in MWI the measurement apparatus and the human looking at the measurement apparatus can all be considered as part of the quantum system. I also agree that relationships/entanglement/correlation develop between these different subsystems in MWI. MWI does a nice job of making these correlations apparent to us. What I disagree with is your statement that "observation still happens". The end result of an observation is the observer having the subjective conscious experience of having seen something. MWI doesn't make the link between the physical state and the observer's experience. $\endgroup$ – jgerber Sep 25 '18 at 7:55
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    $\begingroup$ That is not the physicist's meaning of measurement. Physical quantum measurement is the interaction with a quantum system in order to extract classical information. Any other quantum system can achieve this if (1) it can entangle with the measured system differently for the different basis states being measured and (2) it has enough degrees of freedom (ideally infinite) to ensure decoherence can effectively supress interference terms when it measures a superposition of basis states. $\endgroup$ – Bruce Greetham Sep 25 '18 at 9:13
  • $\begingroup$ If your definition of measurement is that the system apparatus become entangled with the measurement apparatus in a particular way then yes, measurements can happen in the MWI. If you want to call observation the same thing as measurement then yest observation can happen in MWI. In either case, no special status is given to any hermitian operators other than the Hamiltonian which drives the Schrodinger equation. You can express the dynamics in one basis or another and things will look qualitatively different (because the structure of entanglement is governed by the form of the interaction... $\endgroup$ – jgerber Sep 25 '18 at 9:24
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    $\begingroup$ I think I'm getting into bad form with so many comments here but my final note is to emphasize that decoherence by itself doesn't solve the preferred basis problem, it just wags its eyes suggestively at it. See arxiv.org/pdf/quant-ph/0312059.pdf $\endgroup$ – jgerber Sep 25 '18 at 9:29

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