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I have a quantum mechanical system situated within Euclidean space $\mathbb{R}^3$ represented by the state $|\psi\rangle \in \mathcal{H}$, where $\mathcal{H}$ is the corresponding Hilbert space for my system. If I transform my frame of reference in $\mathbb{R}^3$, is this represented on the Hilbert space of states as

  1. a transformation of my state vector $|\psi\rangle \mapsto |\psi'\rangle = U|\psi\rangle$, whilst observable operators remain constant $O\mapsto O$, which I interpret as an active transformation,
  2. a transformation of my observable operators $O\mapsto O' =U^\dagger O U$ whilst states remain constant $|\psi\rangle \mapsto |\psi\rangle$, which I interpret as a passive transformation,
  3. or both of the above at the same time?

When we make a change of reference frame, we expect observables which are not scalar quantities to change (i.e. a shift in displacement from origin, components of momentum). My first guess would be the passive one because I like to view the state vector $|\psi\rangle$ as an abstract vector independent of observer so it should remain invariant. Instead under a frame transformation it is the measuring apparatus that is transforming (i.e. the fixed axes and it’s scale, hence observable operators), but the active one should be equally valid too as it is expectation values $\langle \psi | O | \psi \rangle$ that are the physical things, so I’m happy to accept either 1 or 2.

However if both states and observables transformed, we would find identical eigenvalues, i.e. suppose $O|\psi\rangle = \lambda |\psi\rangle$, if we transformed both state and observable we would find identical eigenvalues: $O'|\psi'\rangle = \lambda |\psi'\rangle$ therefore if this was the case, the states give identical expectation values for all observers which surely is not what we want.

Now in A Course in Modern Mathematical Physics by Peter Szekeres, he sets the scene for Wigner's theorem by explicitly stating observers in different frames observe both a transformed state and have transformed observable operators... What is going on?

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  • $\begingroup$ @DanYand I agree, a frame transformation should leave scalar quantities like the magnitude of vector quantities invariant, such as momentum. However, the components of the momentum will change between frames and that is what I’m discussing here. The eigenvalues of the operator $p_x$ corresponding x-component of momentum will change between frames. If both states and operators evolved then $p_x = p’_x$ which is nonsense $\endgroup$ – Matt0410 Nov 17 '18 at 10:32
  • $\begingroup$ When we rotate the whole universe, do we also rotate the frame the universe is situated within? I see no reason why. If the frame and universe both rotated then nothing observable would happen but if we didn’t rotate the frame we’d observe everything has rotated about the origin so eigenvalues should change $\endgroup$ – Matt0410 Nov 18 '18 at 17:18
  • $\begingroup$ Yes that is what I am confused about - the eigenvalues of frame dependent quantities should change between frames when looking at the same state. Your second to last comment above only tells me that the eigenvalue spectrum of the transformed operator will be the same. I am asking why if both observers observe the same state, surely the eigenvalues they measure for $|\psi \rangle$ vs. the transformed frame state $U |\psi \rangle$ should have different eigenvalues which can only be acheived if we transform operators or states, not both $\endgroup$ – Matt0410 Nov 19 '18 at 21:39
  • $\begingroup$ I haven't tried posting a real answer (just comments) because I'm not sure how to answer without being sure I understand what is meant by "frame". When I'm unsure about the meaning of a word (or notation), my usual policy is to look at the context (such as how the equations are being used) to get more clues; but since I don't have Szekeres' book, I can't do that. All I can do is tentatively suggest that this looks very much like a language issue, not a paradox with quantum mechanics. Hopefully somebody else can step in and help clarify things. $\endgroup$ – Chiral Anomaly Nov 21 '18 at 4:45

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