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Let's consider a quantum system in a state $\left| \psi \right>$ relative to an observer using a coordinate frame $F$. We now introduce a second observer, using a coordinate frame $F'$, related to $F$ by a rotation by an angle $\theta$ about some fixed axis (as an example, the details of the transformation do not really matter). Now in quantum mechanics, we represent the transformation between those two reference frames by a unitary operator $\hat{U}$. This operator can be applied to the state $\left| \psi \right>$ to obtain the state of the system relative to $F'$. On the other hand, if $\hat{A}$ represents some observable relative to $F$, then $\hat{U}^{\dagger}\hat{A}\hat{U}$ represents the same observable relative to $F'$.

Now my question is: When changing frames, do we transform both the states and the observables, or do we only transform one of those? Is it analogous to the Schrödinger and Heisenberg pictures of time evolution, where we choose whether it is the states or operators that evolve in time? If we transform only the states or only the operators, is the choice related to whether we view the transformation as active or passive?

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The "changing frames" language might be confusing the issue. Coordinate systems and observers are two independent concepts. All of the observers in any given scenario can be described using a single coordinate system. The choice of coordinate system doesn't matter at all, aside from mathematical convenience. Some coordinate systems might be more convenient for some observers, but that's the only connection between them. It's a matter of convenience, not necessity.

So, the answer depends on what you mean by "changing frames":

  • If it means changing coordinates, so that the given unitary transformation is being used to represent a mere change of coordinates, then we need to apply it to everything (operators and states) so that it doesn't change any predictions. Changing the coordinate system can't have any physical consequences, because coordinates are just labels.

  • If it means changing the observer, then the answer depends on what kind of model we're using. I'll call them in-practice models and in-principle models.

In practice, solving the Schrödinger equation for a one multi-electron atom is already difficult enough, nevermind solving the Schrödinger equation for a person made of jillions of molecules. So, instead of treating the observer as a physical entity, we usually omit the observer, and instead we artificially specify which observable is being measured. That's what I mean by an in-practice model. If $O$ is an observable and $U$ is a unitary operator that implements a rotation, then $U^\dagger OU$ is a different observable. In any given state, we could choose to measure either $O$ or $U^\dagger OU$, and these should typically give different results. So in this case, we should apply $U$ only to the observable, not to the state. Equivalently, we could apply $U$ only to the state, not the observable. These two options are equivalent, just like the Schrödinger and Heisenberg pictures are equivalent. (This equivalence comes from the fact that applying the same $U$ to both the observable and the state doesn't have any physical consequences.)

In principle, quantum theory allows using a model that includes complex macroscopic observers (like people) as part of the physical quantum system. That's what I mean by an in-principle model. In this type of model, changing the observer means changing the state, because the state describes the whole physical system including the observer. But this isn't a symmetry operation. If a person moves to a different location in the laboratory, then the state — which describes both the laboratory and the person — has changed in some very complicated way. Such a complicated change cannot be described by anything as simple as an overall translation or rotation. We can still describe it as a unitary transformation if we want to, because any change from one given state to another given state can be described by a unitary transformation, but it's not a symmetry.

By the way, even with an in-principle model, we still need to tell the theory which outcome we actually experience when an observable is measured. Using an in-principle model doesn't avoid the need for the usual projective-measurement rules, but it does allow us to defer application of those rules until after the physical process of observation — which the model itself describes — is complete.

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If you are only making a basis change, you have to transform both, otherwise you will end up with inconsistent results. The physics of the system must not depend on the choice of basis. Using $|\psi'\rangle = \hat{U}^\dagger |\psi\rangle$ and $\hat{A}' = \hat{U}^\dagger \hat{A} \hat{U}$, the transformation exactly cancels when calculating $$\langle\psi'|\hat{A}'|\psi'\rangle = \langle\psi|\hat{A}|\psi\rangle$$ as it needs to be.

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  • $\begingroup$ This does not look right. There are two ways of doing a symmetry transformation, an active transformation, and the passive transformation. In active transformation, states are assumed to change under the transformation while operators don't and in the passive transformation, states are left unchanged while operators change. Symmetry transformations do change the expectation values. For example, under a finite space translation, $\langle x\rangle\xrightarrow{T(a)} \langle x\rangle+a$. See the discussion, for example, Griffiths & Schroeter section 6.2.1. @noah $\endgroup$
    – SRS
    Jul 16 at 14:51
  • $\begingroup$ I think, OP is asking about symmetry transformation and your answer is about basis transformation. @noah $\endgroup$
    – SRS
    Jul 16 at 15:02
  • $\begingroup$ All I was trying to say is that symmetry transformations can change the expectation values. For example, under a translation, $\langle x\rangle\to \langle x\rangle+a$. Under parity $\langle x\rangle\to -\langle x\rangle$. And I am not talking about basis transformations but symmetry transformations. It seems that noah's answer is about basis transformation while the OP is asking about symmetry transformations. @Andrew $\endgroup$
    – SRS
    Jul 17 at 5:21
  • $\begingroup$ @SRS OK I see -- I agree! I'll delete my comments. $\endgroup$
    – Andrew
    Jul 17 at 5:43
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The laws of physics should be the same in all reference frames. This means that if your measurement apparatus is measuring a value in one Reference Frame (RF), then it should measure the same thing in another RF. The only way to satisfy this is if you apply the transformation $\hat{U}$ to both the kets and the operators, so that \begin{align} \langle \hat{A} \rangle_{F'} &= \langle \psi \vert \hat{A} \vert \psi \rangle_{F'} \\ &= (\langle \psi \vert\hat{U}^{\dagger}) (\hat{U} \hat{A} \hat{U}^{\dagger})(\hat{U}\vert \psi \rangle)_F \\ &= \langle \psi \vert\hat{A}\vert \psi \rangle_F \\ &= \langle \hat{A} \rangle_F \end{align}

One way to look at this is thinking the transformation on the kets are rotating the particle and the transformation on the operator is rotating the measurement apparatus alongside, so that the measurement result is the same.

As a side note, although I talk about measurement apparatuses, the same conclusion can be reached from more abstract terms. We can say that we expect the expectation value of a given operator to be observer independent. In particular, a given eigenstate of an operator $\hat{A}$ should still be an eigenstate, regardless of the observer, and should have the same eigenvalue for all observers. This is accomplished by the above transformation on both $\hat{A}$ and the eigenstate.

If you want, on the other hand, not describe the system in another reference frame, but make a physical transformation on the system (i.e. rotate a particle by rotating the beamer that is producing it, or evolve the system in time), then you should apply $\hat{U}$ only to one of the objects: either the kets or the operators, not both. The fact that you can choose which one you are applying the transformation to is, as you mentioned, analogous to the situation of the Heisenberg/Schroedinger pictures.

TL;DR: Changes of reference frame transform both objects; Physical transformations on the system transform either, but only one.

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  • $\begingroup$ But way, if the origin of F is 100m away from the origin of F’, and $A=x$, wouldn’t you want $\langle A \rangle_F \neq \langle A \rangle_{F’}$? $\endgroup$
    – Andrea
    Jul 17 at 22:04
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    $\begingroup$ The representation $A = x$ is tied to a reference frame (that of the measurement apparatus, for example). Unless you were to move the measurement apparatus with yourself when going from frame $F$ to frame $F'$, then you would move away from both the measurement apparatus and the particle, such that the expected value of the measurement stays the same. You could consider moving the measurement apparatus with yourself when making the reference frame change, but then you would be making a physical transformation on the system, which fits into the second category explained in my answer. $\endgroup$ Jul 17 at 23:16
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    $\begingroup$ Right on. Great answer $\endgroup$
    – Andrea
    Jul 17 at 23:20
  • $\begingroup$ @Andrea In other words, if you are just making a reference frame change, then $A_{F'} = x -100$ and $\vert x \rangle_{F'} = \vert x+100\rangle_{F} $ in that case, such that $\langle A \rangle_{F'} = x = \langle A \rangle_F$. $\endgroup$ Jul 17 at 23:20
  • $\begingroup$ @Andrea Thanks :) $\endgroup$ Jul 17 at 23:21

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