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If we have an observable $A$, and a unitary operator $\hat U$, one can easily show that both $\hat A$ and $\hat U \hat A \hat U^{\dagger}$ have the same spectrum - in fact, they are called unitarily equivalent.

For example, in Stern-Gerlach experiment, one can easily show that $S_x$ and $S_z$ operators are unitarily equivalent by the rotation operator.

Motivated by the above example, as in that case, $S_x$ and $S_z$ are physically the same operators in the sense that any proper of one should be a proper of the other as well, since one can free to choose a new $z'$ axis in the $x$ axis, i.e the choice of our axis was arbitrary, so in that sense, they are the same operators, by when we use them at the same time, they just correspond to a different dimensions of the system.

Added:

As @JohnForkosh interpreted my intuition in a more clear & correct sense, this equivalent might be captured by the property that $S_x$ and $S_z$ can be measured by the same experimental apparatus, see his comment below.

Question:

Therefore, with this intuition, for any two unitarily equivalent observables, can both be measured by the same experimental apparatus ?

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    $\begingroup$ 1. It is not clear to me here what you mean by "some arbitrary choice", nor what you mean by $S_x$ and $S_z$ being "physically the same operators". They are not the same operators - a state that is an eigenstate of one is not in general an eigenstate of the other! 2. Consider the pre-eminent example of unitarily equivalent operators: Position and momentum, with the Fourier transform being the unitary transformation between them. You'll be hard-pressed to find anyone willing to say that position and momentum are "physically the same". $\endgroup$ – ACuriousMind Jan 22 at 17:46
  • $\begingroup$ @ACuriousMind I didn't know that Fourier transform was a unitary operator, I just say the case $S_x $ and $S_z$, and the from the way how the professor presented, that was the natural question to ask. $\endgroup$ – onurcanbektas Jan 22 at 18:24
  • $\begingroup$ @JohnForkosh That is a good way of interpreting what my intuition asks; thanks a lot. $\endgroup$ – onurcanbektas Jan 24 at 7:37
  • $\begingroup$ @JohnForkosh I have that book, so I'll definitely check out, thanks, but, does that not imply - with what ACuriousMind said - we can measure both position and momentum with the same operator ? To be honest, I don't know how do we measure momentum in QM. $\endgroup$ – onurcanbektas Jan 24 at 9:13

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