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.


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.


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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