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I have a question. Suppose we have a quantum mechanical system, of which we can measure 3 different observables. Each of the observables has an eigenvector spectrum and we can express a state as a linear combination of the eigenvectors (we assume that they form a basis). My question is the following.

The system is in particular state $|\psi\rangle$.
If we want to express it as a linear combination of basis vectors, which basis vectors do we use (we have 3 different types since we have 3 observable)?

Or maybe, if we assume that the observables commute and form a complete set, is the basis made of eigenvectors that belong to all 3 observables?

Now let's say we have an observable that has a discrete basis and another that has a continuous one. Assuming that they commute, will the basis with eigenvectors of both of them be a mix of a discrete and a continuous eigenvector spectrum?

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All representations of the state are equivalent in Quantum mechanics. For a state $|\alpha⟩$ with two observables $A$ and $B$ over the space $\{|a'⟩\}$ and $\{|b'⟩\}$, it's possible to write $$|\alpha⟩= \sum{⟨a'|\alpha⟩|a'⟩}$$ $$|\alpha⟩= \sum{⟨b'|\alpha⟩|b'⟩}$$ Both of these representations are equivalent. This can be seen by considering a unitary transformation, $$|b'⟩= U|a'⟩$$ Without much effort it can be seen that $U$ has the form, $$U=\sum |b'⟩⟨a'|$$ Substituting this transformation, $$|\alpha⟩= \sum{⟨b'|\alpha⟩|b'⟩} = \sum{⟨a'|U^{\dagger}|\alpha⟩U|a'⟩}$$ Hence the expansion coefficients of $|\alpha⟩$ in $b'$ representation is $⟨a'|U^{\dagger}|\alpha⟩$. Also the operator $A$ in $b'$ representation will become $U^{\dagger}AU$. Similar procedure is to be followed for continuous basis. This is irrespective of whether the operators commute or not. If the operators commute then a common set of eigen basis exists for both operators. In the case of compatible observables, the basis are generally written as $|a',b'⟩$.

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    $\begingroup$ @BioPhysicist Thanks for the correction. $\endgroup$ Sep 19, 2020 at 4:38

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