Measuring non-commuting observable at once

Question

Given an Hilbert space $H$ (finite dimensional for sake of clarity), and two non-commuting operators $$A = \sum_a a |a\rangle\langle a|$$ and $$B=\sum_a b |b\rangle\langle b|,$$ is it possible to find an Hilbert space $H'$ and two commuting operators $A'$, $B'$ with the same spectrum of $A$ and $B$, together with a unitary operator $U : H \to H'$, such that, for each $|\psi \rangle$ in $H$, the following hold

$$|\langle \psi | a \rangle |^2 = Tr[P(a) U |\psi \rangle \langle \psi|U^\dagger]$$ for each eigenvalue $a$ of $A$. ($P(a)$ is the projector onto the eigenspace of $A'$ relative to the eigenvalue $a$.)

And analogously for $B$?

If the above is true, the physical interpretation would be that in principle is possible to "measure" two non-commuting operators by suitably enlarging and evolving the system, in the sense that you could afterward measure two commuting operators yielding the same statistics as the non-commuting ones. I guess this is not possible but I could not find an easy proof, any suggestion?

EDIT:

As rightfully noted below, it is wrong to speak of a unitary between two different Hilbert spaces (with different dimension) so I will put the question in a more precise ground.

Given $H$, $A$ and $B$ as above, is it possible to find a space $V$, operators $A'$,$B'$ acting in $H' = H \otimes V$ and with the same spectrum of $A$ and $B$ but with $[A',B']=0$ together with an unitary operator $U : H' \to H'$ such that

$ Tr[P(a) \otimes Id \ \ |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0|] = Tr[P'(a) \ U |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0| U^\dagger] \ \ \forall |\psi\rangle \in H, a \in \sigma(A)$

Where $|0\rangle$ is a fixed state in $V$, $P(a)$ is the projector on the eigenspace of $A$ relative to the eigenvalue $a$ and $P'(a)$ is the same for $A'$. ($\sigma(A)$ denotes the spectrum of $A$).

And analogously for B?

POSSIBLE ANSWER: Inspired (but not totally convinced) by the given answers, I think I have found an un-attackable proof that the above is impossible. As already told below, finding $A'$ and $B'$ with the above request is equivalent to find $A''$ (and $B''$) for which

$Tr[P(a) \otimes Id |\psi\rangle\langle \psi|\otimes |0\rangle\langle 0|] = Tr[P''(a) |\psi \rangle \langle \psi|\otimes|0\rangle\langle 0|]$ (and similar for B).

That is we can absorb the unitary evolution in the definition of $A'$ and $B'$, and so we will. Note though that this does not mean that $A'=U A U^{\dagger}$.

In addition, the above imply that $|a\rangle|0\rangle$ must be eigenstate for $A'$ with eigenvalue $a$. Moreover, if we introduce the basis $|a\rangle|n\rangle$ for the total tensor space, we find that no vector of the form $|a'\rangle|n\rangle$ with $a'\neq a$ can appear in the decomposition of the general eigenvector of $A'$ relative to $a$. (Orthogonality of eigenvectors relative to different eigenvalues). Therefore the general eigenvector of $A'$ relative to $a$ must be of the form $|a\rangle|v\rangle$

Hence the projector $P'(a)$ is forced to be of the form $P(a) \otimes P_V(a)$, with $P_V(a)$ a suitable projector of dimension at least 1 (it must project at least on $|0\rangle$). Of course a similar result hold for $B'$.

Finally we can write $0=[P'(a),P'(b)]|\psi\rangle|0\rangle= ([P(a),P(b)]|\psi\rangle)\ |0\rangle$ for each $|\psi\rangle,a,b$ and this imply that $[A,B]=0$ that is absurd.

Answer 1

If $A^\prime$ and $B^\prime$ commute then there exists a set of mutual eigenvectors of $A^\prime$ and $B^\prime$. For any eigenbasis of $A^\prime$ there exists a unitary transformation $W$ which takes that basis to the mutual eigenbasis of $A^\prime$ and $B^\prime$. Consequently if there is a unitary operation such that $ |\langle \psi | b \rangle |^2 = Tr[P(b) U |\psi \rangle \langle \psi|U^\dagger] $ for one basis there is one which will make the statement true projecting onto the mutual eigenbasis of $A^\prime$ and $B^\prime$.

Working in this basis we can, by assumption, write \begin{equation} |\langle \psi | a \rangle |^2 = Tr[P(a) U |\psi \rangle \langle \psi|U^\dagger] \end{equation} However, since the we are working in a mutual eigenbasis of $A^\prime$ and $B^\prime$ we can also write \begin{align} |\langle \psi | b \rangle |^2 &= Tr[P(b) U |\psi \rangle \langle \psi|U^\dagger] \\ &=Tr[P(a) U |\psi \rangle \langle \psi|U^\dagger]\\& = |\langle \psi | a \rangle |^2\end{align}

Setting $|\psi\rangle = |a\rangle$ for each $a$ implies that $A$ and $B$ have a mutual eigenbasis, contradicting our assumption that $A$ and $B$ did not commute.

Edit: simpler proof \begin{align} 0 &= [A^\prime, B^\prime] \\& = A^\prime B^\prime - B^\prime A^\prime\\ & = U AU^\dagger U BU^\dagger - UBU^\dagger U AU^\dagger\\ & = U\left(AB - BA\right)U^\dagger\\ & = U[A,B]U^\dagger \Rightarrow [A,B] = 0\end{align}

Answer 2

First, note that a unitary transformation can not modify the commutation relations..

$$AB-BA=C$$ Use the fact that $U^\dagger U=U U^\dagger=1$ to get, $$AU U^\dagger B-BU U^\dagger A=C$$ and then multiply by the conjugate transpose from the left and $U$ from the right, $$ U^\dagger AU U^\dagger B U^\dagger- U^\dagger BU U^\dagger AU^\dagger= U^\dagger C U=\\ AB-BA=C$$

EDIT: this also works if you substitute $U$ with any invertible transformation $S$ for which exists $S^{-1}$. Just use $S S^{-1}=1$. If the transformation is not invertible then there is no way you can recover information about $A$ and $B$ from your measurements.

For instance, let's generalize your example,

$$Tr[P(a) \otimes Id \ \ |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0|] = Tr[P'(a) \ S |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0| S^{-1}],$$ so that $S$ is any invertible transformation. Using the property of the trace, $$Tr[P(a) \otimes Id \ \ |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0|] = Tr[ S^{-1} P'(a) \ S |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0|],$$ redefine, $$Tr[P(a) \otimes Id \ \ |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0|] = Tr[ P'_S(a) \ |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0|].$$ This is equivalent to $$[(\langle \psi| \otimes \langle 0| )|a\rangle \otimes |0 \rangle]^2 = [ \langle (\psi| \otimes \langle 0|) S |a'\rangle]^2$$ Since this holds for any $| \psi \rangle$ including $|a \rangle$ and the fact that eigenvectors are normalized it follow $|a \rangle =S |a' \rangle$. Now since $A'$ and $B'$ commute they share the same eigenstates $|a'\rangle$. Therefore by following the a similar procedure starting from, $$ Tr[P(b) \otimes Id \ \ |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0|] = Tr[P'(a) \ S |\psi\rangle \langle \psi| \otimes |0 \rangle \langle 0| S^{-1}]$$we obtain that $|b \rangle = S |a' \rangle$, $|b \rangle = |a \rangle$ which means that that $A$ and $B$ share the same eigenstates which means that they commute.

What's the underlying point? The point is that after we perform a projective measurement on the joint eigenstate, $|a' \rangle$ of $A'$ and $B'$ we want to map the information back to which eigenstates of $A$ and $B$ this corresponds to. However any "map" $M(|a' \rangle)= |a \rangle$ is not a map if it also maps to some different state $M(|a' \rangle) =|b \rangle \neq |a \rangle$ (if $A$ and $B$ do not commute $|b \rangle$ is in general some superposition of eigenstates $| a \rangle).$ That is, it can not give us a certain result about which value of $B$ the measurement obtained or vice versa.