Measuring non-commuting observable at once

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.

• I don't understand. How did you "enlarge" the system? Unitary operators can only rotate the space. – Bubble Sep 20 '14 at 18:35
• I had been a little bit sloppy in this. Consider H' as H tensor V, the unitary above then must work as described if the input state is |psi>|0> for a fixed |0> and any |psi>. I had in mind the same setup of the no-cloning theorem. – giulio bullsaver Sep 20 '14 at 18:41
• So you enlarge the space and the perform a similarity transform? – Bubble Sep 20 '14 at 18:46
• Whether two operators and their spectral families commute or not is preserved by unitary transformations, and the spectral decomposition of self-adjoint operators is unique. So it is not possible to obtain two commuting operators by unitarily transforming two non-commuting operators. – yuggib Sep 20 '14 at 18:51
• Also, remember that you may have very different operators with the same spectrum. Having the same spectrum does not mean two operators are the same from a physical point of view. So, even if you can choose two commuting operators with the same spectrum as the original ones, they would not be unitarily related to the others, and would have different physical meaning. You are simply measuring something else. – yuggib Sep 20 '14 at 19:08

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 $$|\langle \psi | a \rangle |^2 = Tr[P(a) U |\psi \rangle \langle \psi|U^\dagger]$$ 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}

• How do you justify the equality $Tr[P(b)\dotsc]=Tr[P(a)\dotsc]$? It is correct if and only if $A$ and $B$ have the same spectral decomposition, i.e. they are equal. – yuggib Sep 20 '14 at 18:40
• I have edited to clarify this point. We should be able to choose $U$ so that we are projecting onto the mutual eigenbasis of the 2 operators. – By Symmetry Sep 20 '14 at 18:57
• With your notation $P(b)$ is the projection on the eigenvalue $b$ of $B'$, $P(a)$ the analogous for $A'$. They are not the same. There is a projection $P(a,b)=P(a)P(b)=P(b)P(a)$ that projects on the eigenspace of $b$ for $B$ and $a$ for $A'$ but is in general a space smaller than the one where either $P(b)$ or $P(a)$ projects. The equality is then wrong. I think you second proof clarifies your point better, and without mistakes. – yuggib Sep 20 '14 at 19:03
• @BySymmetry, in your edit you assume that A' = U A U*. Could you motivate it? I agree that A' as above defined will have the same statistic on U|psi> as A had on |psi> (Heisenberg picture), but is it the only possible choice of A' with this property? – giulio bullsaver Sep 20 '14 at 19:20
• @giuliobullsaver in fact they are different operators, with different spectral decomposition and different meaning. – yuggib Sep 20 '14 at 19:50

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.