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A pair of non-commuting Observables $\hat{X}$ and $\hat{P}$ does not have a common set of eigenfunctions, i.e., it can not be measured simultaneously. Let us for the sake of simplicity assume that $[\hat{A},\hat{B}]=i\hat{1}$ as for position and momentum observables.

On the other hand, it is well known that using an enlarged Hilbert space, it is possible to define Observables $\hat{A}$ and $\hat{B}$ in this enlarged Hilbert space with $[\hat{A},\hat{B}]=0$ so that $\langle \hat{A} \rangle = \langle \hat{X} \rangle$ and $\langle \hat{B} \rangle = \langle \hat{P} \rangle$ (see, e.g., section 3.3). Since $\hat{A}$ and $\hat{B}$ commute, there exists a common set of eigenfunctions for which $\hat{A} |A B \rangle = A |A B \rangle$ and $\hat{B} |A B \rangle = B |A B \rangle$, respectively. Briefly put, it is possible to define a pair of commuting observables $\hat{A}$ and $\hat{B}$, which can be simultaneously measured in order to infer $\hat{X}$ and $\hat{P}$.

If one assumes that the states in the two Hilbert spaces have to factorize (i.e., that they are separable), then it is possible to derive the uncertainty relation $\Delta \hat{A} \Delta \hat{B} \geq 1$. Thus, a measurement in the enlarged Hilbert space can be performed simultaneously, but at the cost of an additional uncertainty in comparison with the Kennard relation $\Delta \hat{X} \Delta \hat{P} \geq \frac{1}{2}$.

However, according to Robertsons's uncertainty relation, one has $\Delta \hat{A} \Delta \hat{B} \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle| = 0$, with some kind of minimal uncertainty state $|A B \rangle$. Motivated by this statenment, my question is: Is it possible to utilize the common eigenfunctions $|A B \rangle$ to perform a joint measurement of non-commuting observables in an enlarged Hilbert space without uncertainty? If no, why not? If yes, why cant this be exploited for measurements?

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  • $\begingroup$ The enlarged Hilbert space is sth. you add for performing the measurement. Thus, it will always be in a product state with the system you want to measure, and $\Delta\hat A\Delta\hat B\ge 1$ will hold -- as you point out yourself, the minimal uncertainty states are entangled states of system + ancilla. $\endgroup$ – Norbert Schuch Jan 23 '16 at 13:15
  • $\begingroup$ I agree that the usual approach is to have a product state. But why is it not possible to entangle system + ancilla by means of a unitary transformation so that the joint state is a minimal uncertainty state with $\hat{A}\hat{B} = 0$? Or, put differently, how could one prepare such an optimal measurement state $|AB\rangle$? $\endgroup$ – Hendrik Jan 23 '16 at 13:49
  • $\begingroup$ The problem is unitarily invariant. If you use some $U$ to entangle it before, then $\hat A$ and $\hat B$ will be modified accordingly. It simply does not work out. Since the uncertainty relation still holds if you use any POVM measurement, adding ancillas in any conceivable way cannot help. $\endgroup$ – Norbert Schuch Jan 23 '16 at 13:58
  • $\begingroup$ Again, I agree to your statement. However, what meaning does the eigenfunction $|AB\rangle$ have in this case? It seems to exist, but at the same time seems to be impossible to prepare? $\endgroup$ – Hendrik Jan 23 '16 at 14:09
  • $\begingroup$ Precisely, ,the joint eigenfunctions (before the unitary) are entangled states between system and ancilla. (Rather than $X$ and $P$, think of Pauli $X$ and Pauli $Z$, which you can "make commuting" by measuring $X\otimes X$ and $Z\otimes Z$. Their joint eigenstates are Bell states.) $\endgroup$ – Norbert Schuch Jan 23 '16 at 14:36

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