Given the state of a system as $\rho_s$ and that of the ancilla (pointer) as $\rho_a$, the Von-Neumann measurement involves entangling a system with ancilla and then performing a projective measurement on the ancilla. This is often represented as $$[\mathcal{I} \otimes P_i] U(\rho_s \otimes \rho_a)U^{-1} [\mathcal{I} \otimes P_i],$$ where $\mathcal{I}$ is the identity on system space, $P_i$ is the projector corresponding to $i$-th outcome, and $U$ is the combined unitary.

My question: How to choose the form of $U$?

  • $\begingroup$ are you asking given a map $\Phi$ represented as $\Phi(\rho)=\operatorname{tr}_a[U(\rho\otimes \rho_a)U^\dagger]$, how to find the unitary $U$ in this representation? $\endgroup$
    – glS
    Jul 15, 2020 at 23:35
  • $\begingroup$ Actually, a simple example would be sufficient. $\endgroup$
    – Rob
    Jul 16, 2020 at 10:48
  • $\begingroup$ a simple example of what? $\endgroup$
    – glS
    Jul 16, 2020 at 11:28
  • $\begingroup$ An example of U, that would lead to a valid measurement. $\endgroup$
    – Rob
    Jul 16, 2020 at 13:39
  • $\begingroup$ I am asking the following: Given the combined state $\rho_s \otimes \rho_a$, give an example (or a general form) of $U$, such that the quantity I wrote in my question, represents a valid measurement. $\endgroup$
    – Rob
    Jul 16, 2020 at 16:44

1 Answer 1


No such representation exists, unless the underlying spaces were trivial to begin with.

To see this let us first simplify notation by defining $\rho_{sa}:=U(\rho_s\otimes\rho_a)U^{-1}$ (which is obviously a state again). Now the question is: given $\{P_i\}_{i\in I}$ such that each $P_i$ is a positive semi-definite projection and $\sum_{i\in I}P_i={\bf1}_a$ when is $\{({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i)\}_{i\in I}$ a valid POVM? Our goal is to show that this is true (if and) only if both the system and the ancilla are one dimensional, hence trivial.

If $\{({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i)\}_{i\in I}$ were a valid POVM, then necessarily $\sum_{i\in I}({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i)={\bf1}_{sa}$. Using the projective property $P_i^2=P_i$, taking the trace yields \begin{align*} {\rm tr}({\bf1}_{sa})&={\rm tr}\Big(\sum_{i\in I}({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i) \Big)\\ &=\sum_{i\in I}{\rm tr}\big(({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i)\big)\\ &=\sum_{i\in I}{\rm tr}\big(({\bf1}_s\otimes P_i)({\bf1}_s\otimes P_i)\rho_{sa}\big)\\ &=\sum_{i\in I}{\rm tr}\big( ({\bf1}_s\otimes P_i)\rho_{sa} \big)\\ &={\rm tr}\Big( \Big({\bf1}_s\otimes \sum_{i\in I}P_i\Big)\rho_{sa} \Big)\\ &={\rm tr}(\rho_{sa})=1\,. \end{align*} In the last step we used that $\sum_{i\in I}P_i={\bf1}_a$ because $\{P_i\}_{i\in I}$ is a measurement, as well as that $\rho_{sa}$ is a state. But ${\rm tr}({\bf1}_{sa})$ is the product of the dimension of the system and the dimension of the ancilla, so this product being $1$ forces that both the system and the ancilla are of dimension $1$---hence they trivial, as we wanted to show.


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