# Representing a von Neumann measurement as $[\mathcal{I} \otimes P_i] U(\rho_s \otimes \rho_a)U^{-1} [\mathcal{I} \otimes P_i]$, how do we choose $U$?

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 $$U$$?

• 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?
– glS
Jul 15, 2020 at 23:35
• Actually, a simple example would be sufficient.
– Rob
Jul 16, 2020 at 10:48
• a simple example of what?
– glS
Jul 16, 2020 at 11:28
• An example of U, that would lead to a valid measurement.
– Rob
Jul 16, 2020 at 13:39
• 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.
– Rob
Jul 16, 2020 at 16:44

## 1 Answer

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.