2
$\begingroup$

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 $U$?

$\endgroup$
5
  • $\begingroup$ 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
    Commented Jul 15, 2020 at 23:35
  • $\begingroup$ Actually, a simple example would be sufficient. $\endgroup$
    – Rob
    Commented Jul 16, 2020 at 10:48
  • $\begingroup$ a simple example of what? $\endgroup$
    – glS
    Commented Jul 16, 2020 at 11:28
  • $\begingroup$ An example of U, that would lead to a valid measurement. $\endgroup$
    – Rob
    Commented 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
    Commented Jul 16, 2020 at 16:44

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.