0
$\begingroup$

Consider a (possibly unphysical) non-linear transformations of bi-partite quantum states,

$$\mathcal{N} (a A + b B) \neq a \mathcal{N}(A) + b \mathcal{N} (B)$$ for some density matrices $A,B \in \mathcal{S}(\mathcal{H} \otimes \mathcal{H}^{\prime})$ and some $a+b=1$. Suppose that

$$Tr_{\mathcal{H}^{\prime}} \left[ \mathcal{N} (a A + b B) \right] = a Tr_{\mathcal{H}^{\prime}} [\mathcal{N}(A)] + b Tr_{\mathcal{H}^{\prime}} [\mathcal{N} (B)] $$ hence the reduced dynamics on $\mathcal{H}$ is linear.

Can we also have that the reduced dynamics is linear on $\mathcal{H}^{\prime}$? $$Tr_{\mathcal{H}} \left[ \mathcal{N} (a A + b B) \right] = a Tr_{\mathcal{H}} [\mathcal{N}(A)] + b Tr_{\mathcal{H}} [\mathcal{N} (B)] $$ or is this just nonsense and, if linear on $\mathcal{H}$, it must be nonlinear on $\mathcal{H}^{\prime}$?

$\endgroup$

1 Answer 1

1
$\begingroup$

Sure this is possible. Just put the non-linearity on the matrix elements of the output which do not appear in either reduced density matrix.

For instance, for two qubits, the entries on the anti-diagonal do not appear in the reduced density matrix. If the non-linearity lives in those entries, and the other entries are linear, you get an examples of such a map which is linear in either reduced state, but not as a whole.

$\endgroup$
3
  • $\begingroup$ Thanks for the clear answer! I have a follow up question: consider a map of the kind you described, where the nonlinearity is hidden in the correlations but cannot be witnessed at the level of local mmts (without "sharing notes"). My intuition tells me that there always exist a global unitary which, when applied to the transformed state, shifts the nonlinearity in at least one of the reduced states. Is this true? $\endgroup$
    – bb2002
    Commented Feb 15, 2023 at 10:00
  • $\begingroup$ For example, the two-qubit states $| \Phi^+ \rangle \langle \Phi^+|$ and $\mathbb{I}/4$ share the same reduced states. But applying $U$ mapping Bell's basis states to computational basis states, the reduced states will no longer coincide in the two scenarios. $\endgroup$
    – bb2002
    Commented Feb 15, 2023 at 10:26
  • 1
    $\begingroup$ @bb2002 Yes - there will always be a states $|\phi\rangle$ which will have non-zero overlap with the non-linear elements, i.e. its expectation value will be non-linear. Then you can rotate it onto |00>. $\endgroup$ Commented Feb 15, 2023 at 10:48

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.