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Assume that for some $k < n$, there is some pure state $|\psi\rangle$ and another mixed state $\rho$ on n qubits such that for all subset $K$ of $k$ qubits, we have:

$\text{Tr}_\bar{K}(|\psi\rangle \langle\psi|) = \text{Tr}_\bar{K}\big(\rho\big)$

Does it always imply the existence of a pure state $|\psi'\rangle$ on n qubits such that $|\psi\rangle \neq |\psi'\rangle$ and:

$\text{Tr}_\bar{K}(|\psi\rangle\langle\psi|) = \text{Tr}_\bar{K}(|\psi'\rangle\langle\psi'|)$

for all such $K$?

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  • $\begingroup$ Right. This is implied by the assumption that k < n. $\endgroup$ – Yonathan Touati Nov 30 '16 at 10:37
  • $\begingroup$ To be clear, $\bar K$ is the complement of $K$? If so, it would be good to edit the answer to make this explicit. Also, note that double-dollar notation ($$...$$) does work here. $\endgroup$ – Emilio Pisanty Dec 1 '16 at 0:25
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This is a great question :D

We can call all pure states $|\psi\rangle$ for which there do not exist any pure states $|\psi'\rangle$ with $Tr_{\bar{K}}(|\psi\rangle\langle\psi|)=Tr_{\bar{K}}(|\psi'\rangle\langle\psi'|)$ "uniquely determined among pure states" (UDP).

Similarly we can call all pure states $|\psi\rangle$ for which there do not exist any mixed or pure states $\rho$ with $Tr_{\bar{K}}(|\psi\rangle\langle\psi|)=Tr_{\bar{K}}(\rho)$ "uniquely determined among all states" (UDA).

Your question amounts to asking, does not being UDA imply not being UDP. We can reformulate this by asking does being UDP imply being UDA.

In recent work my colleagues and I have addressed this question in the negative. There do exist states $|\psi\rangle$ that are not UDA (ie there exist $\rho$ such that $Tr_{\bar{K}}(|\psi\rangle\langle\psi|)=Tr_{\bar{K}}(\rho)$) but that are also UDP (ie there are no $|\psi'\rangle$ such that $Tr_{\bar{K}}(|\psi\rangle\langle\psi|)=Tr_{\bar{K}}(|\psi'\rangle\langle\psi'|)$).

The example we consider in our paper is the following 4 qubit family of states:

$|\psi\rangle=c_0|w_0\rangle+c_2|w_2\rangle+c_4|w_4\rangle$

With $|w_i\rangle \approx P_{sym} (|0\rangle^{\otimes 4-i}|1\rangle^{\otimes i})$ where $P_{sym}$ is the projector onto the symmetric subspace. For certain parameter ranges (namely when $c_i\in \mathbb{R}$, $c_2\neq0, c_0\neq-c_4$, $(c_2^2/2-c_0^2)(c_2^2/2-c_4^2)\neq c_2^2(c_0+c_4)^2/3$) we can prove that this state is UDP. What's more, we can show that if the 2-party reduced density matrix of our state is separable then the state is not UDA. This can be seen by noting that if $\rho_2=Tr_{2}(\psi)$ is separable then it can be expressed as $\rho_2=\sum_i P_i \sigma_i \otimes\sigma_i$ which implies the existence of a state $\rho_4=\sum_i P_i \sigma_i^{\otimes 4}$ which shares the same reduced density matrices as $\psi$. Using this fact we can find regions in the parameter ranges where $\psi$ is both UDP and not UDA

So the answer to your question is, curiously, no.

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  • $\begingroup$ A closely related question can be found here: physics.stackexchange.com/questions/225378/… $\endgroup$ – Joel Klassen Dec 1 '16 at 0:53
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    $\begingroup$ It would be nice if you could e.g. give some example (so one does not have to go and read the paper). $\endgroup$ – Norbert Schuch Dec 1 '16 at 7:25
  • $\begingroup$ BTW, is there a relation to pure N-representability? My take for constructing a counterexample would have been to start from a state which is N-representable but not pure N-representable. $\endgroup$ – Norbert Schuch Dec 1 '16 at 7:27
  • $\begingroup$ Is this the lowest dimension example possible? $\endgroup$ – Vendetta Feb 14 '17 at 13:43
  • $\begingroup$ Vendetta, can you clarify what you mean by lowest dimensional example? $\endgroup$ – Joel Klassen Feb 15 '17 at 17:10

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