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Imagine that I have a Bell state of two qubits. If I can produce many copies (always of the same state) but I am allowed only to measure one of the qubits, I would be able to tell that the two qubits are entangled, because no matter in which basis I choose to measure the outcome probabilities are 50%. However I cannot tell exactly which of the four Bell states are the qubits in.

Generalizing this concept, are there cases when from the repeated measurement of $N-1$ qubits, is it possible to reconstruct the pure state of the $N$-entangled qubits? Or there is always some amplitude that remains unknown?

I guess for this to happen there would need to be a $N$ qubit pure state $|\Psi\rangle$, with density matrix $\rho_{N}=|\Psi\rangle\langle|\Psi|$, such that $\rho_{N-1}=$Tr$_{1}$($\rho$) is unique to that state (where Tr$_{1}$ is the trace over qubit 1).

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  • $\begingroup$ You don't actually know that they're entangled from the 50/50 random measurements. That would also happen if you were just given random pure states. It's simultaneous correlation in anticommuting bases that tells you entanglement is present. $\endgroup$ Mar 8, 2023 at 23:31
  • $\begingroup$ @CraigGidney if you know that there is a pure state, and that there are two particles that start always in the same state, wouldn't you be able to say that the two particles are entangled? Or is there another pure state that provides the same reduced result as the Bell states? $\endgroup$
    – Mauricio
    Mar 9, 2023 at 0:06
  • $\begingroup$ Yes, if you're promised that the system is pure, then 50/50 random results in all bases of a qubit indicate that qubit must be in an entangled state. $\endgroup$ Mar 9, 2023 at 0:21
  • $\begingroup$ @CraigGidney, exactly that's the case I am considering. $\endgroup$
    – Mauricio
    Mar 9, 2023 at 0:24

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There is always some amplitude that remains unknown. Let $|\Psi\rangle$ be the pure state of all $N$ qubits. The measurement statistics on all qubits except qubit $A$ are optimally described by the reduced density matrix $$ \rho=\text{Tr}_A(|\Psi\rangle\langle\Psi|). $$ However, for all single-qubit unitaries $U_A$, it is true that $$ \text{Tr}_A(U_A\otimes I|\Psi\rangle\langle\Psi|U_A^\dagger\otimes I)=\rho_A. $$ This is a standard property of the partial trace. From a physics perspective, this can be understood as imagine the qubit $A$ were thousands of light years away from everything else. Somebody could apply any $U_A$ they wanted and that could not have any effect on the measurement results on the rest of the system.

Hence, we conclude that there is always a local unitary ambiguity in the determination of $|\Psi\rangle$.

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Suppose for contradiction that you are able to infer the final qubit's value, and its value is $|0\rangle$. But now suppose Eve comes along and tweaks the preparation of the entangled state so that it has an additional rotation of that final qubit, by $\theta$ degrees around the X axis.

Nothing you can do depends on this rotation, because it commutes with all gates and measurements you're able to apply. So, even if you're told that Eve is going to specifically only rotate the qubit around the X axis by some unknown amount, you can't tell if the qubit is $R_X(0^\circ)|0\rangle$ or $R_X(45^\circ)|0\rangle$ or $R_X(30^\circ)|0\rangle$ or etc. There's a degree of freedom in the system, which Eve is using and that you have no access to, that corresponds to an arbitrary rotation of the final qubit. Therefore in general you cannot infer the state of the last qubit from the state of the n-1 other qubits.

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  • $\begingroup$ I do not follow this argument by contradiction so easily. I cannot assume that the state of the final qubit is $|0\rangle$, what I can try to infer is the whole state. For example if the system is $|00\rangle+|11\rangle$, the final qubit is not in any given state. $\endgroup$
    – Mauricio
    Mar 9, 2023 at 0:41
  • $\begingroup$ @Mauricio The $|0\rangle$ was just an example. Ultimately there is a rotation on the final qubit, which affects the state, that commutes with all actions you can take. Therefore you cannot infer the whole state. $\endgroup$ Mar 9, 2023 at 6:37

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