I have asked two similar questions 1, 2 but have not received a definite answer, so here I have narrowed the issue down to one hopefully more easily addressed aspect of the big question. Unfortunately I do not have a quantum computation background, and don't yet understand the usual notation used in the field. A link to an article that provides an answer that I can gradually decipher would be sufficient.

Given an elementary entangled state of N qubits, is there a quantum operation that produces the complement of the entangled state?

By "elementary entangled state of N qubits", I mean a state in which all of the N qubits are entangled with each other such that measuring the value of any one qbit will determine the values of all the other qubits.

By "complement", I mean a superposition of all states that are orthogonal to the elementary entangled state, where the "weights" of all the independent components of the superposition are the same.

The closest I've been able to come on my own to answering the question is to see that the hypothetical operation would not lose any information and would be reversible. And, it seems that simply adding an ancillary bit to identify a state as the complement of another state would almost be good enough (except that the components of that state wouldn't be very easily accessible for downstream operations).

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    $\begingroup$ Consider the state $\vert 11\rangle$. What is the "complement"? Is it an equal superposition of $\vert 00\rangle, \vert 01\rangle$ and $\vert 10\rangle$ or an equal superposition of $\vert 00\rangle, \vert \Psi^+\rangle$ and $\vert\Psi^ -\rangle$, where $\vert \Psi^+\rangle = \frac{1}{\sqrt{2}}( \vert 01\rangle + \vert 10\rangle)$ and $\vert \Psi^-\rangle = \frac{1}{\sqrt{2}}( \vert 01\rangle - \vert 10\rangle)$? Both are "complements" according to your definition but are distinct states. $\endgroup$
    – rnva
    Commented Oct 15, 2020 at 19:50
  • $\begingroup$ Perhaps my understanding is way off, but I think there are four possible defined states of two qbits: |00>, |01>, |10> and |11>. In that case the complement of |11> per my definition would be the normalized version of (|01> + |00> + |10>) -- or any other representation of the same superposition using a different basis. $\endgroup$
    – S. McGrew
    Commented Oct 15, 2020 at 20:49
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    $\begingroup$ "There are four possible defined states of two qubits" - No, this is not true. Any superposition of those states is also a well-defined state. If you meant to say that those four states form a basis, then note that the four Bell states also form a basis for two qubits. The point is that there is a freedom in choosing the "complement" that you must remove in order for the required output state to be well-defined $\endgroup$
    – rnva
    Commented Oct 16, 2020 at 14:40
  • $\begingroup$ Do you mean that there is freedom in choosing the basis with which to represent the complement? It "feels like" the complement shound be the same regardless of how it is represented. But I guess I wouldn't be asking if I understood this point. $\endgroup$
    – S. McGrew
    Commented Oct 16, 2020 at 15:43
  • $\begingroup$ Yes - it is a subspace and you can choose any basis you like for it. An equal superposition of those basis vectors yields a different state for different choice of basis vectors e.g. $\frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle) \neq \frac{1}{\sqrt{2}}(\vert +\rangle + \vert -\rangle)$ $\endgroup$
    – rnva
    Commented Oct 16, 2020 at 16:43


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