Quantum computing is not my field, so answers understandable to a layman will be most useful. Please forgive any incorrect terminology in my question!

Assume that a set of the states of N qubits exists as a superposition of M entangled qubit states, such that the possible states in the superposition comprise a subset B of the universal set comprising all possible states of N unentangled qubits.

If a quantum NOT operation is applied to each of the N qubits in B to obtain a new superposition R, what is the result? Is it the set complementary to B? I suspect it should be the set $R=B^C$, containing a superposition of 2^N - M different possible states.

  • $\begingroup$ If you let your state be the W state $\frac{1}{\sqrt{3}}(\vert 001\rangle+\vert 010\rangle +\vert 100\rangle)$ then applying a $\mathsf{NOT}$ gate to each of the three qubits would convert the state to $\frac{1}{\sqrt{3}}(\vert 110\rangle+\vert 101\rangle +\vert 011\rangle)$, right? This does not correspond to the uniform superposition over the $5$ basis states of the compliment of $W$, because you are missing $\vert 000\rangle$ and $\vert 111\rangle$. $\endgroup$
    – Mark S
    Sep 26, 2020 at 22:49
  • $\begingroup$ I think that makes sense. Is there a different operation that does produce the complement of B? $\endgroup$
    – S. McGrew
    Sep 26, 2020 at 22:58
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    $\begingroup$ I upvoted the question because I thought that was where you were going/what you were looking for, and I had to think about it for a bit but I believe the answer is "not likely". If there were, I think you could leverage it to have some really fast algorithms to solve certain problems way faster than we think possible. For example, you could evaluate a boolean function $f$ having only a single satisfying instance, measure the output of $f$ to collapse on the uniform superposition of unsatisfying assignments, and then do your operation to get back to the single satisfying assignment. $\endgroup$
    – Mark S
    Sep 26, 2020 at 23:03
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    $\begingroup$ Also you may even be in an eigenstate of the $\mathsf{NOT}$ operations. For example, suppose your superposition is a Bell state $\frac{1}{\sqrt{2}}(\vert 00\rangle+\vert 11\rangle)$. A $\mathsf{NOT}$ operation (aka an $X$ gate) on both qubits puts you in the state $\frac{1}{\sqrt{2}}(\vert 11\rangle+ \vert 00\rangle)$, e.g. back to where you started, and not in any complement of the basis. $\endgroup$
    – Mark S
    Sep 26, 2020 at 23:17
  • $\begingroup$ You read my mind. Super fast algorithms. But I'd like a definitive reason for why finding the complement can't be done. $\endgroup$
    – S. McGrew
    Sep 27, 2020 at 0:48

1 Answer 1


To roll up some of the comments thread, initially we can consider letting our state have $N=2$ qubits entangled in one of the Bell states corresponding to a uniform superposition of the positive sum of $M=2$ of the four "universal set" basis states on $2$ qubits, say:

$$\vert\Phi^+\rangle=\frac{1}{\sqrt{2}}(\vert 00\rangle+\vert 11\rangle).$$

Flipping both qubits together brings us to:

$$\frac{1}{\sqrt{2}}(\vert 11\rangle+\vert 00\rangle);$$

i.e. the same state.

Alternatively we could consider acting on another Bell state, say:

$$\vert\Phi^-\rangle=\frac{1}{\sqrt{2}}(\vert 00\rangle-\vert 11\rangle).$$

However, such a mapping provides:

$$\frac{1}{\sqrt{2}}(\vert 11\rangle-\vert 00\rangle)=-\vert\Phi^-\rangle,$$

which is the same up to a global phase.

Thus, the bit-flip/$\mathsf{CNOT}$ operation does not simply move from a superposition of a subset of the basis states to the corresponding complementary subset of basis states; indeed, the states may already be in an eigenstate of the $\mathsf{CNOT}$ operation(s).

The OP's idea of partitioning a superposition into two sets and "flipping between" them runs against the BBBV theorem, which limits how easy it can be to find quick solutions to blackbox problems; it also might run against the no cloning theorem, which limits the ability to copy unknown states.

Another quick way to see the same is to consider a state such as the uniform superposition over all basis states - there the "complementary" set is null; thus, it would not make sense to flip between two complimentary sets.

  • $\begingroup$ I can see that the quantum NOT won't form the complementary set of superimposed states. Also, I can see that if the "B" superposition survives the operation, the no cloning theorem will be violated. That suggests, though, that if an operation does exist that can form the complement, it must necessarily "consume" the original "B" superposition. I will ask a new question about that. $\endgroup$
    – S. McGrew
    Sep 27, 2020 at 22:09
  • $\begingroup$ Sure, but what is the "complement" of the uniform superposition over all basis states? How would such an operation work? The complement of the universe is the null set. $\endgroup$
    – Mark S
    Sep 27, 2020 at 22:29
  • $\begingroup$ I'm not sure. Guess you can't say a qbit is a superposition of nothing! But maybe any single definite state would, in the quantum context, serve as the complement of the universal superposition. $\endgroup$
    – S. McGrew
    Sep 27, 2020 at 22:59

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