In a $3$ - qubit system, the $GHZ_{3}$ - state can be constructed in the following manner:

$|\psi \rangle _{1} = (H \otimes I \otimes I)(|0\rangle \otimes |0\rangle \otimes |0\rangle) = |+\rangle \otimes |0\rangle \otimes |0 \rangle = \frac{1}{\sqrt{2}}[|000\rangle + |100\rangle]$

To flip the two 0's in second and third qubit, note that a $CNOT$ can be applied such that the control is first qubit and target is second qubit to give $|\psi \rangle _{2} = \frac{1}{\sqrt{2}}[|000\rangle + |110\rangle]$. Then, allowing for second qubit to be the control and third qubit to be target, the state $|\psi \rangle _{3} \equiv |GHZ_{3} \rangle = \frac{1}{\sqrt{2}}[|000\rangle + |111\rangle]$ is arrived at.

However, 6 other similar $GHZ_{3}$ - like state exists. They are ${|000\rangle, |001\rangle, |010\rangle, |011\rangle, |100\rangle, |110\rangle}$.

Is it possible to go from $|GHZ_{3}\rangle$ to each of the above $6$ states? If so, how?

  • 1
    $\begingroup$ Are you asking what unitary will transform $|GHZ_3\rangle$ to e.g. $|010\rangle$? You already have $U$ such that $U|000\rangle = |GHZ_3\rangle$. So $(I\otimes X \otimes I)U^\dagger |GHZ_3\rangle = |010\rangle$. $\endgroup$
    – forky40
    Oct 8 at 16:00
  • 1
    $\begingroup$ @forky40 does $(I \otimes X \otimes I) U^{\dagger} |GHZ_{3}\rangle$ follows from a theorem or is this a trial and error method? edit: Ah got it. The hermiticity of the unitary matrix is the key to "eliminating" the superposition because unitary matrix have inverse! $\endgroup$
    – Physkid
    Oct 9 at 0:15


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