# constructing $6$ GHZ-like states from the GHZ state

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?

• 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 $(I\otimes X \otimes I)U^\dagger |GHZ_3\rangle = |010\rangle$. Oct 8 at 16:00
• @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! Oct 9 at 0:15