I am trying to understand the section on the Wikipedia page for GHZ states entitled "Pairwise entanglement". In this section, it is claimed that measuring the third particle in a GHZ state in the X basis can leave behind a maximally entangled Bell state. However, there is no source cited and I am having some trouble understanding this, so I was wondering if anyone could perhaps explain this further.

Also, how would the measurement outcome of this X-basis measurement affect the state produced?


1 Answer 1


As mentioned in the article, you can rewrite the GHZ state as \begin{align} \frac{1}{\sqrt{2}} (|000\rangle + |111)&= \frac{1}{2\sqrt{2}}(|000\rangle + |111 \rangle + \overbrace{|001\rangle + |110 \rangle - |001\rangle - |110\rangle}^{=0} + |000\rangle + |111\rangle ) \\ &= \frac{1}{2\sqrt{2}}\left[(|000\rangle + |111 \rangle + |001\rangle + |110 \rangle) + (|000\rangle - |001\rangle - |110\rangle + |111\rangle ) \right] \\ &= \frac{1}{2\sqrt{2}} \left[\left( |00\rangle + |11\rangle \right) \otimes(|0\rangle + |1\rangle)+ \left( |00\rangle - |11\rangle \right) \otimes(|0\rangle - |1\rangle) \right] \\ &= \frac{1}{2} \left[\left( |00\rangle + |11\rangle \right) \otimes |+\rangle + \left( |00\rangle - |11\rangle \right) \otimes |-\rangle\right] \end{align}

where I've used $|\pm\rangle \equiv \frac{1}{\sqrt{2}}(|0\rangle \pm |1\rangle)$ instead of the article's $|L\rangle, |R\rangle$. So you can see that if you measure qubit 3 in the X-basis (where measurement of "+" results in outcome "0" and "-" gives "1") you will end up with either the Bell state $|00\rangle + |11\rangle$ if you measure "0", or the state $|00\rangle - |11\rangle$ if you measure "1". In the latter case you can apply a $Z$ operation to qubit 1 or 2 to recover the Bell state.

  • $\begingroup$ Thank you very much, that's helped a lot! I think I was partly confused by their |𝐿⟩ and |𝑅⟩ notation, but that has made it clear $\endgroup$
    – clundin
    Jan 2, 2021 at 2:11
  • $\begingroup$ No problem; $|L\rangle$ and $|R\rangle$ are a lot more common to see in the context of photonic quantum computing/protocols where the qubit is encoded in the polarization of light, "Left" or "Right". $\endgroup$
    – forky40
    Jan 2, 2021 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.