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?
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.
