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This exercise ask me to explain what happens when the first qubit is measured in the diagonal base ($|+\rangle,|-\rangle$), considering this state:

$|GHZ\rangle=\dfrac{1}{\sqrt{2}} (|000\rangle+|111\rangle)$

How can I proceed?

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We know that

$$ \begin{gather} |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}} \\ |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}} \end{gather} $$

Thus, we can rewrite the $GHZ$ state as

$$ \begin{align} |GHZ\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle|00\rangle+|1\rangle|11\rangle\right) \\ &=\frac{1}{2}\left(|+\rangle|00\rangle+|-\rangle|00\rangle+|+\rangle|11\rangle-|-\rangle|11\rangle \right) \\ &= \frac{1}{2}|+\rangle\left( |00\rangle+|11\rangle \right)+\frac{1}{2}|-\rangle\left(|00\rangle-|11\rangle\right) \end{align} $$

Using this final representation of the state, it is pretty easy to see the answer to your question. There are two terms in which the first qubit is in the state $|+\rangle$ each with amplitude $1/2$, therefore the probability of finding it in this state is $|1/2|^2+|1/2|^2=1/2$. And the same thing applies with $|-\rangle$, i.e., the probability of finding the first qubit in this state is $|1/2|^2+|-1/2|^2=1/2$.

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    $\begingroup$ Thank you so much! Clear & concise. $\endgroup$
    – user18257
    Sep 11 at 8:19

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