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?

  • $\begingroup$ Can you state what are the basis states for the diagonal basis as given on your exercise? This may help avoid confusions $\endgroup$
    – epelaez
    Sep 10, 2021 at 20:58
  • 1
    $\begingroup$ I edited the question. Thanks $\endgroup$
    – user18257
    Sep 10, 2021 at 21:07
  • $\begingroup$ related: quantumcomputing.stackexchange.com/questions/15370/… $\endgroup$
    – forky40
    Sep 11, 2021 at 0:10

1 Answer 1


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 $|1/2|^2+|-1/2|^2=1/2$.

  • 1
    $\begingroup$ Thank you so much! Clear & concise. $\endgroup$
    – user18257
    Sep 11, 2021 at 8:19

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