I'm following this notebook written using amazon-braket, and there's a statement I don't understand

... To reiterate, the following output is expected: $$\left|\mathrm{GHZ}\right> = \frac{1}{\sqrt{2}}\left(\left|0,0,0\right> + \left|1,1,1\right>\right) = \left[\frac{1}{\sqrt{2}},0,0,0,0,0,0,\frac{1}{\sqrt{2}}\right], $$ for which $\color{red}{\left<ZZZ\right>=0}$ and $\left<111|\mathrm{GHZ}\right>=\frac{1}{\sqrt{2}}$.

The highlighted expression is the one I'm having troubles with. Isn't that expectation value $1/2$ instead?

\begin{eqnarray} \left<\mathrm{GHZ}|ZZZ|\mathrm{GHZ}\right> &=& \frac{1}{2}\left( \left<000|ZZZ|000\right> + \left<000|ZZZ|111\right> + \left<111|ZZZ|000\right> + \left<111|ZZZ|111\right> \right) \\ &=& \frac{1}{2}\left(0 + 0 + 0 + 1 \right) \\ &=& \frac{1}{2} \end{eqnarray}



1 Answer 1


Your off-diagonal contributions are correct, since $Z$ is diagonal in this basis, but your diagonal contributions seem to be taking the values of the kets to be the values of the $Z$ operator's diagonal. The $Z$ operator written in the basis of $\{ |0\rangle, |1\rangle \}$ is $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$. So $\langle iii|ZZZ|iii \rangle$ is $(+1)^3$ for $i=0$ and $(-1)^3$ for $i=1$, yielding the stated result of 0.

  • $\begingroup$ I'm such an idiot, I was assuming $\{|0\rangle, |1\rangle \}$ were the eigenstates of $Z$, $Z|z\rangle = z|z\rangle$, thanks for your help! $\endgroup$
    – caverac
    Nov 22, 2021 at 13:26
  • 2
    $\begingroup$ Oh, but $|0\rangle$ and $|1\rangle$ are the eigenstates of $Z$! However, the corresponding eigenvalues are not equal to the labels of the states. Instead, we have $Z|z\rangle=(-1)^z|z\rangle$. $\endgroup$ Nov 22, 2021 at 17:04

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