1
$\begingroup$

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}

Thanks!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 '21 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 '21 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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