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The GHZ state is defined as $|\mathrm{GHZ}\rangle = \frac{|000\rangle + |111\rangle}{\sqrt{2}}$. Is there a name for the phase flipped GHZ state, i.e. $Z_1(|\mathrm{GHZ}\rangle)=\frac{|000\rangle - |111\rangle}{\sqrt{2}}$ ?

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    $\begingroup$ I don't think so. You could call the two states as $|\mathrm{GHZ}_{\pm}\rangle$. The important think (perhaps) is that both these states are in the same entanglement class (the two entanglement classes being the one of GHZ and W states, for example, see this). $\endgroup$ – keisuke.akira Aug 10 '20 at 22:31
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As @keisuke.akira already mentioned, and as far as I am aware, there is not really an agreed upon name for such a specific state.

I'd like to add one more thing: it might even be a bad idea to give that state (identified as $Z_{1}|\mathrm{GHZ}\rangle$) a specific name, because (as you may be well aware) $Z_{1}|\mathrm{GHZ}\rangle = Z_{2}|\mathrm{GHZ}\rangle = Z_{3}|\mathrm{GHZ}\rangle$. This could create some unwarranted ambiguity.

If you identify the state as $\frac{1}{\sqrt{2}}(|000\rangle - |111\rangle)$ you don't have this problem, of course.

Also, it may be of interest to note that the $|\mathrm{GHZ}\rangle$ state (generalized to $n$ qubits) is often used in anonymous quantum networking, precisely because of the fact that such a local phase flip has a completely non-local effect, thereby hiding the identity of the party applying the phase flip.

Now, this may be really out of focus, but there's one other thing that might be of interest to you. You might now that the $\mathrm{GHZ}$ state is (up to local Cliffords) equivalent to a graph state. Moreover, in this review paper they introduce (Eq. $(28)$, page $16$) the set of states $\{|W\rangle = Z_{W}|G\rangle\}$, where $|G\rangle$ is a graph state. This set forms an orthonormal basis, and if the $|\mathrm{GHZ}\rangle$ state were a graph state your state $Z_{1}|\mathrm{GHZ}\rangle$ would be an element of this basis, namely $|W_{100}\rangle$. However, the $\mathrm{GHZ}$ state is only equivalent to a graph state, so it doesn't really work. (If it were to work, we would have $|W_{100}\rangle = |W_{010}\rangle = |W_{001}\rangle$, which is clearly not possible for a basis!)

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