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In the picture above, the last layer with dot-shape, is called 1D ladder of controlled Z gates. It is stated in this paper to show the barren plateau issue. I want to know which is the control qubit, the higher one or the lower one?


Controlled Z gate is symmetric with respect to the two qubits.

Its action in the computational basis is to leave the states $|00\rangle$, $|01\rangle$, $|10\rangle$ unchanged and to flip the phase of the state $|11\rangle$. Consequently, it does not matter which qubit we designate as control: it flips the phase only if both qubits are in state $|1\rangle$.

You can also see this from its matrix

$$ CZ = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}. $$

Swapping qubits corresponds to swapping the two middle columns and the two middle rows and in the case of CZ this transformation does not change the matrix.

  • $\begingroup$ I notice it's a stupid question, sorry. But I just can't think it out a few minutes ago... $\endgroup$ – narip Dec 30 '20 at 1:34
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    $\begingroup$ Admittedly the name "controlled-Z" does suggest that one of the qubits serves a different role than the other! I suppose the symmetric symbol is trying to rectify the confusing name... $\endgroup$ – Adam Zalcman Dec 30 '20 at 1:44

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