I'm reading a Hamiltonian simulation example proposed in this paper. From their notation, the operator $D_y$ (sometimes it's called $H_y$) serves the function to diagonalize the Pauli matrix $\sigma_y(Y)$ (the corresponding circuit is illustrated below): $$ D_y\ (or\ H_y)=HSX=\frac{1}{\sqrt{2}}\begin{bmatrix} i & 1\\ -i & 1 \end{bmatrix} \quad\quad\quad [A] $$

However, unlike $D_x$, which is the Hadamard gate, I found $D_y$ is sometimes written in different ways, like in this answer by @Craig Gidney:

$$ D_y\ (or\ H_{YZ}) = \frac{Y+Z}{\sqrt{2}} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -i\\ i & -1 \end{bmatrix} \quad\quad\quad [B] $$

and in this answer by @Davit Khachatryan:

$$ D_y\ (or \ 'Y'_{not\ pauli\ here}) = U_2(0,\pi/2) =\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -i\\ 1 & i \end{bmatrix} \quad\quad\quad [C] $$

Thus I'm wondering are those different versions of $D_y$ the same thing? Are they essentially all belong to $U_2$ gate?

Also, when should we use two $U_3$ gates outside of the CNOT gates to perform the time-evolution simulation (like in this case, the answer from @KAJ226)?



1 Answer 1


A and C are technically the same thing, they only differ by a global phase, which makes no difference.

B is technically different, but serves the same purpose. Note that if $UYU^\dagger=Z$, then $$R_zUYU^\dagger R_z^\dagger=R_zZR_z^\dagger=ZR_zR_z^\dagger=Z$$ so if the only effect you're interested in is the diagonalisation of $Y$, there is a freedom of an arbitrary $Z$ rotation $R_z=e^{i\theta Z/2}$ in the definition of your unitary.

In the end, we have the relations $$ D^C=-iD^A=S^\dagger D^B. $$

  • $\begingroup$ Thanks for the answer! Are all of the definitions serve the same purpose to diagonalize $Y$? Could you explain a bit more about how the definition of B is different from that of A? Thanks:) $\endgroup$
    – ZR-
    Nov 16, 2020 at 14:47
  • 1
    $\begingroup$ It's just a different matrix that achieves the same result of diagonalising $Y$. Which should you pick? whichever you can implement more easily. $\endgroup$
    – DaftWullie
    Nov 16, 2020 at 14:58

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