4
$\begingroup$

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)?

Thanks!!!

$\endgroup$
2
$\begingroup$

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 $$ D^C=-iD^A=S^\dagger D^B. $$

$\endgroup$
2
  • $\begingroup$ 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 '20 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 '20 at 14:58

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.