2
$\begingroup$

I have the following operators acting on two qubits, denoted as $(1)$ and $(2)$:

$$T_1=\displaystyle\exp\left(-i\frac{\pi}{4}Z\otimes Z\right)\cdot R_z^{(1)}\left(\frac{\pi}{2}\right)\cdot R_z^{(2)}\left(\frac{\pi}{2}\right),\quad T_2=R_y^{(2)}\left(-\frac{\pi}{2}\right)\cdot C_Z\cdot R_y^{(2)}\left(\frac{\pi}{2}\right).$$

I should prove that up to global phases $T_1=C_Z$ and $T_2=C_X$ (the controlled $Z$ and $X$ gates).

For convenience, I omit the labels for the qubits, which should be clear by the ordering. Using the usual expansion for rotations, I ended up finding

$$T_1=\frac{1}{2\sqrt 2}(\mathbb I\otimes\mathbb{I}-iZ\otimes Z)(\mathbb I\otimes \mathbb I-iZ\otimes\mathbb I)(\mathbb I\otimes \mathbb I-i\mathbb I \otimes Z)=\frac{1+i}{\sqrt 2}\begin{pmatrix} -Z \\ & \mathbb I\end{pmatrix},\\ T_2=\frac{1}{2}(\mathbb I\otimes \mathbb I+i\mathbb I\otimes Y)C_Z(\mathbb I\otimes \mathbb I-i\mathbb I\otimes Y)=\begin{pmatrix} \mathbb I & \\ & X\end{pmatrix}.$$ So, I've checked the calculations a few times and it looks like there is something wrong with the first operator. Here's the catch: there was a mistake in the original $T_2$ I was given that I had to change in order to find the correct sequence to obtain a $C_X$, and there might be a similar problem with $T_1$ (unless I am of course making some mistake somewhere). Does anyone see what the issue may be?

(As a reference, $T_1$ comes from the interaction shift in NMR computing, while $T_2$ from Wineland's experiment with trapped ions).

$\endgroup$
3
$\begingroup$

I agree with your calculation for $T_1$ and $T_2$ starting from your expansion of the operations (the second set of equations). Since there's a couple of possible different conventions for the definition of, for example $R_z(\theta)$, I cannot be certain that the expansion itself is correct. Indeed, it requires relatively minor modification to make things work. For example, if you replace both of the $\pi/2$ rotation angles in the statement for $T_1$ with $-\pi/2$ rotation angles, it all works happily.

It may assist you to know the way that I do this calculation. For $T_1$, you've got 3 matrix exponentials, all of which are using diagonal operators. This means they all commute and hence can be grouped together as a single exponential. $$ T_1=e^{-i\pi Z_1Z_2/4}e^{i\pi Z_1/4}e^{i\pi Z_2/4}=e^{i\pi(Z_1+Z_2-Z_1Z_2)/4} $$ So I can just concentrate on the diagonal elements of these, $$ Z_1+Z_2-Z_1Z_2=\text{diag}(1,1,1,-3) $$ I can add (or subtract) identity matrices as this only contributes a global phase $$ Z_1+Z_2-Z_1Z_2-I=\text{diag}(0,0,0,-4). $$ So when I multiply by $\pi/4$, I'll get a $\pi$ for the final element which, exponentiated, gives the -1 I need.

$\endgroup$

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.