Is $M = a \mathbb{I} - ib \sigma_Z$ a valid representation in terms of logic gates?


I have a matrix $M= \begin{pmatrix} a - ib & 0 \\ 0 & a + ib \end{pmatrix}$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$. I need to represent this matrix in terms of the quantum logic gates. Is the following representation valid \begin{equation} M = a \mathbb{I} - ib \sigma_Z, \end{equation}

where $ \mathbb{I} $ and $\sigma_Z$ are the identity matrix and Pauli-z gate, respectively. Can this be implemented in a lab?


Posted 2019-08-26T11:55:45.690

Reputation: 283

1Does $a^2 + b^2 = 1$? What 'fundamental logic gates' are you using? – Niel de Beaudrap – 2019-08-26T13:03:42.913

@NieldeBeaudrap, thanks. But $a^2+b^2 \ne 1$. I am talking about the quantum logic gates. Maybe the word fundamental is confusing in my question. – Rob – 2019-08-27T03:37:41.230

3if $a^2+b^2\neq1$ then the matrix is not unitary, and as such, non-reversible. Depending on your definition of "quantum logic gate", this might not qualify as one. Are you sure that's what you want? This aside, sure, you can write $M$ using that formula with Pauli operators – glS – 2019-08-27T07:13:57.563



For this matrix to be a unitary, you need, as Neil mentioned, that $a^{2} + b^{2} = 1$. If that is the case, then the matrix $M$ is a rotational matrix along the $z$-axis. In other words (and how it shows on the wikipedia page you linked), it is a phase gate $R_{\phi}$, or also known as $R_{z}(\theta)$.

Since a global phase can always be omitted, we can write: \begin{equation} M = \begin{bmatrix}a - ib & 0 \\ 0 & a + ib\end{bmatrix} = (a - ib)\begin{bmatrix}1 & 0 \\ 0 & \frac{a + ib}{a - ib}\end{bmatrix} = (a - ib)\begin{bmatrix}1 & 0 \\ 0 & \frac{a^{2} - b^{2} - 2iab}{a^{2} + b^{2}}\end{bmatrix} \hat{=} \begin{bmatrix}1 & 0 \\ 0 & a^{2} - b^{2} - 2iab\end{bmatrix}, \end{equation} where $\hat{=}$ indicates that it has the same action on a qubit (this is the global phase, practically). So from that you can easily compute the $\phi$ as $2\tan(\frac{b}{a})$.


Posted 2019-08-26T11:55:45.690

Reputation: 3 248

Thanks. Are you sure $\phi = 2 tan(b/a)$ and not tan^{-1}(2ab/(a^2-b^2)) ? – Rob – 2019-08-28T11:52:24.597


If $A$ is an operator, and $A^2=\mathbb{I}$, then you can show that

$$e^{-ixA}=\cos(x)\mathbb{I}-i\sin(x)A $$

Just write down $e^{-ixA}$ as a series and separate the even and odd indices. Since $\sigma_z^2=\mathbb{I}$, provided that $a^2+b^2=1$ (hence this transformation is a valid quantum gate) then you can write in polar coordinates $a=\cos(\theta)$, $b=\sin(\theta)$, and your gate is

$$e^{-i\theta\sigma_x} $$

which is clearly a rotation around the $z$ axis in the Bloch sphere. As others have mentioned is $a^2+b^2\neq 1$ then this isn't a quantum gate, as it is not unitary.


Posted 2019-08-26T11:55:45.690

Reputation: 845