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

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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 $$$$M = a \mathbb{I} - ib \sigma_Z,$$$$

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

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

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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: $$$$M = \begin{bmatrix}a - ib & 0 \\ 0 & a + ib\end{bmatrix} = (a - ib)\begin{bmatrix}1 & 0 \\ 0 & \frac{a + ib}{a - ib}\end{bmatrix} = (a - ib)\begin{bmatrix}1 & 0 \\ 0 & \frac{a^{2} - b^{2} - 2iab}{a^{2} + b^{2}}\end{bmatrix} \hat{=} \begin{bmatrix}1 & 0 \\ 0 & a^{2} - b^{2} - 2iab\end{bmatrix},$$$$ where $$\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})$$.

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

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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.