[NOTES/QCQI-04002] Two Qubit Gates

\newcommand{\DD}[2][]{\frac{d^2 #1}{d^2 #2}}$ 
$\newcommand{\matrixelement}[3]{\langle#1|#2|#3\rangle}$
$\newcommand{\PP}[2][]{\frac{\partial^2 #1}{\partial #2^2}}$
$\newcommand{\dd}[2][]{\frac{d#1}{d#2}}$
$\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}$
$\newcommand{\average}[2]{\langle#1|#2|#1\rangle}$
$\newcommand{\ket}[1]{\langle #1\rangle}$

 1 Two Qubit Gates

Gates acting on two qubits $\ket{x}\ket{y}$ are known as two qubits.  Two qubit $\ket{x}\ket{y}$, more generally $\ket{\psi_1}\ket{\psi_2}$ are elements of $\mathcal{H}\otimes\mathcal{H} \equiv \mathcal{H}^{(2)}$  CNOT, or controlled NOT, gate is a two qubit gate. It action on $\ket{x}\ket{y}$ leaves the first qubit $\ket{x}$, called control qubit, unchanged and flips the second qubit $\ket{y}$ if the control bit is $\ket{1}$ \text{CNOT}~\ket{0}\ket{0} = \ket{0}\ket{0} \quad&\text{CNOT}~\ket{0}\ket{1}=\ket{0}\ket{1}\\ \text{CNOT}~\ket{1}\ket{0}=\ket{1}\ket{1} \quad&\text{CNOT}~\ket{1}\ket{1}=\ket{1}\ket{0} Here $x,y\rightarrow\{0 \text{\,or\,} 1\}$. More generally, using the notation $\ket{x}\ket{y}\equiv\ket{xy}$, we have  \alpha\ket{00}+\beta\ket{01}+\gamma\ket{10}+\delta\ket{11} \overset{CNOT}{\longrightarrow}~ \alpha\ket{00}+\beta\ket{01} + \gamma\ket{11}+\delta\ket{10} 

Matrix representation

$$ CNOT = \begin{bmatrix}~ I_2&\vline&0\\ \hline0&|&\sigma_x~\end{bmatrix} $$ Digramatically \noindent$\ket{x}$\,\text{------}$\bullet$\text{------}\,\,$\ket{x}$ \hspace{0.3in}$|$ $\ket{y}$\,\text{------}$\oplus$\text{------} $\ket{x+y}$\qquad where $x,y=0,1$\\ ~~~~~~~CNOT gate

Exercise/Example

<ol class="itemlist">
<li> The figure shows a Hadamard gate on the first qubit followed by a CNOT gate on $\ket{00}$. It will give $$ \ket{00} ~\overset{H}{\rightarrowtail}~ {\ket{0}+\ket{1}\over\sqrt{2}}\oplus \ket{0}~ \overset{CNOT}{\text{---------}} ~{\ket{00}+\ket{11}\over\sqrt{2}} $$ Exercise find action of the above network on $\ket{01},\ket{10},\ket{11}$.\\ </li>
<li> Frequently one needs NOT operation on a single qubit\\ \begin{equation} \ket{0}\stackrel{NOT}{\rightarrowtail} \ket{1},\qquad \ket{1}\stackrel{NOT}{\rightarrowtail} \ket{0}\\ \end{equation} \noindent This operation can be achieved by adding a control bit $\ket{1}$ and using CNOT gate as follows. \begin{equation} \ket{1}\ket{0}\stackrel{NOT}{\rightarrowtail} \ket{1}\ket{1},\qquad \ket{1}\ket{1}\stackrel{NOT}{\rightarrowtail} \ket{1}\ket{0}\\ \end{equation} </li>
</ol>

More 2 qubit gates

The action of Hadamard gate followed by a CNOT gate on two qubits is as follows. \begin{align*} \ket{00} = \ket{0}\ket{0} \stackrel{H}{\rightarrowtail}~{\ket{0}+\ket{1}\over\sqrt{2}}~\ket{0}~\stackrel{CNOT}{\rightarrowtail} ~{\ket{00}+\ket{11}\over\sqrt{2}}\\ \ket{10} = \ket{1}\ket{0} \stackrel{H}{\rightarrowtail}~{\ket{0}-\ket{1}\over\sqrt{2}}~\ket{0}~\stackrel{CNOT}{\rightarrowtail} ~{\ket{00}-\ket{11}\over\sqrt{2}}\\ \ket{x_1x_2}=\ket{x_1}\ket{x_2}\longrightarrow ??\qquad \text{where}~ x_1,x_2\in \{0,1\} 

2 qubit controlled phase gate

 \begin{array}{l} \ket{x}\,\text{------}\bullet\text{------} \ket{y}\,\text{------}\fbox{V}\text{------} \end{array}\right\}~ \ket{0}\ket{y}~~~~\text{does~nothing~ to $\ket{y}$} \left.\begin{array}{l} \ket{1}\,\text{------}\bullet\text{------}\\ \ket{y}\,\text{------}\fbox{U}\text{------} \end{array}\right\}~ \ket{1}(U\ket{y})~~~~\text{applies unitary operator $u$ only $\ket{y}$} \end{align*} In a controlled gate the control bit(s) do not change $$\ket{x}\ket{1} \stackrel{CNOT}{\rightarrowtail} \ket{x}\ket{x\oplus1}$$ \noindent so $\ket{x}\ket{0} \rightarrowtail \ket{x}\ket{x}$.

CNOT gate

has representation \begin{equation*} \text{CNOT} = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{pmatrix} \end{equation*} in computational basis $\{\ket{00},\ket{01}, \ket{10}, \ket{11}\}$

CV gate is the gate given by the matrix $$V=\begin{pmatrix}1&0\\ 0& i\end{pmatrix}$$ \noindent CV gate is a controlled phase gate where the phase matrix is $V$. Note that $V$ is unitary and $V^4=I$.

Example: A CNOT gate can be built from $H$ and CV gates