Here, we will describe how to construct some simple two-qubit states using some primitive gates.
In what follows, a two-qubit quantum circuit consisting of one $CNOT$ gate and two Hadamard $H$ gates that computes the following unitary transformation will be described:
\[U=\begin{pmatrix}
1&0&0&0 \\
0&1&0&0\\
0&0&1&0 \\
0&0&0&-1
\end{pmatrix}\]
Consider the for computational basis states and their corresponding representation as column vectors:
\[\ket{00}=\begin{pmatrix}
1\\0\\0\\0
\end{pmatrix},
\ket{01}=\begin{pmatrix}
0\\1\\0\\0
\end{pmatrix},
\ket{10}=\begin{pmatrix}
0\\0\\1\\0
\end{pmatrix},
\ket{11}=\begin{pmatrix}
0\\0\\0\\1
\end{pmatrix}.\]
Observe that $U$ leaves the first three basis states unchanged and only multiplies the last by a factor of $-1$:
\[U\ket{00}=\ket{00},
U\ket{01}=\ket{01},
U\ket{10}=\ket{10},
U\ket{11}=-\ket{11}.\]
Consider the following circuit consisting of a Hadamard gate acting on the second qubit, followed by a controlled-NOT gate, and then with a second Hadamard gate acting on the second qubit again.
The action of this circuit on each of the four basis states is described below:
\[\ket{00} \overset{I\otimes H}\mapsto\super{\ket{00}+\ket{01}}\overset{CNOT}\mapsto\super{\ket{00}+\ket{01}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{00}+\ket{01}+\ket{00}-\ket{01})=\ket{00},\]
\[\ket{01} \overset{I\otimes H}\mapsto\super{\ket{00}-\ket{01}}\overset{CNOT}\mapsto\super{\ket{00}-\ket{01}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{00}+\ket{01}-\ket{00}+\ket{01})=\ket{01},\]
\[\ket{10} \overset{I\otimes H}\mapsto\super{\ket{10}+\ket{11}}\overset{CNOT}\mapsto\super{\ket{11}+\ket{10}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{10}-\ket{11}+\ket{10}+\ket{11})=\ket{10}, \]
\[\ket{11} \overset{I\otimes H}\mapsto\super{\ket{10}-\ket{11}}\overset{CNOT}\mapsto\super{\ket{11}-\ket{10}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{10}-\ket{11}-\ket{10}-\ket{11})=-\ket{11}.\]
Thus, the action of the circuit is identical to the unitary transformation $U$ expressed as the matrix since the four basis states are mapped similarly.
Now, define the one-qubit gates $H$ and $S$ as
\[H=\frac{1}{\sqrt{2}}\begin{pmatrix}
1&1\\
1&-1
\end{pmatrix}\]
and
\[S=\begin{pmatrix}
1&0\\
0&i
\end{pmatrix}\]
In each case, the $4\times4$ matrix corresponding to the two-qubit controlled gate is given.
Denote the following gate by $c_1H_2$:
Observe how the $c_1H_2$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_1H_2}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_1H_2}\mapsto\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},\]
\[\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}\overset{c_1H_2}\mapsto\super{\ket{10}+\ket{11}}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0\\0\\1\\1\end{pmatrix},\]
\[\ket{11}=\begin{pmatrix} 0\\0\\0\\1\end{pmatrix}\overset{c_1H_2}\mapsto\super{\ket{10}-\ket{11}}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0\\0\\1\\-1\end{pmatrix}.\]
The matrix representation of the gate $c_1H_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1H_2=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
0&0&\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}
\end{pmatrix}.\]
Denote the following gate by $c_2H_1$:
Observe how the $c_2H_1$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_2H_1}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_2H_1}\mapsto\super{\ket{01}+\ket{11}}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0\\1\\0\\1\end{pmatrix},\]
\[\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}\overset{c_2H_1}\mapsto\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}, \ \ \ \ \
\ket{11}=\begin{pmatrix} 0\\0\\0\\1\end{pmatrix}\overset{c_2H_1}\mapsto\super{\ket{01}-\ket{11}}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0\\1\\0\\-1\end{pmatrix}.\]
Then, matrix representation of the gate $c_1H_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1H_2=\begin{pmatrix}
1&0&0&0\\
0&\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\
0&0&1&0\\
0&\frac{1}{\sqrt{2}}&0&\frac{-1}{\sqrt{2}}
\end{pmatrix}.\]
Denote the following gate by $c_1S_2$:
First note that $S\ket{0}=\ket{0}$ and $S\ket{1}=i\ket{1}$. Now observe how the $c_1S_2$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_1S_2}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_1S_2}\mapsto\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},\]
\[\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}\overset{c_1S_2}\mapsto\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{11}=\begin{pmatrix} 0\\0\\0\\1\end{pmatrix}\overset{c_1S_2}\mapsto\ket{11}=\begin{pmatrix} 0\\0\\0\\i\end{pmatrix}.\]
The matrix representation of the gate $c_1S_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1S_2=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&i
\end{pmatrix}.\]
Denote the following gate by $c_2S_1$:
Observe how the $c_2S_1$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_2S_1}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_2S_1}\mapsto\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},\]
\[\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}\overset{c_2S_1}\mapsto\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{11}=\begin{pmatrix} 0\\0\\0\\1\end{pmatrix}\overset{c_2S_1}\mapsto\ket{11}=\begin{pmatrix} 0\\0\\0\\i\end{pmatrix}.\]
Note that the action of $c_2S_1$ is identical to that of $c_1S_2$ from case (iii), and so the matrix representation of $c_2S_1$ is also the same:
\[c_2S_1=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&i
\end{pmatrix}.\]
Here, the $4\times4$ matrix corresponding to the following quantum circuit will be given:
where $S$ is defined as it was above, and the last two-qubit gate denoted by $SWAP$ transposes the two qubits (more precisely, the $SWAP$ gate maps $\ket{00}\mapsto\ket{00}, \ket{01}\mapsto\ket{10}, \ket{10}\mapsto\ket{01},\ \text{and}\ \ket{11}\mapsto\ket{11}$).
Denote the overall action of this circuit by the unitary operator $U$. In order to express the action of $U$ as a matrix, first observe how $U$ acts on the four computational basis states:
\[\begin{array}{rcl}
\ket{00} &\overset{H\otimes I}\mapsto&\super{\ket{00}+\ket{10}} \\
&\overset{c_1S_2}\mapsto&\super{\ket{00}+\ket{10}} \\
&\overset{I\otimes H}\mapsto&\frac{1}{2}(\ket{00}+\ket{01}+\ket{10}+\ket{11}) \\
&\overset{SWAP}\mapsto&\frac{1}{2}(\ket{00}+\ket{10}+\ket{01}+\ket{11})
\end{array}
\implies U\begin{pmatrix}1\\0\\0\\0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\1\\1\\1\end{pmatrix},\]
\[\begin{array}{rcl}
\ket{01} &\overset{H\otimes I}\mapsto&\super{\ket{01}+\ket{11}} \\
&\overset{c_1S_2}\mapsto&\super{\ket{01}+i\ket{11}} \\
&\overset{I\otimes H}\mapsto&\frac{1}{2}(\ket{00}-\ket{01}+i\ket{10}-i\ket{11}) \\
&\overset{SWAP}\mapsto&\frac{1}{2}(\ket{00}-\ket{10}+i\ket{01}-i\ket{11})
\end{array}
\implies U\begin{pmatrix}0\\1\\0\\0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\i\\-1\\-i\end{pmatrix},\]
\[\begin{array}{rcl}
\ket{10} &\overset{H\otimes I}\mapsto&\super{\ket{00}-\ket{10}} \\
&\overset{c_1S_2}\mapsto&\super{\ket{00}-\ket{10}} \\
&\overset{I\otimes H}\mapsto&\frac{1}{2}(\ket{00}+\ket{01}-\ket{10}-\ket{11}) \\
&\overset{SWAP}\mapsto&\frac{1}{2}(\ket{00}+\ket{10}-\ket{01}-\ket{11})
\end{array}
\implies U\begin{pmatrix}0\\0\\1\\0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\-1\\1\\-1\end{pmatrix},\]
\[\begin{array}{rcl}
\ket{11} &\overset{H\otimes I}\mapsto&\super{\ket{01}-\ket{11}} \\
&\overset{c_1S_2}\mapsto&\super{\ket{01}-i\ket{11}} \\
&\overset{I\otimes H}\mapsto&\frac{1}{2}(\ket{00}-\ket{01}-i\ket{10}+i\ket{11}) \\
&\overset{SWAP}\mapsto&\frac{1}{2}(\ket{00}-\ket{10}-i\ket{01}+i\ket{11})
\end{array}
\implies U\begin{pmatrix}1\\0\\0\\0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\-i\\-1\\i\end{pmatrix}.\]
Then the matrix constructed that contains the four output vectors as its columns gives the matrix representation of the unitary operation $U$ given by the circuit:
\[U=\frac{1}{2}\begin{pmatrix}
1&1&1&1\\
1&i&-1&-i\\
1&-1&1&-1\\
1&-i&-1&i
\end{pmatrix}.\]
In what follows, a two-qubit quantum circuit consisting of one $CNOT$ gate and two Hadamard $H$ gates that computes the following unitary transformation will be described:
\[U=\begin{pmatrix}
1&0&0&0 \\
0&1&0&0\\
0&0&1&0 \\
0&0&0&-1
\end{pmatrix}\]
Consider the for computational basis states and their corresponding representation as column vectors:
\[\ket{00}=\begin{pmatrix}
1\\0\\0\\0
\end{pmatrix},
\ket{01}=\begin{pmatrix}
0\\1\\0\\0
\end{pmatrix},
\ket{10}=\begin{pmatrix}
0\\0\\1\\0
\end{pmatrix},
\ket{11}=\begin{pmatrix}
0\\0\\0\\1
\end{pmatrix}.\]
Observe that $U$ leaves the first three basis states unchanged and only multiplies the last by a factor of $-1$:
\[U\ket{00}=\ket{00},
U\ket{01}=\ket{01},
U\ket{10}=\ket{10},
U\ket{11}=-\ket{11}.\]
Consider the following circuit consisting of a Hadamard gate acting on the second qubit, followed by a controlled-NOT gate, and then with a second Hadamard gate acting on the second qubit again.
The action of this circuit on each of the four basis states is described below:
\[\ket{00} \overset{I\otimes H}\mapsto\super{\ket{00}+\ket{01}}\overset{CNOT}\mapsto\super{\ket{00}+\ket{01}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{00}+\ket{01}+\ket{00}-\ket{01})=\ket{00},\]
\[\ket{01} \overset{I\otimes H}\mapsto\super{\ket{00}-\ket{01}}\overset{CNOT}\mapsto\super{\ket{00}-\ket{01}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{00}+\ket{01}-\ket{00}+\ket{01})=\ket{01},\]
\[\ket{10} \overset{I\otimes H}\mapsto\super{\ket{10}+\ket{11}}\overset{CNOT}\mapsto\super{\ket{11}+\ket{10}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{10}-\ket{11}+\ket{10}+\ket{11})=\ket{10}, \]
\[\ket{11} \overset{I\otimes H}\mapsto\super{\ket{10}-\ket{11}}\overset{CNOT}\mapsto\super{\ket{11}-\ket{10}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{10}-\ket{11}-\ket{10}-\ket{11})=-\ket{11}.\]
Thus, the action of the circuit is identical to the unitary transformation $U$ expressed as the matrix since the four basis states are mapped similarly.
Now, define the one-qubit gates $H$ and $S$ as
\[H=\frac{1}{\sqrt{2}}\begin{pmatrix}
1&1\\
1&-1
\end{pmatrix}\]
and
\[S=\begin{pmatrix}
1&0\\
0&i
\end{pmatrix}\]
In each case, the $4\times4$ matrix corresponding to the two-qubit controlled gate is given.
Denote the following gate by $c_1H_2$:
Observe how the $c_1H_2$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_1H_2}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_1H_2}\mapsto\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},\]
\[\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}\overset{c_1H_2}\mapsto\super{\ket{10}+\ket{11}}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0\\0\\1\\1\end{pmatrix},\]
\[\ket{11}=\begin{pmatrix} 0\\0\\0\\1\end{pmatrix}\overset{c_1H_2}\mapsto\super{\ket{10}-\ket{11}}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0\\0\\1\\-1\end{pmatrix}.\]
The matrix representation of the gate $c_1H_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1H_2=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
0&0&\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}
\end{pmatrix}.\]
Denote the following gate by $c_2H_1$:
Observe how the $c_2H_1$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_2H_1}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_2H_1}\mapsto\super{\ket{01}+\ket{11}}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0\\1\\0\\1\end{pmatrix},\]
\[\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}\overset{c_2H_1}\mapsto\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}, \ \ \ \ \
\ket{11}=\begin{pmatrix} 0\\0\\0\\1\end{pmatrix}\overset{c_2H_1}\mapsto\super{\ket{01}-\ket{11}}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0\\1\\0\\-1\end{pmatrix}.\]
Then, matrix representation of the gate $c_1H_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1H_2=\begin{pmatrix}
1&0&0&0\\
0&\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\
0&0&1&0\\
0&\frac{1}{\sqrt{2}}&0&\frac{-1}{\sqrt{2}}
\end{pmatrix}.\]
Denote the following gate by $c_1S_2$:
First note that $S\ket{0}=\ket{0}$ and $S\ket{1}=i\ket{1}$. Now observe how the $c_1S_2$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_1S_2}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_1S_2}\mapsto\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},\]
\[\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}\overset{c_1S_2}\mapsto\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{11}=\begin{pmatrix} 0\\0\\0\\1\end{pmatrix}\overset{c_1S_2}\mapsto\ket{11}=\begin{pmatrix} 0\\0\\0\\i\end{pmatrix}.\]
The matrix representation of the gate $c_1S_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1S_2=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&i
\end{pmatrix}.\]
Denote the following gate by $c_2S_1$:
Observe how the $c_2S_1$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_2S_1}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_2S_1}\mapsto\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},\]
\[\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}\overset{c_2S_1}\mapsto\ket{10}=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{11}=\begin{pmatrix} 0\\0\\0\\1\end{pmatrix}\overset{c_2S_1}\mapsto\ket{11}=\begin{pmatrix} 0\\0\\0\\i\end{pmatrix}.\]
Note that the action of $c_2S_1$ is identical to that of $c_1S_2$ from case (iii), and so the matrix representation of $c_2S_1$ is also the same:
\[c_2S_1=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&i
\end{pmatrix}.\]
Here, the $4\times4$ matrix corresponding to the following quantum circuit will be given:
where $S$ is defined as it was above, and the last two-qubit gate denoted by $SWAP$ transposes the two qubits (more precisely, the $SWAP$ gate maps $\ket{00}\mapsto\ket{00}, \ket{01}\mapsto\ket{10}, \ket{10}\mapsto\ket{01},\ \text{and}\ \ket{11}\mapsto\ket{11}$).
Denote the overall action of this circuit by the unitary operator $U$. In order to express the action of $U$ as a matrix, first observe how $U$ acts on the four computational basis states:
\[\begin{array}{rcl}
\ket{00} &\overset{H\otimes I}\mapsto&\super{\ket{00}+\ket{10}} \\
&\overset{c_1S_2}\mapsto&\super{\ket{00}+\ket{10}} \\
&\overset{I\otimes H}\mapsto&\frac{1}{2}(\ket{00}+\ket{01}+\ket{10}+\ket{11}) \\
&\overset{SWAP}\mapsto&\frac{1}{2}(\ket{00}+\ket{10}+\ket{01}+\ket{11})
\end{array}
\implies U\begin{pmatrix}1\\0\\0\\0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\1\\1\\1\end{pmatrix},\]
\[\begin{array}{rcl}
\ket{01} &\overset{H\otimes I}\mapsto&\super{\ket{01}+\ket{11}} \\
&\overset{c_1S_2}\mapsto&\super{\ket{01}+i\ket{11}} \\
&\overset{I\otimes H}\mapsto&\frac{1}{2}(\ket{00}-\ket{01}+i\ket{10}-i\ket{11}) \\
&\overset{SWAP}\mapsto&\frac{1}{2}(\ket{00}-\ket{10}+i\ket{01}-i\ket{11})
\end{array}
\implies U\begin{pmatrix}0\\1\\0\\0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\i\\-1\\-i\end{pmatrix},\]
\[\begin{array}{rcl}
\ket{10} &\overset{H\otimes I}\mapsto&\super{\ket{00}-\ket{10}} \\
&\overset{c_1S_2}\mapsto&\super{\ket{00}-\ket{10}} \\
&\overset{I\otimes H}\mapsto&\frac{1}{2}(\ket{00}+\ket{01}-\ket{10}-\ket{11}) \\
&\overset{SWAP}\mapsto&\frac{1}{2}(\ket{00}+\ket{10}-\ket{01}-\ket{11})
\end{array}
\implies U\begin{pmatrix}0\\0\\1\\0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\-1\\1\\-1\end{pmatrix},\]
\[\begin{array}{rcl}
\ket{11} &\overset{H\otimes I}\mapsto&\super{\ket{01}-\ket{11}} \\
&\overset{c_1S_2}\mapsto&\super{\ket{01}-i\ket{11}} \\
&\overset{I\otimes H}\mapsto&\frac{1}{2}(\ket{00}-\ket{01}-i\ket{10}+i\ket{11}) \\
&\overset{SWAP}\mapsto&\frac{1}{2}(\ket{00}-\ket{10}-i\ket{01}+i\ket{11})
\end{array}
\implies U\begin{pmatrix}1\\0\\0\\0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\-i\\-1\\i\end{pmatrix}.\]
Then the matrix constructed that contains the four output vectors as its columns gives the matrix representation of the unitary operation $U$ given by the circuit:
\[U=\frac{1}{2}\begin{pmatrix}
1&1&1&1\\
1&i&-1&-i\\
1&-1&1&-1\\
1&-i&-1&i
\end{pmatrix}.\]