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}.