Constructing simple two qubit states

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:
\[U=\frac{1}{2}\begin{pmatrix}
1&1&1&1\\
1&i&-1&-i\\
1&-1&1&-1\\
1&-i&-1&i
\end{pmatrix}.\]

Entangled vs. Seperable states

If possible, each two-qubit state given below will be expressed as a product of two one-qubit states. Otherwise, it will be shown that such a factorization is impossible (in this case, the qubits are said to be entangled).


Case (i):
$\ket{\psi}=\frac{1}{2}\ket{00}+\frac{i}{2}\ket{01}-\frac{1}{2}\ket{10}-\frac{i}{2}\ket{11}$

Consider the one-qubit states $\ket{\phi_1}=\frac{1}{\sqrt2}\ket{0}-\frac{1}{\sqrt2}\ket{1}$ and $\ket{\phi_2}=\frac{1}{\sqrt2}\ket{0}+\frac{i}{\sqrt2}\ket{1}$. Then expanding the tensor product of the two states $\ket{\phi_1}\otimes\ket{\phi_2}=\ket{\phi_1}\ket{\phi_2}$ yields:
\[\begin{array}{rcl}\ket{\phi_1}\ket{\phi_2}&=&(\frac{1}{\sqrt2}\ket{0}-\frac{1}{\sqrt2}\ket{1})(\frac{1}{\sqrt2}\ket{0}+\frac{i}{\sqrt2}\ket{1}) \\
\\
&=&\frac{1}{\sqrt2}\frac{1}{\sqrt2}\ket{0}\ket{0}+\frac{1}{\sqrt2}\frac{i}{\sqrt2}\ket{0}\ket{1}-\frac{1}{\sqrt2}\frac{1}{\sqrt2}\ket{1}\ket{0}-\frac{1}{\sqrt2}\frac{i}{\sqrt2}\ket{1}\ket{1} \\
\\
&=&\frac{1}{2}\ket{00}+\frac{i}{2}\ket{01}-\frac{1}{2}\ket{10}-\frac{i}{2}\ket{11}.
\end{array}\]
Thus, $\ket{\psi}=\ket{\phi_1}\ket{\phi_2}$ can be expressed as the tensor product of two one-qubit states and is therefore a separable state.


Case (ii):
$\ket{\psi}=\frac{1}{2}\ket{00}+\frac{1}{2}\ket{01}+\frac{1}{2}\ket{10}-\frac{1}{2}\ket{11}$
It will be shown that $\ket{\psi}$ is an entangled state. For the sake of contradiction, suppose that $\ket{\psi}$ can be written as the tensor product of two one-qubit states expressed in their general form $\ket{\phi_1}=\alpha\ket{0}+\beta\ket{1}$ and $\ket{\phi_2}=\alpha'\ket{0}+\beta'\ket{1}$, where $\alpha,\beta,\alpha',\beta'\in\mathbb{C}$. Then,
\[\begin{array}{rcl}
\ket{\psi}&=&\ket{\phi_1}\ket{\phi_2} \\
\\
&=&(\alpha\ket{0}+\beta\ket{1})(\alpha'\ket{0}+\beta'\ket{1}) \\
\\
&=&\alpha\alpha'\ket{00}+\alpha\beta'\ket{01}+\beta\alpha'\ket{10}+\beta\beta'\ket{11}
\end{array}\]
Now compare these amplitudes to those actually provided in $\ket{\psi}$:
\[\alpha\alpha'=\frac{1}{2}, \alpha\beta'=\frac{1}{2}, \beta\beta'=\frac{1}{2}, \beta\beta'=\frac{-1}{2}.\]
Since these products are all nonzero numbers and $\alpha,\beta,\alpha',\beta'\in\mathbb{C}$, this implies that $\alpha,\beta,\alpha',\beta'\neq 0$. Consider the following relationships which must also hold:
\[\begin{array}{rl}
&\alpha\alpha'-\alpha\beta'=\frac{1}{2}-\frac{1}{2}=0\\
\implies&\alpha(\alpha'-\beta')=0 \\
\implies&\alpha'=\beta',
\end{array}\]
and likewise
\[\begin{array}{rl}
&\beta\alpha'+\beta\beta'=\frac{1}{2}+\frac{-1}{2}=0\\
\implies&\beta(\alpha'+\beta')=0 \\
\implies&\alpha'=-\beta',
\end{array}\]
but then $\beta'=-\beta'$ implies that $\beta'=0$ which contradicts the fact that $\beta'\neq0$. Therefore, the two-qubit state $\ket{\psi}$ cannot be expressed as the tensor product of two one-qubit states, and must be entangled.


Case (iii):
$\ket{\psi}=\frac{9}{25}\ket{00}+\frac{12}{25}\ket{01}+\frac{12}{25}\ket{10}-\frac{16}{25}\ket{11}$

Consider the two identical one-qubit states $\ket{\phi_1}=\ket{\phi_2}=\frac{3}{5}\ket{0}+\frac{4}{5}\ket{1}$. Then expanding the tensor product of the two states $\ket{\phi_1}\otimes\ket{\phi_2}=\ket{\phi_1}\ket{\phi_2}$ yields:
\[\begin{array}{rcl}\ket{\phi_1}\ket{\phi_2}&=&(\frac{3}{5}\ket{0}+\frac{4}{5}\ket{1})(\frac{3}{5}\ket{0}+\frac{4}{5}\ket{1}) \\
\\
&=&\frac{3}{5}\frac{3}{5}\ket{0}\ket{0}+\frac{3}{5}\frac{4}{5}\ket{0}\ket{1}+\frac{4}{5}\frac{3}{5}\ket{1}\ket{0}+\frac{4}{5}\frac{4}{5}\ket{1}\ket{1} \\
\\
&=&\frac{9}{25}\ket{00}+\frac{12}{25}\ket{01}+\frac{2}{25}\ket{10}+\frac{16}{25}\ket{11}.
\end{array}\]
Thus, $\ket{\psi}=\ket{\phi_1}\ket{\phi_2}$ can be expressed as the tensor product of two one-qubit states and is therefore a separable state.