Depicting the Transformation O-gate

The following figure depicts the transformation gate $O_i$:

Let’s say, $i=0$. In that case, we’ll apply the function $f_0$. Per definition, $f_0(x)=0$. When we insert this into the above equation, we can see the output of $O_i$ is equal to its input:

$$O_0(|x\rangle\otimes|y\rangle)=|x\rangle\otimes|y\oplus|f_0(x)\rangle=|x\rangle\otimes|y\oplus|0\rangle=|x\rangle\otimes|y\rangle$$

We can safely state that not changing a state is reversible.

When $i=1$, we apply the function $f_1$, which returns 0 for $x=0$ and 1 for $x=1$. Thus, $f_1(x)=x$.

$$O_1(|x\rangle\otimes|y\rangle)=|x\rangle\otimes|y\oplus f_1(x)\rangle=|x\rangle\otimes|y\oplus x\rangle$$

The truth table of the term $|x\rangle\otimes|y\oplus x\rangle$ shows that it is reversible.

When we apply $f​_2$, it returns 1 for $x=0$ and 0 for $x=1$. We can then say $f​_2$ $f_2(x)=x\oplus1$.

$$O_2(|x\rangle\otimes|y\rangle)=|x\rangle\otimes|y\oplus f_2(x)\rangle=|x\rangle\otimes|y\oplus x \oplus 1\rangle$$

The truth table discloses that the term $|x\rangle\otimes|y\oplus x \oplus 1\rangle$ is reversible, too.

Finally, $f_3$ always returns 1.

$$O_3(|x\rangle\otimes|y\rangle)=|x\rangle\otimes|y\oplus|f_3(x)\rangle=|x\rangle\otimes |y\oplus 1\rangle)$$

The output is like the input, but with a reversed $y$.

The truth table shows the following:

• $O_i$ is a valid two-qubit gate for all $i$
• The output of $O_i$

The following truth table shows how the $O_i$-gate transforms pairs of qubits in the basis states.

As usual, when we only look at qubits in the basis states, there’s nothing special with a quantum circuit. However, things get interesting when the qubits are in a state of superposition.

Let’s input the states $|+\rangle$ ...