Question

The CONTROLLED SWAP ( $\mathbf{C}\mathbf{-}\mathbf{SWAP}$) gate takes as input$\mathbf{3}$ qubits and swaps the second and third if and only if the first qubit is a $\mathbf{1}$.

1. Show that each of the$\mathbf{NOT}\mathbf{,}\mathbf{CNOT}\mathbf{and}\mathbf{C}\mathbf{-}\mathbf{SWAP}$ gates are their own inverses.
2. Show how to implement anrole="math" localid="1658207684748" $\mathbf{AND}$ gate using a$\mathbf{C}\mathbf{-}\mathbf{SWAP}$ gate, i.e., what inputs$\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{c}$ would you give to a$\mathbf{C}\mathbf{-}\mathbf{SWAP}$ gate so that one of the outputs is $\mathbf{a}\mathbf{\wedge }\mathbf{b}$?
3. How would you achieve fanout using just these three gates? That is, on input $\mathbf{a}$and$\mathbf{0}$ , output$\mathbf{a}$ and$\mathbf{a}$ .
4. Conclude therefore that for any classical circuit $\mathbf{C}$there is an equivalent quantum circuit$\mathbf{Q}$ using just NOT and C-SWAP gates in the following sense: if$\mathbf{C}$ outputs $\mathbf{Y}$on input $\mathbf{x}$, then $\mathbf{Q}$outputs$\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{⟩}$ on input $\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{⟩}$. (Here$\mathit{z}$ is some set of junk bits that are generated during this computation.)
5. Now show that that there is a quantum circuit ${\mathbf{Q}}^{-1}$ that outputs$\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{⟩}$ on input$\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}⟩$ .
6. Show that there is a quantum circuit$\mathbf{Q}\mathbf{\text{'}}$ made up of$\mathbf{NOT}\mathbf{,}\mathbf{CNOT}\mathbf{and}\mathbf{C}\mathbf{-}\mathbf{SWAP}$gates that outputs$\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{0}⟩$ on input $\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}⟩$.

Short answer

1. The inverse of $\mathrm{NOT},\mathrm{CNOT}\mathrm{and}\mathrm{C}-\mathrm{SWAP}$gates is the same gates itself.
2. AND gate is implemented with a C-SWAP gate by giving the input of$\mathrm{c}$ as $0$.
3. The fanout can be achieved by giving the input for third qubit ( $\mathrm{c}$) as$1$ with the represented input for second qubit ( $b$) as $0$.
4. Any classical circuit can be implemented using the C-SWAP gate and NOT gate.
5. A quantum circuit that outputs$|\mathrm{x},0,0⟩$ on input $|\mathrm{x},\mathrm{y},\mathrm{z}⟩$is there.
6. A quantum circuit is there that is made of NOT, CNOT and C-SWAP gates.

Step-by-step solution

Define quantum circuit

Circuits that perform Fourier transform faster than classical computers are called Quantum Circuits.A quantum operation is similar to operations of gates like AND or NOT gate.It works on a single qubit or two qubits.

Quantum operation on input$0$ is:

Quantum operation on input$1$ is:

Calculate the inverse of NOT, CNOT and C-SWAP gates

(a)

The gates NOT, CNOT and C-SWAP are its own inverse.

NOT Gate:

NOT gate performs the inverse operation $\mathrm{NOT}\left(\mathrm{a}|0⟩\right)=|1⟩$.

accordingly, taking inverse two times gives the equal output as proven:

$\begin{array}{c}\mathrm{NOT}\left(\mathrm{NOT}\left(|\mathrm{x}⟩\right)\right)=\mathrm{NOT}\left(|\mathrm{y}⟩\right)\\ =\mathrm{x}\end{array}$

CNOT Gate:

The CNOT gate works with $2$qubits and its miles indicated by using CNOT.

First qubit is the “manipulate qubit” and second qubit is the “target qubit”.

If the primary qubit is $0$then it produces the identical entry as output.

in any other case, it flips the second qubit of input

The output of CNOT gate for the bits $00$and $10$is shown beneath:

$\begin{array}{l}\mathrm{CNOT}\left(|00⟩\right)=|00⟩\\ \mathrm{CNOT}\left(|10⟩\right)=|11⟩\end{array}$

CNOT gate performs the inverse operation.

NOT gate can be connected to the second qubit because the first qubit is a control bit and it cannot be changed.

Thus, taking the inverse twice provides the identical output as shown:

$\begin{array}{c}\mathrm{CNOT}\left(\mathrm{CNOT}\left(|\mathrm{x}⟩\right)\right)=\mathrm{CNOT}\left(|\mathrm{y}⟩\right)\\ =\mathrm{x}\end{array}$

C-SWAP Gate:

The C-SWAP gate works with three qubits.

Here, first qubit is the “manipulate qubit” and second and $\frac{1}{3}$qubit are the “goal qubit”. If the first qubit is 1 then it swaps the second qubit and $0.33$ qubit.

CNOT gate plays the inverse operation.

NOT gate is added to the second qubit and third qubit since the first qubit is a manage bit and it cannot be changed.

For that reason, taking inverse twice affords the equal output as shown:

$\begin{array}{c}\mathrm{C}-\mathrm{SWAP}(\mathrm{C}-\mathrm{SWAP}\left(|\mathrm{x}⟩\right))=\mathrm{C}-\mathrm{SWAP}\left(|\mathrm{y}⟩\right)\\ =\mathrm{x}\end{array}$

Thus, the inverse of$\mathrm{NOT},\mathrm{CNOT}\mathrm{and}\mathrm{C}-\mathrm{SWAP}$ gates is the same gates.

Implement AND gate using C-SWAP gate

(b)

AND gate is implemented with a C-SWAP gate by giving the input of third qubit ($c$) as .

The truth table of C-SWAP is shown:

Input	Output
A	b	c	a’	b’	c’
0	0	0	0	0	0
0	0	1	0	0	1
0	1	0	0	1	0
0	1	1	0	1	1
1	0	0	1	0	0
1	0	1	1	1	0
1	1	0	1	0	1
1	1	1	1	1	1

The highlighted rows show the output of$\mathrm{a}\wedge \mathrm{b}$ in the column${\mathrm{c}}^{\text{'}}$ for the inputs$\mathrm{a}$ and $b$. It is shown in the following diagram:

Therefore, the AND gate is implemented using a C-SWAP gate by giving the input$c$ as $0$.

Obtain the fanout

(c)

Fanout is achieved by using C-SWAP alone by giving the input for third qubit ( $\mathrm{c}$) as with the represented input for second qubit ($b$) as .

The truth table of C-SWAP is shown:

Input	Output
A	B	c	a’	b’	c’
0	0	0	0	0	0
0	0	1	0	0	1
0	1	0	0	1	0
0	1	1	0	1	1
1	0	0	1	0	0
1	0	1	1	1	0
1	1	0	1	0	1
1	1	1	1	1	1

Hence, a fanout is achieved using NOT, CNOT and C-SWAP gates.

 Obtain the equivalent quantum circuit

(d)

The section (b) implements AND gate using C-SWAP gate and adding NOT gate to the output of C-SWAP gate performs the inverse operation.Thus, any classical circuit can be implemented using the C-SWAP gate and NOT gate.

The truth table of C-SWAP is shown:

Input	Output
X	Y	z	x’	y’	z’
0	0	0	0	0	0
0	0	1	0	0	1
0	1	0	0	1	0
0	1	1	0	1	1
1	0	0	1	0	0
1	0	1	1	1	0
1	1	0	1	0	1
1	1	1	1	1	1

The circuit is:

Therefore, any classical circuit is implemented using the C-SWAP gate and NOT gate.

Obtain Quantum Circuit of Q'

(e)

The truth table of C-SWAP is shown:

Input	Output
X	Y	z	x’	y’	z’
0	0	0	0	0	0
0	0	1	0	0	1
0	1	0	0	1	0
0	1	1	0	1	1
1	0	0	1	0	0
1	0	1	1	1	0
1	1	0	1	0	1
1	1	1	1	1	1

The highlighted rows show the outputfor the inputs.

It is shown in the following circuit Q’:

Therefore, a quantum circuit that outputs $|\mathrm{x},0,0⟩$on input$|\mathrm{x},\mathrm{y},\mathrm{z}⟩$ is made.

Obtain the quantum circuit

(f)

The truth table of C-SWAP is shown:

Input	Output
X	Y	Z	x'	y'	z'
0	0	0	0	0	0
0	0	1	0	0	1
0	1	0	0	1	0
0	1	1	0	1	1
1	0	0	1	0	0
1	0	1	1	1	0
1	1	0	1	0	1
1	1	1	1	1	1

The resulting circuit is shown:

Therefore, a quantum circuit is made of NOT, CNOT and C-SWAP gates.