Quantum computing - form a SWAP gate from x3 controlled-NOT gates

Click For Summary
SUMMARY

Three controlled-NOT (CNOT) gates can be combined to create a SWAP gate in a two-qubit system. The correct matrix representation for the CNOT gate is crucial for achieving the desired SWAP gate output. The matrices involved include the identity and the X gate, specifically CNOT(2->1) which is represented as I⊗|0><0| + X⊗|1><1|. The final SWAP gate matrix is confirmed to be accurate as 1 & 0 & 0 & 0; 0 & 0 & 1 & 0; 0 & 1 & 0 & 0; 0 & 0 & 0 & 1.

PREREQUISITES
  • Understanding of quantum gates, specifically controlled-NOT gates.
  • Familiarity with matrix representation of quantum operations.
  • Knowledge of qubit systems and their manipulation.
  • Basic principles of quantum mechanics and linear algebra.
NEXT STEPS
  • Study the matrix multiplication of quantum gates in detail.
  • Learn about the implementation of SWAP gates in quantum algorithms.
  • Explore the properties and applications of controlled-NOT gates in quantum circuits.
  • Investigate the role of identity and X gates in quantum computing.
USEFUL FOR

Quantum computing students, researchers in quantum algorithms, and professionals working on quantum circuit design will benefit from this discussion.

SK1.618
Messages
8
Reaction score
0

Homework Statement



Show that three controlled-NOT gates (for a 2 qubit system) can be combined to form a SWAP gate. The control qubit alternates between the 2 qubits for each consecutive c-NOT gate. (The diagram is Figure 5 of the following notes: http://www-inst.eecs.berkeley.edu/~cs191/fa07/lectures/lecture9_fa07.pdf )

Homework Equations



The explicit matrix form of a controlled-NOT gate is

\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{matrix}

The Attempt at a Solution



Multiply the following 3 matrices, representing c-NOTs with alternating control gate:

\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{matrix}

\begin{matrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{matrix}

\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{matrix}

The correct answer should be (for a SWAP gate):

\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{matrix}

But this is not what I get. I think there may be a problem with my matrix representation of the second c-NOT gate in the series.
 
Physics news on Phys.org
turns out, as you guessed, the matrix representation of the CNOT(2->1) gate is incorrect.

CNOT(2->1) = I[itex]\otimes[/itex]|0><0| + X[itex]\otimes[/itex]|1><1|

=
1 0 0 0
0 0 0 1
0 0 1 0
0 1 0 0
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K