The discussion revolves around the use of swap gates in Quantum Fourier Transform (QFT) circuits, specifically addressing the reasons for their inclusion in the circuit design as presented in Nielsen & Chuang's work. Participants explore the implications of qubit order in relation to endian-ness and its effect on frequency representation.
Participants generally agree on the role of swap gates in addressing endian-ness in QFT circuits, but there is some disagreement regarding the specific frequency representations associated with qubit states, indicating unresolved aspects of the discussion.
The discussion includes potential confusion regarding the correct frequency values associated with specific qubit states, highlighting the need for clarity in definitions and interpretations of output in QFT circuits.
jimmycricket said:I'm currently working through Nielsen & Chuang's section on the circuit design for implementing the QFT. I'm confused as to why swap gates are used in the model to swap the order of qubits. Heres what I'm looking at http://www.johnboccio.com/research/quantum/notes/QC10th.pdf page 247 figure 5.1
Can anybody help?
Strilanc said:Here's a blog post about the QFT with some puzzles in a circuit simulator.
The basic reason is that the QFT's output has the qubits in the reverse order of what you typically want. It switches you from big-endian bits to little-endian bits (or vice versa), so that the amplitude of the state 10010 corresponds to the 7/N frequency instead of the 18/N frequency (or vice-versa).
You don't have to swap them back into the right order, it just makes it easier to think about the QFT when used as a tool because you're keeping the endian-ness consistent. It makes the result match the usual definition of the discrete Fourier transform, where F(1) corresponds to the lowest non-zero frequency (instead of what F(N-1) corresponds to).
Do you mean the 9/N frequency not 7/N. Am I right in saying the qubits in your example reversed would read 01001=9?Strilanc said:Here's a blog post about the QFT with some puzzles in a circuit simulator.
The basic reason is that the QFT's output has the qubits in the reverse order of what you typically want. It switches you from big-endian bits to little-endian bits (or vice versa), so that the amplitude of the state 10010 corresponds to the 7/N frequency instead of the 18/N frequency (or vice-versa).
jimmycricket said:Do you mean the 9/N frequency not 7/N. Am I right in saying the qubits in your example reversed would read 01001=9?