Why Are Swap Gates Used in Quantum Fourier Transform Circuits?

[jimmycricket]

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]

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?

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).

 

[jimmycricket]

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).

Thats cleared that up, Thanks

 

[jimmycricket]

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).

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]

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?

Yeah, my mistake.