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I am currently reading Shor's algorithm on my own and I come across a paper via this link. It shows the circuit for implementing Shor's alogrithm. Here it depicts that taking a QFT circuit on the number a on the qubits b that are already in fourier space would result in the addition of a and b in Fourier space.

enter image description here

The addition is quite intuitive. However, when it comes the subtraction, I am stuck.Why taking an inverse QFT circuit would result in the $b-a$ or $2^{n+1} - (a-b)$? Wouldn't it return $a+b$ as the second stage output is $\Phi(a+b)$?

enter image description here

devance
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In the paper they mention that a thick bar drawn on the left of a gate is their notation for its inverse. So in the second circuit they have drawn the inverse of $\phi ADD(a)$, i.e. $\phi ADD(-a)$.

$QFT^{-1}$ is there to un-Fourier transform the result, not to actually perform the substraction.

user96233
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