Question

What is the QFT modulo M of $\overline{)\mathbf{j}\mathbf{>}}$

Short answer

The QFT modulo M of $\overline{)j>}$is

$QFT\left(j\right)=\left(\frac{1}{\sqrt{M}}\sum _{j=0}^{M-1}{W}^{jI}\overline{)j>}\right)$

Step-by-step solution

Step 1:Discrete Fourier Transform Formula

Quantum Fourier Transform:

Performing linear transformation on the “quantum bits” is called “quantum Fourier transform” (QFT).

It is similar to the “discrete Fourier transform” (DFT) where it works on the quantum state’s vector amplitude.

The “classical DFT” works on the vector of $\left(a_0,K,a_{N-1}\right)$and map it to the vector$\left(b_0,K,b_{N-1}\right)$.

It is defined by the following formula,$b_I=\left(\frac{1}{\sqrt{M}}\sum _{j=0}^{M-1}{a}_j{w}^{jI}\right)$

QFT modulo of j>

Here, the value of w is w= $e^{\frac{2\pi i}{M}}$and it is the $N^{th}$root of unity.

Similar to DFT, QFT works on the quantum state $\sum _{j=0}^{M-1}{a}_j\overline{)j>}$and map it to the quantum state.

$\sum _{j=0}^{M-1}{b}_j\overline{)j>.}$

It is defined by the following formula.

localid="1658904981124" $\overline{)j>}=\left(\frac{1}{\sqrt{M}}\sum _{j=0}^{M-1}{w}^{jI}\overline{)J>}\right)$Therefore,

$QFT\left(j\right)=\left(\frac{1}{\sqrt{M}}\sum _{j=0}^{M-1}{w}^{jI}\overline{)j>}\right)$