Question

$$\begin{gathered} \mathbf{Convocation}\mathbf{-}\mathbf{Multiplication}\mathbf{}\mathbf{.}\mathbf{}\mathbf{Suppose}\overline{)\mathbf{\alpha }\mathbf{⟩}\mathbf{=}\mathbf{\sum }_{\mathbf{j}}{\mathbf{a}}_{\mathbf{j}}\overline{)\mathbf{j}\mathbf{⟩}\mathbf{}\mathbf{by}\mathbf{}\mathbf{I}\mathbf{}\mathbf{to}\mathbf{}\mathbf{get}\mathbf{}\mathbf{the}\mathbf{}\mathbf{superposition}\mathbf{}\overline{)\mathbf{\alpha }}}}\mathbf{\text{'}}\mathbf{⟩}\mathbf{=}\mathbf{\sum }_{\mathbf{j}}{\mathbf{\alpha }}_{\mathbf{j}}\overline{)\mathbf{j}\mathbf{+}\mathbf{I}\mathbf{⟩}\mathbf{.}} \\ \mathbf{If}\mathbf{}\mathbf{the}\mathbf{}\mathbf{QFT}\mathbf{}\mathbf{of}\mathbf{}\overline{)\mathbf{\alpha }\mathbf{⟩}}\mathbf{}\mathbf{is}\mathbf{}\overline{)\mathbf{\beta }\mathbf{⟩}\mathbf{,}\mathbf{}\mathbf{show}\mathbf{}\mathbf{that}\mathbf{}\mathbf{the}\mathbf{}\mathbf{QFT}\mathbf{}\mathbf{OF}\mathbf{}\mathbf{\alpha }\mathbf{\text{'}}}\mathbf{}\mathbf{IS}\mathbf{}\mathbf{\beta }\mathbf{\text{'}}\mathbf{,}\mathbf{}\mathbf{Where}\mathbf{}\mathbf{\beta j}\mathbf{\text{'}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{j}}{\mathbf{\omega }}^{\mathbf{ij}}\mathbf{.}\mathbf{}\mathbf{Conclude}\mathbf{}\mathbf{that}\mathbf{}\mathbf{if}\mathbf{}\overline{)\mathbf{\alpha }\mathbf{\text{'}}\mathbf{⟩}\mathbf{=}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{{\displaystyle \frac{\mathbf{M}}{\mathbf{K}\mathbf{-}\mathbf{1}}}}\sqrt{\frac{\mathbf{K}}{\mathbf{M}}}\overline{)\mathbf{jk}\mathbf{+}\mathbf{I}\mathbf{⟩}\mathbf{,}}} \\ \mathbf{then}\mathbf{}\overline{)\mathbf{\beta }\mathbf{\text{'}}\mathbf{⟩}\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{k}}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{k}\mathbf{-}\mathbf{1}}}{\mathbf{\omega }}^{\frac{\mathbf{IjM}}{\mathbf{K}}}\overline{)\frac{\mathbf{jM}}{\mathbf{k}}\mathbf{⟩}}\mathbf{.} \end{gathered}$$

Short answer

It can be proved that the QFT of $\alpha \text{'}$is$\beta \text{'}$ by offsetting / and applying QFT.

Step-by-step solution

Explain Convolution-Multiplication.

$$\begin{gathered} \mathrm{Consider}\mathrm{that}\mathrm{the}\overline{)\mathrm{\alpha }⟩=\sum _{\mathrm{j}}{\mathrm{a}}_{\mathrm{j}}}\overline{)\mathrm{j}}⟩\mathrm{by}\mathrm{I}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{superposition}\overline{)\mathrm{\alpha }\text{'}}⟩=\sum _{\mathrm{j}}{\mathrm{\alpha }}_{\mathrm{j}}\overline{)\mathrm{j}+\mathrm{I}⟩.} \\ \mathrm{If}\mathrm{the}\mathrm{QFT}\mathrm{of}\overline{)\mathrm{\alpha }⟩\mathrm{is}\overline{)\mathrm{\beta }⟩.}} \end{gathered}$$

Show that the QFT of α' is β'.

Consider that the $i^{th}$coefficient against $\overline{)\beta ⟩and}\overline{)\beta \text{'}⟩are,}$

$$\begin{gathered} {\beta }_i=\frac{1}{\sqrt{M}}\sum _j{\omega }^{ij}{a}_j......\left(1\right) \\ {\beta }_i^{\text{'}}=\frac{1}{\sqrt{M}}\sum _j{\omega }^{i\left(j+I\right)}{a}_{j.........\left(2\right)} \end{gathered}$$

By comparing Equations (1) and (2),

${\beta }_i^{\text{'}}={\beta }_i{\omega }^{il}......\left(3\right)$ $Consider,\overline{)\alpha =\rangle \sum _{jj=0}^{\frac{M}{K-1}}}\sqrt{\frac{k}{M}}\overline{)jk⟩,accordingtotheclaim}\overline{)\beta ⟩=\frac{1}{\sqrt{k}}\sum _{j=0}^{k-1}}\overline{)\frac{jM}{K}⟩.}$

After offsetting $\overline{)\alpha ⟩}$by I ,the result is

$\overline{)\alpha \text{'}\rangle =\sum _{j=0}^{\frac{M}{K-1}}}\sqrt{\frac{K}{M}}\overline{)jk+I⟩.....\left(4\right)}$

Perform QFT to get $\overline{)\beta \text{'}}\rangle .$According the Equation (3), the value of the $i^{th}$coefficient is $i=j\left(\frac{M}{K}\right)at\overline{)\beta \rangle .}$at .

Hence,

$\overline{)\beta \text{'}}\rangle =\frac{1}{\sqrt{k}}\sum _{j=0}^{k-1}{\omega }^{\frac{IjM}{K}}\overline{)\frac{jM}{K}⟩......\left(5\right)}$

Therefore, It is proved that the QFT of $\alpha \text{'}$is $\beta \text{'}$by the Equations (4) and (5).