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Deutsch-Josza algorithm

  1. Jan 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider a general function ##f:\{0,1\}^n \rightarrow \{0,1\}## (not necessarily constant or balanced). The function gives ##f(x)=0## for a portion##z## of the different possible inputs ##x\in\{0,1\}^n##, and ##f(x)=1## for the rest of the inputs. Consider that we construct the Deutsch-Josza algorithm for this function. Calculate the probability, that the algorithm gives the answer "The function is constant" and the probability that it gives "The function is balanced".

    ##f(x)## is balanced= the number of inputs for which the function takes values ##0## and ##1## are the same .
    ##f(x)## is balanced= ##f(x)=0## for all inputs or ##f(x)=1## for all inputs.

    2. Relevant equations
    At the end of the Deutsch-Josza algorithm, the wave function is in the sate
    [tex]
    |\psi\rangle=\sum_{y\in \{0,1\}^n} \left( \frac{1}{2^n} \sum_{x\in \{0,1\}^n} (-1)^{f(x)+x\cdot y} \right)
    [/tex]
    where ##x\cdot y= \sum_{k=1}^n x_k y_k \;(\mod 2)##

    3. The attempt at a solution
    The amplitude associated with the classical state ##|0^n\rangle## is

    [tex]
    \alpha=\frac{1}{2^n}\sum_{x\in \{0,1\}^n} (-1)^{f(x)}
    [/tex]

    Thus
    [tex]\begin{align*}
    P(constant)&=\lvert \frac{1}{2^n} \sum_{x\in \{0,1\}^n, i=1}^z (-1)^{f_0(x)}+\frac{1}{2^n} \sum_{x\in \{0,1\}^n, j=1}^{2^n-z} (-1)^{f_1(x)}\rvert^2\\
    &=\lvert \frac{1}{2^n} \left(z-2^n+z\right)\rvert^2=\lvert\frac{2z-2^n}{2^n}\rvert^2
    \end{align*}
    [/tex]


    However, I struggle finding the expression for ##P(balanced)##.

    My attempt is finding the classical state ##|1^n\rangle##:

    [tex]\begin{align*}
    P(balanced)&= \lvert( \frac{1}{2^n} \sum_{x\in \{0,1\}^n} (-1)^{f(x)+\sum_{k=1}^n x_k}\rvert^2\\
    &=\lvert \frac{1}{2^n}\sum_{x\in \{0,1\}^n, i=1}^z (-1)^{\sum_{ k=1}^n x_k} + \frac{1}{2^n}\sum_{x\in \{0,1\}^n, j=1}^{2^n-z} (-1)^{1+\sum_{k=1}^n x_k}\rvert^2

    \end{align*}[/tex]

    This should be ##1## for ##z=\frac{2^n}{n}## and ##0## for ##z=2^n##. However, I don't think this result is correct.

    Thank you so much!
     
  2. jcsd
  3. Jan 26, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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