In a book for quantum information, I found the following expression:

$$\sum_{x_0\in\{0,1\}^n}4\big\|\langle x_0|\phi^k\rangle|x_0\rangle\big\|^2=1$$

If so, would not it be 4 as a result? Now, I would just like to know how the result comes out to be 1? I already have a guess, but I can not describe it good enough, or put into the right words.

For information: $|\phi^k\rangle$ is defined as follows $|\phi^k\rangle=U_KU_{k-1}...U_2U_1|\phi\rangle$, where $U_K$ means a unitary transformation.

Can somebody explain to me clearly, or at least comprehensibly, why the result of this expression is 1?

I hope the question is understandable. I hope that the title is ok, if not let me know and I will improve! If you need more information, let me know.

  • $\begingroup$ What is known about the initial state $|\psi\rangle$? $\endgroup$ – Siddharth Bhat Mar 31 '19 at 13:33
  • $\begingroup$ There is only stated: We assume that we are in the state $|\phi\rangle$. I suppose that means a superposition. So a normalized state. I edited the question, ist was $\phi$ not $\psi$ $\endgroup$ – P_Gate Mar 31 '19 at 13:44
  • $\begingroup$ That seems wrong. Consider $|\phi\rangle = 0^n$. Then the given expression will have value 4, not 1. $\endgroup$ – Siddharth Bhat Mar 31 '19 at 13:45
  • $\begingroup$ @Siddhartha Bhat,I think so too, but why is that? How did you figure this out, thats I would be interested in. $\endgroup$ – P_Gate Mar 31 '19 at 13:46
  • 1
    $\begingroup$ Because the $\langle x_0|0^n\rangle = 0$ if $x_0 \neq 0^n$, and $\langle 0^n|0^n\rangle = 1$. So the final vector will be $4\lVert 1 |0^n \rangle \rVert^2 = 4 \cdot 1$ $\endgroup$ – Siddharth Bhat Mar 31 '19 at 14:19

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