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I am currently learning from Nielsen and Chuang and I am currently learning about Deutsch-Jozsa algorithm. However, I am stumped with the mathematics of the algorithm at the following section: The section

I understand intuitively that it works very similarly with Deutsch algorithm where we could measure a global state of the function with only 1 measurement, but I couldn't do it mathematically. Why is the amplitude simply so without including the $x \cdot z$ factor? I am also having a hard time on the summation over $x$ on the amplitudes. Why is the resulting amplitude for the constant case is $\pm 1$ and 0 for the balanced case?

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Why is the amplitude simply so without including the $x \dot z$ factor?

When you calculate the amplitude of the $|0\rangle^{\otimes n}$ state, you have $z = 0$ (the integer representation of the state you're looking at), so $x \dot z = 0$ for any $x$.

Why is the resulting amplitude for the constant case is ±1 and 0 for the balanced case?

For the constant case, if $f(x) = 0$, each of the terms $(-1)^{f(x)}/2^n = 1/2^n$, and there are $2^n$ of them, so they add up to 1. If $f(x) = 1$, each of the terms is $-1/2^n$, and they add up to -1, with the phase difference habitually discarded.

For the balanced cases, exactly half of the terms evaluate to $1/2^n$, and the other half to $-1/2^n$, so they cancel each other out, and the resulting sum is 0.

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  • $\begingroup$ Thank you for the reply. I would like to clarify some things. Is z only 0 because initially it was $|0⟩^{⊗n}$? So if let's just say I use $|1⟩^{⊗n}$ The factor in question should persist right? $\endgroup$
    – Michael Teguh Laksana
    May 30, 2021 at 4:25
  • $\begingroup$ z is 0 because we're looking at the amplitude of the basis state |z⟩ = |0...0⟩ in the resulting state $|\psi_3\rangle$. If we look at the amplitude of any other component, xz will remain, yes $\endgroup$
    – Mariia Mykhailova
    May 30, 2021 at 5:15

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