In the chapter about the Grover algorithm, it is suggested that the gate which executes the phase shift is given in the following form:

enter image description here

Now I have looked at this gate in detail and come to the conclusion that this "only" negates the state $|00\rangle$, but not all others, this can be shown for different inputs $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$ (the gate changes only $|00\rangle$ to -$|00\rangle$) or see here. I'm sorry, because in the chapter it says a few pages earlier: $|x\rangle$ becomes $-|x\rangle$ and $|0\rangle$ remains to $|0\rangle$ (picture left section). But that does not correspond to the circuit (picture right section). That's why I ask myself, what is right now? I want to do this a bit further, we say the Grover operator is:

$$H^{\otimes n}(2|0\rangle\langle 0|-I)H^{\otimes n}$$

That means I have several Hadamard transformations before and then do the phase shift, followed by Hadamard transformations. I think that you can see this in $(2|0\rangle\langle 0|-I)$ (matrix with 1 in the first row, all other rows -1), that $|x\rangle$ becomes $-|x\rangle$ and $|0\rangle$ stays $|0\rangle$. But that does not correspond to the circuit (picture right).

  • $\begingroup$ An overall minus sign (global phase), makes no difference no the final outcome. So it doesn't matter if you make $(2|0\rangle\langle 0|-I)$ or $-(2|0\rangle\langle 0|-I)$ $\endgroup$ – DaftWullie Mar 26 at 9:48
  • $\begingroup$ I almost thought so. So that means, if I would use the loop in the form (top right), then probably my solution amplitude would be negative, right? But that does not make any difference because a minus in front of the state does not change, because the measurement is calculated in the amount $| \alpha |^2$, ok? $\endgroup$ – user4961 Mar 26 at 9:52
  • $\begingroup$ Yes, exactly. Indeed, given you'll be repeating this operation, if you happen to repeat it an even number of times, the sign would disappear anyway. $\endgroup$ – DaftWullie Mar 26 at 11:01

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