# N&C quantum circuit for Grover's algorithm

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

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).

• 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 $-(2|0\rangle\langle 0|-I)$ – DaftWullie Mar 26 at 9:48
• 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? – user4961 Mar 26 at 9:52
• 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. – DaftWullie Mar 26 at 11:01