@Vogel612 I'm not sure what you mean exactly by the notation x∗ and ↔?

@Vogel612 In this case though, I suppose you could say that, when f(x) = 1, it does a bit flip, but this bit flip is a basis-dependent operation, so when you apply it to |->, what you get is -|-> - but then if you have some state |a->, you get -|a->, so it's acting on 'both', if that makes sense?

If the system isn't entangled, this just gives a global phase, which you can't measure

But if the system is entangled, you're operating on a specific state, less so a specific register

I mean, there's this whole thing about how oracles abstract away a lot (often too much) detail, so I wouldn't worry too much about this - the oracle is physically acting on both registers at once, so any operation it does can change either one of the registers, or both

Like, if we take the simplest case, where x has a bit-length of 1 and f(0) = 0, f(1) = 1. We start with the state |+-> = (1/2)(|00>-|01>+|10>-|11>), you can say that the oracle is a CNOT that takes the RHS and bit flips when the first bit is 1 to give (1/2)(|00>-|01>-|10>+|11>). You could also consider the CNOT in the Hadamard basis, which is looks completely different - if the 2nd bit is |->, it swaps the first bit between |+> and |-> yet these are the exact same operation

@Mithrandir24601 those two notation issues are me being not used to tex anymore and messing up $\Leftrightarrow$ and $x^*$ respectively

and $x^*$ denotes the state that is supposed to be searched / 😀found

@Vogel612 Ahhh, I see! Makes sense