Why is an oracle qubit necessary in Grover's algorithm?

Question

I am a bit confused about the necessity of an oracle qubit in Grover's algorithm.

My question is, does it depend on how you implement your oracle whether you need an oracle qubit or not? Or, it there any reason for an oracle qubit? (such as, there exist some problems that cannot be solved without an oracle qubit, or it's easier to think about the problem with an oracle qubit, or it's a convention, etc)

Many resources introduce Grover's algorithm with an oracle qubit, but I found there are some cases that you do not need an oracle qubit.

For example, here are two implementations of Grover's algorithm in IBM Q simulator. One is using an oracle qubit, and the other is not. In both cases, I would like to find |11> from a space of |00>, |01>, |10>, and |11>. In both cases, oracle successfully flips |11> to -|11>.

・With an oracle qubit (Link to IBM Q simulator)

・Without an oracle qubit (Link to IBM Q simulator)

Answer 1

From the perspective of defining the quantum circuit, the oracle qubit is not strictly necessary. For example, in Grover's search, you might normally define the action of the oracle as $$ U|x\rangle|y\rangle=|x\rangle|y\oplus f(x)\rangle, $$ where $f(x)$ returns 1 if $x$ is the marked item. However, we always use this in a particular way, inputting $(|0\rangle-|1\rangle)/\sqrt{2}$ on the oracle qubit. This has the net effect of just implementing a phase on the marked item. In other words, it is entirely equivalent to the implementation of a new unitary $$ \tilde U|x\rangle=(-1)^{f(x)}|x\rangle $$

However, where it makes a difference is the practical reality. Searching for an item, we will actually need some sort of circuit that recognises the marked item, based on the input of $x$. At that point, it's far easier to think about outputting the answer onto the oracle bit, rather than somehow directly building the unitary that gives the phase without using the oracle qubit. Indeed, I suspect if I asked you to design a generic version $\tilde U$, you'd come up with $U$ with the extra qubit as the solution.