Advantage of Hadamard gate over rotation about the X axis for creating superpositions

Question

When I look at most circuits (admittedly small sample as I'm a beginner), the Hadamard gate is used a lot to prepare a superposition from say the $\mid0\rangle$ state.
But upon a little reflection, we can prepare a superposition using a $\dfrac{\pi}{2}$ rotation about the X axis.

I do know that a successive application of the Hadamard gate yields the initial state back (for any state).
If we have one of $\mid0\rangle$ or $\mid1\rangle$, we can recover them using a succession of said rotation followed by a NOT gate (Pauli-X).

So why is the Hadamard gate preferred to create superpositions when it uses more gates (rotation about Z then rotation about X then rotation about Z again)?
If it is because the Hadamard gate allows recovery of any initial state, why is that property so important? (Even when not actually used when I look at the examples I see.)

Answer 1

It's mostly about simplicity and adopted convention. In the end, this is basically the same question as "why should I pick a universal set of gates A rather than a universal set B?" (see here). Experimentalists would pick the universal set they have available. Theorists just pick something that they like to work with, and eventually a convention is adopted. But it doesn't matter which convention they adopt because any universal set is easily converted into any other universal set, and it is (or should be) understood that the quantum circuits describing algorithms are not what you actually want to run on a quantum computer: you need to recompile them for the available gate set and optimise based on the available architecture (and this process is unique to each architecture).

You could use operations such as $\sqrt{X}$, but they are a little bit more fiddly because of all the imaginary numbers that appear. Or there's $\sqrt{Y}$ which gives an even more direct comparison to $H$, avoiding imaginary numbers.

One of the main purposes of $H$ in a quantum circuit is to prepare uniform superpositions: $H|0\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$. But $\sqrt{Y}$ also does this: $\sqrt{Y}|1\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$. When you start combining multiple Hadamards on unknown input states (i.e. the Hadamard transform), it has a particularly convenient structure $$ H^{\otimes n}=\frac{1}{\sqrt{2^n}}\sum_{x,y\in\{0,1\}^n}(-1)^{x\cdot y}|x\rangle\langle y|. $$

The Hadamard gives you some very nice inter-relations (reflecting basis changes between pairs of mutually unbiased bases), $$ HZH=X\qquad HXH=Z \qquad HYH=-Y. $$ It also enables relations between controlled-not and controlled phase, and between controlled-not in two different directions (swapping control and target). There are similar relations for $\sqrt{Y}$: $$ \sqrt{Y}Z\sqrt{Y}^\dagger=YZ=iX \qquad \sqrt{Y}X\sqrt{Y}^\dagger=YX=-iZ\qquad \sqrt{Y}Y\sqrt{Y}^\dagger=Y $$ Part of this looking (slightly) nicer is because, as stated in the question, $H^2=\mathbb{I}$.

One way that many courses introduce the basic idea of quantum computation, and interference, is to use the Mach-Zehnder interferometer. This consists of two beam splitters which, mathematically, should be described by $\sqrt{X}$ (or $\sqrt{Y}$ would do). Indeed, this is important for a first demonstration because of course these operations are "square root of not", which you can prove is logically impossible classically. However, once that initial introduction is over, theorists will often substitute the beam splitter operation for Hadamard, just because it makes everything slightly easier.

Answer 2

I think the major advantages of the Hadamard gate are "usability" stuff, as opposed to fundamental mathematical stuff. It's just easier to remember and simpler to apply.

1. The Hadamard gate's marix is real and symmetric. Makes it easy to remember.
2. Hadamard is its own inverse. Makes it easy to optimize in circuits. Any two Hs that meet cancel out; whereas $\sqrt{X}$ tends to meet $\sqrt{X}$ as often as $\sqrt{X}^{-1}$ leaving behind $X$ operations.
3. Hadamard's effect on operators is easy to remember: swap X for Z. Whereas for $\sqrt{X}$ style operations you need to remember a right hand rule. If you pass a hadamard over a CZ, it turns into a CNOT. If you pass a $\sqrt{Y}$ over a CZ, whether you get a CNOT or a CNOT+Z depends on whether you went left-to-right or right-to-left.
4. In the surface code you need twist defects or distilled states to do $\sqrt{X}$ gates. Hadamard operations need neither (though the twists are more efficient...).
5. The Hadamard is unique. There are two values $M$ such that $M^2 = X$, and so you need an agreed upon convention for which one $\sqrt{X}$ is.

PS: it would be better to compare a Hadamard to a 90 degree rotation about the Y axis, not the X axis, because the Hadamard operation is equivalent $\sqrt{Y}$ up to Pauli operations ($H \propto Z \cdot \sqrt{Y}$).

Answer 3

Any Hermitian quantum gate $U$ is "self-recovering". This is because $U$ is unitary, and $$UU^{\dagger}=U^{\dagger}U=I$$ If $U$ is also Hermitian, then $U=U^{\dagger}$ and $$UU=I$$

Hadamard gate prepares $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ superposition from $|0\rangle$ state. If you need this superposition, you use Hadamard. If you need a different superposition, $\alpha|0\rangle + \beta|1\rangle$ with some $\alpha$ and $\beta$, you need a different gate or a sequence of gates; Hadamard gate has no advantage here.