DaftWullie
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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. There's just a conceptual simplicity of doing as much as possible with realOr there's $$\sqrt{Y}$$ which gives an even more direct comparison to $$H$$, avoiding imaginary numbers. There's

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 fact that 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), $$HZH=X\qquad HXH=Z \qquad HYH=-Y$$$$HZH=X\qquad HXH=Z \qquad HYH=-Y.$$ It also enables relations between controlled-not and controlled phase, and between controlled-not in two different directions (swapping control and target).

Of course, there There are similar relations for $$\sqrt{X}$$, but they're not quite so nice$$\sqrt{Y}$$: $$\sqrt{X}Z\sqrt{X}^\dagger=XZ=-iY \qquad \sqrt{X}X\sqrt{X}^\dagger=X\qquad \sqrt{X}Y\sqrt{X}^\dagger=XY=iZ$$$$\sqrt{Y}Z\sqrt{Y}^\dagger=YZ=iX \qquad \sqrt{Y}X\sqrt{Y}^\dagger=YX=-iZ\qquad \sqrt{Y}Y\sqrt{Y}^\dagger=Y$$ Part of this looking (slightly) nicer is also 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.

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.

DaftWullie
• 50.4k
• 2
• 37
• 108

It's mostly about simplicity. You could use operations such as $$\sqrt{X}$$, but they are a little bit more fiddly because of all the imaginary numbers that appear. There's just a conceptual simplicity of doing as much as possible with real numbers. There's also the fact that Hadamard gives you some very nice inter-relations, $$HZH=X\qquad HXH=Z \qquad HYH=-Y$$ It also enables relations between controlled-not and controlled phase, and between controlled-not in two different directions (swapping control and target).

Of course, there are similar relations for $$\sqrt{X}$$, but they're not quite so nice: $$\sqrt{X}Z\sqrt{X}^\dagger=XZ=-iY \qquad \sqrt{X}X\sqrt{X}^\dagger=X\qquad \sqrt{X}Y\sqrt{X}^\dagger=XY=iZ$$ Part of this looking nicer is also 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 $$\sqrt{X}$$. 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.

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.