# $H = e^{i\pi/4} \sqrt{iNOT}$?

In the paper Valley qubit in Gated MoS$$_2$$ monolayer quantum dot, a description of how a $$NOT$$ gate would be performed on a qubit in the described device is given.

The authors say that in the described implementation the operation performed is an $$iNOT$$ gate, and that the Hadamard operation can be implemented by performing half of the $$iNOT$$ Rabi transition. Particularly, the authors say that $$H = e^{i\pi/4}\sqrt{iNOT}$$, (actually they say $$H = e^{i\pi/4}\sqrt{NOT}$$, but I assume this is a typo).

Most generally, my question is: Does $$H = e^{i\pi/4}\sqrt{iNOT}$$? I cannot work it out.

My confusion may stem from my lack of understanding concerning why the implemented operation corresponds to $$iNOT$$ instead of simply $$NOT$$. My understanding is that $$iNOT = i\sigma_x$$. I'd appreciate any insight you have on this as well. Thank you.

It cannot be the case that $$H=e^{i\pi/4}\sqrt{iNOT}$$. Whatever your interpretation of $$iNOT$$ (I'd agree with your definition), just square the thing. $$H^2=I$$, the identity, and so it is certainly not the case that $$I=e^{i\pi/2}iNOT.$$
It is true, however, that if you perform the sequence that would give you an operation such as $$NOT$$ or $$iNOT$$, and you only evolve for half the time, that gives you something that achieves an equivalent result to the Hadamard. It's usually referred to as a beam-splitter. For example, if a theorist writes about a Mach-Zehnder interferometer, they usually write down a sequence of Hadamard - phase - Hadamard, whereas an experimentalist will use beam splitter - phase - beam splitter. It achieves the same practical task (e.g. identify if the phase is 0 or $$\pi$$) although the interpretation of the results is different as which output is which gets switched.