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.


1 Answer 1


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.


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