Doing maths with controlled-half NOTs

Question

In Quantum Computation with the simplest maths possible there is a section titled "Doing maths with a controlled-half NOT" which covers a reversible-(N)AND circuit with controlled-half NOTs.

What would the unitary matrix for a controlled-half NOT be?
How could a reversible-XNOR gate be constructed with controlled-half NOTs?
How would a half-adders, full adders & ripple carry adders be constructed from controlled-half NOTs?

score 4 · Accepted Answer · edited Jul 17 '18 at 15:00

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This is the gate that I would call controlled-square-root-of-not. Bit more of a mouthful, I know, but perhaps conveys more accurately what it's doing. The point is that it's a unitary $U$ such that $U^2$ is the controlled-not. There are probably a few ways of writing down such a thing, but, for example $$ U=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{e^{i\pi/4}}{\sqrt{2}} & \frac{e^{-i\pi/4}}{\sqrt{2}} \\ 0 & 0 & \frac{e^{-i\pi/4}}{\sqrt{2}} & \frac{e^{i\pi/4}}{\sqrt{2}} \end{array}\right) $$

There's a trivial answer to your other questions. Take the normal circuits for each of these things built out of (n)and gates, and substitute the reversible (n)and circuits in their place. Of course, there may be optimisations to improve things slightly...

edited Jul 17 '18 at 15:00

Sanchayan Dutta

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answered Jul 16 '18 at 08:17

DaftWullie

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C-sqrt-of-not! Nice. I am not however interested in using (n)and gates. Xor & xnor(which as I understand can be composed of xor gates) are the only classic gates that interest me. I can ask the xnor & rpca questions separately if it's too much for one answer? – user820789 Jul 16 '18 at 20:01
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It’s more that it’s really not worth anyone’s bother coming up with answers for you. (You ask a lot of questions which are essentially novel research questions, but researchers are busy doing their research, not yours. You need to do it, or at least be seen to have tried) I’ve given you one root to an answer (all classical gates can be decomposed in terms of nand, so now you can decompose them in terms of this gate). At first guess, writing those gates using cnot, and then replacing with two copies of the square root is probably the best. – DaftWullie Jul 17 '18 at 05:20
I don't mean to distract from anyone's research; I generally ask a question here only when I get stuck in my own (a lot of questions stemming from a lot of research). I figure asking a group of experts & others that are learning accelerates my learning exponentially (at which point hopefully I can provide more answers as well). Also, apologies if I've done anything to seem ungrateful. You blew me away w/ your answer to my 1st question (& many others) to the point that I have been really fired up in my research ever since (more & more each day it seems)! – user820789 Jul 17 '18 at 05:48
I would like to add that it may have been more accurate to say xor is the only classical gate I have some understanding of. Also, the only reason I didn't accept your answer after reading it is that I'm holding out for the source of the info to weigh in. After reading your comment, perhaps it is naive of me to think they will take the time. – user820789 Jul 17 '18 at 05:58
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What i’m Really saying is that either the strategies i’m Suggesting are (essentially) optimal, or it’s possible to get a quantum speedup, in which case that’s a huge result, not to be wasted here! – DaftWullie Jul 17 '18 at 06:08
I would like to be able to understand why it's either/or & in the latter case where such a result would not be wasted. Thank you again for taking the time to answer & respond! – user820789 Jul 17 '18 at 06:16
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well, either you can do it classically (replacing a classical representation of the circuit with elementary steps built out of your root-cnot) or there's something better. There's no way to tell in advance whether there is something better or not. You just keep trying until you find it, or you give up. Of course, for xor, there's no way that you'll be able to use fewer than 2 root-cnot gates (and we know 2 do the job) and similarly for xnor it seems unlikely that you could do it with fewer than 4 (and 4 is enough because you can make not using cnot controlled off a $|1\rangle$ state). – DaftWullie Jul 17 '18 at 07:01
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I had never heard I'd half-not before, is it just square root of not? – user1271772 No more free time Jul 17 '18 at 09:39
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@user1271772 If I'm reading the blog article correctly, yes. I hadn't heard it called that either. – DaftWullie Jul 17 '18 at 14:17
cc @user1271772 After doing more research & thinking, I'm curious to find out more about nth roots of not (nth roots of swap may be interesting as well - will look into). I was able to locate two references: Fault Models For Quantum Mechanical Switching Networks (Section 6 pg.8) & NMR Quantum Computation (PDF Section 7.3) - neither is very in depth. Considered asking as a new question (or updating this one) thought I'd check in here first though. – user820789 Jul 18 '18 at 02:37
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That’s very much a separate question. Perhaps if I ask it (give me a couple of hours) that will let me ask it so that I can give the answer I think you need rather than the answer you might think you want. You can always edit the question if it doesn’t serve your purpose. – DaftWullie Jul 18 '18 at 05:42
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@meowzz The n'th root of NOT is just a 360/n degree phase gate framed by Hadamard gates so that it rotates around the X axis instead of the Z axis. – Craig Gidney Jul 18 '18 at 16:19

Doing maths with controlled-half NOTs

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