logoalt Hacker News

SyzygyRhythmtoday at 8:39 AM2 repliesview on HN

I was skimming the paper and came to this: > This transformation is like an AND gate - it ignores the index qubit and places the flag qubit in the state |1> if and only if either of the original components had the state |1> for the flag qubit.

Shouldn't that be an OR gate? Not only does the description above say "if and only if either of the original components had the state |1>", which is an OR, but the truth table listed above shows the same thing for the flag qubit.

Of course, one could say it's an AND on the |0> states, which is just De Morgan's law, but that's pretty awkward phrasing.


Replies

anonymousiamtoday at 5:05 PM

Demorgan's theorem says AND and OR are equivalent, and only depend upon the polarity of the bits. So if "state |1>" is a binary zero, AND is the proper logical operator.

show 1 reply
slwvxtoday at 2:53 PM

Are you sure you're looking at the right paper? I don't find the sentence you mention in the paper.

show 1 reply