What's the square root of NOT?

Question

Most concise answer given here gets 10 points. Links given only get 0 points, and saying that this question makes no sense deserves -10 points. Check quantum computing.

Psst...."NOT", not "naught"

Touche, I can't stop people from getting their two points' worth

Folks! I'm asking about a quantum logic gate.

For those that think this is a silly question, please check this American Scientist article:

http://www.americanscientist.org/template/AssetDetail/assetid/24592?&print=yes

Kudos to Ben for a great answer in spite of the fact he says he knows nothing about quantum computing. He's on the right track.

Answer 1

I know nothing about Quantum Computing, however, I know a little about Boolean Arithmetic, so I will take a shot at this.

The only way this makes sense to me is, when NOT is a function of two bits, and returns two bits. That is NOT(A,B) = (~A,~B), where ~A is the not of the single bit A. If this is the case then if I define SQRTNOT to be the function of two bits such that NOT(A,B) = SQRTNOT(SQRTNOT(A,B)). In fact, I can come up with two such functions, given by the truth table below:

A, B : 0,0 0,1 1,1 1,0
SQRTNOT1 : 0,1 1,1 1,0 0,0
SQRTNOT2 : 1,0 0,0 0,1 1,1

Here is a description of the functions:
SQRTNOT1(A,B) = (B, ~A)
SQRTNOT2(A,B) = (~B, A)

Note that SQRTNOT1(SQRTNOT1(A,B)) = SQRTNOT2(SQRTNOT2(A,B)) = NOT(A,B) = (~A,~B)

Not sure if this is what you had intended, but it makes some sense to me =)

Answer 2

How about the following idea:

I'll denote the Boolean algebra in consideration by B. Let us look at the 2 x 2 matrices with values in B. Embed B diagonally, via the inclusion a goes to
( a, 0 )
( 0, ~a )
Call this inclusion phi for the time being.

Now, NOT acts on a matrix coordinate wise, so NOT( phi(a) ) = phi( NOT( a ) ) and NOT commutes with this embedding.

Consider the operator N defined by taking
( a, b )
( c, d )

to the matrix
( b, d )
( a, c )

Now consider the action of N^2 on the embedded phi( B ).

N^2 takes the matrix
( a, 0 )
( 0, ~a )

to the matrix
( ~a, 0 )
( 0, a )

which is NOT a, under our embedding.

Answer 3

Sorry, I don't know the answer either, however, you might consider that as the average IQ here is around 100 and the average apparent IQ is far lower, I doubt highly that you'll be able to get a concise answer out of this.

Most of the people viewing this think in terms of applied and abstract arithmetic and algebra. Though your question seems to be based in this field, 'NOT' has no numeric value and thus cannot have a valid square root.

I don't know of too many places where people who are well versed in quantum mechanics and quantum computing will answer such a question, but I doubt it's here.

Something tells me that your 10 points will go unrewarded.

Answer 4

NOT^(1/2)

Answer 5

Is asking for the square root of NOT any different from asking what the derivative of OR is, or the integral of AND? Or how about mundane things like the log of an apple or the sine of a lamp?

NOT is a logical operator and the answer either doesn't exist or there's a different definition of square root that I (nor many other people) are aware of.

NOT usually preceeds and is applied to a boolean expression, negating it as a result. In a sense it's equivalent to (-1) in terms of finding the additive inverse, so seeing as NOT is similar to (-1), then the square root of NOT would be i.

But again, that's purely on an arbitrary guess based on what NOT is similar to. f(x) = sqrt(x) usually takes in a real number, though it can take in a complex number as well. Perhaps there's a definition of square root that takes in the boolean negation too.

Answer 6

Check Wikipedia

Answer 7

An input is put through two NOT gates and the output is NOT the input (e.g. 0 being put through two NOT gates and ending up 1). The square root of NOT is the state after going through just one of these gates.

Answer 8

0 x 0 = 0 therefore, square root of 0 is 0.

Answer 9

well, if itz some kind of riddle then....

NOT canNOT be a product of two similar numbers....

unless ure askin the sq root of NOT^2... then itz NOT...

JK....

Answer 10

Not....u cant multiply not by itself to get another number except not!