As you go forward and read more advanced algorithms, it’s likely you’ll encounter different notation for the same objects. Sometimes this is due to different authors’ preferences, but
other times it’s because  one author who changes notation in different contexts to emphasize distinct aspects of the problem. Today we’ll consolidate the notation and explain all
the possible choices, so you have one reference to come back to if you get lost.
I should by preface this by saying that we’ll be talking only about basis kets, not general kets. We’re going to describe – in some cases review -- different ways to express the few
CBS kets in H-n, not the infinity of vectors in that space. With that qualification, we begin with the familiar representation of CBS kets in H-n.  We’ve met different ways to express them.
That notation remains in force.  It’s the most common.  These CBS kets can be aligned with a totally different vector space.  The finite vector space of n-tuples whose values are either 0
or 1, and that’s our second environment.
Now, this is not a Hilbert space, just an oddball vector space.  The most notable weirdness of it is that its vectors are added using mod-2 addition applied component-wise. We designate
this space by Z_2 all raised to the N. The third and final environment we want to associate with the CBS kets of H-n is one we’ve actually seen. We’ve been calling it the “encoded
form” of the CBS tensors. So instead of writing ket-11010, we might write ket-26, which is the decimal form of 11010. This environment is called Z-subscript 2^n.  You have to watch
those parentheses to distinguish it from the vector space Z_2 all to the N.
The only thing new here is a note about the addition operation, which is not integer addition. There'll will be times when we want to add these encoded numbers inside the
kets.  It’ll usually appear at the output of a quantum oracle’s B register. So, we formally introduce the group of integers Z-sub 2^N, with this strange addition. It’s the same addition as
in the vector space I just described –- mod-2 -- applied to each of the numbers’ binary digits. Somehow it seems more palatable when we apply it to two vectors than when we’re
using it on two encoded decimal integers, which is why I want you to be aware of this third environment.  A likely mistake would be to add 2 + 3 inside the ket, and get the incorrect
answer, 5, rather than the correct answer, 1. A consequence of these different notational systems is that the nth order Hadamard operator can look like this one day,
and like this another …
and … like this on a third.
