U2L4b.txt

#
# File:   content-mit-8370x-subtitles/U2L4b.txt
#
# Captions for course module
#
# This file has 36 caption lines.
#
# Do not add or delete any lines.  If there is text missing at the end, please add it to the last line.
#
#----------------------------------------

So what I'm going to do is try to work it out
for some very small examples.
And once we've seen how it works, for n equals 1.
That's one qubit, then we'll say something about general n,
then we'll--
n as a power of two, then we'll talk
about how it works for two qubits,
and I'll show you exactly how it works.
And then I'll draw the circuit for general n
and explain why it works.
So what is this for any case, 1?

n equals 1.
1 over root n that's root 2 times 1 1 1.
And here we have e to the minus pop
to pi i over 2, which is just e to the pi i.
Which is what minus 1?
Minus 1 [INAUDIBLE].

So yesterday, we saw Simon's algorithm and the transform
we used to make Simon's algorithm work was comparable
was h to the n.
Different than I think.
So this is the Fourier transform.

1 over the group.

Over the group of binary strings of length that--

so you see Simon's algorithm was the Fourier transform just
over a different group.
So today, we're doing the Fourier transform over
the cyclic group mod n.
So that's the Fourier transform for n equals 1.