U1L3n.txt

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What is E to the I theta minus I theta over 2 sigma-y?
Well, there are two ways of doing this.
Actually, there's probably more than two ways of doing this.
Two ways to do it.

Diagonalize.

So you find U so U sigma-y U-dagger
is equal to a diagonal.

Exponentiate D. So that gives you E to the minus I theta D
over 2.
And you can do this just by taking
each of the diagonal elements of D and raising it and--
you know, exponentiating it to get E to the minus I
theta D sub I over 2.
And then undiagonalize.
Reverse U. So you now have U-dagger E
to the minus I theta D over 2 U. But I
want to do it the other way just to show you
how the other way works.

OK E to the I theta over 2 sigma-y is equal to I.
What we're going to do is just write this out
in terms of a Taylor series.
Minus theta over 2 sigma-y plus I squared theta over 2
squared sigma-y squared.
And I think we need 1/2 factorial.

Plus 1/3 factorial times I times theta over 3, theta over 2
cubed sigma-y cubed plus dot-dot-dot.

So what we want to do is I want to say the even terms are I
minus 1/2 theta over 2 squared.
Sigma-y squared is I because you take any of the following
matrices and you square them and you
get I. Plus 1/4 factorial theta over 2
to the fourth times I, dot-dot-dot.
And the R terms, well, all the--
sigma-y is sigma-y cubed, so this is sigma-y--

OK, so this is minus I theta over 2,
plus I cubed over 3 factorial theta over 2 cubed, et cetera.
And this is just cosine of theta over 2,
and this is minus I sine theta over 2,
so this is cosine theta over 2 times I minus I sine
theta over 2 times sigma-y, which
is equal to cosine theta over 2 minus sine theta over 2,
sine theta over 2, cosine theta over 2.
Gosh.
Now it seems to me I've gotten one of the--

Assuming I didn't get this sign wrong, but you can see,
this is just a rotation of angle theta over 2 and the x-z plane.

So this is how you get rotations of theta
around the z-axis, the y-axis, and the x-axis.