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U1L5c.txt
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U1L5c.txt
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#
# File: content-mit-8370x-subtitles/U1L5c.txt
#
# Captions for course module
#
# This file has 114 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
Before we go on, I think I need to tell you
some facts about matrices.
So a unitary matrix satisfies finds
U dagger equals U inverse.
A Hermitian matrix satisfies H is equal to H dagger.
And sigma x, sigma y, sigma z are both unitary and Hermitian.
So if we take minus i, i and 1 minus 1,
so if we take the transpose of this
and take its complex conjugate, we get the same thing here.
We get the same thing here.
And we get the same thing here.
And that's what it means for this to be Hermitian.
So this is a dagger and not a plus.
It's a dagger and not a transpose the way
it looked like.
I'm sorry I meant to write dagger.
But I didn't draw the--
I didn't draw this upper thing quite as well as I should.
OK, so the fact we need, both unitary and Hermitian matrices
of dimension d have d eigenvectors.
Dimension d have d eigenvectors.
And these are, well, they're all orthogonal.
And you can-- and since eigenvectors aren't--
the length of eigenvectors is not specified,
you can assume that there are all units.
So you can assume that they're all orthonormal.
So H is equal--
and Hermitian M is equal to summation i equals 1
through d lambda i b sub i.
And a unitary is i equals 1 through d e to the i theta
b sub i, b sub i, where in both the cases,
the b sub i's form a basis.
And lambda i is of course the eigenvalues.
And e to the i theta are again the eigenvalues.
So this-- well, I will prove it.
Am I speaking loud enough for the back of the class to hear?
Oh, good.
So I will prove it in a little while.
Suppose proof, well, I will use the fact that there always
exist one eigenvector--
exists one eigenvector.
And over here, we need to use the fact
that we're using matrices over the complex numbers
because this isn't necessarily true for matrices
if you don't allow complex numbers.
And so that means M v equals lambda i v. So or U v equals
e to the i theta v.
So the eigenvalue for unitary matrices
has to be e to the i theta because, remember,
unitary matrices take norm one vectors to norm one vectors.
So if you have an eigenvalue, this has to be unit length.
So it has to be e to the i theta.
For Hermitian matrices, it has to be real, lambda i.
Let's call that lambda.
Lambda is real since v M v, complex conjugate,
is equal to v M adjoint v because what
we do is flip this around and take the adjoint of the middle.
And that's just equal to v M v. And this was lambda.
So now, what we do is change the basis.
I guess M prime equals U M U dagger So this
is how you change bases.
So first, you know, basis vector is v. Now,
if v is an eigenvector, we know that M times 1, 0, 0
is equal to lambda 1, 0, 0, 0, 0.
And what that means is that this entry is lambda,
and all these other entries are 0.
But M is Hermitian, so that means that all of these entries
also have to be 0.
OK?
And now, we have a smaller matrix,
which is this, which also has to be Hermitian.
And we can use induction and show
that we can find a unitary transformation, which
changes the bases so that M is diagonal with real elements.
So the exact same proof applies to unitary,
except that U equals e to the i theta 0, 0, 0, 0, 0.
And we know that the columns have length 1.
And the row vectors have length 1.
So that means all of these other things
have to be 0 because row vectors in unitary U are norm 1.
And again, we can apply induction and get that U is--
that we can diagonalize U so that it has e to the i
thetas along the diagonal.
Yeah?
Does theta have to be the same for every eigenvector.
No.
OK.
Theta is going to be different.
So this is going to be e to the i theta 1, e to the i theta 2,
and e to the i theta n.
And again, this lambda doesn't have to be the same.
It can be lambda 1, lambda 2, through lambda n.
And it's clear this is a unitary matrix.
And it's clear that this is a Hermitian matrix.