WEBVTT
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PROFESSOR: So here comes the
point that this quite fabulous
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about Hermitian operators.
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Here is the thing that it
really should impress you.
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It's the fact that any,
all Hermitian operators
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have as many eigenfunctions and
eigenvalues as you can possibly
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need, whatever that means.
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But they're rich.
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It's a lot of those states.
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What it really means is that the
set of eigenfunctions for any
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Hermitian operator--
whatever Hermitian operator,
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it's not just for some
especially nice ones--
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for all of them you
get eigenfunctions.
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And these eigenfunctions,
because it has vectors,
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they are enough to span
the space of states.
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That is any state can be
written as a superposition
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of those eigenvectors.
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There's enough.
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If you're thinking finite
dimensional vector spaces,
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if you're looking at
the Hermitian matrix,
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the eigenvectors
will provide you
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a basis for the vector space.
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You can understand anything
in terms of eigenvectors.
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It is such an important theorem.
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It's called the spectral
theorem in mathematics.
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And it's discussed in
lots of detail in 805.
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Because there's
a minor subtlety.
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We can get the whole
idea about it here.
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But there are a couple
of complications
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that mathematicians
have to iron out.
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So basically let's
state we really
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need, which is the following.
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Consider the collection of
eigenfunctions and eigenvalues
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of the Hermitian operator q.
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And then I go and say, well,
q psi 1 equal q 1 psi 1 q psi
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2 equal q2 psi 2.
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And I actually don't
specify if it's
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a finite set or an infinite set.
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The infinite set, of course,
is a tiny bit more complicated.
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But the result is true as well.
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And we can work with it.
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So that is the set up.
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And here comes the claim.
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Claim 3, the
eigenfunctions can be
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organized to satisfy the
following relation, integral dx
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psi i of x psi j of x
is equal to delta ij.
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And this is called
orthonormality.
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Let's see what this all means.
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We have a collection
of eigenfunctions.
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And here it says
something quite nice.
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These functions are like
orthonormal functions,
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which is to say each
function has unit norm.
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You see, if you
take i equal to j,
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suppose you take psi 1 psi 1,
you get delta 1 1, which is 1.
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Remember the [INAUDIBLE] for
delta is 1 from the [INAUDIBLE]
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are the same.
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And it's 0 otherwise.
psi 1 the norm of psi 1
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is 1 and [INAUDIBLE]
squared [INAUDIBLE]
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psi 1, psi 2, psi3, all of
them are well normalized.
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So they satisfied this thing
we wanted them to satisfy.
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Those are good states.
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psi 1, psi 2, psi 3,
those are good states.
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They are all normalized.
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But even more, any two
different ones are orthonormal.
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This is like the 3
basis vectors of r3.
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The x basic unit vector, the y
unit vector, the z unit vector,
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each one has length 1, and
they're all orthonormal.
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And when are two
functions orthonormal?
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You say, well, when vectors are
orthonormal I know what I mean.
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But orthonormality for functions
means doing this integral.
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This measures how different one
function is from another one.
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Because if you have
the same function,
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this integral and this
positive, and this all adds up.
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But for different
functions, this
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is a measure of the inner
product between two functions.
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You see, you have the dot
product between two vectors.
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The dot product of two functions
is an integral like that.
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It's the only thing
that makes sense
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So I want to prove
one part of this,
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which is a part that is doable
with elementary methods.
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And the other part is a
little more complicated.
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So let's do this.
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And consider the case if
qi is different from qj,
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I claim i can prove
this property.
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We can prove this
orthonormality.
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So start with the integral
dx of psi i star q psi j.
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Well, q out here at psi j is qj.
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So this is integral dx
psi i star qj psi j.
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And therefore, it's equal to qj
times integral psi i star psi
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j.
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I simplified this by
just enervating it.
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Because psi i and psi
j are eigenstates of q.
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Now, the other thing I
can do is use the property
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that q is Hermitant and move
the q to act on this function.
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So this is equal to integral
dx q i psi i star psi j.
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And now I can keep
simplifying as well.
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And I have dx.
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And then I have the complex
conjugate of qi psi i psi i,
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like this, psi j.
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And now, remember q is an
eigenvalue for Hermitian
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operator.
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We already know it's real.
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So q goes out of the
integral as a number.
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Because it's real,
and it's not changed.
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Integral dx psi i star psi j.
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The end result is that we've
shown that this quantity is
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equal to this second quantity.
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And therefore moving this--
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since the integral is the same
in both quantities, this shows
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that q i minus qj, subtracting
these two equations,
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or just moving one to one side,
integral psi i star psi j dx
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is equal to 0.
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So look what you've proven
by using Hermiticity,
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that the difference
between the eigenvalues
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times the overlap between
psi i and psi j must be 0.
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But we started
with the assumption
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that the eigenvalues
are different.
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And if the eigenvalues are
different, this is non-zero.
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And the only possibility
is that this integral is 0.
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So this implies
since we've assumed
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that qi is different than qj.
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We've proven that psi i
star psi j dx is equal to 0.
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And that's part of
this little theorem.
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That the eigenfunctions
can be organized
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to have orthonormality
and orthonormality
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between the different points.
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My proof is good.
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But it's not perfect.
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Because it ignores one
possible complication,
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which is that here we wrote the
list of all the eigenfunctions.
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But sometimes something
very interesting
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happens in quantum mechanics.
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It's called degeneracy.
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And degeneracy
means that there may
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be several eigenfunctions
that are different but have
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the same eigenvalue.
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We're going to find that soon--
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we're going to find,
for example, states
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of a particle that move in
a circle that are different
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and have the same energy.
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For example, a particle moving
in a circle with this velocity
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and a particle moving in a
circle with the same magnitude
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of the velocity in
the other direction
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are two states that
are different but have
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the same energy eigenvalue.
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So it's possible that this
list not all are different.
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So suppose you have like three
or four degenerate states,
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say three degenerate states.
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They all have the
same eigenvalue.
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But they are different.
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Are they orthonormal or not?
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The answer is-- actually
the clue is there.
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The eigenfunctions can
be organized to satisfy.
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It would be wrong if you say
the eigenfunctions satisfy.
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They can be
organized to satisfy.
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It means that, yes, those ones
that have different eigenvalues
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are automatically orthonormal.
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But those that have
the same eigenvalues,
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you may have three
of them maybe,
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they may not necessarily
be orthonormal.
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But you can do linear
transformations of them
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and form linear combinations
such that they are orthonormal.
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So the interesting
part of this theorem,
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which is the more difficult
part mathematically,
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is to show that when
you have degeneracies
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this still can be done.
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And there's still enough
eigenvectors to span the space.