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PROFESSOR: This
is very important.

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This is the beginning of
the uncertainty principle,

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the matrix formulation
of quantum mechanics,

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and all those things.

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I want to just tabulate the
information of matrices.

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We have an analog,
so we have operators.

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And we think of
them as matrices.

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Then in addition to operators,
we have wave functions.

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And we think of them as vectors.

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The operators act on the
wave functions or functions,

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and matrices act on vectors.

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We have eigenstate
sometimes and eigenvectors.

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So matrices do the same thing.

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They don't necessarily commute.

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There are very many
examples of that.

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I might as well give
you a little example

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that is famous in the
theory of spin, spin 1/2.

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There is the Pauli matrices.

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Sigma 1 is equal to 1, 1, 0, 0.

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Sigma 2 is 0 minus i, i 0, and
sigma 3 is 1 minus 1, 0, 0.

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And a preview of
things to come--

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the spin operator is
actually h bar over 2 sigma.

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And you have to think of sigma
as having three components.

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That's where it is.

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Spins will be like that.

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We won't have to deal
with spins this semester.

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But there it is, that spin 1/2.

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Somehow these matrices
encode spin 1/2.

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And you can do simple things,
like sigma 1 times sigma 2.

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0, 1, 1, 0 times
0 minus i, i, 0.

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Let's see if I can
get this right.

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i, 0, 0 minus i.

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And you can do sigma 2 sigma
1 0 minus i, i 0, 0, 1, 1,

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0 equals minus i, 0, 0, i, i.

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So I can go ahead here.

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And therefore, sigma 1
commutator with sigma 2

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is equal to sigma 1, sigma
2 minus sigma 2, sigma 1.

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And you can see that they're
actually the same up to a sign,

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so you get twice.

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So you get 2 times
i 0, 0 minus i.

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And this is 2i times
1 minus 1, 0, 0.

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And that happens to
be the sigma 3 matrix.

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So sigma 1 and sigma 2
is equal to 2i sigma 3.

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These matrices
talk to each other.

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And you would say, OK,
these matrices commute

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to give you this matrix.

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This thing commutes to give you
a number so that surely it's

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a lot easier.

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You couldn't be more wrong.

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This is complicated,
extraordinarily complicated

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to understand what this means.

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This is very easy.

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This is 2 by 2 matrices
that you check.

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In fact, you can write
matrices for x and p.

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This correspondence is
not just an analogy.

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It's a concrete fact.

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You will learn-- not too much
in this course, but in 805--

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how to write matrices
for any operator.

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They're called matrix
representations.

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And therefore, you could ask
how does the matrix for x look.

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How does the matrix for p look?

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And the problem
is these matrices

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have to be infinite dimensional.

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It's impossible to
find two matrices whose

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commutator gives you a number.

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Something you can prove in
math is actually not difficult.

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You will all prove it through
thinking a little bit.

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There's no two matrices that
commute to give you a number.

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On the other hand,
very easy to have

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matrices that commute to
give you another matrix.

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So this is very strange and
profound and interesting,

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and this is much simpler.

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Spin 1/2 is much simpler.

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That's why people do
quantum computations.

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They're working with
matrices and simple stuff,

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and they go very far.

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This is very difficult. x
and p is really complicated.

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But that's OK.

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The purpose of this
course is getting

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familiar with those things.

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So I want to now generalize
this a little bit more

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to just give you the
complete Schrodinger

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equation in three dimensions.

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So how do we work
in three dimensions,

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three-dimensional physics?

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There's two ways
of teaching 804--

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it's to just do everything
in one dimension, and then

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one day, 2/3 of the way
through the course--

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well, we live in
three dimensions,

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and we're going to
add these things.

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But I don't want to do that.

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I want to, from the
beginning, show you

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the three-dimensional
thing and have

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you play with
three-dimensional things

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and with one-dimensional
things so that you don't get

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focused on just one dimension.

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The emphasis will be in
one dimension for a while,

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but I don't want you to
get too focused on that.

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So what did we have
with this thing?

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Well, we had p equal
h bar over i d dx.

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But in three dimensions,
that should be the momentum

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along the x direction.

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We wrote waves like that with
momentum along the x direction.

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And py should be
h bar over i d dy,

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and pz should be h
bar over i d dz--

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momentum in the x,
y, and z direction.

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And this corresponds
to the idea that if you

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have a wave, a de Broglie
wave in three dimensions,

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you would write this--

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e to the i kx minus
omega t, i omega t.

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And the momentum would be
equal to h bar k vector,

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because that's how
the plane wave works.

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That's what de
Broglie really said.

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He didn't say it
in one dimension.

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Now, it may be easier to write
this as p1 equal h bar over i d

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dx1, p2 h bar over i d dx2,
and p3 h bar over i d dx3

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so that you can say that all
these three things are Pi

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equals h bar over i d dxi--

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and maybe I should put pk,
because the i and the i

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could get you confused--

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with k running from 1 to 3.

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So that's the momentum.

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They're three momenta,
they're three coordinates.

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In vector notation,
the momentum operator

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will be h bar over i
times the gradient.

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You know that the gradient is
a vector operator because d dx,

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d dy, d dz.

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So there you go.

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The x component of the
momentum operators,

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h bar over i d dx, or
d dx1, d dx2, d dx3.

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So this is the
momentum operator.

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And if you act on this wave
with the momentum operator,

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you take the gradient,
you get this--

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so p hat vector.

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Now here's a problem.

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Where do you put the arrow?

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Before or after the hat?

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I don't know.

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It just doesn't look
very nice either way.

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The type of notes I think
we'll use for vectors

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is bold symbols so there will
be no proliferation of vectors

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there.

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So anyway, if you have this
thing being the gradient acting

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on this wave function, e
to the i kx minus i omega

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t, that would be h
over i, the gradient,

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acting on a to the i kx
vector minus i omega t.

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And the gradient acting on
this-- this is a vector--

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actually gives you a vector.

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So you can do
component by component,

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but this gives you i k vector
times the same wave function.

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So you get hk, which is the
vector momentum times the wave

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function.

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So the momentum operator
has become the gradient.

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This is all nice.

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So what about the
Schrodinger equation

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and the rest of these things?

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Well, it's not too complicated.

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We'll say one more thing.

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So the energy operator,
or the Hamiltonian,

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will be equal to p vector
hat squared over 2m plus

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a potential that depends on
all the coordinates x and t,

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the three coordinates.

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Even the potential is radial,
like the hydrogen atom,

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is much simpler.

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There are conservation laws.

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Angular momentum works nice.

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All kinds of beautiful
things happen.

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If not, you just
leave it as x and p.

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And now what is p hat squared?

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Well, p vector hat squared
would be h bar over--

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well, I'll write this-- p vector
hat dotted with p vector hat.

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And this is h over i
gradient dotted with h over i

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gradient, which is minus
h squared Laplacian.

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So your Schrodinger
equation will

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be ih bar d psi dt is equal to
the whole Hamiltonian, which

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will be h squared over 2m.

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Now Laplacian plus v
of x and t multiplied

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by psi of x vector and t.

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And this is the full
three-dimensional Schrodinger

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equation.

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So it's not a new invention.

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If you invented the
one-dimensional one,

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you could have invented the
three-dimensional one as well.

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The only issue was recognizing
that the second dx squared now

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turns into the full Laplacian,
which is a very sensible thing

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to happen.

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Now, the commutation relations
that we had here before--

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we had x with p is
equal to ih bar.

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Now, px and x failed to
commute, because d dx and x,

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they interact.

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But px will commute with y.

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y doesn't care
about x derivative.

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So the p's failed to commute.

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They give you a number with
a corresponding coordinate.

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So you have the i-th component
of the x operator and the j-th

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component of the p operator--

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these are the components--

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give you ih bar delta
ij, where delta ij

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is a symbol that gives
you 1 if i is equal to j

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and gives you 0 if i
is different from j.

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So here you go.

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X and px is 1 and 1.

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Delta 1, 1 is 1.

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So you get ih bar.

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But if you have x
with py or p2, you

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would have delta
1, 2, and that's 0,

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because the two indices
are not the same.

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So this is a neat way of
writing nine equations.

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Because in principle,
I should give you

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the commutator of x
with px and py and pc,

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00:15:25,000 --> 00:15:29,530
y with px, py, pc,
and z with px, py, pc.

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You're seeing that,
in fact, x just

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talks to px, y talks
to py, z talks to pz.

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So that's it for the
Schrodinger equation.

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Our goal is going to be to
understand this equation.

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So our next step is
to try to figure out

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the interpretation of this psi.

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We've done very nicely by
following these things.

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We had a de Broglie wave.

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00:16:01,950 --> 00:16:03,180
We found an equation.

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00:16:03,180 --> 00:16:05,590
Which invented a free
Schrodinger equation.

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00:16:05,590 --> 00:16:08,990
We invented an interacting
Schrodinger equation.

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But we still don't know what
the wave function means.