WEBVTT

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PROFESSOR: So we
continue today our study

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of electromagnetic fields,
and quantum mechanics,

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and particles in those
electromagnetic fields.

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So last time, we
described what we

00:00:17.430 --> 00:00:20.970
must do in order to
couple a particle

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to an electromagnetic field.

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And the rule was oddly simple.

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You get the
Schrodinger equation,

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and the Hamiltonian is now
changed into this form.

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If you had a free
particle, you're

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accustomed to have
p squared over 2m.

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That has changed, and you
have a new term, with p,

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minus q over c, A squared.

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Thank you.

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So now our task is to understand
how this is compatible,

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and what are the implications
of these changes.

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They're pretty
significant changes.

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This expression,
p, minus q over c,

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A, is what used to be just p.

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And that used to be
just the kinetic energy.

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And now it looks a little
more strange, in fact.

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So in terms of
electromagnetic potentials,

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we know that the
potentials are not unique.

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There are this gauge
transformations

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that establish that these
are completely equivalent

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potentials-- so two
potentials, A and phi,

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and A prime and phi prime,
are physically equivalent.

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They define the same
electromagnetic field

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if they are related in this way.

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We then mention
that therefore you

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have an issue with the
Schrodinger equation.

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You would want the physics
to remain invariant

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in their gauge transformations.

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So you could write the
Schrodinger equation

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for the new gauge potentials,
with A prime and phi prime.

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And you could compare with
the Schrodinger equation

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with the old
potentials, A and phi.

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And you could ask, OK, is
it solved by the same wave

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function?

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And the answer is no.

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The wave function must change.

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But there is a way to get one
way function from the other.

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Here is the formula.

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Psi prime and psi are related by
this factor, this function, U,

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which involves exponential of
the gauge parameter, multiplied

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by a couple of
physical constants,

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including the charge,
q, of the particle.

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Now we're going to use the
notation, q, all the time.

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Many books put E in there.

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And that's the charge
of the electron.

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And then you always wonder,
is that E, or minus E,

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or which sign is it.

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So here, q is the
charge of the particle.

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If you have an
electron, q is minus E.

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But let's leave it q open there,
so you can work with arbitrary

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charges in any circumstance.

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So this Hamiltonian is
what we have to understand.

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They want to make a
couple more comments

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about what this Hamiltonian is.

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And we must think a little about
the gauge invariance as well.

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So there it is.

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A Hamiltonian contains
this term squared.

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So let's just ask
ourselves, what

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does that term really imply.

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Well, there's 1 over 2m.

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So for the
Hamiltonian, 1 over 2m.

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And we have this factor squared.

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So following the
careful things we're

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accustomed to do in
quantum mechanics,

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I would put P squared here,
minus q over c, P dot A,

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minus q over c, A dot P, plus
q squared over c squared,

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A squared, plus q phi.

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That's how the
Hamiltonian looks.

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But you have to be careful.

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Here, this term we kind
of understand what it is.

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It's vector potential squared--

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no problem.

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Here is p squared.

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We're accustomed to that.

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That's a Lapacian.

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We have two terms that
could be a little strange.

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A dot P, probably is all clear.

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A is dotted with a vector
momentum, which is an operator.

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It's the gradient.

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So that's going to differentiate
whatever is to the right.

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The question is,
what this P dot A?

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And P dot A must be thought
in the operator way,

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that P acts on everything
to the right, including A.

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So if this Hamiltonian is to
act in a wave function, that

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doesn't mean that the P
operator just acts on A.

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It acts on A and
everything to the right.

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So P dot A, if you think
of it as a derivative,

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is h bar over i, divergence
of A, plus A dot P. You see,

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this P is h bar over i gradient,
but it's acting on everything.

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So you could put up psi
to the right of here,

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and it would act on
everything there.

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So maybe in order to
justify this equation,

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you should just write h bar over
i gradient acting on A times

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a wave function, psi.

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And go through this and see
that at the end of the day,

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this P dot with
A means act on A,

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but then you still have to act
on everything to the right.

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So that's this term here.

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So this term is
pretty important.

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Sometimes your A satisfies what
is called the Coulomb gauge,

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in which delta of A is zero,
but in general it doesn't.

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So this helps clarify what
the Hamiltonian really is.

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It would be a mistake
to say that these two

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terms are the same.

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And it would be also a mistake
to say that this thing is just

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the divergence of A.
Both are wrong things.

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So the whole Hamiltonian,
now, is P squared over 2m.

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Then you have twice--

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well, let's have this term.

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So you have plus, the i
goes to the numerator,

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as i h bar q, over 2m
c, divergence of A.

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And that's just the
divergence of A, the function.

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It's not any more a
differential operator.

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It's just acted on it.

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Then you have this term twice.

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So minus q over mc, A dot
P, plus this term, q squared

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over 2mc squared, A
squared, plus q phi.

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So that's the Hamiltonian.

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Let me just make
sure I got it right--

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P squared over 2m, plus
[INAUDIBLE],, minus qAP--

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OK, so that's right.

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OK, so this is our Hamiltonian.

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If you just want to
write it very explicitly.

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It's not generally all
that useful to have

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the explicit form.

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But in some examples-- there
will be at least one example

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where it is nice to know.

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And in fact, the H formula
on the top of the blackboard

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hides, in a nice way, a
little bit the complexity

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of this whole coupling.

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Yes.

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AUDIENCE: Is this
for scalar particles?

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PROFESSOR: Yes, this is
for a scalar particle.

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So we're taking a particle,
at this moment, without spin.

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For spin particles, there
would be a little extra term

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sometimes.

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So let's say a couple more
words about this thing,

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in particular the
gauge invariance.

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So we now have seen
what the Hamiltonian is.

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Let's think of the
gauge invariance.

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How could you establish
this gauge invariance?

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So an identity
that is very useful

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takes the following form.

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This differential operator--
h bar over i gradient, the p,

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minus q over c, A
prime, times U--

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and you could put the
psi here if you wish--

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is in fact equal to U,
h bar over i gradient,

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minus q over c, A, psi.

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So I call this a very
remarkable identity.

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And look what's happening here.

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This is useful for
this kind of equation.

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If you have a psi
prime which is U psi,

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the factor U interacts very
nicely with this operator.

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In fact, as you move U
from the right to the left,

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the gauge potential
goes from A prime to A.

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A rather nice thing about
this derivative-- it's

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as if this derivative had
a very special symmetry

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property that U can
be moved across,

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almost as if U was a constant.

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But of course it's not.

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But when you move it
across, the only effect

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is to change the A
prime to A. So a gauge

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transformation on the A.

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That is the reason this
derivative, equipped

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with an extra term, is sometimes
called a gauge covariant

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derivative.

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That is, it transforms nicely
under gauge transformations.

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It does a nice job.

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I write this equation because,
in part of the exercises,

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you will be asked to show that
this statement about gauge

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transformations is correct, that
if this equation-- top-- holds,

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the bottom equation holds with
the replacements indicated

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here.

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And this identity makes the
task of proving the equation--

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the gauge invariance--
very simple.

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Because you can imagine
this psi prime--

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put U psi-- and then
getting the U out,

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so the psi prime
just will become psi,

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and the U going
through these factors

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and simplifying very nicely.

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So this is a simple
equation to show.

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For example, I can
take the left-hand side

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and see what happens.

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So the first term,
you have a gradient

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acting on the product
of two functions.

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And that's just the
gradient acting on each.

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So h over i, gradient
of U, times psi, plus U,

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h over i, gradient of psi.

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That's the first term.

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The second term is minus
q over c, A, times U psi--

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because A prime is A
plus gradient of lambda.

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So you're going to
have minus q over c,

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gradient of lambda, U psi.

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So I wrote the first line.

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Now we can take the
gradient of U. U is here.

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When we take the gradient
of U, what do we get?

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h bar over i, now
gradient of U--

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you differentiate
the exponentials,

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so you take the gradient
of what is in the exponent.

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So it's iq over h bar
c, gradient of lambda,

00:15:00.570 --> 00:15:04.270
times U itself, because
you're differentiating

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an exponential, times psi.

00:15:10.820 --> 00:15:15.720
And then you have all
these other terms.

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I'll couple this term.

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These two terms here are plus U
h bar over i, gradient of psi,

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minus q over c.

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U and A commute, because A is
a function of x and t, and U

00:15:43.050 --> 00:15:47.140
is a function of lambda, which
is also a function of x and t.

00:15:47.140 --> 00:15:49.090
So there's no momentum here.

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These two things commute, so
U can be moved to the left--

00:15:53.950 --> 00:15:59.780
A psi minus this term.

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And if we've done
the arithmetic right,

00:16:06.960 --> 00:16:11.690
the first and last term
should cancel, and they do.

00:16:11.690 --> 00:16:14.320
That i cancels,
the h bar cancels,

00:16:14.320 --> 00:16:18.500
there's q over c times that,
and there's minus that term,

00:16:18.500 --> 00:16:21.020
so these two cancel.

00:16:21.020 --> 00:16:24.110
And this is precisely
the right-hand side.

00:16:29.840 --> 00:16:32.435
So this covariant
derivative is very nice.

00:16:35.470 --> 00:16:38.210
There's something about the
Schrodinger equation of course

00:16:38.210 --> 00:16:44.090
that, in a sense, it's all
made of covariant derivatives.

00:16:44.090 --> 00:16:48.080
Let's look at that.

00:16:48.080 --> 00:16:51.590
So any version of the
Schrodinger equation--

00:16:51.590 --> 00:16:54.760
here is the typical version
of the Schrodinger equation.

00:16:54.760 --> 00:17:00.550
So recall that the vector
potential in general

00:17:00.550 --> 00:17:03.610
can be thought of
as a four-vector.

00:17:03.610 --> 00:17:06.609
You may or may not have seen
this in the literal dynamics.

00:17:06.609 --> 00:17:11.260
But it's just like time
and x form a four-vector.

00:17:11.260 --> 00:17:15.640
The scalar potential
and the vector potential

00:17:15.640 --> 00:17:19.460
form a four-vector.

00:17:19.460 --> 00:17:26.305
So the four-vector with index
mu, 0, 1, 2, 3 form this.

00:17:29.180 --> 00:17:32.700
And then, if you look at
the Schrodinger equation,

00:17:32.700 --> 00:17:34.710
what do we have?

00:17:34.710 --> 00:17:45.000
We have i h bar, d
psi, dt, minus q phi--

00:17:45.000 --> 00:17:47.660
I'm bringing the last term
on the right-hand side

00:17:47.660 --> 00:17:48.680
to the left.

00:17:48.680 --> 00:18:00.980
So it's minus q times phi, which
is minus A0 psi is equal to 1

00:18:00.980 --> 00:18:10.040
over 2m, this i h
bar over i gradient,

00:18:10.040 --> 00:18:16.080
minus q over c, A,
squared, and psi.

00:18:19.380 --> 00:18:20.790
And now look at this.

00:18:23.340 --> 00:18:26.250
It can be written as follows.

00:18:26.250 --> 00:18:48.070
Minus c times h over i, d over d
of ct, minus q over c, A0, psi,

00:18:48.070 --> 00:18:50.480
equal the same thing
on the right-hand side.

00:18:50.480 --> 00:18:54.500
So I've rewritten
the left-hand side

00:18:54.500 --> 00:19:00.250
in a slightly different
way, all the terms.

00:19:00.250 --> 00:19:06.132
So I put an extra
c, this d, dct.

00:19:06.132 --> 00:19:07.570
That canceled the c.

00:19:07.570 --> 00:19:11.750
The h bar, the i went
to the denominator.

00:19:11.750 --> 00:19:17.140
And now this all looks like
this covariant derivative.

00:19:17.140 --> 00:19:20.410
Look at this covariant
derivative. h over i, d dx,

00:19:20.410 --> 00:19:27.460
minus q over c, A. And here
it is, h over i, d dx0--

00:19:27.460 --> 00:19:29.470
because 0 is component
of the [INAUDIBLE],,

00:19:29.470 --> 00:19:32.110
minus q over c, A0.

00:19:32.110 --> 00:19:35.260
So the whole
Schrodinger equation

00:19:35.260 --> 00:19:38.810
is built with this
funny derivatives-- d

00:19:38.810 --> 00:19:44.510
dx minus the vector
potential added in the net.

00:19:44.510 --> 00:19:46.820
These are the
covariant derivatives.

00:19:46.820 --> 00:19:51.140
These are nice operators.

00:19:51.140 --> 00:19:54.260
You see, the
operator P is always

00:19:54.260 --> 00:19:58.580
called the canonical momentum--

00:19:58.580 --> 00:20:01.785
canonical momentum.

00:20:05.570 --> 00:20:11.090
And this canonical momentum is
a momentum such that x with P,

00:20:11.090 --> 00:20:14.675
if you put the hat, is i h bar.

00:20:17.890 --> 00:20:27.400
But this canonical momentum,
P, is not mass times velocity,

00:20:27.400 --> 00:20:29.380
not at all.

00:20:29.380 --> 00:20:34.740
This canonical momentum
is a little unintuitive.

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It's the one that
generates translation.

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The one that is mass
times velocity is really

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this whole combination,
is mass times velocity.

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Because if it's mass times
velocity, this term, 1 over 2m,

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the mass squared times
velocity squared,

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that gives you kinetic energy.

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So we have to be aware that
the canonical momentum is not

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necessarily the simplest,
most intuitive object.