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PROFESSOR: One of
the things you will

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be thinking about
in the homework

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is the definition of
gauge-invariant observables.

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So an operator O is a
gauge-invariant observable--

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gauge environment
observable-- if, see,

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under a gauged
transformation, O can change.

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The operator can change under
a gauged transformation.

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You'd say, oh, but isn't it
supposed to be gauge-invariant?

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But, well, it's not
quite the operator itself

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that is gauge-invariant.

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It is its expectation value.

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So you should have that
psi prime of O prime psi

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prime is equal to psi O psi.

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So the operator may change.

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And that's OK.

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The wave function
will change as well.

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So the question is that
if you put the new wave

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function and the new operator
and you find the expectation

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value, do you get
the same thing?

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That's what you really can ask.

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You cannot ask more than that.

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And you will see very funny
things when you do this.

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You will see that
this operator, P,

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the momentum operator
that you like,

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that you find has been
intuitive for you so far,

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is not gauge-invariant.

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You cannot compute the
expectation value of it

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and expect something physical.

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It's not gauge-invariant.

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The thing that is
gauge-invariant is P minus q

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over c A--

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is gauge-invariant.

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So it's some small
calculations you will do.

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In order to get the
gauge-invariant operator,

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you do, in general, have
to put in a dependence.

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That's the funny thing.

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So this thing will not
be gauge-invariant.

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And, you know,
it's almost obvious

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that P is not gauge-invariant.

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Why?

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Because P, under a
gauge transformation--

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well, the gauge
transformation, say,

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change A, change phi
prime, change psi.

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But it doesn't say change P. So
in the gauge transformation, P

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goes to P.

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But then let's look at
psi prime P psi prime.

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Is it equal to psi P psi?

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Well, here you
have that psi prime

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is psi times e to the minus
i q lambda over h bar c.

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You have P here.

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And e to the i q
lambda over h bar c.

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Psi prime-- no, psi.

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That's the left-hand side.

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And this P is such that
it takes derivatives.

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And lambda depends on space.

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So this is going to
give you something.

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This is not going to cancel.

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This is not going to cancel,
and therefore the left-hand side

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is not going to be equal
to the right-hand side.

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The faces don't go
across each other.

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So therefore, there it is.

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P is no good.

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Not intuitive, not
physical, not observable.

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It's a very strange
thing that once you

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put electromagnetic fields,
P, the canonical generator

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of translation, loses
its privileged status.

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Not anymore we can
think of P so easily.

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And look, however.

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If you had here P
minus qA, P minus qA

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is precisely the kind
of thing that you

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can move the face
across, and you

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will see that then it works.

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You get something to simplify,
and it's all very nice.

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So gauge invariance makes
for funny things to happen.

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I want to do an example that
illustrates quantization,

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another pretty surprising thing.

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So we have quite a few
things to do today.

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So let me try to do this
magnetic fields example.

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Magnetic field on a torus.

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So we've talked about
what is a circle.

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And we say a circle is the
line with the identification

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x equal x plus L. So the point 0
is identified with the point L.

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And that's a circle.

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This line on this point
is the same as this one,

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so you return.

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A torus that we use--
so, this is a circle--

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a torus, which is
something that we sometimes

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think of like this,
is, on the other hand,

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an identification of
the following kind.

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Here is the x-axis.

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Here is the y-axis.

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Here is Lx.

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Here is Ly.

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And we say that any point xy is
identified with x plus Ly Lx,

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and every point xy is identified
with xy plus Ly, which

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is to say that any point
with some value of x

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is identified with a point
with x increasing by Lx.

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So this line is supposed
to be glued to this line.

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And this line is supposed
to be glued to that line.

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So this is identified
and this is identified.

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Perhaps the most intuitive
way of thinking of this

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is take this strip, take
these two sides, glue them.

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Now you have a cylinder.

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And now you're supposed to
glue this end to the first end.

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So you glue them like that, and
you form something like this.

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So here is a torus.

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Now we try to put the
magnetic field on this torus.

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So you could say,
well, this is x and y.

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There is a torus.

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There's the z-direction.

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Let's put the
magnetic field that

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goes through the torus
in the z-direction.

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B z.

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B in the z-direction.

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And we'll put a constant
one, B constant.

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So it's time-independent
and space-independent.

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And then you say, OK, let's
look at my Maxwell's equations.

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Divergence of B equals 0.

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B is constant.

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Good.

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Curl of B is related to current.

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There's no current.

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Plus dE dt,
displacement current.

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There's no electric field.

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Good equation.

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Curl of E is minus dB dt, no E.

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There's no problem, obviously.

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So B, for any
constant value of B,

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it satisfies
Maxwell's equations.

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So any constant
value of B should

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be an allowed magnetic field.

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Or so we would think,
because it actually is not.

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So why does it go
wrong, this intuition?

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We'll see.

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But let's put an assumption.

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The assumption is that we're
doing quantum mechanics,

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and there exists a
particle with charge.

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And I'll call it this time q.

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I'll just--

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So if we're doing
quantum mechanics,

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and there exists a
particle of charge q,

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I need potentials to describe
the quantum mechanics.

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I need an A field
to describe this.

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So I know B is consistent,
and it makes sense,

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and we can use it,
if and only if I

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can find a vector potential.

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So our task is to find a
vector potential on this torus.

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Well, at first sight that
doesn't sound too difficult.

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You can say, all right, let's
find the vector potential.

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That's easy to find an A such
that curl of A is equal to B.

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So B z is dx A y minus dy A x.

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So let's simplify my life.

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Let's take A y to
be equal to B z,

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and let's call this magnetic
field B0, the constant value

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B0.

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Constant.

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So A y would be equal to B0
times x, and A x would be 0.

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OK.

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That's it.

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Look.

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A y is equal to B0 x.

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So I take the
derivative dx of A y.

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I get B0.

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dy of A x is 0.

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Perfect.

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All done, we would say.

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You found the A y.

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But there is a problem with this
[? A i. ?] What is the problem?

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If this is a torus,
it means these points

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are the same as these points.

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So I should have,
the vector potential

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here must be the same as
the vector potential there.

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Because it is, after
all, the same point.

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You're gluing the surface.

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But, OK, this point and
this point differ by y.

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And this vector potential
doesn't depend on y.

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So, phew.

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That's good.

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A is well-defined because it
has the same value anywhere here

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as anywhere there.

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So I can write it as A y of
x, y plus L y at any point x.

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And y plus L y is the
same thing as A y of xy.

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So that's very good.

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But we're going to run
into trouble in the second

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now, because the A y should be
the same here and here, too.

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Because that's also the
same points in the torus.

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This line is
identified with this.

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So when I change x by L
x, A y should not change.

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And here it seems like A y at x
plus L x, and the same point y,

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is not the same as A y at xy.

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And this is a torus, so this
point, x plus L x and y and y

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should be the same.

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So this vector potential is
not the same here as here.

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So, actually, this vector
potential doesn't look good.

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You thought you could write
the magnetic field in terms

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of a vector potential,
but this vector potential

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doesn't seem to
live on the torus.

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It's just not a
well-defined magnetic vector

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potential in the torus.

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It doesn't have the same
value here and here.

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So in principle,
we're not through.

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We're not out of the woods here.

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We have not been able to find
a good vector potential yet.

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But here, the gauge
transformations

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come a little to the rescue.

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When you work with
vector potentials,

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you don't really need
that the vector potential

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be the same here as here.

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It is enough if the
vector potential here

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and here, which
is the same point,

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they differ by a
gauge transformation.

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So the vector potential
is a subtle object.

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Here is the torus.

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In one part of the
torus it has a formula.

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In the other part, it may
look like it's not consistent.

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But you can use
another formula related

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by a gauge transformation.

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So this vector
potential here is OK

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if I manage to show that
what I get on this side

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is just the gauge transformation
of what I was getting here.

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So physically
these are the same.

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And yes, you have a unique
configuration on the torus

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if the vector potential here is
gauge-equivalent to the vector

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potential there.

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So let's try to do that.

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So, gauge transformations
to the rescue.

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OK.

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A y of x plus Lxy.

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I want it to be a
gauge transformation

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of the field at A y xy.

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So remember, A prime was
A plus gradient of lambda.

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So I'm going to write this
as A at the same point, which

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is xy--

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it's physically the same point--

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plus the gradient of lambda, in
this case would be dy lambda.

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OK.

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Now we have to find
the gauge parameter.

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To show that they're
gauge-equivalent here and here,

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I must find the gauge parameter.

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That's the lambda.

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So let's do the formula here.

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This is B0.

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There's a formula
here, B0 times x.

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So x plus Lx times x plus
Lx is equal to B0 times

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x plus dy lambda.

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So I cancel here and I get
B0 Lx is equal to dy lambda.

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OK.

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All right.

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So, well, it seems that
we're succeeding again,

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against all odds.

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And we could take lambda, for
example, to be B0 Lx times y.

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And that would be
our gauge parameter.

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00:17:54,800 --> 00:17:55,420
OK.

254
00:17:55,420 --> 00:17:57,310
So are we happy already?

255
00:17:57,310 --> 00:17:59,635
Have we shown that
everything is working?

256
00:18:02,390 --> 00:18:05,150
Well, there's still a problem.

257
00:18:05,150 --> 00:18:12,410
This time the gauge parameter
doesn't seem to be well-defined

258
00:18:12,410 --> 00:18:15,710
on the torus as
well when I change--

259
00:18:15,710 --> 00:18:18,740
you know, we didn't have
problems with y here.

260
00:18:18,740 --> 00:18:20,980
We had problems with x.

261
00:18:20,980 --> 00:18:24,440
When we tried to fix it with
x, we find the gauge parameter.

262
00:18:24,440 --> 00:18:29,450
But the gauge parameter
doesn't seem to have--

263
00:18:29,450 --> 00:18:31,760
it does have problems with y.

264
00:18:31,760 --> 00:18:35,540
It's not the same at this
point and at this point.

265
00:18:35,540 --> 00:18:40,610
Because at those points,
y differs by L y,

266
00:18:40,610 --> 00:18:44,990
and the gauge parameter,
again, doesn't quite

267
00:18:44,990 --> 00:18:49,000
seem to live on the torus.

268
00:18:49,000 --> 00:18:51,760
It seems like an endless
amount of complication.

269
00:18:51,760 --> 00:18:56,560
But we're near the
end of the road now.

270
00:18:56,560 --> 00:19:02,320
The thing that comes to our
rescue is that lambda, in fact,

271
00:19:02,320 --> 00:19:07,140
doesn't quite have to
be that well-defined.

272
00:19:07,140 --> 00:19:07,660
Why?

273
00:19:07,660 --> 00:19:12,190
Because the gauge transformation
of the charged particle

274
00:19:12,190 --> 00:19:23,550
says that you need to do i
Q over hc lambda times psi.

275
00:19:23,550 --> 00:19:26,600
That's how you do
gauge transformations.

276
00:19:26,600 --> 00:19:30,480
And in fact, this, remember,
this is the U thing.

277
00:19:30,480 --> 00:19:37,020
And even though we write
A prime as A D lambda,

278
00:19:37,020 --> 00:19:39,330
this term can be
written in terms

279
00:19:39,330 --> 00:19:45,760
of U. U is the master quantity
that has all the information.

280
00:19:45,760 --> 00:19:56,820
In fact, D lambda is roughly
U minus 1 gradient of U,

281
00:19:56,820 --> 00:19:58,770
up to a series of factors.

282
00:19:58,770 --> 00:20:02,220
So everything
depends on U. So we

283
00:20:02,220 --> 00:20:06,210
don't have to worry too
much if lambda is not

284
00:20:06,210 --> 00:20:08,700
well-defined on the torus.

285
00:20:08,700 --> 00:20:11,610
What has to be
well-defined on the torus

286
00:20:11,610 --> 00:20:17,100
is U, the exponential of lambda.

287
00:20:17,100 --> 00:20:25,690
And that, it's a saver, because
U, now, is e to the i q.

288
00:20:25,690 --> 00:20:33,990
Lambda is B0 Lxy over h bar c.

289
00:20:33,990 --> 00:20:40,830
That's U. And U will be
well-defined on the torus

290
00:20:40,830 --> 00:20:49,120
if, when you change y plus y
plus L y, this doesn't change.

291
00:20:49,120 --> 00:20:56,110
And for that, this whole thing,
when you change y by L y,

292
00:20:56,110 --> 00:21:00,550
must change by a factor of 2 pi.

293
00:21:00,550 --> 00:21:04,710
So if-- you see, you're
living dangerously with this.

294
00:21:04,710 --> 00:21:07,420
It just doesn't want
to exist on a torus.

295
00:21:07,420 --> 00:21:12,430
But if this happens to
be equal to 2 pi n--

296
00:21:12,430 --> 00:21:23,670
so if you have q B0 Lx Ly
over hc equal to 2 pi n,

297
00:21:23,670 --> 00:21:25,580
everything is OK.

298
00:21:25,580 --> 00:21:31,720
That thing, when you
change y for y plus L y,

299
00:21:31,720 --> 00:21:35,350
this whole exponent
changes by 2 pi n.

300
00:21:35,350 --> 00:21:37,620
And i times 2 pi n is 1.

301
00:21:37,620 --> 00:21:42,940
So U is well-defined
on the torus.

302
00:21:42,940 --> 00:21:44,540
So we're almost there.

303
00:21:44,540 --> 00:21:45,770
Here it is.

304
00:21:45,770 --> 00:21:51,550
B0-- there is a quantization.

305
00:21:51,550 --> 00:22:05,330
B0 times the area of the
torus is equal to 2 pi n--

306
00:22:05,330 --> 00:22:17,310
pi n-- hc over q n.

307
00:22:17,310 --> 00:22:23,670
So you cannot have
arbitrary magnetic field.

308
00:22:23,670 --> 00:22:28,760
The magnetic field is such
that the flux is quantized.

309
00:22:28,760 --> 00:22:31,910
So here is the
illustration of an example

310
00:22:31,910 --> 00:22:35,270
I wanted to mention
from the beginning,

311
00:22:35,270 --> 00:22:39,860
that if you have a magnetic
field that solves Maxwell's

312
00:22:39,860 --> 00:22:41,570
equations, you're not done.

313
00:22:41,570 --> 00:22:43,310
You have to find potentials.

314
00:22:43,310 --> 00:22:45,900
And sometimes there are
funny things happening.

315
00:22:45,900 --> 00:22:48,200
And in particular,
here is the funny thing

316
00:22:48,200 --> 00:22:51,470
that has happened
here, is that in order

317
00:22:51,470 --> 00:22:55,790
to have a well-defined vector
potential on the torus,

318
00:22:55,790 --> 00:22:59,600
you've been forced to
quantize the magnetic field.

319
00:22:59,600 --> 00:23:05,960
And B0 times A is
equal to phi, the flux.

320
00:23:10,250 --> 00:23:15,500
And this flux is a multiple
of this quantity, which

321
00:23:15,500 --> 00:23:19,220
is sometimes called
the quantum of flux,

322
00:23:19,220 --> 00:23:22,400
the least possible flux.

323
00:23:22,400 --> 00:23:24,020
n, the flux quantum.

324
00:23:30,390 --> 00:23:36,660
So this flux quantum, it is
a famous number, in fact.

325
00:23:36,660 --> 00:23:38,870
I have it somewhere here.

326
00:23:46,750 --> 00:23:48,160
Somewhere in my notes.

327
00:23:48,160 --> 00:23:50,050
Yes, here it is.

328
00:23:50,050 --> 00:23:56,230
Phi 0 for electrons,
for q equal to e--

329
00:23:56,230 --> 00:24:02,470
phi 0-- these are
units you seldom use.

330
00:24:02,470 --> 00:24:12,560
But it's about 2.068 times
10 to the minus 15 webers.

331
00:24:15,250 --> 00:24:17,110
Who knows what a weber is?

332
00:24:20,200 --> 00:24:32,420
Or 2.068 times 10 to
the minus 7 Maxwells.

333
00:24:35,520 --> 00:24:38,880
Anybody know what a Maxwell is?

334
00:24:38,880 --> 00:24:42,310
OK, a weber is a unit of flux.

335
00:24:42,310 --> 00:24:45,590
It's tesla times meter squared.

336
00:24:45,590 --> 00:24:48,040
That's far too big.

337
00:24:48,040 --> 00:24:53,070
The Maxwell, it's a
more natural thing.

338
00:24:53,070 --> 00:24:59,880
Maxwell is gauss times
centimeters squared.

339
00:24:59,880 --> 00:25:02,460
So, you know, the magnetic
field of the Earth

340
00:25:02,460 --> 00:25:04,350
is about half a gauss.

341
00:25:04,350 --> 00:25:07,060
A centimeter square
you can imagine.

342
00:25:07,060 --> 00:25:10,230
And that's the value of
the flux quantum, which

343
00:25:10,230 --> 00:25:15,020
will have a role later for us.