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PROFESSOR: Before
we get started,
let me ask you guys if
you have any questions,
pragmatic or otherwise,
about the course so far.
Seriously?
To those of you
reading newspapers,
I encourage you to find
a slightly different time
to do so.
I really encourage you find
a slightly different time
to do so, thanks.
So far we've done basic
rules of quantum mechanics.
We've done solids.
We understand a lot
about electrons in atoms,
the periodic table, and
why diamond is transparent.
One thing we did along the
way is we talked about spin.
We found that when we looked
at the angular momentum
commutation
relations, these guys.
We found that the
commutators are the same.
For these commutators, we can
get total angular momentum, l l
plus 1 times h bar
squared for l squared.
And h bar m for a constant and
integer m for angular momentum
in a particular direction, which
we conventionally called z.
But we also found that there
were half integer values
of the total spin
and of the spin
in a particular direction
which, for example, with Spin
s as 1/2, 3/2, 5/2, et cetera.
And we discovered that riding a
wave function on a sphere that
interpreted these
states as states
with definite probability to
be at a particular position
on a sphere, a function of
theta and phi, was inconsistent.
In order for the wave function
to satisfy those properties
that have those eigenvalues
and, in particular, half integer
eigenvalues, we discovered
that the wave function
had to be doubly valued.
And thus it would equal to
minus itself, it equaled 0.
So the rest of
today and tomorrow
is going to be an exploration
of spin, or tomorrow--
next lecture, is going to
be an exploration of spin
and the consequences of
these spin 1/2 states.
Exactly what they
are, we already
saw that they're important
for understanding
the structure of
the periodic table.
So we know they're there.
And they're present
from the experiment,
the Stern-Gerlach
experiment that we've
discussed many times.
But before I get
onto that, I want
to improve on the last
experiment we did.
So in particular, in
this last experiment
we talked about the effective
mass of an object interacting
with a fluid or an
object interacting
with its environment.
And why the mass of the
object that's moving
is not the same as
the mass of the object
when you put it on a balance.
And that's this basic
idea of renormalization.
And we demonstrated that.
I demonstrated that
with a beaker of water.
So I had a beaker of water
and pulled a ping pong ball
under water.
We calculated that it
should accelerate upward
when released at 20 times
the acceleration of gravity,
depending on the numbers
you use, very, very rapidly.
And in fact it went
glug, glug, glug,
but it wasn't a terribly
satisfying experiment
because it's very hard
to get the timing right.
And the basic issue there is
that the time scales involved
were very short.
How long did it
take for a ping pong
ball to drop from
here to the surface?
Not much time.
And to rise through the water,
not a whole lot of time.
So I did that experiment,
it was sort of comical.
But I wanted to improve on it.
So over the weekend, I
went down to my basement
and tweaked the
experiment a little bit.
And I called up a
couple of my friends,
and we did an improved
version of this experiment.
So this is a diver, I
think this one is Kathy.
Oh, shoot, we need to
turn off the lights.
Sorry.
I totally forgot.
good You'll never
see this if we don't.
You can do it.
There we go.
All right.
So here you see my friend, I
think this one's, actually,
is it Kathy?
I'm not sure.
It's hard to tell when
they have their marks on.
And we're in the tank at
the New England Aquarium,
and she's going to perform
this experiment for us.
And we're going to film
it, as you can see,
the bubbles moving
rather slowly,
with a high speed camera filming
at 1,200 frames per second
with which we'll be able to
analyze the data that results.
The camera cost as
much as a nice house.
And it's not mine, but
it's important to have
friends who trust you.
OK, so here we are.
You might notice something
in the background.
Before we get
started, I just want
to emphasize that one
should never take casually
the dangers of
doing an experiment.
When you plan an experiment
you must, ahead of time,
design the experiment, design
the experimental parameters.
We designed the lighting.
We designed everything.
We set it up, but
there are always
variables you haven't
accounted for.
And a truly great
experimentalist
is one who has taken account
of all the variables.
And I just want to
emphasize that I'm not
a great experimentalist.
So here, for
example, is a moment.
I probably should have
thanked the sharks,
it just occurred to me.
Anyone who wants to take this
experimental data, which,
as you can probably guess,
filmed for different purposes,
I just manage to get the
ping pong ball into the tank.
Anyone who wants to get this
and actually take the data,
come to me, I'll give
you the raw footage.
And you can read
off the positions,
and hopefully for
next lecture we'll
have the actual plot
of the acceleration
as a function of time.
So with that said and done,
the moral of the story
is you have to account
for all variables.
The other moral of the story is
that you saw this very vividly.
In the motion of the ping pong
ball up, when it was released,
there is that very rapid
moment of acceleration
when it bursts up
very, very rapidly.
But it quickly slows
in its acceleration.
Its acceleration slows down.
In fact, a slightly
funny thing happens.
If you look
carefully, and again,
anyone who wants
this can get access
to the video, what you'll
see is that the ping pong
ball accelerates and then
it sort of slows down.
It literally
decreases in velocity,
accelerates and slows down.
Can anyone think what's
going on in that situation?
AUDIENCE: Boundary
layer formation.
PROFESSOR: Sorry?
AUDIENCE: Boundary
layer formation.
PROFESSOR: Good.
Say that in slightly more words.
AUDIENCE: It's
starting to pick up
more and more water [INAUDIBLE].
PROFESSOR: Yeah.
Exactly. so what's going on
is as this guy starts slowly
moving along, it's pulling
along more and more water,
each bit of water
around it is starting
to drag along the
layers of water nearby,
and it builds up
a sheath of water.
Now that water starts
accelerating, and the ping pong
ball and the bubble of water
that it's dragging along
need to come to equilibrium
with each other.
They need to settle down
smoothly to a nice uniform
velocity.
But it takes a while for
that equilibrium to happen.
And what actually happens
is that the ping pong
ball drives up.
It pulls up the water.
Which then drags with
the ping pong ball.
So you can see that in
the acceleration, which
is oscillatory with a
slight little oscillation.
So it's a damped but
not overdamped harmonic
oscillator motion.
Any other questions
about the effective mass
of an electron and solid
before moving-- Yeah?
AUDIENCE: Why does
it speed up after?
That explains why it
slows down because it's
forming that sheath--
PROFESSOR: Right.
Why it speeds back up is it's
sort of like a slingshot.
As this guy gets going a little
ahead of the pack of water,
the pack of water has
an extra driving force
on top of the buoyancy.
It has the fact that it's
a little bit displaced.
So it catches up, but it's
going slightly greater velocity
than it would be if there
were uniform velocity.
So this guy catches up
with the ping pong ball.
OK so this is an of course
of course in fluid mechanics,
but I guess we don't
actually need this anymore.
Picking up on spin.
So the commutation relations
for spin are these.
And as we saw last time,
we have spin states.
We have we can construct towers
of states because from the sx
and s1 we can build s plus-minus
is equal to sx plus-minus sy.
Sorry.
i, thank you.
So we can build towers of
states using the raising
and lowering operators
as plus-minus.
And those states need to
end, they need to terminate.
So we find that the spin can
have totalling momentum of s
squared h bar
squared l l plus 1.
And S in some
particular direction,
which we conventionally
called z, is h bar m.
I don't want to call this l.
I want to call this s.
For orbital angular
momentum this would be l
and this would be an integer.
But for spinning
angular momentum
these are all the
states we could build,
all the towers we could build,
which were 2n plus 1 over 2,
which were not expressible
in terms of wave functions,
functions of a
position on a sphere.
These are all the
1/2 integer states.
So s could be 1/2,
3/2, 5/2, and so on.
And then m sub s is going
to go from minus s to s
in integer steps, just like
m for the orbital angular
momentum, l.
So I want to talk
about these states
in some detail over the rest of
this lecture and the next one.
So the first thing to talk
about is how we describe spin.
In principle, this is easy.
What we want, is
we want to describe
the state of a particle that
carries this intrinsic angular
momentum spin.
So that's easy.
The particle sits at some
point, but the problem
is it could be sitting at some
point with angular momentum
with spin in the z direction,
say, plus h bar over 2.
And let's focus on the
case s is equal to 1/2.
So I'll be focusing,
for the lecture,
for simplicity on
the case, total spin
is 1/2, which is the
two-state ladder,
but all of this
generalizes naturally.
In fact, that's a very good
test of your understanding.
So for s as 1/2,
we have two states,
which I will conventionally
call the up in the z direction
and the down in the z direction.
And I will often
omit the subscript.
If I omit the subscript
it's usually z
unless from context you see
that it's something else.
So the wave function tells
us the state of the system.
But we need to know now
for a spinning particle
whether it's in the spin 1/2
up or spin 1/2 down state.
And so we could write that
as there's some amplitude
that it's in the plus 1/2
state of x and at position x.
And there's some amplitude
that it's at position x
and it's in the minus 1/2
state or the down state.
And we again need that the
total probability is one.
Another way to say this
is that the probability
that we find the particle to be
at x with plus or minus h bar
upon 2 being the spin
in the z direction.
So at x, spin in the z direction
is h bar upon 2 plus or minus
is equal to norm squared of
psi plus or minus of x squared.
So this is one way you
could talk about spin,
and you could develop the
theory of spin nicely here.
But it's a somewhat
cumbersome formalism.
The formulation I
want to introduce
is one which involves
matrices and which
presages the study
of matrix mechanics
which you'll be using in 805.
So instead, I want to
take these two components,
and what we see already is
that we can't use a single wave
function to describe a
particle at spin 1/2.
We need to use two functions.
And I want to organize
them in a nice way.
I'm going to write
them as psi is
equal to-- and I'll call
this capital psi of x--
is a two component vector,
or so-called spinner,
psi up of x and psi down of x.
So it's a two component object.
It's got a top and
a bottom component.
And notice that its conjugate,
or its adjoint, psi dagger,
is going to be equal
to psi up, complex
conjugate psi down, a row
vector, or a row spinner.
And for normalization we'll
need that the total probability
is 1 which says that
psi capital dagger
psi, psi capital with
psi is equal to 1.
But this is going to be
equal to the integral dx.
And now we have to take the
inner product of the two
vectors.
So integral dx of psi up
squared plus psi down squared.
Cool?
Yep?
AUDIENCE: What's the
coordinate x representing here?
PROFESSOR: The coordinate x is
just representing the position.
So what I'm saying
here is I have,
again, we're in one dimension
just for simplicity,
it's saying, look if I have a
particle that carries spin 1/2,
it could be anywhere.
Let's say it's at position x.
So what's the amplitude at
position x and spinning up,
and I'm not going to
indicate spinning down.
OK?
I like my coffee.
So that's what the of x
indicates, and I've just
been dropping the of x.
So there's some probability
that it's at any given point
and either spin up or spin down.
Now, again, it's important,
although I'm going to do this,
and we conventionally do
this spin up and down,
this spin is
pointing in a vector
space that's two dimensional.
It's either plus 1/2 or
minus 1/2 h bar upon 2 h bar.
So it's not like
the spin is an arrow
in three dimensional
space that points.
Rather, what it is, it's
saying, if I measure
the spin along some axis, it
can take one of two values.
And that was shown in the
Stern-Gerlach experiment, where
if we have a gradient of
the magnetic field, dbz,
in the z direction, and this
is our Stern-Gerlach box,
and we send an electron
in, the electron always
comes out in one
of two positions.
OK.
Now, this is not saying there is
a vector associated with this,
that the electron has a
angular momentum vector that
points in some
particular direction.
Rather, it's saying that
there are two possible values,
and we're measuring
along the z direction.
Cool?
So it's important not
to make that mistake
to think of this as some
three dimensional vector.
It's very explicitly a vector in
a two dimensional vector space.
It's not related to
regular rotations.
Yeah?
AUDIENCE: Where you write
psi of x equals psi plus 1
plus psi [INAUDIBLE].
PROFESSOR: Yes.
AUDIENCE: Do we need a
1 over square root of 2
in front of that thing?
PROFESSOR: Yeah, I haven't
assumed their normalization,
but each one could be
independently normalized
appropriately.
AUDIENCE: So [INAUDIBLE].
PROFESSOR: Right.
The whole thing has to
be properly normalized,
and writing it this way,
this is just the [INAUDIBLE].
Good.
So there's another
nice bit of notation
for this which is
often used, which
is probably the most
common notation.
Which is to write psi
of x is equal to psi up
at x times the vector 1, 0 plus
psi down of x times the vector
0, 1.
So this is what I'm going
to refer to as the up
vector in the z direction.
And this is what
I'm going to refer
to as the down in the
z direction vector.
And that's going to allow me to
write all the operations we're
going to need to study
spin in terms of simple two
by two matrices.
Yeah?
AUDIENCE: Will those two
psi's not be the same?
PROFESSOR: Yeah, in
general, they're not.
So for example, here's a
situation, a configuration,
a quantum system could be in.
The quantum system could
be in the configuration,
the particle is here
and it's spinning up.
And it could be in
the configuration
the particles over here,
and it's spinning down.
And given that it could be
in those two configurations
it could also be in
the superposition
over here and up and
over here and down.
Right?
So that would be different
spatial wave functions
multiplying the
different spin wave
functions, spin part
of the wave function.
Make sense?
OK.
So these are not
the same function.
They could be the same function.
It could be that you could be
either spin up or spin down
at any given point with
some funny distribution,
but they don't need
to be the same.
That's the crucial thing.
Other questions?
Yeah.
AUDIENCE: I know
that [INAUDIBLE]
is a way to call them.
So like [INAUDIBLE]
something, are they
anti-parallel, perpendicular,
or are they something?
PROFESSOR: Yeah.
So here's what we can say.
We know that an electron
which carries sz
is plus h bar upon 2,
and a state corresponding
to minus h bar upon 2 are
orthogonal because those
are two different eigenvectors
of the same operator, sz.
So these guys are orthogonal.
Up in the z direction
and down to the z
directions are orthogonal.
And thank you for
this question, it's
a good way to think
about how wrong it
is to think of up and down being
up and down in the z direction.
Are these guys orthogonal?
These vectors in space?
AUDIENCE: No.
PROFESSOR: No, they happen
to be parallel with a minus 1
in the overlap, right?
So sz as up sc as down
are orthogonal vectors,
but this is clearly
not sz as down, right?
So the direction that they're
pointing, the up and down,
should not be thought
of as a direction
in three-dimensional space.
AUDIENCE: It's just [INAUDIBLE].
PROFESSOR: It's just
a different thing.
What it does tell you, is if
you rotate the system by a given
amount, how does the phase
of the wave function change.
But what it does tell you is how
the spin operations act on it.
In particular, sz acts with
a plus 1/2 or minus 1/2.
They're just different states.
Yeah?
AUDIENCE: Is this
similar to what
happens when
polarizations [INAUDIBLE]?
PROFESSOR: It's similar to
the story of polarizations
except polarizations are
vectors not spinners.
It's similar in the sense
that they look and smell
like vectors in
three-dimensional space,
but they mean slightly--
technically-- slightly
different things.
In the case of
polarizations of light,
those really are honest
vectors, and there's
a sharp relationship
between rotations in space.
But that's a sort of quirk.
They're both spin and vectors.
These are not.
OK so here's a notation
I'm going to use.
Just to alert you,
a common notation
that people use
in Dirac notation
is to say that the wave function
is equal to psi up at x times
the state up plus psi down
at x times the state down.
So for those of you who speak
Dirac notation at this point,
then this means the
same thing as this.
For those of you who don't,
then this means this.
What I want to do is I want to
develop a theory of the spin
operators, and I
want to understand
what it means to be
a spin 1/2 state.
Now, in particular, what I mean
by develop a theory of the spin
operators, if I was talking
about four orbital angular
momentum, say the
orbital angular momentum
in the z direction, I know
what operator this is.
If you hand me a wave function,
I can act on it with lz
and tell you what the result is.
And that means that I can
construct the eigenfunctions.
And that means I can construct
the allowed eigenvalues,
and I can talk
about probabilities.
Right?
But in order to
do all that I need
to know how the operator acts.
And we know how
this operator acts.
It acts as h bar upon i dd
phi where phi is the angular
coordinate around the equator.
And so given any wave function,
a function of x, y, and z
or r, theta, and phi, I
can act with this operator,
know how the operator acts.
And it's true, that again, lx
with lx is equal to i h bar lz.
It satisfies the same
time computation relation
as the spin.
But we know the spin operators
cannot be expressed in terms
of derivatives along a sphere.
I've harped on that many times.
So what I want to know
is what's the analog
of this equation for spin?
What is a representation
of the spin operators
acting on the spinners,
acting on states
that carry half integer spin?
We know it's not going
to be derivatives.
What is it going to be?
Everyone cool with the goal?
In order to do that,
we need to first decide
just some basic definition
of spin in the z direction.
So what do the angular
momentum operators do?
Well whatever else is true of
the spin in the z direction
operator, sz acting on a state
up is equal to h bar upon 2 up.
And sz actually
on a state down is
equal to h bar
minus upon 2 down.
And similarly s
squared on up or down--
Oh, by the way, I'm going
to relatively casually
oscillate between the notations
up and plus, up or down,
and plus or minus.
So sometimes I will write plus
for up and minus for down.
So I apologize for
the sin of this.
s squared on plus, this
is the plus 1/2 state,
is equal to h bar squared
ss plus 1, but s is 1/2,
so 1/2 times 1/2 plus 1 is 3/4
h bar squared times 3 over 4.
And we get the same
thing up, ditto down.
Because s squared acts the same
way on all states in a tower.
Going up and down through
a tower of angular momentum
states, raising and
lowering, does not
change the total
angular momentum
because s plus and s
minus commute with s
squared because they have
exactly the same commutation
relations as the
angular momentum.
That's an awesome sound.
So I want to know
what these look like
in terms of this vector
space notation, up and down.
And for the moment I'm
going to dispense entirely
with the spatial dependence.
I'm going to treat
the spatial dependence
as an overall constant.
So we're equally likely
to be in all positions.
So we can focus just on
the spin part of the state.
So again I want to replace
up by 1, 0 and down by 0, 1.
And I want to think
about how this looks.
So what this looks like is
sz acting on [INAUDIBLE] 1,
0 is equal to h
bar upon 2, 1, 0.
And sz on 0, 1 is
h bar upon 2, 0, 1.
And s squared on 1, 0 is equal
to 3 h bar squared on 4, 1, 0.
And ditto for 0, 1.
Yeah?
AUDIENCE: That [INAUDIBLE]
should have a line over it.
PROFESSOR: Oh, thank you.
Yes, it really should.
AUDIENCE: So how do
you get 3/4 there?
PROFESSOR: 3/4, good.
Where that came from
is that remember
that when we constructed the
eigenfunctions of l squared,
l squared acting
on a state lm was
equal to h bar
squared l l plus 1.
l plus 1 on [? file m. ?] Now
if we do exactly the same logic,
which we actually
did at the time.
We did in full generality
whether the total angular
momentum was an integer
or a half integer.
We found that if
we took, I'm just
going to use for the half
integer states the symbol s,
but it's exactly the
same calculation.
s squared on phi and again
sm sub s is equal to h bar
squared s s plus
1 phi s m sub s.
OK and so for s
equals 1/2 then s,
s plus 1, is equal to 1/2
times 3/2, which is 3/4.
Yeah?
AUDIENCE: [INAUDIBLE]
s equals [INAUDIBLE].
Isn't that [INAUDIBLE]?
PROFESSOR: Ah, but remember,
does l go negative?
Great.
Does s go negative?
No. s is just
labeling the tower.
So s is 0, it's 1, it's 2.
And so for example, here, these
are states where the s is 1/2
and the s in the z direction
can be plus 1/2 or minus 1/2.
s is 3/2 and then s in
the z direction can be m
is 3/2, 1/2, minus
1/2, minus 3/2.
OK.
Other questions?
OK, yeah.
AUDIENCE: What about the
lowering and raising of those?
PROFESSOR: Good, we're going
to have to construct them,
because, what are they?
Well, they lower
and raise, so we're
going to have to build the
states that lower and raise.
AUDIENCE: And would lowering
on the up will give you down,
but raising on up--?
PROFESSOR: Awesome.
So what did it mean
that we had towers?
Let me do that back here.
So the question is,
look, we're going
to have to use the raising
and lowering operators
at the end of the
day, but what happens
if I raise the bottom
state-- what if I raise down?
I'll get up.
And what happens if I raise up?
You get 0.
That's the statement
that the tower ends.
On the other hand, if s is 3/2,
what happens if I raise 1/2?
I get 3/2.
And if I raise 3/2, I get
nothing, I get 0, identically.
And that's the statement
that the tower ends.
So for every tower
labeled by s, we
have a set of states labeled by
m which goes from minus s to s
in integer steps.
The raising operator
raises us by 1,
the lowering
operator lowers by 1.
The lowering operator
annihilates the bottom state,
the raising operator
kills the top state.
Cool?
Yeah.
AUDIENCE: Do states
like 3/2 and 5/2
have anything akin
to up and down?
PROFESSOR: Yeah,
so-- do they have
anything akin to up and down.
Up and down is just a name.
It doesn't really
communicate anything other
than it's shorthand for
spin in the z direction 1/2.
So the question could
be translated as,
are there convenient
and illuminating names
for the spin 3/2 states?
And I don't really know.
I don't know.
I mean, the states exist.
So we can build
nuclear particles that
have angular momentum 3/2,
or 5/2, all sorts of things.
But I don't know
of a useful name.
For the most part,
we simplify our life
by focusing on the 1/2 state.
And as you'll discover in
8.05, the spin 1/2 states,
if you know them
very, very well,
you can use everything you
know about them to construct
all of the spin 5/2 8/2--
well, 8/2 is stupid, but-- 9/2,
all those from the spin 1/2.
So it turns out spin
1/2 is sort of Ur--
in a way that can be
made very precise,
and that's the theory
of Lie algebras.
Yeah.
AUDIENCE: Can you
just elaborate on what
you meant by, you can't really
think of spin as an angular
momentum vector?
PROFESSOR: Yeah.
So OK, good.
So the question is, elaborate a
little bit on what you mean by,
spin can't be thought of as
an angular momentum vector.
Spin certainly can be thought
of as an angular momentum,
because the whole point here
was that if you have a charged
particle and it carries spin,
then it has a magnetic moment.
And a magnetic
moment is the charge
times the angular momentum.
So if it carries spin,
and it carries charge,
and thus it carries
magnetic moment--
that's pretty much what we
mean by angular momentum.
That's as good a
diagnostic as any.
Meanwhile, on top of satisfying
that experimental property,
this, just as a set of
commutation relations,
these commutation relations
are the commutation relations
of angular momentum.
It just turns out that we can
have states with total angular
momentum little s, which is 1/2
integral-- 1/2, 3/2, et cetera.
Now, the things that I want
to emphasize are twofold.
First off, something I've
harped on over and over again,
so I'll attempt to limit
my uses of this phrase.
But you cannot think of these
states with s is 1/2 as wave
functions determining
position on a sphere.
So that's the first
sense in which
you can't think of it as
equivalent to orbital angular
momentum.
But there's a
second sense, which
is that up and down should
not be thought of as spin
in the z direction being up and
spin in the z direction being
down meaning a vector in
three dimensions pointing up
and a vector in three
dimensions pointing down,
because those states
are orthogonal.
Whereas these two
three-dimensional vectors
are not orthogonal--
they're parallel.
They have a non-zero
inner product.
So up and down, the names we
give these spin 1/2 states,
should not be confused with
pointing up in the z direction
and down in the z direction.
It's just a formal name we give
to the plus 1/2 and minus 1/2
angular momentum in
z direction states.
Does that answer your question?
AUDIENCE: Yeah,
so, when you make
a measurement of
the value of spin--
so perhaps you do a
Stern-Gerlach experiment--
and you get what
the spin is, can you
not then say, all right, this
is spin plus 1/2, spin minus?
It's as z is positive
1/2 as z is minus 1/2?
PROFESSOR: Yeah, absolutely.
So if you do a
Stern-Gerlach experiment,
you can identify those
electrons that had spin plus 1/2
and those that had
spin minus 1/2,
and they come out
in different places.
That's absolutely true.
I just want to emphasize
that the up vector does not
mean that they're
somehow attached
to the electronic vector that's
pointing in the z direction.
Good.
Yeah.
Go ahead, whichever.
AUDIENCE: How do we verify that
uncharged particles have spin?
PROFESSOR: Yeah, that's
an interesting question.
So the question
is, how do we know
if an uncharged
particle has spin?
And there are many
ways to answer
this question,
one of which we're
going to come to later
which has to do with Bell's
inequality, which is a
sort of slick way to do it.
But a very coarse
way is this way.
We believe, in a deep
and fundamental way,
that the total angular
momentum of the universe
is conserved, in
the following sense.
There's no preferred
axis in the universe.
If you're a cosmologist,
just stay out
of the room for the
next few minutes.
So there's no preferred
axis in the universe
and the law of physics should
be invariant under rotation.
Now, if you take a system
that has a bunch of particles
with known angular momentum--
let me give you an example.
Take a neutron.
A neutron has spin 1/2.
Wait, how did I know that?
We can do that experiment by
doing the following thing.
We can take a neutron
and bind it to a proton
and see that the resulting
object has spin 1.
So let me try to think of a way
that doesn't involve a neutron.
Grant me for the moment that
you know that a neutron has
spin 1/2.
So let's just imagine that we
knew that, by hook or by crook.
We then do the
following experiment.
We wait.
Take a neutron, let
it sit in empty space.
When that neutron decays,
it does a very cool thing.
It decays relatively
quickly into your proton
and an electron that you see.
You see them go flying away
the proton has positive charge,
and the electron has negative
charge and it goes flying away.
But you've got a problem.
Because you knew that
the neutron had spin 1/2,
which is [INAUDIBLE].
And then you decay one into
a proton and an electron.
And the total angular
momentum there
is 1/2 plus 1/2
or 1/2 minus 1/2.
It's either 1 or 0.
And you've got a problem.
Angular momentum
hasn't been conserved.
So what do you
immediately deduce?
That another particle must
have also been emitted
that had 1/2 integer
angular momentum
to conserve angular momentum.
And it couldn't carry any
charge because the electron
and the proton were neutral,
and the neutron is neutral.
So things like
this you can always
deduce from conservation
of angular momentum
one way or the other.
But the best way
to do it is going
to be some version of
addition of angular momentum
where you have some object
like an electron and a proton
and you allow them
to stick together
and you discover it
has total spin 1.
Yeah.
We can talk about that in
more detail afterwards.
That's a particularly nice
way to do the experiment.
Yeah.
AUDIENCE: Angular momentum
[INAUDIBLE] weird vector
since if you reflect your
system through [INAUDIBLE].
How does that work?
PROFESSOR: Yeah, OK, good.
I don't want to get into
this in too much detail,
but it's a really good question,
so come to my office hours
and ask or go to
recitations and ask.
It's a really good question.
The question is this--
angular momentum
has a funny property under
parity, under reflection.
So if you look in a mirror this
way, here's angular momentum
and it's got--
right-hand rule, it's
got angular momentum up--
if I look in the mirror,
it's going this way.
So it would appear to have
right angular momentum down.
That's what it looks like
if you reflect in a mirror.
Other direction.
So that's a funny property
of angular momentum.
It's also a true property
of angular momentum.
It's fine.
And what about spin,
is the question.
Does spin also have this
funny property under parity,
is that basically the question?
Yeah, and it does.
And working out exactly
how to show that
is a sort of
entertaining exercise.
So again, it's beyond
the scope of the lecture,
so come ask me in office hours
and we can talk about that.
Yeah, one more.
AUDIENCE: For orbital angular
momentum, say for l equals 1,
we had states like m
equals plus 1 and minus 1.
PROFESSOR: Yes.
AUDIENCE: And we did think
of those as angular momentum
vectors.
PROFESSOR: Absolutely.
AUDIENCE: But those states are
also orthogonal, are they not?
PROFESSOR: Yeah, those
states are also orthogonal.
AUDIENCE: So even though
the angular momentum vectors
aren't orthogonal,
they're still--
it's just a different sense.
PROFESSOR: That's exactly right.
So again, even in the case
of integer angular momentum,
you've got to be
careful about talking
about the top state
and the bottom state
corresponding to pointing
in some direction,
because they're
orthogonal states.
However, they do correspond to
a particular angular momentum
vector in three
dimensional space.
They correspond to a
distribution on the sphere.
So there's a sense in
which they do correspond
to real rotations, real
eigenfunctions on a sphere,
and there's also a sense
in which they don't,
because they're
still orthogonal.
That's exactly right.
So let me move on.
I'm going to stop
questions at this point.
So good.
So these are the
properties that need
to be satisfied
by our operators.
And it's pretty easy to
see in this basis what
these operators must be.
Sz has eigenvectors
1, 0 and 0, 1.
So Sz should be equal to h
bar upon 2 1, 0, 0, minus 1.
So let's just check
this on 1, 0 gives me 1,
0, so it gives me the same
thing back times h bar upon 2.
Cool.
And acting on 0, 1,
or the down state,
we get h bar upon
2 times 0 minus 1.
That could be minus 1.
Oh, sorry, 1, 0 gives me 0 and
0 minus 1 on 1 gives me minus 1,
which is 1 with a minus sign.
That's a minus sign.
So this works out like a champ.
And S squared, meanwhile,
is equal to-- well,
it's got to give me h
bar squared times 3/4
for both of these vectors.
So h bar squared--
and meanwhile, these
are eigenstates-- h bar squared
times 3/4 times 1, 0, 0, 1.
So we know one other
fact, which was brought up
just a minute ago, which
was that if we take S plus
and you act on the state
0, 1, what should you get?
If you raise your 1-- 1, 0.
Great.
So we also worked out the
normalization coefficient
on the problem set.
And that normalization
coefficient turns out to be 1.
And let's be careful--
we've got an h
bar, for dimensional
analysis reasons.
So meanwhile, S minus,
similarly, on 0,
1, is equal to 0.
And S plus on 0, 1,
is equal to-- oh,
sorry, we already did that.
We want S minus on 1, 0.
Let's see-- S minus on--
we want S plus on 1,
0 is equal to-- right, 0.
And S minus on 1, 0 is
equal to h bar 0, 1.
OK, so putting
all this together,
you can pretty quickly
get that S plus
is equal to-- we
need an h bar and we
need it to raise the lower
one and kill the top state.
So on 1, 0, what does S plus do?
That gives us 0 that gives us 0.
Good.
And on the lowered
state, 0, 1, that
gives me a 1 up top
and that gives me
a 0 downstairs, so it
works out like this.
So we've got h bar.
Similarly, S minus is equal
to h bar times 0, 0, 1, 0.
So we've got these
guys-- so much
from just the definitions
of raising and lowering.
And by taking
inner products, you
can just derive those
two lines from these.
But notice that Sx is equal
to S plus plus S minus upon 2,
and Sy is equal to S plus
minus S minus upon 2i.
So this tells us that Sx is
equal to h bar upon 2 times
S plus-- we're going to
get a 1 here-- plus S
minus-- we're going to
get a 1 here-- 0, 1, 1, 0.
And Sy is equal to, again,
upon 2i times h bar, h
bar upon 2i, times S plus, which
is going to give me 1 and minus
S minus which is going
to give me minus 1, 0, 0.
But we can pull this i in,
so 1/i is like minus i.
So minus i times minus
1 is going to give me i
and minus i times 1
is going to give me i.
So--
AUDIENCE: Shouldn't
it be minus i?
PROFESSOR: Sorry?
Yeah, did I write i?
That should've been minus i.
Thank you.
So now we have a nice
representation of these spin
operations, of the
spin operators.
And explicitly we have that
Sx is equal to h bar upon 2 0,
1, 1, 0, Sy is equal to h
bar upon 2 0, minus i, i, 0.
And Sz is equal to h bar
upon 2 1, 0, 0, minus 1.
So why is Sz the only
one that's diagonal?
Is it something
special about Sz?
AUDIENCE: I mean, we've chosen
z as the axis along which
to project S squared.
PROFESSOR: Exactly.
So the thing that's
special about Sz
is that at the very
beginning of this,
we decided to work in a
basis of eigenstates of Sz,
with definite values of Sz.
So if they have definite
values, then acting with Sz
is just going to
give you a number.
That's what it is to
be a diagonal matrix.
You act on a basis vector,
you just get a number out.
So we started out by working
in the eigenbasis of Sz.
And as a consequence, we
find that Sz is diagonal.
And this is a general
truth that you'll
discover in matrix
mechanics when
you work in the
eigenbasis of an operator,
that operator is represented
by a diagonal matrix.
And so we often say, rather
than to work in an eigenbasis,
we often say, to diagonalize.
Yeah.
AUDIENCE: Are the
signs right for Sy?
Because if we had h bar over
2i, and as we initially had a 1
in the top right and minus
1 in the bottom left,
shouldn't we just multiply by i?
PROFESSOR: I'm pretty
sure-- so originally, we
had a downstairs i, right?
So let's think about
what this looked like.
This was 1 and minus 1, right?
Agreed?
So this is minus 1 over i.
So we pull in the i.
So that we go from minus
1 to minus 1 over i.
And we go from 1 to 1 over i.
And I claim that one
over i is minus i.
AUDIENCE: Oh, OK.
PROFESSOR: OK?
And minus 1 over i is i.
That cool?
Good.
OK, so this is i and minus i.
I always get that
screwy, but it's
useful to memorize
these matrices.
You might think it's a little
silly to memorize matrices.
But these turn out to
be ridiculously useful
and they come up all the time.
This is called sigma x.
This is called sigma y.
And this is called sigma z.
And different people decide
whether to put the 1/2 in there
or not, the h bar
does not go in there.
Some people put in the 2, some
people don't put in the 1/2,
it's a matter of taste.
Just be careful and be
consistent, as usual.
And these are called the
Pauli matrices because A,
we really like Pauli and
B, Pauli introduced them.
Although he didn't actually
introduce them in some sense--
this mathematical
structure was introduced
ages and ages and ages ago.
But physicists cite
the physicist, not
the mathematician.
OK I'm not saying that's good.
I'm just saying it happens.
So notice a
consequence of these.
An important consequence
of these-- the whole point
here was to build a
representation of the spin
operators.
Now whatever else the
spin operators do,
they had better satisfy
that computation relation,
otherwise they're not
really spin operators.
That's what we mean by
being spin operators.
So let's check.
Is it true that Sx commutator
with Sy is equal to i h bar Sz?
So this is a question mark.
And let's check.
Let's do the commutator.
From the Sx, we're going
to get an h bar upon 2.
From the Sy, from each Sy, we're
going to get a factor of h bar
upon 2, so I'll just write that
as h bar upon 2 squared-- times
the commutator of two
matrices-- 0, 1, 1,
0 commutator with
0, minus i, i, 0.
This is equal to h bar
squared upon 4 times--
let's write this out.
The first term is going to be
this matrix times this matrix.
That's going to be, again,
a matrix-- 0, 1, 0, 1.
So that first one is a 1.
0, 1, minus i-- oh,
sorry, that's an i.
0, 1, that's an i,
that's a minus i.
So 0, 1, 0, i gives me an i.
0, 1, minus i, 0 gives me a 0.
Second row-- 1, 0,
0, i gives me 0.
And 1, 0, minus i,
0 gives me minus i.
And then the second term is
the flipped order, right?
The commutator term.
So we get minus the commutator
term, which is going to be 0,
minus i, 0, 1.
That gives me minus i.
0, minus i, 1, 0,
that gives me 0.
Bottom row-- i, 0, 0, 1-- 0.
And i, 0, 1, 0 give me i.
OK, so notice what
we get this is
equal to h bar squared upon 4.
And both of those matrices
are the same thing.
Those matrices are both i, 0,
0, minus i with minus i, 0, 0,
i, giving us i, 0, 0, minus
i times 2 from the two terms.
The 2's cancel,
and this gives me
h bar squared upon 2
times i, 0, 0, minus i.
But this is also known as--
pulling out an i and an h bar--
times h bar upon 2
1, 0, 0, minus 1.
This is equal to i h bar Sz.
So these matrices represent
the angular momentum
commutators quite nicely.
And in fact, if you
check, all the commutators
work out beautifully.
So quickly, just
as a reminder, what
are the possible
measurable values
of Sz for the spin 1/2 system?
What possible
values could you get
if you measured Sz,
spin in the z direction?
Yeah, plus or
minus h bar upon 2.
Now what about the eigenvectors?
What are they-- of Sz?
In this notation,
there are these states.
There's one eigenvector,
there's the other.
But let's ask the same
question about Sx.
So for Sx, what are the
allowed eigenvalues?
Well, we can answer
this in two ways.
The first way we can answer
this is by saying look,
there's nothing deep about z.
It was just the stupid
direction we started with.
We could have started by working
with the eigenbasis of Sx
and we would've found
exactly the same story.
So it must be plus or
minus h bar upon 2.
But the reason you make
that argument is A,
it's slick and B,
it's the only one
you can make without knowing
something else about how
Sx acts.
But now we know what Sx is.
Sx is that operator.
So now we can ask, what are the
eigenvalues of that operator?
And if you compute the
eigenvalues of that operator,
you find that there
are two eigenvalues
Sx is equal to h
bar upon 2 and Sx
is equal to minus h bar upon 2.
And now I can ask, well,
what are the eigenvectors?
Now, we know what
the eigenvectors
are because we
can just construct
the eigenvectors of Sx plus.
And if you construct the
eigenvectors of Sx plus--
should I take the time?
How many people want me to
do the eigenvectors of Sx
explicitly?
Yeah, that's kind
of what I figured.
OK, good.
So the eigenvectors of Sx
are, for example, on 1,
1 is equal to-- well, Sx
on 1, 1, the first term,
that 0, 1, is going
to give me a 1, the 1,
0 is going to give me a 1.
So this is h bar upon 2
coefficient of Sx on 1, 1.
And Sx on 1, minus 1 is going
to give me h bar upon 2 minus 1,
minus 1, because
all Sx does is swap
the first and second components.
So it gives me minus
1, takes it to the top,
but that's just an
overall minus sign.
So again, we have the correct
eigenvalues, plus and minus
h bar upon 2, and now
we know the eigenvector.
So what does this tell us?
What does it tell us that
up in the x direction
is equal to 1 over root 2 if
I normalize things properly.
Up in the z direction plus
down in the z direction.
That's what this is telling me.
This vector is equal to up
in the z direction-- that's
this guy-- plus down
in the z direction.
But this isn't
properly normalized,
and properly normalizing it
gives us this expression.
So what does this tell us?
This tells us that if we happen
to know that the system is
in the state with angular
momentum, or spin,
angular momentum in the x
direction being plus 1/2, then
the probability to measure
up in the z direction
in a subsequent
measurement is 1/2.
And the probability to
measure down is 1/2.
If you know it's up
in the x direction,
the probability of measuring
up in the z or down in the z
are equal.
You're at chance.
You're at even odds.
Everyone agree with that?
That's the meaning
of this expression.
And similarly, down in the
x direction is equal to 1
over root 2, up
in the z direction
minus down in the z direction.
And we get that from here.
This state is explicitly, by
construction, the eigenstate
of Sx as we've constructed Sx.
And we have a natural
expression in terms
of the z eigenvectors
up and down.
That's what this
expression is giving us.
It gives us an expression
of the Sx eigenvector
in the basis of Sz eigenvectors.
So for example, this tells
you that the probability
to measure up in the
z direction, given
that we measured down
in the x direction
first-- so this is the
conditional probability.
Suppose I first measured
down in the x direction,
what's the probability
that I subsequently
measure up in the z direction?
This is equal to--
well, it's the norm
squared of the
expansion coefficient.
So first down in the x
direction and the probability
that we're up in the z direction
is 1 upon root 2 squared.
Usual rules of
quantum mechanics--
take the expansion coefficient,
take its norm squared,
that's the probability-- 1/2.
And we can do exactly
the same thing for Sy.
So let's do the
same thing for Sy
without actually working
out all the details.
So doing the same thing for
Sy, up in the y direction
is equal to 1 upon root 2 times
up in the z direction plus i
down in the z direction.
And down in the y direction
is equal to 1 upon root 2
up in the z direction minus
i down in the z direction.
And I encourage you to
check your knowledge
by deriving these
eigenvectors, which
you can do given our
representations of Sy.
Now here's a nice thing that
we're going to use later.
Consider the following.
Consider the operator
S theta, which
I'm going to define as cosine
theta Sz plus-- oh, sorry.
I'm not even going
to write it that way.
So what I mean by S
theta is equal to-- take
our spherical directions and
consider an angle in the-- this
is x, let's say, y, x, z--
consider an angle theta down
in the zx plane, and we
can ask, what is the spin
operator along the
direction theta?
What's the angular momentum
in the direction theta?
If theta, for example, is
equal to pi/2, this is Sx.
So I'm just defining
a spin operator,
which is the angular momentum
along a particular direction
theta in the xz plane.
Everyone cool with that?
This is going to turn out
to be very useful for us,
and I encourage you to
derive the following.
And if you don't derive
the following, then
hopefully it will be
done in your recitations.
Well, I will chat with your
recitation instructors.
And if you do
this, then what are
we going to get for the
eigenvalues of S theta?
What possible eigenvalues
could S theta have?
[? AUDIENCE: None. ?]
PROFESSOR: Good.
Why?
AUDIENCE: Because it can't
have any other redirection
[INAUDIBLE] no matter
where you start.
PROFESSOR: Fabulous.
OK.
So the answer that
was given is that it's
the same, plus or
minus h bar upon 2
as Sz, or indeed as Sx or
Sy, because it can't possibly
matter what direction you
chose at the beginning.
I could have called
this direction theta z.
How do you stop me?
We could have done that.
We didn't.
That was our first
wave answering.
Second wave answering is what?
Well, construct the
operator S theta
and find its eigenvectors
and eigenvalues.
So I encourage you to do that.
And what you find
is that, of course,
the eigenvalues are plus
or minus h bar upon 2
and up at the angle theta is
equal to cosine of theta upon 2
up in the z direction
plus sine of theta
upon 2 down in the z direction.
And down theta is
equal to cosine theta
upon 2 up in the z direction
minus sine of theta
over 2 down in the z direction.
Oops, no.
I got that wrong.
This is sine, and
this is minus cosine.
Good.
That makes more sense.
OK.
So let's just sanity check.
These guys should be
properly normalized.
So if we take the norm
squared of this guy,
the cross terms vanish
because up z and down z
are orthogonal states.
So we're going to get a co
squared plus the sine squared.
That's 1.
So that's properly normalized.
Same thing for this guy.
The minus doesn't
change anything
because we norm squared.
Now, I'll check that
they're orthogonal.
If we take this guy dotted into
this guy, everything's real.
So we get a cosine
from up up, we're
going to get a cosine sine.
And from down down, we're going
to get a minus cosine sine.
So that gives us 0.
So these guys are
orthogonal, and they
satisfy all the nice
properties we want.
So this is a good check--
do this-- of your knowledge.
And if we had problem
sets allowed this week,
I would give you this in your
problem set, but we don't.
Yeah, OK.
Suppose that I measure.
So I want to use these
states for something.
Suppose that I
measure Sz and find
Sz is equal to 1/2 h
bar plus h bar upon 2,
OK, at some moment in time.
First question is easy.
What's the state of the system
subsequent to that measurement?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Man, you
all are so quiet today.
What's the state of the system
subsequent to measurement
that Sz is plus h bar upon 2?
AUDIENCE: Up z.
PROFESSOR: Up z.
Good.
So our state psi is
up z upon measurement,
OK, after measurement.
And I need new chalk.
OK.
Now, if I measure Sx, so 1, 2,
measure Sx, what will I get?
AUDIENCE: [INAUDIBLE]
PROFESSOR: OK.
What values will I observe
with what probabilities?
Well, first off, what
are the possible values
that you can measure?
AUDIENCE: Plus or
minus h bar over 2.
PROFESSOR: Right, the
possible eigenvalues,
so which is plus or minus
h bar upon 2, but with what
probability?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Exactly.
So we know this from
the eigenstate of Sx.
If we know that we're
in the state up z,
we can take linear
combinations of this guy
to show that up z is
equal to 1 over root 2.
So let's just check.
If we add these two
together, up x and down x,
if we add them
together, we'll get 1/2,
1/2, 2/2, a root 2 times up
z, is up x plus down x, up x
plus down x, dividing
through by the 1/2.
So here we've expressed
up in the z direction
in a basis of x and y, which
is what we're supposed to do.
So the probability that
I measure a plus 1/2
as x is equal to
plus 1/2 h bar upon 2
is equal to 1 over
root 2 squared, so 1/2.
And ditto, Sx is equal
to minus h bar upon 2.
Same probability.
OK?
What about Sy?
Again, we get one of two values.
Oops, plus or
minus h bar upon 2.
But the probability of
measuring plus is equal to 1/2,
and the probability that we
measure minus 1/2 is 1/2.
That's h bar upon 2.
OK?
So this should look familiar.
Going back to the
very first lecture,
hardness was spin
in the z direction,
and color was spin in
the x direction, I guess.
And we added, at one
point, a third one,
which I think I
called whimsy, which
is equal to spin
in the y direction.
OK?
And now all of the
box operation that we
used in that very first
lecture, you can understand
is nothing other than
stringing together
chains of Stern-Gerlach
experiments
doing Sx, Sy, and Sz.
So let's be explicit about that.
Let's make that concrete.
Actually, let's do that here.
So for example, suppose
we put a random electron
into an Sz color box.
OK.
Some are going to
come out up, and some
are going to come out
down in the z direction.
And if we send this into
now an Sx color box,
this is going to give us
either up in the x direction
or down in the x direction.
And what we'll get
out is 50-50, right?
OK.
So let's take the ones
that came out down.
And if we send those back
into an Sz, what do we get?
AUDIENCE: 50-50.
PROFESSOR: Yeah, 50-50,
because down x is 1 over 2 up z
plus 1 over 2 down z.
What was going on in that
very first experiment, where
we did hardness,
color, hardness?
Superposition.
And what superposition
was a hard electron?
It was 1 over a 2,
white plus black.
We know precisely which
superposition, and here it is.
OK?
And now, let's do that
last experiment, where
we take these guys, Sx down, and
I'm turning this upside down,
beam joiners.
We take the up in the x
direction and the down
in the x direction, and
we combine them together,
and we put them back
into Sz, what do we get?
Well, what's the state?
1 over root 2 down x
plus 1 over root 2 up x?
AUDIENCE: [INAUDIBLE]
PROFESSOR: It's up z.
Up z into an Sz
box, what do we get?
Up z with 100%.
That's white with 100%, or
I'm sorry, hard with 100%.
AUDIENCE: [INAUDIBLE]
down z with a down z.
PROFESSOR: Sorry.
AUDIENCE: You put in down.
PROFESSOR: Oh, I put in down.
Shoot, I'm sorry.
Well, yeah, indeed,
I meant to put in up.
Yes, down z with
100% confidence.
If we remove the mirror, 50-50.
If we add in the mirror, 100%.
And the difference
is whether we're
taking one component
of the wave function,
or whether we're superposing
them back together.
All right?
Imagine we know we have
a system in this state,
and I say, look, this component
is also coincidentally very far
away, and I'm going
to not look at them.
So of the ones that I look
at, I have 1 over root 2 up x.
But if I look at
the full system,
the superimposed system,
that adds together
to be an eigenstate of Sz.
These are our color boxes.
Yeah?
AUDIENCE: But how
do you know, when
you put the two beams
to the beam joiner,
it serves to add their
two states together?
PROFESSOR: Yeah, excellent.
This is a very subtle point.
So here we have to decide what
we mean by the beam splitter.
And what I'm going to
mean by the beam splitter,
by the mirrors
and beam joiners--
so the question is, how do
we know that it does this
without changing
the superposition?
And what I want to do
is define this thing
as the object that takes the
two incident wave functions,
and it just adds them together.
It should give me the
direct superposition
with the appropriate
phases and coefficients.
So if this was plus up x,
then it stays plus up x.
If it's minus up x, it's
going to be minus up x.
Whatever the phase is of
this state, when it gets here
it just adds together
the two components.
That's my definition
of that adding box.
AUDIENCE: So realistically,
what does that look like?
PROFESSOR: Oh, what?
You think I'm an
experimentalist?
[LAUGHTER]
Look, every time I
try an experiment,
I get hit by a shark.
OK?
[LAUGHTER]
Yeah, no.
How you actually implement
that in real systems
is a more complicated story.
So you should
direct that question
to Matt, who's a very
good experimentalist.
OK.
So finally, let's go back to
the Stern-Gerlach experiment,
and let's actually run the
Stern-Gerlach experiment.
I guess I'll do that here.
So let's think about what the
Stern-Gerlach experiment looks
like in this notation, and
not just in this notation,
in the honest language of spin.
And I'm going to do a slightly
abbreviated version of this
because you guys can
fill in the details
with your knowledge
of 802 and 803.
OK.
So here's the
Stern-Gerlach experiment.
We have a gradient in
the magnetic field.
This is the z direction.
And we have a gradient
where B in the z direction
has some B0, a
constant plus beta z.
OK?
So it's got a constant
piece and a small gradient.
Everyone cool with that?
It's just the magnetic field
gets stronger and stronger
in the z direction,
there's a constant,
and then there's a rate of
increase in the z direction.
And I'm going to send
my electron through.
Now remember, my electron has
a wave function, psi electron,
is equal to-- well, it's
got some amplitude to be up,
a up z, plus some amplitude
to be down, b down z.
And if this is a
random electron,
then its state is
going to be random,
and a and b are going to
be random numbers whose
norm squared add up to
1, proper normalization.
Cool?
So here's our random
initial state,
and we send it into
this region where
we've got a magnetic
field gradient.
And what happens?
Well, we know that the
energy of an electron that
carries some angular momentum
is a constant, mu naught,
times its angular
momentum dotted
into any ambient magnetic field.
Whoops, sorry,
with a minus sign.
This is saying that
magnets want to anti-align.
Now, in particular,
here we've got
a magnetic field
in the z direction.
So this is minus mu 0 Sz Bz.
And Bz was a constant, beta
0-- sorry, B0 plus beta z.
So the energy has two terms.
It has a constant term,
which just depends on Sz,
and then there's a
term that depends
on z as well as depending
on Sz [INAUDIBLE].
So we can write this as a--
so Sz, remember what Sz is.
Sz is equal to h bar
upon 2, 1, 0, 0, minus 1.
So this is a matrix with
some coefficient up a--
do I want to write this out?
Yeah, I guess I don't really
need to write this out.
But this is a matrix, and
this is our energy operator.
And it acts on any given state
to give us another state back.
It's an operator.
OK.
And importantly,
I want this to be
only-- this is in some
region where we're
doing the experiment, where we
have a magnetic field gradient.
Then outside of this region,
we have no magnetic field
and no magnetic field gradient.
So it's 0 to the left
and 0 to the right.
So as we've talked about
before, the electron
feels a force due to this
gradient to the magnetic field.
The energy depends on
z, so the derivative
of the energy with
respect to z, which
is the force in the z
direction, is non-zero.
But for the moment that's
a bit of a red herring.
Instead of worrying about
the center of mass motion,
let's just focus on
the overall phase.
So let's take our initial
electron with this initial wave
function a up z plus
b down z, and let's
note that in this time-- so
what does this matrix look like?
OK.
So fine, let's actually
look at this matrix.
So Sz is the matrix 1, 0, 0,
minus 1 times h bar upon 2.
So we have h bar
upon 2 minus mu 0.
And then B0 plus beta
z, which I'll just
write as B, which is equal
to some constant C, 0, 0,
some constant minus C.
Everyone agree with that?
Where a constant, I
just mean it's a number,
but it does depend on z,
because the z is in here.
Now, here's the nice thing.
When we expand the energy on
up z, this is equal to C of z
up z, because there's
our energy operator.
And energy on down z is equal
to minus Cz of z down to z.
So up and down in
the z direction
are still eigenfunctions
of the energy operator.
We've chosen an
interaction, we've
chosen a potential which
is already diagonal,
so the energy is diagonal.
It's already in its eigenbasis.
So as a consequence, this is the
energy, energy of the up state
is equal to, in the z
direction, is equal to plus C.
And energy in the down state,
energy of the down state,
is equal to minus C of z.
Everyone agree with that?
So what this magnetic
field does it
splits the degeneracy of
the up and down states.
The up and down
states originally
had no energy splitting.
They were both zero energy.
We turn on this magnetic
field, and one state
has positive energy,
and the other state
has negative energy.
So that degeneracy
has been split.
Where did that original
degeneracy come from?
Why did we have a degeneracy
in the first place?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Spherical symmetry.
And we turn on the magnetic
field, which picks out
a direction, and we
break that degeneracy,
and we lift the splitting.
OK?
So here we see that the
splitting is lifted.
And now we want to ask, how
does the wave function evolve
in time?
So this was our
initial wave function.
But we know from the Schrodinger
equation that psi of t
is going to be equal to--
well, those are already
eigenvectors-- so a e to the
minus i e up t upon h bar up z
plus b e to the minus i e
down t upon h bar down z.
All right?
But we can write this as equals
a e to the i mu 0 b0 t over 2
plus i mu 0 beta t z upon 2 up
z plus b same e to the minus
i--oops, that should be,
oh, no, that's plus--
that's e to the i
mu 0 b0 t over 2.
And if we write this
as two exponentials,
e to the minus i mu 0
beta t z upon 2-- oh no,
that's a minus,
good, OK-- down z.
OK.
So what is this telling us?
So what this tells us
is that we start off
in a state, which has
some amplitude to be up
and some amplitude
to be down, a and b.
And at a later
time t, sine of t,
what we find after we run
it through this apparatus
is that this is the amplitude.
What do you notice?
Well, we notice two things.
The first is that
the system evolves
with some overall energy.
So the phase rotates as usual.
These are energy
eigenstates, but the amount
of phase rotation depends on z.
So in particular, this is e
to the i some number times z.
And what is e to the
i some number times z?
If you know you have a state
that's of the form e to the i
some number, which I will, I
don't know, call kz times z,
what is this telling
you about the system?
This is an eigenstate
of what operator?
AUDIENCE: Momentum.
PROFESSOR: Momentum
in the z direction.
It carries what momentum?
AUDIENCE: h bar kz.
PROFESSOR: h bar kz, exactly.
So what about this?
If I have a system
in this state,
what can you say about its
momentum in the z direction?
AUDIENCE: [INAUDIBLE]
PROFESSOR: It's non-zero.
Right here's a z.
These are a bunch of constants.
It's beta, t, mu 0, 2, i.
So it's non-zero.
In fact, it's got minus
some constant times z.
Right?
So this is a state
with negative momentum.
Everyone agree?
It's got momentum z
down in the z direction.
What about this guy?
Momentum up.
So the states that are up in
the z direction get a kick up.
And the states that are
down in the z direction
get a kick down.
They pick up some momentum
down in the z direction.
Yeah?
AUDIENCE: Where did
that big T come from?
PROFESSOR: Big T should
be little t, sorry.
Sorry, just bad
handwriting there.
That's just t.
Yeah?
AUDIENCE: So in this
case, the eigenvalues--
it's a function of z?
PROFESSOR: Mhm.
AUDIENCE: Is that allowable?
PROFESSOR: Yeah.
So that seems bad, but remember
what I started out doing.
Oop, did I erase it?
Shoot, we erased it.
So we started out saying,
look, the wave function is up
in the z direction, or
really rather here, up
in the z direction
times some constant.
Now, in general, this
shouldn't be a constant.
It should be some function.
So this should be a function
of position times up
in the z direction.
This is a function position
times down in the z direction.
So what we've done here is we
said, under time evolution,
that function changes
in time, but it
stays some linear
combination up and down.
So you can reorganize this as--
the time evolution equation
here is an equation for
the coefficients of up z
and the coefficient of down z.
AUDIENCE: OK.
PROFESSOR: Other questions?
OK.
So the upshot of all this is
that we run this experiment,
and what we discover is that
this component of the state
that was up z gets a kick
in the plus z direction.
And any electron that
came from this term
in the superposition
will be kicked up up z.
And any electron that came
from the superposition down z
will have down.
Now, what we really mean is
not that an electron does this
or does that, but rather
that the initial stage
of an electron that's here,
with the superposition of z
up and down, ends up in the
state as a superposition of up
z being up here and
down z being down here.
OK?
An electron didn't do one,
it didn't do the other.
It ends up in a superposition,
so the state at the end.
So what the Stern-Gerlach
experiment has done,
apparatus has done,
is it's correlated
the position of the
electron with its spin.
So if you find the amplitude,
to find it up here and down
is very small.
The amplitude to find it up
here and up is very large.
Similarly, the amplitude, to
find it down here and up is 0.
And the amplitude, to find it
down here and down is large.
Cool?
So this is exactly what
we wanted from the boxes.
We wanted not to do
something funny to the spins,
we just wanted to correlate
the position with the spin.
And so the final state is a
superposition of these guys.
And which superposition?
It's exactly this superposition.
So these calculations are
gone through in the notes
that are going to be posted.
OK.
So questions at this point?
Yeah?
AUDIENCE: And so when
you put the electron
through the
Stern-Gerlach device,
does that count as a measurement
of the particle's angular
momentum?
PROFESSOR: Excellent.
When I put the electrons through
the Stern-Gerlach device,
do they come out with
a definite position?
AUDIENCE: No.
PROFESSOR: No.
At the end of the
experiment, they're
in a superposition of either
being here or being here.
And they're in a
superposition of either
being up spin or down spin.
Have we determined the
spin through putting it
through this apparatus?
No.
We haven't done any measurement.
The measurement comes when
we now do the following.
We put a detector here
that absorbs an electron.
We say, ah, yeah, it got hit.
And then you've measured
the angular momentum
by measuring where it came out.
If it comes out down here,
it will be in the positive.
So this is a nice example of
something called entanglement.
And this is where we're
going to pick up next time.
Entanglement says the following.
Suppose I know one
property of a particle,
for example, that it's up here,
or suppose I'm in the state psi
is equal to up, so up here
and up in the z direction plus
down there and down
in the z direction.
OK?
This means that, at the moment,
initially, if I said, look,
is it going to be up or down
with equal probabilities, 1
over root 2 with
equal amplitudes,
if I measure spin in
the z direction, what
value will I get?
What values could
I get if I measure
spin in the z direction?
Plus or minus 1/2.
And what are the
probabilities that I
measure plus 1/2 or minus 1/2?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Even odds, right?
We've done that experiment.
On the other hand, if
I tell you that I've
measured it to be up
here, what will you then
deduce about what its spin is?
AUDIENCE: Plus.
PROFESSOR: Always plus 1/2.
Did I have to do the
measurement to determine
that it's plus 1/2.
No, because I already
know the state,
so I know exactly what I will
get if I do the experiment.
I measure up here, and
then the wave function
is this without the 1 over
root 2, upon measurement.
But as a consequence, if I
subsequently measure up z,
the only possible value
is up in the z direction.
Yeah?
And this is called entanglement.
And here it's entanglement of
the position with the spin.
And next time, what
we'll do is we'll
study the EPR experiment,
which says the following thing.
Suppose I take two electrons,
OK, take two electrons,
and I put them in the state one
is up and the other is down,
or the first is up and
the second is down.
All right?
So up, down plus down, up.
OK?
I now take my two electrons, and
I send them to distant places.
Suppose I measure one of them
to be up in the x direction.
Yeah?
Then I know that the other
one is down in that direction.
Sorry.
If I measure up in
the z direction,
I determine the other one
is down in the z direction.
But suppose someone over here
who's causally disconnected
measures not spin
in the z direction,
but spin in the x direction.
They'll measure one of two
things, either plus or minus.
Now, knowing what we know,
that it's in the z eigenstate,
it will be either plus 1/2 in
the x direction or minus 1/2
in the x direction.
Suppose I get both measurements.
The distant person over here
does the measurement of z
and says, aha, mine is up, so
the other one must be down.
But the person over here doesn't
measure Sz, they measure Sx,
and they get that it's plus Sx.
And what they've done
as a result of these two
experiments, Einstein,
Podolsky, and Rosen say,
is this electron has been
measured by the distant guy
to be spin z down, but by this
guy to be measured spin x up.
So I know Sx and Sz definitely.
But that flies in the face of
the uncertainty relation, which
tells us we can't have spin
z and spin x definitely.
Einstein and Podolsky
and Rosen say
there's something missing
in quantum mechanics,
because I can do
these experiments
and determine that this
particle is Sz down and Sx up.
But what quantum
mechanics says is
that I may have done
the measurement of Sz,
but that hasn't
determined anything
about the state over here.
There is no predetermined value.
What we need to do is
we need to tease out,
we need an experimental
version of this tension.
The experimental
version of this tension,
fleshed out in the
EPR experiment,
is called Bell's inequality.
We studied it in the
very first lecture,
and we're going to show it's a
violation in the next lecture.
See you guys next time.
[APPLAUSE]
