The following
content is provided under a Creative
Commons license. Your support will help MIT
OpenCourseWare continue to offer high-quality
educational resources for free. To make a donation or to
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. So today, we'll continue our
kind of review that included, of course, the last lecture,
the variational principle that's supposed to be new stuff
you didn't see in 804. And today, as we
continue, we'll talk about position and momentum for
about 30 minutes or 40 minutes, and then begin
the study of spin. That will be spin-1/2 with
a Stern-Gerlach experiment and the mathematics
that comes out of it. Now, we will talk about the
Stern-Gerlach experiment in quite some detail
so that you can appreciate what
was going on there. And then we will extract a few
of the mathematical lessons that this experiment tells
us about quantum mechanics. Immediately after
that, which will be probably middle of next
lecture, we will pivot. And as we learn this
mathematics that the Stern-Gerlach experiment
is telling us or asking us for, we will go in some detail
on the necessary mathematics for quantum mechanics. We'll talk about vector
spaces, linear operators, Hermitian operators, unitary
operators, [INAUDIBLE], matrix representations,
all kinds of things. That probably will be about two
weeks, three lectures at least. So it will be a nice study. And in that way,
people that don't have a background
in linear algebra will feel more comfortable with
what we're going to be doing. And I think even
for the people that have a background
in linear algebra, you will gain a new
appreciation about the concepts that we meet here. So today, we begin, therefore,
with position and momentum, and these are operators
in quantum mechanics. And they have letters
to denote them. x, we put a hat with it,
that's a position operator. p, we put a hat on it. And the position and momentum
operators don't commute. And the commutator
is given by ih bar. Now, we have been dealing
so far with wave functions. Our wave functions, where
these functions of x and t, they represent the dynamics
of your system, the dynamics of your particle as
it moves in time. But time, as you are seeing
in quantum mechanics, is a little bit of a spectator. It's an arena where
things happen. But the operators, and most
of the interesting things, are going on without
reference to time. Time evolution, you
have an expansion of a wave function
in terms of energy, eigenstates, at a given time. And then you can
evolve it easily with the way we've
learned, adding e to the minus i et over h bar
for each energy eigenstate. So time will play no role here. So when I talk about the
wave function, at this moment you could put the
time, but we will talk about the wave functions
that have no time dependence. So, say, a psi of
x wave function. So this psi of x may be the true
wave function at time equals 0, or you could just simply
think of it as the psi of x. Now, this wave function
means that we're treating x in a
particular way, and we say that we're working in the
x representation, the position representation. Now, this means that
we have an easy way to figure out what
this operator does when it acts on this function. So what it acts
on this function, it will give you
another function, and the definition of
this is that the position operator acting on
the function psi of x is defined to be
another function, which is the function
x times psi of x. Well, we're talking about these
wave functions and operators on wave functions. And a recurrent theme
in quantum mechanics is that we will think
of wave functions, sometimes we call them states. Sometimes we call them vectors. And we basically think of
wave functions as vectors. And things that act
on wave functions are the things that
act on vectors. And the things that
act on vectors, as you know in
mathematics, is matrices. So we're compelled, even
at this early stage, to get a picture of
how that language would go if we're talking
about these things. So how do we think of a
wave function as a vector? And how do we think
of x as a matrix? So there's a way to do that. It will not be totally
precise, but it's clear enough. So suppose you have
a wave function, and we're interested in
its values from 0 up to a. This wave function is a
function of x between 0 and a. So it's the psi of x
for x between a and 0. That's all the information. What we're going to
do is we're going to divide this thing,
this line, this segment, into a lot of pieces. And we're going to
say, look, instead of writing a function like
sine of x or cosine of x, let's just give the
values and organize them as if this will be a
vector of many components. So let's divide this
in sizes epsilon, such that N times
epsilon is equal to a. So there are N of
these intervals. So we think of psi as a
vector whose first component is psi at 0. The second is psi at epsilon. The third is psi at 2 epsilon. And the last one is
psi at N epsilon. And depending on how much
accuracy you want to work with, you take epsilon smaller and
larger, keeping a constant. And this would be like
summarizing all the information of a function in a vector. Now, that's intuitively a
nice way to think of it. May look, with your background
in classical physics, a little strange that we
sort of put the value at 0 along the x-axis, first
component, the value at epsilon along the y, the value
of 2 epsilon along the z. But we need more axes. So you need many axes here. In this case, this is a
N plus 1 column vector. It has N plus 1
entries, because 0 up to N, that's N plus 1 entries. But that's a fine way
of thinking of it. Not exact because
we have an epsilon. In this way of thinking
about the wave function, we can then ask, what does
the matrix x hat look like? So x hat is an operator,
and it acts this way. So here is how it looks like. We would think of x hat as an
N plus 1 times N plus 1 matrix. And its entries
are 0 everywhere, except in the
diagonal, where they are 0 epsilon, 2
epsilon, up to N epsilon. And here is a big 0 and a big 0. This, I claim, is the way you
should think of the x operator if you thought of the wave
function the way we wrote it. And how do we check that? Well, x operator acting on psi
should be this acting on that. And then, indeed,
we see that if x hat is acting on psi of
x, what do we get? Well, it's easy to multiply a
diagonal matrix times a vector. Here you get 0 times psi of 0. You get a vector, so let
me make this thinner. Then I get epsilon times psi
of epsilon, 2 epsilon times psi of 2 epsilon, up to N
epsilon times psi of N epsilon. And indeed, that matrix looks
like the matrix associated with this wave
function because here is the value at 0 of
this wave function. Here is the value at epsilon of
this wave function, and so on. So this has worked
out all right. We can think of the wave
function as a column vector, and then the position operator
as this vector as well. Now, given that we know how
the x operator is defined, we can also think
easily about what is the expectation value
of x on a wave function. Something that you
really know, but now maybe becomes a little clearer. Here you're supposed
to do psi star of x times the x operator
acting on psi of x. But we have the
definition of this, so this is, as you imagine, dx--
and I should put primes maybe, well, I don't have to
put primes-- dx psi star of x x psi of
x, which is what you would have done anyway. Well, given that we've started
with this, we can ask also, is there eigenstates
of the x operator? Yes, there are. but then fortunately,
are a bit singular. So what should be
an eigenstate of x? It's some sort of state. Intuitively, it has a definite
value of the position. So it just exists
for some value of x. So it's naturally thought
as a delta function. So let me define a
function, psi sub x0 of x. So it's a function
of x labeled by x0, and define it to be
delta of x minus x0. So I claim that is an
eigenstate of x hat. x hat on psi x0 of x is
equal, by definition, to x times psi x0 of x, which
is x times delta of x minus x0. And when you multiply
a function of x times a delta
function in x, it is possible to evaluate
the function that is being multiplied
by the delta function at the place where the
delta function fires. It has the same effect
on integrals or anything that you would do. So here, this is equal to
x0 times delta x minus x0. You evaluate the x at x0. And finally, this is x0 times
that function psi x0 of x. And therefore, you've shown
that this operator acting on this function reproduces
the function-- that's the definition of
eigenstate as an operator-- and the eigenvalue is the number
that appears here, and it's x0. So this function
is an eigenstate of x hat with
eigenvalue, e.v., x0. The only complication
with this eigenfunction is that it's not normalizable. So it doesn't
represent the particle. It can be used to
represent the particle, but it's a useful function. You can think of it
as something that can help you do physics,
and don't insist that it represents a particle. So this is the
story for position. And the position gets actually
more interesting as soon as you introduce the
dual quantity, momentum. So what is momentum here? So momentum is an operator, and
this operator must be defined. Now, you had a
shorthand for it in 804, which is p hat equal
h bar over i d dx. And this shorthand
means actually that, in what we call the
position representation where we're using wave functions
that depend on x, well, the momentum is given
by this operator. And the story of why
this was the case was sort of something
that was elaborated on in 804, the work
of de Broglie, that saw that the
wavelength of the wave has to do with the
momentum of a wave. And finally, people
understood that this would measure the
momentum of the wave. So this is the operator. And therefore, in the
representation that we're working-- representation
is a word that has a lot of precise meaning, but now
I'm just using it in the sense that, well, we're working
either with x's or with p's. And we're working with x's. That's why p looks like
something to do with x. So what is p hat
on a wave function? Well, that's what this means. It's another wave
function obtained by taking the x derivative. So that's the definition of
it acting on a wave function. The one thing that must
be verified, of course, is that this definition
is consistent or implies this commutation relation. So you've defined
it as an operator. x, we've defined
it as an operator. But most of us think
that it doesn't look like an operator
is multiplying. But it is an operator. So this one does look
like an operator. It's a differential operator. And you can try to see
if this equation is true. And the way to test
these commutators is something that,
again, I don't think is unfamiliar to you,
but let's go through it, is that you try to evaluate
this product of operators acting on a wave function. And if things work
out well, we'll see you should get ih bar
times that wave function. If that is the case,
you say, OK, I've proven that equation, because
it's an operator equation. The left-hand side
of that equation is the product in different
orders of two operators, therefore it's an operator. The right-hand side
is another operator. It's the operator multiplied by
ih, anything that you'll get. Well, if this is an
operator identity, the operator on the left
must be equal to the operator on the right, which just means
that, acting on anything, they must give the same answer. So if I managed to prove
that this is equal to this, I've proven that for
anything that is the answer. And therefore, I can
write the top one. And let me just do
it, even though this may be kind of familiar
to many of you. It's good to do this
slowly once in your life. So let's go through this. So this says x operator
p operator on psi minus p operator of
x operator on psi. When you have several operators,
like ABC acting on psi, this really means
let C act on psi, and then let B act on C psi,
and then let A act on that. The operators act one by one. The closest one acts first. So here I'm supposed
to let B act on psi, but that means that thing. So now x is acting
on h over i d psi dx. On this one, I have
p acting on x psi, because that's
what x hat psi is. Here, this is multiplication
by x of a function of x. So this is just h
over i x d psi dx. And here, I have h over i d
dx of this whole thing x psi. And you can see that
when you act here, you act first on the x,
and you get something. And then you act on the psi,
and you get this same term. So the only contribution
here is equal to minus h over i, the d dx on x
times psi, which is ih bar psi, which is
what I wanted to show. So this is true. And therefore, you could
say that this definition is consistent with your
definition of x, and they represent
this operator. One more thing you could try
to do, and it's fun to do it, is we had a matrix
representation for x. Can I think of p as a matrix? How would you do it? What kind of matrix
would p look like? Well, yes? AUDIENCE: You just generate
a finite difference equation. PROFESSOR: You could
do it, exactly, with taking finite differences. So for example, if
you think that you want to produce the wave
function psi prime at 0, psi prime at epsilon,
psi prime, that's what the derivative
gives you, you'll write this as 1 over
epsilon, say, psi at epsilon minus psi at 0. That's the derivative
at 0 roughly. It would be psi at 2 epsilon
minus psi at 0 over 2. And you could build it. You could build it. I'm not going to do it. You may want to do
it and try and see how the derivative
operator looks as a matrix. And then if you really want
to spend some time thinking about it, you could try
to see if this matrix and this matrix commute
to give the right answer. And as you try it,
you will figure out all kinds of funny
things that we will talk about
later in the course. So you can represent the
momentum operator as a matrix indeed, and there are
interesting things to say about it, and
it's a good subject. So let's continue
with the momentum and ask for eigenstates
of the momentum. So eigenstates of
p, you know them. They're e to the ipx things. So let's write them with some
convenient normalization. This is an [INAUDIBLE]
wave function that depends on x
with momentum p. And we'll write it,
as a definition, as e to the ipx over h bar, and
I'll put it a 2 pi h bar here. It's kind of a
useful normalization. Then p hat on psi p of x,
well, p hat is supposed to take h over i d dx, and
take h over i d dx of psi p. And h over i cancels
the i over h. When you take the d
dx, you get p out, and you get the
same wave function. So indeed, you get
p times psi p of x. So indeed, this is the
eigenstate of the momentum operator, and it has momentum p. Well, what is the use of this? Well, say you have a
representation, what we call the position
representation, of the wave function
and operators. Let us think now of the
momentum representation. So what does all that mean? Well, there is the Fourier
transform operation in which we have psi of p. Well, let me write it
this way, actually. I'll write any psi
of x physically can be represented as a sum
of momentum eigenstates. Therefore, that's
Fourier's theorem, minus infinity to infinity
dp e to the ipx over h bar square root of 2 pi
h psi tilde of p. That's Fourier transformation,
defines psi tilde of p. And Fourier's theorem is
the fact that not only you can do that, but you can invert
it so that psi tilde of p can also be written as an
integral, this time over x from minus infinity to infinity
e to the minus ipx over h bar, also 2 pi h bar psi of x. So let's ponder this equation
for a couple of minutes. Well, as a physicist,
you think of this, well, this is telling me that
any wave function could be written as a superposition
of momentum eigenstates. Here are the
momentum eigenstates. And for each value
of momentum, you have some coefficient here
that tells me how much of that momentum eigenstate I have. Now, here is the opposite one. Psi tilde of p and psi of
x are related in this way. So these coefficients, if
you want to calculate them, you calculate them this way. But now let's think of it as
a change of representation. The physics is
contained in psi of x. All what you wish to know about
this physical system in quantum mechanics is there in psi of x. But it's also there in
psi of p, because they contain the same information. So there are different ways of
encoding the same information. What is the relation
between them? This, we thought
of it as a vector, vector in position space, an
infinite dimensional space that is talking about positions. This is another vector
in momentum space. Think of it now
the infinite line. So this is an infinite vector
with all those points little by little, from minus infinity
to plus infinity, all of them there, gigantic vector. And here is another
gigantic vector with p from minus infinity to infinity. And in between,
there's an integral. But now, with your picture
of quantum mechanics, you see an integral, but
you also see a matrix. And what is this matrix? Think of this as some
sort of psi sub p. And this as some
sort of matrix, px psi x, in which if you have a
product-- you'll remember when you multiply matrices,
a matrix on a vector, you sum over the second index. That's the product for matrix. And then the first
index is the index here. So here is what it,
more or less, is like. Psi tilde of p
[? subtend ?] by this, and this matrix depends on two
labels, p and x, and it's that. So it's a matrix full of phases. So how do you pass from the
coordinate representation of the information, a vector
of all values of the wave function in all positions? By multiplying with this
matrix of phases that is here, and it gives you
this representation. So different representations
means using different vectors to represent the physics. And this vector is
a very nice one. And because of these properties
of the momentum operator and all these things, this
vector is also a very nice one. And there's an
integral transform or some sort of infinite matrix
product that relates them. And we shouldn't be
uncomfortable about it. That's all fine. So we say that we
have, for example, psi of x as one representation
of the state and psi tilde of p as another representation
of the same physics. We can do one more thing
here, If I continue. We can take that boxed equation
on the blackboard up there and act with h bar over
i d dx on psi of x. So that is equal to
h i d dx, and I'll write what psi of x is, is
minus infinity to infinity dp e to the ipx over
h bar square root of 2 pi h bar psi tilde of p. Now, when we act on this, as
you know, h bar over i d dx just acts on this and
produces the factor of p. So this is equal to minus
infinity to infinity dp e to the ipx over h bar over
square root of 2 pi h bar p times psi tilde of p. So look at this equation again. This double arrow is
to mean that there are equivalent physics in them. They have the same information. It's the same data encoded
in a different way. And that different
way, this arrow is Fourier transformation. And this Fourier transformation
is explained here. So now you have Fourier
transformation the same way. So here we have--
what we've learned is that h over i d dx of psi is
represented in momentum space by p psi tilde of p. And this was p hat
acting on psi of x. So the corresponding thing
in momentum space of p hat acting on psi of x is p
multiplying psi tilde of p, which is to say that we can
think of the abstract operator p hat acting on psi tilde of
p as just p psi tilde of p. So in momentum space,
the operator p hat acts in a very easy way. In coordinate space,
it takes derivatives. In momentum space,
it's multiplicative. So in position space,
x is multiplicative. But in momentum space, x
would not be multiplicative. x would also be a derivative. So I leave it for
you as an exercise to show that or convince
yourself in several ways, that x hat is really i h bar d
dp in p space, in i h bar d dp. All right. So that's really all I wanted to
say about position and momentum operators at this moment. They will come back when we'll
introduce bra-ket notation in detail. We'll revisit this a little. But the main concepts
have been illustrated. Are there questions? We're about to leave
this, so if you have any questions
at this moment. Yes? AUDIENCE: Could
you explain again how you used this
[INAUDIBLE] h bar over i d dx assign to [INAUDIBLE]? PROFESSOR: Right. So the question was, why did
I associate these things? So it really goes back here to
what the meaning of this arrow is. The meaning of this arrow
is Fourier transformation. So this psi tilde and psi of
x are related in this way. That's Fourier
transformation, and that's what we mean by this arrow. It also means that whatever
physics you have here, you have it there. So really, when you have
something acting on a state, for example, if you have some
operator acting in here, well, you get a new wave function. And there should be
one on the right that corresponds to it, that has
the same information as the one in which you've
acted with something. So what we claim here is that,
also in the sense of Fourier transformation or having the
same information, h bar over i, the derivative of psi,
is encoded by this. So we say, thinking
abstractly, what is this? This is the momentum operator. Therefore, I'm going to say that
the momentum operator really is the same momentum
operator, whether it acts on wave functions that you show
them to mean this way or wave functions that, because
you're in another mood, you decide to give them
to me in momentum space. So as you change your mood, the
operator takes different forms but is doing the same thing. It's totally reversible. It's acting on that-- you
see, the operator is always the same, but you give me the
data in two different ways, then the operator has to do
the thing in a different way. So that's what it means
that the operator has different representations. In the [INAUDIBLE]
representation, it looks like a derivative. In the momentum representation,
it looks like multiplying. Other questions? Yes? AUDIENCE: So by saying
that they sort of represent [INAUDIBLE] to the
same positions, does that mean that h bar over
i p e to the xi and p psi p are like the same [INAUDIBLE]? PROFESSOR: That h bar over
d dx psi and p-- yeah. They are the same
data, the same state represented in different ways. Yeah. All right. So time for a change. We're going to talk about
Stern-Gerlach and spin. Now, spin will keep us busy the
biggest chunk of this semester. So it will be spin-1/2, and
we're really going to go into enormous detail on it. So this is just the
beginning of the story that will be elaborated
at various stages. So at this moment, I will
talk about this experiment that led to the
discovery of spin, and if you try to invent
the theory that describes this experiment, what you
would possibly begin doing. And then we go through
the mathematics, as I mentioned to you, for maybe
a week and a half or two weeks, and then return to the
spin with more tools to understand it well. So the subject is the
Stern-Gerlach experiment, Stern-Gerlach experiment. So the Stern-Gerlach experiment
was done in Frankfurt, 1922. It was an experiment
that, in fact, people were extraordinarily confused. It was not clear why
they were doing it. And for quite a
while, people didn't understand what
they were getting, what was happening with it. In fact, Pauli had thought
that the electron has like two degrees of
freedom and didn't know what it was, those
two degrees of freedom. Kronig suggested that
it had to do somehow with the rotation
of the electron. Now, Pauli said that's nonsense. How can an electron rotate
and have angular momentum because it has a rotation? It would have to rotate
so fast, even faster than the speed of light to
have the angular momentum, and then this little ball
that would be the electron would disintegrate. And it made no sense to him that
there would be such a thing. So Kronig didn't publish this. Then there were another two
people, Uhlenbeck and Goudsmit, at the same time,
around 1925, had the same idea, angular
momentum of this particle. And their advisor was
Ehrenfest, and said it doesn't make too much sense,
but you should publish it. [LAUGHTER] And thanks to their
publishing, they are given credit for discovering
the spin of the electron. And Pauli, a couple
of years later, decided, after all, I was wrong. Yes, it is spin, and
it's all working out. And 1927, five years after
the experiment basically, people understood
what was going on. So what were these
people trying to do? First, Stern and Gerlach
were atomic physicists, and they were just
interested in measuring speeds of thermal
motion of ions. So they would send
beams of these ions and put magnetic fields and
deflect them and measure their velocities. And eventually,
they were experts doing this kind of thing. And they heard of Bohr, that
said that the electron has angular momentum and is going
around the proton in circles, so it might have
angular momentum. They said, oh, if it has
angular momentum because it's going around the proton,
maybe we can detect it. And when they did the
experiment, they got something. And they said, well,
we're seeing it. But it was not that. They were not seeing the
orbital angular momentum of the electron because that
electron in these silver atoms actually has no
angular momentum, as we will see, no
orbital angular momentum. It only has spin. So they were actually
seeing the spin. So it was a big confusion. It took some time. Basically, they took the
beam, and they split it with a magnetic field,
and the clean split was something nobody understood. So they called it
space quantization, as of it's separated in space. Space is quantized. A pretty awful name, of course. There's nothing quantized
about space here. But it reflects that when
you don't know what's really happening, your names
don't come out too well. So what we have to
understand here, our goal today is
to just see what's happening in that experiment,
quantify a bit the results, and then extract the quantum
mechanical lessons from it. So let us begin with
the important thing. You don't see the spin directly. What you see is
magnetic moments. So what's that? So what are magnetic moments? Magnetic moments, mu, is the
analog, the magnetic analog of an electric dipole. A mu is called a
magnetic dipole. You say it has a
magnetic moment. And the magnetic moment is
given by I times the area. What does that mean? Well, a precise discussion
would take some time. But roughly, you
can simplify when you think of a loop that is
in a plane, in which case there's an area
associated to it. And if the loop is
this one, the area is defined as the normal
vector to the oriented loop. So an oriented loop
has an area vector. And the orientation
could be focused the direction of the current. There is some area. And the magnetic moment
is given by this thing. It points up in
the circumstances when this current
goes like that. So that's a magnetic moment. A little bit of units. The way units work out is
that mu B-- magnetic moments and magnetic fields
have units of energy. So magnetic moments
you could define as energy, which is
joules, divided by tesla, or ergs divided by
gauss, because mu B has units of energy. So how do magnetic
moments originate in a charge configuration? Well, you can simply have
a little current like that. But let's consider a
different situation in which you have
a ring of charge, a ring of charge of some radius
R. It has a total charge Q, and it has a linear
charge density lambda. It's uniform, and it's
rotating with some velocity v. If you wish, it also
has a mass M. There are all kinds of [? parameters. ?]
How many? Mass, charge,
radius, and velocity. Here we go. We have our solid ring
of charge rotating, and we want to figure out
something quite fundamental, which is the origin
of this principle. We said, you really
never see spins directly. You never see this intrinsic
angular momentum directly. You see magnetic moments. But then actually
what happens is that there's a
universal relation between magnetic moments
and angular momentum. This is a key
concept in physics. Maybe you've seen it before. Maybe you haven't. Probably you might
have seen that in 802. So how does that go? Let's calculate the
magnetic moment. So the current is the
linear charge density times the velocity. The linear charge density
is the total charge divided by 2 pi R
times the velocity. Now the area, to give
the magnetic moment, we'll have mu is equal
to I times the area. So it would be this
Q times 2 pi R v times the area, which
would be pi R squared. So the pi's cancel,
and we get 1/2 QvR. OK. 1/2 QvR, and that's
fine and interesting. But OK, depends on the radius,
depends on the velocity. So here is the magnetic moment
is supposed to be going up. But what else is going up? The angular momentum of
this thing is also going up. So what is the magnitude
of the angular momentum L? L is angular momentum. Well, it's the mass
times the momentum-- it's the mass momentum
cross R, so MvR. The momentum of R
cross p for each piece, contributes the same, so you
just take the total momentum. This really is 0, but add
them up little by little, and you've got your MvR. So here you have vR, so here
you put 1/2 Q over M MvR. And you discover that mu
is equal to 1/2 Q over M L. So maybe write it better--
Q over 2M L. I'm sorry, this is the normal. The M shouldn't change, M. And I box this relation
because an interesting thing has happened. All kinds of incidentals
have dropped out. Like the velocity
has dropped out. The radius has
dropped out as well. So if I have one
ring with this radius and another ring
with a bigger radius, the relation between
mu and L is the same, as long as it's rotating
with the same speed. So this is actually
a universal relation. It is not just true
for a little ring. It's true for a solid sphere
or any solid object axially symmetric. It would be true. You could consider any object
that is axially symmetric, and then you start considering
all the little rings that can be built. And for every ring,
mu over L is the same, and they all point in
the same direction. Therefore, it's true under
very general grounds. And that is a very
famous relation. So now you could
speculate that, indeed, the reason that a particle may
have a magnetic moment if it's made by a little ball of
charge that is rotating. But that was exactly what
Pauli didn't like, of course. And you would like
to see what's really happening with particles. So when you think of a true
quantum mechanical particle, let's think of a particle
in general, a solid particle rotating. We'll change the name to S
for spin angular momentum. Because that little part,
this is just one particle. We're not thinking of
that little particle going around a nucleus. We're thinking of that
little particle rotating. So this is a little piece
of that little particle that is rotating. So you could ask,
if, for the electron, for example, is
it true that mu is equal to e over 2 mass of
the electron times its spin? So this would be a vindication
of this classical analysis. It might be that it's
related in this way. So actually, it's
not quite true, but let's still improve
this a little bit. In terms of units, we like to
put an h bar here and a 2Me. And put spin here, angular
momentum, divided by h. Because this has no
units, h bar has the units of angular momentum, x times p. It's the same units, so
units of angular momentum. So h bar would be convenient. So that over here, you would
have units of a dipole moment, or magnetic moment,
magnetic moment units. So what does happen
for the electron? Well, it's almost
true, but not quite. In fact, what you get is
that you need a fudge factor. The fudge factor
is that, actually, for elementary
particles, you have a g, which is a constant,
which is the fudge factor, e h bar 2 over M of the
particle S over h bar. Sometimes called
the Lande factor. You must put a number there. Now, the good thing is
that the number sometimes can be calculated and predicted. So when people did
this, they figured out that for the electron the
number is actually a 2. So for the electron, g of
the electron is equal to 2. Now that, you would say,
cannot be an accident. It's twice what you would
predict sort of classically. And the Dirac equation,
the relativistic equation of the electron that
you have not studied yet but you will study soon,
predicts this g equal to 2. It was considered
a great success that that equation gave the
right answer, that people understood that this
number was going to be 2. So for the electron, this is 2. So this quantity is called--
it's a magnetic dipole moment-- is called mu
B for Bohr magneton. So how big is a mu B? It's about 9.3 times 10 to
the minus 24 joules per tesla. AUDIENCE: Professor. PROFESSOR: Yes? AUDIENCE: [INAUDIBLE]. So where exactly does
the fudge factor come in? Is it just merely
because [INAUDIBLE]? PROFESSOR: Right. So the classical
analysis is not valid. So it's pretty invalid, in fact. You see, the picture
of an electron, as of today, is that
it's a point particle. And a point particle
literally means no size. The electron is not a
little ball of charge. Otherwise, it would have parts. So an electron is
a point particle. Therefore, a point particle
cannot be rotating and have a spin. So how does the electron
manage to have spin? That you can't
answer in physics. It just has it. Just like a point particle
that has no size can have mass. How do you have mass
if you have no size? You get accustomed to the idea. The mathematics
says it's possible. You don't run into trouble. So this particle has no size,
but it has an angular spin, angular momentum, as if
it would be rotating. But it's definitely not the
case that it's rotating. And therefore, this 2 confirms
that it was a pointless idea to believe that
it would be true. Nevertheless, kind
of unit analyses or maybe some truth to the
fact that quantum mechanics changes classical mechanics. Turns out that it's
closely related. For the proton, for example, the
magnetic moment of the proton is quite complicated as well
because the proton is made out of quarks that are
rotating inside. And how do you get
the spin of the proton and the magnetic
moment of the proton? It's complicated. The neutron, that has no
charge, has a magnetic moment, because somehow the quarks
inside arrange in a way that their angular momentum
doesn't quite cancel. So for example, the value
for a neutron, I believe, is minus 2.78 or
something like that. It's a strange number. Another thing that is sort of
interesting that is also true is that this mass is
the mass of a particle. So if you're talking
about the magnetic moment of the proton or
the neutron, it's suppressed with respect to
the one of the electron. The electron one is much bigger
because, actually, the mass shows up here. So for a neutron or a
proton, the magnetic moment is much, much smaller. So, in fact, for an
electron then, you would have the following. Mu is equal to minus g, which
is 2, mu B S bar over h. And actually, we
put the minus sign because the electron
has negative charge. So the magnetic moment
actually points opposite. If you rotate this way, the
angular momentum is always up. But if you rotate this
way and you're negative, it's as if the current goes
in the other direction. So this is due to the fact
that the electron is negatively charged. And that's the final expression. So OK, so that's the general
story with magnetic moments. So the next thing is,
how do magnetic moments react when you have
magnetic fields? So that is something
that you can calculate, or you can decide if
you have a picture. For example, if you have a
loop of charge like this, and you have magnetic field
lines that go like this, they diverge a bit. Let me see you use
your right-hand rule and tell me whether
that loop of current will feel a force up or down. I'll give you 30 seconds,
and I take a vote. Let's see how we're
doing with that. And I'll prepare these
blackboards in the meantime. All right. Who votes up? Nobody. Who votes down? Yeah, [INAUDIBLE]. Down, exactly. How do you see down? Well, one way to see this,
look at the cross-section. You would have this
wire here like that. The current is coming in on this
side and going out this way. Here you have the field lines
that go through those two edges, and the magnetic
field is like that. And the force goes like I cross
B. So I goes in, B goes out. The force must be like
that, a little bit of force. In this one, I cross
B would be like that, a little bit of force. Yep. Has a component down because
the field lines are diverging. So what is the force
really given by? The force is given
by the gradient of mu dot B. This is derived in
E&M. I will not derive it here. This is not really the
point of this course. But you can see that
it's consistent. This is saying that the force
goes in the direction that makes mu dot B grow the fastest. Now mu, in this case, is up. So mu dot B is positive, because
mu and the magnetic field go in the same direction. So mu dot b is positive. So the force will be towards
the direction-- that's what the gradient is--
that this becomes bigger. So it becomes bigger here,
because as the field lines come together, that means
stronger magnetic field. And therefore, mu dot B would be
larger, so it's pointing down. If you have a magnetic
field that is roughly in the z direction, there
will be a simplification, as we will see very soon. So what did Stern
and Gerlach do? Well, they were working
with silver atoms. And silver atoms
have 47 electrons, out of which 46 fill up
the levels and equal 1, 2, 3, and 4. Just one lone electron, a 5s
electron, the 47th electron, it's a lonely electron that
is out in a spherical shell, we know now with zero
orbital angular momentum. It's an S state. And therefore, throwing silver
atoms through your apparatus was pretty much the same
thing as throwing electrons, because all these
other electrons are tied up with each other. We know now one has
spin up, one spin down. Nothing contributes, no
angular momentum as a whole. And then you have this
last electron unpaired. It has a spin. So it's like throwing spins. So moreover, throwing spins,
as far as we're concerned, Stern and Gerlach wouldn't care. Because of these relations,
it's throwing in dipole moments. And they would care about
that because magnetic fields push dipole moments up or down. Therefore, what is the
apparatus these people had? It was sort of like
this, with an oven, and you produce
some silver atoms that come out as a gas,
a collimating slit. Then you put axes
here-- we put axes just to know the components
we're talking about. And then there's magnets,
some sort of magnet like this, and the screen over there. So basically, this
form of this magnet that I've tried to draw there,
although it's not so easy, if I would take a cross-section
it would look like this. So the magnetic
field has a gradient. The lines bend a bit,
so there's a gradient of the magnetic field. And it's mostly in the z
direction, so z direction being pointed out here. So there's the magnetic field. The beam then comes here. And the question is, what
do you get on this screen? Now, I have it a little too low. The beam comes there
and goes through there. So the analysis that
we would have to do is basically an
analysis of the forces. And relatively, we
don't care too much. The fact is that
there's basically, because the magnetic field
is mostly in the z direction and varies in z
direction, there will be a force basically
in the z direction. Why is that? Because you take
this, and you say, well, that's roughly mu
z Bz, because it's mostly a magnetic field
in the z direction. And mu is a constant, so it's
basically gradient of Bz. Now, that's a vector. But we're saying also
most of the gradient of Bz is in the z direction,
so it's basically dBz dz. Now, there is some
bending of the lines, so there's a little bit of
gradient in other directions. But people have gone
through the analysis, and they don't matter for
any calculation that you do. They actually average out. So roughly, this gradient
is in the z direction. I'm sorry, the gradient is
supposed to be a vector. So you get a force
in the z direction. And therefore, the thing
that people expected was the following. You know, here comes one atom,
and it has its magnetic moment. Well, they've all been boiling
in this oven for a while. They're very disordered. Some have a z component of
magnetic-- the magnetic moment is pointing like that, so they
have some component, some down. Some are here. They have no component. It's all Boltzmann distributed
all over the directions. Therefore, you're going
to get a smudge like this. Some ones are going to be
deflected a lot because they have lots of z
component of angular momentum or z magnetic moment. Others are going to
be deflected little. So this was the
classical expectation. And the shock was that
you got, actually, one peak here, an empty
space, and another peak there. That was called
space quantization. Stern and Gerlach worked
with a magnetic field that was of about 0.1
tesla, a tenth of a tesla. And in their experiment,
the space quantization, this difference, was
1/5 of a millimeter. So not that big, but
it was a clear thing. It was there. So everybody was confused. They thought it was the orbital
spin, angular momentum that somehow had been measured. At the end of the
day, that was wrong. It couldn't have been that. People understood the Bohr
atom, realized, no, there's no angular momentum there. The idea of the spin
came back, and you would have to do a
calculation to determine what is the value of the spin. So the exact factor took
a while to get it right. But with the idea that mu z is
equal to minus 2 Bohr magenton Sz over h bar, which
we wrote before. Well, mu z, if you
know the strength of your magnetic field, you
can calculate the deflections. You know what mu B is. So therefore, you get
the value for Sz over h. And experiments
suggested that Sz over h was either plus or minus 1/2. And this kind of particle, it
has Sz over h bar equal plus or minus 1/2, is called
the spin-1/2 particle. So again, from this equation,
this can be measured. And you then use this,
and you get this value. So the experiment is
a little confusing. Why did this happen? And how do we think of
it quantum mechanically? Now 804 sort of began
with these kind of things. And you know by now
that what's happening is the following, that somehow,
mathematically, every state is a superposition of a
spin up and a spin down. So every particle
that goes there has half of its brain in the
spin up and half of its brain in the spin down. And then as it goes
through the magnetic field, this thing splits, but each
particle is in both beams still. And they just have
this dual existence until there's a screen
and there's detectors. So they have to decide
what happens, and then either collapses in the
top beam or lower beam. Nothing happens until
you put the screen. That's what we think now
is the interpretation of this experiment. But let's use the
last few minutes to just write this in terms of
boxes and get the right ideas. So instead of drawing
all that stuff, we'll draw a little
box called a z hat box, a Stern-Gerlach apparatus. In comes a beam, out
would come two beams, Sz equal h bar over 2 and
Sz equal minus h bar over 2. And the convention is
that the plus goes up and the minus goes
down, which I think is probably consistent
with that drawing. And that's the
Stern-Gerlach apparatus. It measures Sz, and
it splits the beam. Each particle goes
into both beams until there's a
device that measures and decides where you go. So you can do the
following arrangements. So here's arrangement number 1,
a Stern-Gerlach device with z. Then you block the
lower one and let the top one go as Sz
equal h bar over 2. And then you put another
Stern-Gerlach machine, z hat, that has two outputs. And then you ask,
what's going to happen? And the experiment can be
done and, actually, there's nothing here coming out, and
all the particles come out here with Sz equal h bar over 2. What are we going
to learn from this? In our picture of
quantum mechanics, we're going to think
of this as there are states of the
electron that have-- and I will write
them with respect to z-- they have
plus h bar over 2 and states that have
minus h bar over 2. And what we will think is that
these are really old basis states, that any other state,
even one that points along x, is a superposition of those two. This is a very incredible
physical assumption. It's saying this system is a
2-dimensional complex vector space, two vectors, two
unit, two basis vectors. And from those two, all linear
combinations that are infinite represent all possible
spin configurations. And what is this saying? Well, as we will
translate it into algebra, we will say that, look,
here is a state plus. And when you try to measure,
if it had any minus component, it had nothing. So we will state that as
saying that these states are orthogonal. The minus state and the plus
state have zero overlap. They are orthogonal
basis states. And, for example, well, you
could also do it this way. That would also be 0. And you could also
say that z plus and z plus is 1, because
every state that came in as a plus came out as a plus. They had perfect overlap. So these are two
orthonormal basis vectors. That's what this
seems to suggest. And it's a little
strange, if you think, because there's a
clash between arrows and the notion of
orthonormality. In 3-dimensional
vectors, you think of this vector being
orthogonal to this. But you wouldn't
think of this vector as being orthogonal to that one. And here is the spin is up,
and this is the spin down. And those two are orthogonal. You say, no, they're
anti-parallel. They're not orthogonal. No, they are orthogonal. And that's the endlessly
confusing thing about spin-1/2. So these states, their pictures
of the spins are arrows. But don't think that those
arrows and the dot product give you the orthogonality,
because this is up and down. If you would be doing the
dot product of an up and down vector, you would not get 0. But this is 0. Then you do the
following experiment. So let's do the next one. And the next one is,
again, the z filter. Take this one, block it. Then you put an x filter. And what actually happens
is that you would get states with Sx, now, h bar over 2 and
Sx equal minus h bar over 2, because it's an x filter. The magnetic field is a
line in the x direction. Now, all these things have
Sz equal h bar over 2. And what happens
in the experiment is that 50% of the
particles come out here and 50% come out there. So a spin state
along the x direction has some overlap with a spin
state along the z direction. Normal vectors, a z vector and
an x vector, are orthogonal. Not here for spins. The spin pointing in the z
and the spin pointing in the x are not orthogonal states. They have overlaps. So this means that, for example,
the x plus state and the z plus state have an overlap. This is notations that-- we're
going to be precise later. But the same thing with the x
minus state, it has an overlap, and somehow they're
about the same. Finally, the last experiment
is this, z hat, block again, x hat, but this time block one. So here is a state with Sx
equals minus h bar over 2. Here is a state with
Sz equal h bar over 2. And now you put the
z machine again. And what happens? Well, there's two options. People who were inventing
quantum mechanics no wonder thought about them. Here they could say, look,
I filtered this thing, and now all these electrons
have Sz equal h bar over 2. And now all these electrons have
Sx equal minus h bar over 2. Maybe, actually,
all these electrons have both Sz equal
h over 2 and that because I filtered it twice. So it maybe satisfies both. So if all these electrons
would have Sz equals h over 2 and this,
then you would only get something from the top one. But no, that's not what happens. You get in both. So somehow, the memory of
these states coming from Sz equals h over 2
has been destroyed by the time it
turned into a state with Sx equal minus h over 2. And a state cannot have
simultaneously this and that. That's two properties, because
you get 50% here and 50% there. So we'll discuss next
time a little more about these relations
and how can the states be related, the ones
that we use as the basis vectors and all the
others along x and others that we could build
some other way. All right. See you next week. There's office hours today, 5:00
to 6:00, Monday, 4:30 to 5:30.
