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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Last time we
spoke about the Stern-Gerlach experiment, and how you
could have a sequence of Stern-Gerlach boxes that
allow you to understand the type of states and
properties of the physics having to do with spin-1/2. So the key thing in the
Stern-Gerlach machine was that a beam of silver
atoms, each of which is really like an electron
with a magnetic moment, was placed in an inhomogeneous
strong magnetic field, and that would classically mean
that you would get a deflection proportional to the z-component
of the magnetic moment. What was a surprise
was that by the time you put the screen
on the right side, it really split into
two different beams, as if the magnetic
moments could either be all the way pointing in
the z-direction or all the way down, pointing opposite
to the z-direction, and nothing in between. A very surprising result. So after looking at
a few of those boxes, we decided that we would try
to model the spin-1/2 particle as a two-dimensional
complex vector space. What is the two-dimensional
complex vector space? It's the possible space of
states of a spin-1/2 particle. So our task today to go
into detail into that, and set up the whole
machinery of spin-1/2. So we will do so, even though
we haven't quite yet discussed all the important
concepts of linear algebra that we're going to need. So today, I'm going to
assume that at least you have some vague notions
of linear algebra reasonably well understood. And if you don't, well,
take them on faith today. We're going to go
through them slowly in the next couple
of lectures, and then as you will reread
this material, it will make more sense. So what did we have? We said that the spin states,
or the possible states of this silver atom, that really
correspond to an election, could be described by states z
comma plus and z colon minus So these are the two states. This state we say corresponds
to an angular momentum Sz hat. Sz-- I can put it like that-- of h-bar over 2, and
this corresponds to Sz equals minus h-bar over 2. And those are our two states. The z label indicates that
we've passed, presumably, these atoms through a
filter in the z-direction, so that we know
for certain we're talking about the z-component of
angular momentum of this state. It is positive, and
the values here again have the label z to remind us
that we're talking about states that have been organized using
the z-component of angular momentum. You could ask whether this state
has some angular momentum-- spin angular momentum--
in the x-direction or in the y-direction,
and we will be able to answer that
question in an hour from now. So mathematically, we
say that this statement, that this state, has Sz equals
h-bar over 2 means that there is an operator, Sz hat-- hat for operators. And this operator, we
say, acts on this state to give h-bar over
2 times this state. So when we have a measurement
in quantum mechanics, we end up talking
about operators. So this case is no exception. We think of the operator,
Sz, that acts in this state and gives h-bar over 2. And that same operator,
Sz, acts on the other state and gives you minus h-bar
over 2 times the state. You see, an operator on a
state must give a state. So in this equation, we
have a state on the right, and the nice thing is
that the same state appears on the right. When that happens, you
say that the state is an eigenstate of the operator. And, therefore,
the states z plus, minus are eigenstates of the
operator Sz with eigenvalues-- the number that appears here-- equal to plus, minus h over 2. So the relevant
physical assumption here is the following, that
these two states, in a sense, suffice. Now, what does that mean? We could do the experiment again
with some Stern-Gerlach machine that is along the
x-axis, and say, oh, now we've got states
x plus and x minus and we should add them there. They are also part of the
possible states of the system. Kind of. They are parts of the
possible states of the system. They are possible
states of the system, but we shouldn't add
them to this one. These will be thought
as basis states. Just like any vector
is the superposition of a number times
the x-unit vector plus a number times
the y-unit vector and a number times
the z-unit vector, we are going to
postulate, or try to construct the theory
of spin, based on the idea that all possible spin
states of an electron are obtained by suitable linear
superposition of these two vectors. So, , in fact, what we're
going to say is that these two vectors are the basis of a
two-dimensional vector space, such that every possible state
is a linear superposition. So psi, being any
possible spin state, can be written as some constant,
C1 times z plus plus C2 times z minus where these
constants, C1 and C2 belong to the complex numbers. And by this, we mean that
if any possible state is a superposition like that,
the set of all possible states are the general vectors in a
two-dimensional complex vector space. Complex vector space, because
the coefficients are complex, and two-dimensional, because
there's two basis vectors. Now this doesn't quite
look like a vector. It looks like those
things called kets. But kets are really
vectors, and we're going to make the
correspondence very clear. So this can be called
the first basis state and the second basis state. And I want you to realize
that the fact that we're talking about the complex
vector space really means these coefficients
are complex. There's no claim that the
vector is complex in any sense, or this one. They're just vectors. This is a vector, and
it's not that we say, oh this vector is complex. No. A complex vector space, we
think of as a set of vectors, and then we're allowed
to multiply them by complex numbers. OK, so we have this, and this
way of thinking of the vectors is quite all right. But we want to be more concrete. For that, we're
going to use what is called a representation. So I will use the
word representation to mean some way of exhibiting
a vector or state in a more concrete way. As something that any one
of us would call a vector. So as a matter of notation,
this being the first basis state is sometimes written
as a ket with a 1. Like that. And this being this
second basis state is sometimes written this way. But here is the real
issue of what we were calling a representation. If this is a two-dimensional
vector space, you're accustomed to
three-dimensional vector space. What are vectors? They're triplets of numbers. Three numbers. That's a vector. Column vectors, it's perhaps
easier to think about them. So column vectors. So here's what
we're going to say. We have this state z plus. It's also called 1. It's just a name, but
we're going to represent it as a column vector. And as a column vector,
I'm going to represent it as the column vector 1, 0. And this is why I put
this double arrow. I'm not saying it's
the same thing-- although the really it is-- it's just a way of
thinking about it as some vector in what we
would call canonically a vector space. Yes AUDIENCE: So do the components
of the column vector there have any
correspondence to the actual. Does it have any basis in the
actual physical process going on? Or, what is their connection to
the actual physical [INAUDIBLE] represented here? PROFESSOR: Well, we'll
see it in a second. It will become a little clearer. But this is like saying, I have
a two-dimensional two0 vector space, so I'm going to think of
the first state as this vector. But how do I write this vector? Well, it's the vector ex. Well, if I would write
them in components, I would say, for a vector, I can
put two numbers here, a and b. And this is the a-component
and b-component. So here it is, ex would be 1, 0. And ey would be 0, 1. If I have this notation
then the point a, b is represented by a
and b as a column vector. So at this moment,
it's just a way of associating a vector in
the two-dimensional canonical vector space. It's just the column here. So the other state, minus-- it's also called 2-- will be represented by 0 1 1. 1 And therefore,
this state, psi, which is C1 z plus
plus C2 z minus will be represented as C1 times
the first vector plus C2 times the second vector. Or multiplying, in C1, C2. So this state can be written
as a linear superposition of these two basis
vectors in this way-- you can write it this way. You want to save
some writing, then you can write them with 1 and 2. But as a vector, it's
represented by a column vector with two components. That's our state. Now in doing this,
I want to emphasize, we're introducing the
physical assumption that this will be enough to
describe all possible spin states, which is far from
obvious at this stage. Nevertheless, let's
use some of the ideas from the experiment, the
Stern-Gerlach experiment. We did one example of a box
that filtered the plus z states, and then put it against
another z machine, and then all the states
went through the up. Which is to say that
plus states have no amplitude, no probability
to be in the minus states. They all went through the plus. So when we're going to introduce
now the physical translation of this fact, as saying
that these states are orthogonal to each other. So, this will require the
whole framework, in detail, of bras and kets to
say really, precisely-- but we're going to do
that now and explain the minimum necessary
for you to understand it. But we'll come back to it later. So this physical
statement will be stated as z minus with z plus. The overlap, the
bra-ket of this, is 0. The fact that all particles
went through and went out through the plus output
will state to us, well, these states are
well normalized. So z plus, z plus is 1 Similarly, you could have
blocked the other input, and you would have concluded
that the minus state is orthogonal to the plus. So we also say that these,
too, are orthogonal, and the minus states
are well normalized. Now here we had to
write four equations. And the notation, one
and two becomes handy, because we can summarize
all these statements by the equation Ij
equals delta Ij. Look, if this equation
is 2 with 1 equals 0. The bra 2, the ket 1. This is 1 with 1 is equal to 1. Here is is 1 with 2 is equal to
0, and 2 with 2 is equal to 1. So this is exactly
what we have here. Now, I didn't define for
you these so-called bras. So by completeness, I
will define them now. And the way I will define
them is as follows. I will say that while for
the one vector basis state you associate at 1, 0, you will
associate to one bra, the row vector 1, 0. I sometimes tend to
write equal, but-- equal is all right-- but
it's a little clearer to say that there's arrows here. So we're going to
associate to 1, 1, 0-- we did it before-- but now to the bra, we
think of the rho vector. Like this. Similarly, I can
do the same with 2. 2 was the vector 0, 1. It's a column vector,
so 2 was a bra. We will think of it as
the row vector 0, 1. We're going to do this now
a little more generally. So, suppose you have
state, alpha, which is alpha 1, 1 plus alpha 2, 2. Well, to this, you would
associate the column vector alpha 1, alpha 2. Suppose you have a beta state,
beta 1, 1 plus beta 2, 2. You would associate beta 1,
beta 2 as their representations. Now here comes the
definition for which this is just a special case. And it's a definition
of the general bra. So the general bra here, alpha,
is defined to be alpha 1*, bra of the first, plus
alpha 2*, bra of the second. So this is alpha 1*
times the first bra, which we think of it as 1, 0,
plus alpha 2* times the second bra, which is 0, 1. So this whole thing is
represented by alpha 1*, alpha 2*. So, here we've had a column
vector representation of a state, and the bra is
the row vector representation of the state in which
this is constructed with complex conjugation. Now these kind of
definitions will be discussed in more detail
and more axiomatically early very soon, so that
you see where you're going. But the intuition that
you're going to get from this is quite valuable. So what is the bra-ket? Alpha-beta is the
so-called bra-ket. And this is a number. And the reason for
complex conjugation is, ultimately, that when
these two things are the same, it should give a
positive number. It's like the length squared. So that's the reason for
complex conjugation, eventually. But, for now, you are supposed
to get a number from here. And the a reasonable way
to get a number, which is a definition, is
that you get a number by a matrix multiplication
of the representatives. So you take the representative
of alpha, which is alpha 1*, alpha 2*. And do the matrix product with
the representative of beta, which is beta 1, beta 2. And that's alpha 1*, beta
1 plus alpha 2*, beta 2. And that's the number
called the inner product, or bra-ket product. And this is the true meaning
of relations of this kind. If you're given an
arbitrary states, you compute the inner
product this way. And vectors that satisfy
this are called orthonormal because they're orthogonal
and normal with respect to each other in the
sense of the bra and ket. So this definition,
as you can see, is also consistent with
what you have up there, and you can check it. If you take I with
j, 1, say, with 2-- like this-- you do the inner
product, and you get 0. And similarly for
all the other states. So let's then complete the
issue of representations. We had representations of the
states as column vectors-- two by two column
vectors or row vectors. Now let's talk about this
operator we started with. If this is an operator,
acting on states, now I want to think of
its representation, which would be the way it acts on
these two component vectors. So it must be a two by two
matrix, because only a two by two matrix acts naturally
on two component vectors. So here is the
claim that we have. Claim, that Sz hat
is represented-- but we'll just put equal-- by this matrix. You see, it was an operator. We never talked about matrices. But once we start
talking about the basis vectors as column vectors, then
you can ask if this is correct. So for example, I'm
supposed to find that Sz hat acting
on this state 1 is supposed to be h-bar
over 2 times the state 1. You see? True. Then you say, oh let's put the
representation, h-bar over 2, 1 minus 1, 0, 0. State one, what's
its representation? 1, 0. OK, let's act on it. So, this gives me h-bar over 2. I do the first
product, I get a 1. I do the second
product, I get a 0. Oh, that seems
right, because this is h over 2 times the
representation of the state 1. And if I check this, and as
well that Sz on 2 is equal minus h-bar over 2, 2-- which can also be checked-- I need to check no more. Because it suffices
that this operator does what it's supposed to
do of the basis vectors. And it will do
what it's supposed to do on arbitrary vectors. So we're done. This is the operator
Sx, and we seem to have put together a lot of
the ideas of the experiment into a mathematical framework. But we're not through because
we have this question, so what if you align and
operate the machine along x? What are the possible spin
states along the x-direction? How do you know that
all that the spins state that points along x can be
described in this vector space? How do I know there
exists a number C1, C2 so that this linear
combination is a spin state that points along x. Well, at this moment, you
really have to invent something. And the process of invention
is never a very linear one. You use analogies-- you
use whatever you can-- to invent what you need. So, given that
that's a possibility, we could follow what Feynman
does in his Feynman lectures, of discussing how
to begin rotating Stern-Gerlach machines, and
doing all kinds of things. It's an interesting argument,
and it's a little hard to follow, a little
tedious at points. And we're going to
follow a different route. I'm going to assume that you
remember a little about angular momentum, and I think you
do remember this much. I want to say, well, this
is spin angular momentum. Well, let's compare it with
orbital angular momentum, and see where we are. You see, another way of
asking the question would be, well, what are the
operators Sx and Sy. Where do I get them? Well, the reason I want to
bring in the angular momentum is because there you have Lz,
but you also have Lx and Ly. So angular momentum had
Lz, just like we had here, but also Lx and Ly. Now these spin things look a
lot more mysterious, a lot more basic, because, like Lz,
it was xpy minus ypx. So you knew how this operator
acts on wave functions. You know, it multiplies by
y, takes an x derivative, or it's a dd phi. It has a nice thing, but Sz on
the other hand, there's no x, there's no derivatives. It's a different space. It's working in a
totally different space, in the space of a
two-dimensional complex vector space of column vectors
with two numbers. That's where it acts. I'm sorry there's no dd x,
nothing familiar about it. But that's what we
have been handed. So this thing acts
on wave functions, and thus natural things. Well, the other one
acts on column vectors. Two-by-two-- two
component column vectors, and that's all right. But we also know
that Lz is Hermitian. And that was good,
because it actually meant that this is good observable. You can measure it. Is Sz Hermitian? Well, yes it is. Hermeticity of a matrix-- as we'll discuss it
in a lot of detail, maybe more than you want-- means you can
transpose it complex conjugated, and you
get the same matrix. Well that matrix is Hermitian. So that's nice. That maybe is important. So what other
operators do we have? Lx and Ly. And if we think of Lx as
L1, Ly as L2, and Lz as L3, you had a basic
computation relation. Li with Lj was equal to
i epsilon ijk Lk-hat-- oops i-hbar. And this was called the
algebra of angular momentum. These three operators
satisfy these identities. i and j are here, k is
supposed to be summed over-- repeated in this is our
sum from 1, 2, and 3. And epsilon ijk is totally
anti-symmetric with epsilon 1, 2, 3 equal to plus 1. You may or may not
know this epsilon. You will get some practice
on that very soon. Now for all intents
and purposes, we might as well write
the explicit formulas between Lx, Ly equal i-hbar Lz. Ly Lz equals i-hbar Lx. And Lz Lx-- there
are hats all over-- equal i-hbar Ly. So we had this for
orbital angular momentum, or for angular
momentum in general. So what we're going
to do now is we're going to try to figure out
what are Sx and Sy by trying to find a complete analogy. We're going to declare that S
is going to be angular momentum. So we're going to want that
Sx with Sy will be i-hbar Sz. Sy with Sz will be i-hbar Sx. And finally, Sz with
Sx is i-hbar Sy. And we're going to try that
these things be Hermitian. Sx and Sy. So let me break for a second
and ask if there are questions. We're aiming to
complete the theory by taking S to be angular
momentum, and see what we get. Can we invent
operators Sx and Sy that will do the right thing? Yes. AUDIENCE: What's the
name for the epsilon ijk? I know there's a special name. Levi-Civita? Levi-Civita. Yeah. PROFESSOR: What's the name? AUDIENCE: Levi-Civita tensor. PROFESSOR: That's right. Levi-Civita tensor. It can be used for
cross products. It's very useful
for cross products. It's a really useful tensor. Other questions. More questions about
what we're going to try to do, or this so far. Yes. AUDIENCE: When you use
the term representation, is that like the technical
mathematical term of representation,
like in algebra? PROFESSOR: Yes. It's representation of
operators in vector spaces. So we've used the
canonical vector space with column vectors represented
by entries one and numbers. And then the operators
become matrices, so whenever an operator
is viewed as a matrix, we think of it as
a representation. Other questions. Yes. AUDIENCE: Will we
talk about later why we can make an
analogy between L and S? Or is it [INAUDIBLE]? PROFESSOR: Well you see, this
is a very strong analogy, but there will be
big differences from orbital angular momentum
and spin angular momentum. And basically having
to do with the fact that the eigenvalues
of these operators are plus minus h-bar over 2. And in the orbital
case they tend to be plus minus
integer values of h-bar. So this is a very deep
statement about the algebra of these operators that still
allows the physics of them to be quite different. But this is probably the only
algebra that makes sense. It's angular momentum. So we're going to try to
develop that algebra like that, as well here. You could take it
to be an assumption. And as I said, an experiment
doesn't tell you the unique way to invent the mathematics. You try to invent the
consistent mathematics and see if it coincides
with the experiment. And this is a very natural
thing to try to invent So what are we facing? We're facing a slightly
nontrivial problem of figuring out these operators. And they should be Hermitian. So let's try to think of
Hermitian two-by-two matrices. So here is a Hermitian
two-by-two matrix. I can put an arbitrary
constant here because it should be invariant on
their transposition, which doesn't change this diagonal
value in complex conjugation. So c should be real. d should be real. For the matrix to be
Hermitian, two-by-two matrix, I could put an a here. And then this a would have
to appear here as well. I can put minus ib, and then
I would have plus ib here. So when I transpose a complex
conjugate, I get this one. So this matrix with abc
and d real is Hermitian. Hermiticity is some sort
of reality condition. Now, for convenience, I
would put a 2c and a 2d here. It doesn't change
things too much. Now to look at what
we're talking about. We're talking about this
set of Hermitian matrices. Funnily, you can think of
that again as a vector space. Why a vector space? Well, we'll think about
it, and in a few seconds, it will become clear. But let me just try
to do something here that might help us. We're trying to identify
Sx and Sy from here so that this commutation
relations hold. Well, if Sx and Sy have anything
to do with the identity matrix, they would commute
with everything and would do nothing for you. So, I will remove from
this matrices then trying to understand something having
to do with the identity. So I'll remove a
Hermitian matrix, which is c plus d
times the identity-- the two-by-two identity matrix. This is a Hermitian
matrix, as well. And I can remove it, and then
this matrix is still Hermitian, and this piece that I've removed
doesn't change commutators as they appear on
the left hand side. So if you have an
Sx and an xy here, and you're trying
to do a computation, it would not contribute,
so you might as well just get rid of them. So if we remove this,
we are left with-- you're subtracting c
plus d from the diagonal. So here you'll have c minus d. Here you'll get b minus c,
a minus ib, and a plus ib. And we should keep searching for
Sx and Sy among these matrices. But then you say,
look, I already got Sz, and that was Hermitian. And Sz was Hermitian,
and it had a number, and the opposite number on
the other diagonal entry. If Sx and Sy have a little
bit of Sz, I don't care. I don't want these to
be independent matrices. I don't want to
confuse the situation. So if this thing has something
along Sz, I want it out. So since precisely this number
is opposite to this one, I can add to this matrix
some multiple of Sz and kill these things
in the diagonal. So add the multiple
and Sz multiple, and we finally get this matrix. 0, a minus ib, a plus ib, and 0. So we've made quite
some progress. Let's see now what we have. Well, that matrix could be
written as a times 0, 1, 1, 0 plus b times 0, minus i, i, 0. Which is to say that it's
this Hermitian matrix times a real number, and
this Hermitian matrix times a real number. And that makes sense because
if you take a Hermitian matrix and multiply by a real number,
the matrix is still Hermitian. So this is still Hermitian
because these are real. This is still Hermitian
because a is real, and if you add Hermitian
matrices, it's still Hermitian. So in some sense, the set
of Hermitian matrices, two-by-two Hermitian matrices,
is a real vector space with four basis vectors. One basis vector is this,
another basis vector is this, the third basis
vector is the Sz part, and the fourth basis
vector is the identity that we subtracted. And I'm listing the
other two that we got rid of because physically
we're not that interested given that we want Sx and Sz. So, Sx and Sy. But here it is. These four two-by-two
matrices are sort of the linearly
independent Hermitian matrices. You can think of them as
vectors, four basis vectors. You multiply by real numbers,
and now you add them, and you got the most
general Hermitian matrix. So this is part of the subtlety
of this whole idea of vector spaces of matrices, which
can be thought of as vectors sometimes, as well. So that's why these
matrices are quite famous. But before we just discuss
why they are so famous, let's think of this. Where we're looking
for Sx and Sy, and we actually seem to
have two matrices here that could do the job, as
two independent Hermitian two-by-two matrices. But we must add a little
extra information. We don't know what the scale is. Should I multiply this
by 5 and call that Sx? Or this by 3? We're missing a
little more physics. What is the physics? The eigenvalues of Sx should
also be plus minus h over 2. And the eigenvalues of Sy should
also be plus minus h over 2. Just like for Sz. you could have started the
whole Stern-Gerlach things thinking of x,
and you would have obtained plus minus h over 2. So that is the
physical constraint. I have to figure
out those numbers. Maybe Sx is this one,
as y is this one. And you can say, oh,
you never told us if you're going to get
the unique answer here. And yes, I did tell
you, and you're not going to get a unique answer. There are some sign notations
and some other things, but any answer is
perfectly good. So once you get an answer,
it's perfectly good. Of course, we're going
to get the answer that everybody likes. And the convention is that
happily that everybody uses this same convention. Questions. AUDIENCE: So I have
a related question, because at the beginning we
could have chosen the top right entry to be a plus ib and the
bottom left to be a minus ib and that would have yielded
a different basis matrix. PROFESSOR: Right, I would have
called this plus and minus. Yes. AUDIENCE: Are we going to show
that this is the correct form? PROFESSOR: No, it's
not the correct form. It is a correct form, and it's
equivalent to any other form you could find. That's what we can show. In fact, I will
show that there's an obvious ambiguity here. Well, in fact, maybe I can
tell it do you, I think. If you let Sx go to minus
Sy, and Sy goes to plus Sx, nothing changes in
these equations. They become the same equations. You know, Sx would become
minus Sy, and this Sx-- this is not changed. But, in fact, if you
put minus Sy and Sx as the same commutator
then this one will become actually
this commutator, and this one will become that. So I could change
whatever I get for Sx, change it from minus
Sy, for example, and get the same thing. So there are many
changes you can do. The only thing we need
is one answer that works. And I'm going to write,
of course, the one that everybody likes. But don't worry about that. So let's think of eigenvectors
and eigenvalues now. I don't know how much
you remember that, but we'll just take it at
this moment that you do. So 0, 1, 1, 0 has
two eigenvalues, and lambda equals 1,
with eigenvector 1 over square root of 2, 1, 1. And a lambda equals minus
1 with eigenvector 1 over square root
of 2, 1, minus 1. The other one, it's
equally easy to do. We'll discuss eigenvectors
and eigenvalues later. Minus i, i, 0, 0, plus a
lambda equals one eigenvector, with components 1 over
square root of 2, 1, and i. I'm pretty sure it's 1 and i. Yes, and a lambda
equals minus 1, with components 1 over
square root 2, 1, minus i. Now I put the 1 over
square root of 2 because I wanted them
to be normalized. Remember how you're supposed
to normalize these things. You're supposed to take the
row vector, complex conjugate, and multiply. Well, you would get 1
for the length of this, 1 for the length of this. You would get one for
the length of this, but remember, you have
to complex conjugate, otherwise you'll get 0. Also, you will get one
for the length of this. So these are our eigenvalues. So actually, with eigenvalues
lambda equals 1 and minus 1 for these two, we're
in pretty good shape. We could try Sx to be h-bar
over 2 times 0, 1, 1, 0. And Sy to be h-bar over
2, 0, minus i, i, 0. These would have the
right eigenvalues because if you multiply
a matrix by a number, the eigenvalue gets
multiplied by this number, so the plus minus 1s become
plus minus h-bar over 2. But what are we was
supposed to check? If this is to work,
we're supposed to check these commutators. So let's do one, at least. Sx commutator with Sy. So what do we get? h-bar over
2, h-bar over 2-- two of them-- then the first matrix, 0, 1,
1, 0 times 0, minus i, i, 0, minus 0, minus i, i,
0 times 0, 1, 1, 0. Which is h-bar over 2 times
h-bar over 2, times i, 0, 0, minus i, minus minus i, 0, 0, i. And we're almost there. What do we have? Well, we have h-bar
over 2, h-bar over 2. And we've got 2i and minus 2i. So this is h-bar over 2 times
h-bar over 2, times 2i times 1, minus 1. And this whole thing is i
h-bar, and the other part is h-bar over 2, 1, minus 1,
which is i h-bar as z-hat. Good, it works. You know, the only
thing that could have gone wrong-- you could
have identified 1 with a minus, or something like that It
would have been equally good. Once you have these
operators, we're fine. So one has to check that the
other ones work, and they do. I will leave them
for you to check. And therefore, we've
got the three matrices. It's a very important result-- the Sx Sy, and Sz. I will not rewrite them, but
they should be boxed nicely, the three of them together,
with that one there on top of the blackboard. And of course by construction,
they're Hermitian. They're famous enough
that people have defined the following object. Si is defined to
be h-bar over 2, sigma i, the power
of the matrix sigmas. And these are Pauli matrices,
sigma one is 0, 1, 1, 0. Sigma two is 0, minus i, i, 0. And sigma three is equal
to 1, minus 1, 0, 0. OK, so in principle-- yes, question. AUDIENCE: Is it
at all significant that the Pauli matrices are
all squared [INAUDIBLE]?? PROFESSOR: Yes,
it is significant. We'll use it, but at this
moment, it's not urgent for us. We'll have no application
of that property for a little while,
but it will help us do a lot of the algebra
of the Pauli matrices. AUDIENCE: [INAUDIBLE]
eigenvalues, right? PROFESSOR: Sorry? AUDIENCE: Doesn't that follow
from the eigenvalue properties that we've [INAUDIBLE]
plus or minus one. Because those were both squared. [INAUDIBLE] PROFESSOR: That's right. I think so. Our eigenvalues--
yes, it's true. That the fact that the
eigenvalues are plus minus 1 will imply that these
matrices squared themselves. So it's incorporated
into our analysis. The thing that I
will say is that you don't need it in the
expression of the commutators. So in the
commentators, it didn't play a role to begin with. Put it as an extra condition. Now what is the next thing
we really want to understand? Is that in terms
of plain states, we now have the answer for most
of the experiments we could do. So in particular,
remember that we said that we would have Sx,
for example, having states x plus minus, which
are h-bar over 2 plus minus, x comma plus minus. The states along the
x-direction referred like that would be the
eigenstates of the Sx operator. But we've calculated the
states of the Sx operator-- they're here. The Sx operator is h-bar
over 2 times this matrix. And we have those things. So the plus eigenvalue
and the minus eigenvalue will just show up here. So let me write them, and
explain, in plain language, what these states are. So the eigenstate
with lambda equal 1-- that would correspond to
h-bar over two-- so the x plus corresponds to this vector. So what is that state? It's that vector which, if
you want more explicitly, it's the z plus, plus z minus. This is the state 1 over
square root of 2, 1, 1. The x minus is z
plus, minus z minus. As you see on that
blackboard, it's 1 minus 1. So here it is. The states that you were looking
for, that are aligned along x-- plus x or minus x--
are not new states that have you to add
to the state space. They are linear combinations
of the states you've got. We can invert this formula
and write, for example, that z plus is 1 over
square root of 2, x plus, plus 1 over square
root of 2, x minus. And z minus is 1 over square
root of 2, x plus, minus 1 over square root of-- minus the square root of 2
is already out, I'm sorry-- minus x minus. So actually, this answers
the question that you had. For example, you
put a z plus state, and you put an x filter-- what amplitude do you
have to find a state in the x plus,
given that you start with a state on the z plus? Well, you put an
x plus from here. You get 1 from this
and 0 from this one because the states
are always orthogonal. The states are orthogonal-- you should check that. And therefore, this is
1 over square root of 2. If you ask for x minus
with respect to z plus, that's also 1 over
square root of 2. And these are the
amplitudes for this state to be found in this, for this
state to be found in them. They're equal. The probabilities are 1/2. And that's good. Our whole theory
of angular momentum has given us something
that is perfectly consistent with the
Stern-Gerlach experiment, and it gives you
these probabilities. You can construct in the
same way the y states. So the y states
are the eigenstates of that second matrix, Sy,
that we wrote on the left. So this matrix is Sy,
so its eigenstates-- I'm sorry, Sy is there. Sy is there. The eigenstates are
those, so immediately you translate that to say
that Sy has eigenstates y plus minus, whose eigenvalues
are plus minus h-bar over 2y plus minus. And y plus is equal 1 over
square root of 2, z plus-- and look at the
first eigenvector-- plus iz minus. And, in fact, they can
put one formula for both. Here they are. So, it's kind of
neat that the x1s were found by
linear combinations, and they're orthogonal. Now, if you didn't
have complex numbers, you could not form
another linear combination of this orthogonal. But thanks to these
complex numbers, you can put an i there--
there's no i in the x ones-- and the states are
orthogonal, something that you should check. So again, you can
invert and find the z states in
terms of y, and you would conclude that the
amplitudes are really the same up to signs, or
maybe complex numbers, but the probabilities
are identical. So we've gotten a long way. We basically have
a theory that seems to describe the whole result of
the Stern-Gerlach experiment, but now your theory
can do more for you. Now, in the last
few minutes, we're going to calculate
the states that are along arbitrary directions. So here I produced a
state that is along the x-direction plus, and
along the x-direction minus. What I would like to construct,
to finish this story, is a state that is along
some arbitrary direction. So the state that points
along some unit vector n. So here is space, and here's a
unit vector n with components nx, , ny, and nz. Or you can write the vector n
as nx ex plus ny ey plus nz ez. And I would like
to understand how I can construct, in general,
a spin state that could be said to be in the n direction. We have the ones
along the z, x, and y, but let's try to get something
more general, the most general one. So for this, we think of the
triplet of operators S, which would be Sx, Sy, and Sz. Now you can, if you wish,
write this as Sx-hat ex vector, plus Sy-hat ey vector,
plus Sz hat ez vector. But this object, if
you write it like that, is really a strange object. Think of it. It's matrices, or operators,
multiplied by unit vectors. These vectors have nothing
to do with the space in which the matrices act. The matrices act in an abstract,
two-dimensional vector space, while these vectors are sort
of for accounting purposes. That's why we sometimes
don't write them, and say we have a triplet. So this product
means almost nothing. They're just sitting together. You could put the e to the
left of the x or to the right. It's a vector. You're not supposed to put
the vector inside the matrix, either. They don't talk to each. It's an accounting procedure. It is useful
sometimes; we will use it to derive identities
soon, but it's an accounting procedure. So here's what I want to define. So this is a crazy thing, some
sort of vector valued operator, or something like that. But what we really
need is what we'll call S-hat n, which
will be defined as n dot S. Where we take
naively what a dot product is supposed to mean. This component times
this component, which happens to be an operator. This times this,
this times that. nx Sx plus ny Sy, plus nz Sz. And this thing is
something very intuitive. It is just an operator. It doesn't have anymore
a vector with it. So it's a single operator. If your vector points
in the z-direction, nx and ny z, and you have Sz
because it's a unit vector. If the vector points in the
x-direction, you get Sx. If the vector points in the
y-direction, you get Sy. In general, this we
call the spin operator in the direction
of the vector n-- spin operator in
the direction of n. OK, so what about
that spin operator? Well, it had eigenvalues plus
minus h-bar over 2 along z, x, and y-- probably does still
have those eigenvalues-- but we have to make
this a little clearer. So for that we'll
take nx and ny and nz to be the polar
coordinate things. So this vector is
going to have a theta here on the azimuthal
angle phi over here. So nz is cosine theta. nx and ny have sine theta. And nx cosine phi, and
this one has sine phi. So what is the
operator Sn vector hat? Well, it's nx times Sx. So, I'll put a h-bar
over 2 in front, so we'll have nx sigma x,
or sigma1, plus ny sigma2, , plus nz sigma3. Remember the spin operators are
proportional h-bar over 2 times the sigmas-- so sigma1, sigma2, sigma3. And look what we
get. h-bar over 2. Sigma1 has an nx here, nx. Sigma2 has minus iny plus iny. And sigma3, we
have a nz minus nz. So this is h-bar over 2, nz is
cosine theta, nx minus iny-- you'd say, oh it's a
pretty awful thing, but it's very simple-- nx minus iny is sine theta
times e to the minus i phi. Here it would be sine
theta, e to the i phi, and here we'll have
minus cosine theta. So this is the whole
matrix, Sn-hat, like that. Well, in the last
couple of minutes, let's calculate the
eigenvectors and eigenvalues. So what do we get? Well, for the
eigenvalues, remember what is the computation of
an eigenvalue of a matrix. An eigenvalue for
matrix a, you write that by solving the determinant
of a minus lambda 1 equals 0. So for any matrix a,
if we want to find the eigenvalues
of this matrix, we would have to write
eigenvalues of Sn-hat. We have to ride the
determinant of this, minus lambda i, so the
determinant of h-bar over 2 cosine theta, minus lambda,
minus h-bar over 2 cosine theta, minus lambda. And here, it's sine theta, e
to the minus i phi, sine theta e to the i phi, the
determinant of this being 0. It's not as bad as it looks. It's actually pretty simple. These are a plus b, a minus b. Here the phases cancel out. The algebra you can read in the
notes, but you do get lambda equals plus minus h-bar over 2. Now that is fine, and we
now want the eigenvectors. Those are more non-trivial, so
they need a little more work. So what are you supposed to
do to find an eigenvector? You're supposed to take this
a minus lambda i, acting on a vector, and put
it equal to zero. And that's the eigenvector. So, for this case,
we're going to try to find the eigenvector n plus. So this is the one that
has Sn on this state-- well, I'll write it
here, plus minus h over 2, n plus minus here. So let's try to
find this one that corresponds to the eigenvalue
equal to plus h-bar over 2. Now this state is C1 times z
plus, plus C2 times z minus. These are our basis states,
so it's a little combination. Or it's C1, C2. Think of it as a matrix. So we want the
eigenvalues of that-- the eigenvector for
that-- so what do we have? Well, we would have Sn-hat
minus h-bar over 2 times 1, on this C1, C2 equals 0. The eigenvector equation
is that this operator minus the eigenvalue
must give you that. So the h-bars over
2, happily, go out, and you don't really need
to worry about them anymore. And you get here cosine
theta minus 1, sine theta e to the minus i phi,
sine theta e to the i phi, and minus cosine theta
minus 1, C1, C2 equals 0. All right, so you
have two equations, and both relate C1 and C2. Happily, and the reason
this works is because with this eigenvalue that
we've used that appears here, these two equations
are the same. So you can take
either one, and they must imply the same
relation between C1 and C2. Something you can check. So let me write one of them. C2 is equal to e to the i
phi, 1 minus cosine theta over sine theta C1. It's from the first line. So you have to remember, in
order to simplify these things, your half angle identities. Sorry. 1 minus cosine theta is 2
sine squared theta over 2, and sine theta is 2 sine theta
over 2 cosine theta over 2. So this becomes e to the
i phi sine theta over 2, over cosine theta over 2, C1. Now we want these things
to be well normalized, so we want C1 squared plus
C2 squared equal to 1. So, you know what C2
is, so this gives you C1 squared times 1 plus-- and C2 you use this, when you
square the phase goes away-- sine squared theta over
2, cosine squared over 2 must be equal to 1. Well, the numerator
is 1, so you learn that C1 squared is equal to
cosine squared theta over 2. Now you have to take
the square root, and you could put an i
or a phase or something. But look, whatever
phase you choose, you could choose C1 to
be cosine theta over 2, and say, I'm done. I want this one. Somebody would say, no
let's put the phase, e so to the i pi over 5. So that doesn't look good,
but four or even worse, this phase will show
up in C2 because C2 is proportional to C1. So I can get rid of it. I only should put it
if I really need it, and I don't think I need
it, so I won't put it. And you can always
change your mind later-- nobody's going to take
your word for this. So, in this case, C2 would
be sine theta over 2, e to the i phi. It's nice, but it's [INAUDIBLE]. And therefore, we got
this state n plus, which is supposed to be
cosine theta over 2, z plus, and plus sine theta over
2, e to the i phi, z minus. This is a great result. It gives
the arbitrarily located spin state that point
in the n-direction. As a linear superposition
of your two basis states, it answers conclusively
the question that any spin state
in your system can be represented in this
two-dimensional vector space. Now moreover, if I take
that theta equals 0, I have the z-axis, and it's
independent of the angle phi. The phi angle becomes
singular at the North Pole, but that's all right. When theta is equal to
0, this term is 0 anyway. And therefore, this goes,
and when theta is equal to 0, you recover the plus state. Now you can calculate
the minus state. And if you follow exactly the
same economical procedure, you will get the
following answer. And I think, unless
you've done a lot of eigenvalue calculations,
this is a calculation you should just redo. So the thing that you get,
without thinking much, is that n minus is equal to
sine theta over 2 plus, minus cosine theta over 2,
e to the i phi minus. At least some way of solving
this equation gives you that. You could say, this is
natural, and this is fine. But that is not
so nice, actually. Take theta equal to pi-- no, I'm sorry. Again, you take
theta equal to 0. Theta equal to 0-- this is supposed to be the
minus state along the direction. So this is supposed to
give you the minus state. Because the vector n is
along up, in the z-direction, but you're looking at
the minus component. So theta equals 0. Sure, there's no plus,
but theta equals 0, and you get the minus state. And this is 1, and
phi is ill-defined-- it's not so nice, therefore-- so, at this moment, it's
convenient to multiply this state by e to the
minus i phi, times minus 1. Just multiply it by
that, so that n minus is equal to minus sine theta
over 2, e to the minus i phi, plus, plus cosine
theta over 2, minus. And that's a nice
definition of the state. When theta is equal
to 0, you're fine, and it's more naturally
equivalent to what you know. Theta equal to 0 gives you
the minus state, or z minus. I didn't put the zs
here, for laziness. And for theta equal to
0, the way the phase phi doesn't matter. So it's a little nicer. You could work with this
one, but you might this well leave it like that. So we have our general states,
we've done everything here that required some linear
algebra without doing a review of linear
algebra, but that's what we'll start
to do next time.
