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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Last time we
talked about the spin operator pointing in some
particular direction. There were questions. In fact, there was
a useful question that I think I want to begin
the lecture by going back to it. And this, you received
an email from me. The notes have an extra section
added to it that is stuff that I didn't do
in class last time, but I was told in fact some
of the recitation instructors did discuss this
matter And I'm going to say a few words about it. Now, I do expect you
to read the notes. So things that you will need for
the homework, all the material that is in the notes is
material that I kind of assume you're familiar with. And you've read it
and understood it. And I probably don't cover
all what is in the notes, especially examples
or some things don't go into so much detail. But the notes should really be
helping you understand things well. So the remark I want to make
is that-- there was a question last time that better that
we think about it more deliberately in which we saw
there that Pauli matrices, sigma 1 squared was equal
to sigma 2 squared equal to 2 sigma 3 squared
was equal to 1. Well, that, indeed,
tells you something important about the
eigenvalues of this matrices. And it's a general fact. If you have some matrix M
that satisfies an equation. Now, let me write an equation. The matrix M squared plus alpha
M plus beta times the identity is equal to 0. This is a matrix equation. It takes the whole matrix,
square it, add alpha times the matrix, and then beta
times the identity matrix is equal to 0. Suppose you discover that
such an equation holds for that matrix M. Then,
suppose you are also asked to find eigenvalues
of this matrix M. So suppose there is a vector--
that is, an eigenvector with eigenvalue lambda. That's what having
an eigenvector with eigenvalue lambda means. And you're supposed to calculate
these values of lambda. So what you do here
is let this equation, this matrix on the left,
act on the vector v. So you have M squared plus
alpha M plus beta 1 act on v. Since the matrix
is 0, it should be 0. And now you come and
say, well, let's see. Beta times 1 on v. Well,
that's just beta times v, the vector v. Alpha M on v, but M on
v is lambda v. So this is alpha lambda v. And M squared
on v, as you can imagine, you act with another M here. Then you go to this side. You get lambda Mv, which
is, again, another lambda times v. So M squared
on v is lambda squared v. If acts two times on v. Therefore, this is 0. And here you have, for
example, that lambda squared plus alpha lambda
plus beta on v is equal to 0. Well, v cannot be 0. Any eigenvector-- by definition,
eigenvectors are not 0 vectors. You can have 0 eigenvalues
but not 0 eigenvectors. That doesn't exist. An eigenvector that
is 0 is a crazy thing because this would
be 0, and then it would be-- the eigenvalue
would not be determined. It just makes no sense. So v is different from 0. So you see that lambda squared
plus alpha lambda plus beta is equal to 0. And the eigenvalues, any
eigenvalue of this matrix, must satisfy this equation. So the eigenvalues
of sigma 1, you have sigma 1 squared, for
example, is equal to 1. So the eigenvalues,
any lambda squared must be equal to
1, the number 1. And therefore, the
eigenvalues of sigma 1 are possibly plus or minus 1. We don't know yet. Could be two 1's, 2 minus
1's, one 1 and one minus 1. But there's another nice
thing, the trace of sigma 1. We'll study more the
trace, don't worry. If you are not that
familiar with it, it will become
more familiar soon. The trace of sigma
1 or any matrix is the sum of elements
in the diagonal. Sigma 1, if you remember,
was of this form. Therefore, the trace is 0. And in fact, the traces of any
of the Pauli matrices are 0. Another little theorem
of linear algebra shows that the
trace of a matrix is equal to the sum of eigenvalues. So whatever two
eigenvlaues sigma 1 has, they must add up to 0. Because the trace is 0
and it's equal to the sum of eigenvalues. And therefore, if the
eigenvalues can only be plus or minus 1,
you have the result that one eigenvalue
must be plus 1. The other eigenvalue
must be minus 1, is the only way you
can get that to work. So two sigma 1 eigenvalues of
sigma 1 are plus 1 and minus 1. Those are the two eigenvalues. So in that section
as well, there's some discussion about properties
of the Pauli matrices. And two basic properties
of Pauli matrices are the following. Remember that the spin
matrices, the spin operators, are h bar over 2 times
the Pauli matrices. And the spin operators had the
algebra for angular momentum. So from the algebra
of angular momentum that says that Si Sj is equal
to i h bar epsilon i j k Sk, you deduce after plugging
this that sigma i sigma j is 2i epsilon i j k sigma k. Moreover, there's
another nice property of the Pauli matrices having
to deal with anticommutators. If you do experimentally
try multiplying Pauli matrices,
sigma 1 and sigma 2, you will find out that if you
compare it with sigma 2 sigma 1, it's different. Of course, it's not the same. These matrices don't commute. But they actually-- while
they fail to commute, they still fail to
commute in a nice way. Actually, these are
minus each other. So in fact, sigma 1 sigma 2 plus
sigma 2 sigma 1 is equal to 0. And by this, we mean
that they anticommute. And we have a brief
way of calling this. When this sign was a minus,
it was called the commutator. When this is a plus, it's
called an anticommutator. So the anticommutator of sigma
1 with sigma 2 is equal to 0. Anticommutator defined
in general by A, B. Two operators is AB plus BA. And as you will
read in the notes, a little more analysis
shows that, in fact, the anticommutator of
sigma i and sigma j has a nice formula,
which is 2 delta ij times the unit matrix, the
2 by 2 unit matrix. With this result, you
get a general formula. Any product of two operators,
AB, you can write as 1/2 of the anticommutator plus
1-- no, 1/2 of the commutator plus 1/2 of the anticommutator. Expand it out, that
right-hand side, and you will see
quite quickly this is true for any two operators. This has AB minus BA
and this has AB plus BA. The BA term cancels and the
AB terms are [INAUDIBLE]. So sigma i sigma j
would be equal to 1/2. And then they put down
the anticommutator first. So you get delta ij
times the identity, which is 1/2 of the
anticommutator plus 1/2 of the commutator, which
is i epsilon i j k sigma k. It's a very useful formula. In order to make those
formulas look neater, we invent a notation in which
we think of sigma as a triplet-- sigma 1, sigma 2, and sigma 3. And then we have vectors,
like a-- normal vectors, components a1, a2, a3. And then we have a dot
sigma must be defined. Well, there's an
obvious definition of what this should
mean, but it's not something you're accustomed to. And one should pause
before saying this. You're having a normal
vector, a triplet of numbers, multiplied by a
triplet of matrices, or a triplet of operators. Since numbers commute
with matrices, the order in which you
write this doesn't matter. But this is defined
to be a1 sigma 1 plus a2 sigma 2 plus a3 sigma 3. This can be written as ai
sigma i with our repeated index convention that you sum
over the possibilities. So here is what you're
supposed to do here to maybe interpret
this equation nicely. You multiply this
equation n by ai bj. Now, these are numbers. These are matrices. I better not change this
order, but I can certainly, by multiplying that way, I
have ai sigma i bj sigma j equals 2 ai bj delta ij
times the matrix 1 plus i epsilon i j k ai bj sigma k. Now, what? Well, write it in terms
of things that look neat. a dot sigma, that's a matrix. This whole thing is a matrix
multiplied by the matrix b dot sigma gives you-- Well, ai bj delta ij, this
delta ij forces j to become i. In other words, you can replace
these two terms by just bi. And then you have ai bi. So this is twice. I don't know why I have a 2. No 2. There was no 2 there, sorry. So what do we get here? We get a dot b, the dot product. This is a normal dot product. This is just a number
times 1 plus i. Now, what is this thing? You should try to remember
how the epsilon tensor can be used to do cross products. This, there's just one
free index, the index k. So this must be
some sort of vector. And in fact, if you try the
definition of epsilon and look in detail what this
is, you will find that this is nothing but
the k component of a dot b. The k-- so I'll write it here. This is a cross b sub k. But now you have a cross
b sub k times sigma k. So this is the same as
a cross b dot sigma. And here you got a pretty nice
equation for Pauli matrices. It expresses the general
product of Pauli matrices in somewhat geometric terms. So if you take, for
example here, an operator. No. If you take, for
example, a equals b equal to a unit vector,
then what do we get? You get n dot sigma squared. And here you have
the dot product of n with n, which is 1. So this is 1. And the cross product of two
equal vectors, of course, is 0 so you get
this, which is nice. Why is this useful? It's because with this identity,
you can understand better the operator S hat
n that we introduced last time, which was n
dot the spin triplet. So nx, sx, ny, sy, nz, sc. So what is this? This is h bar over
2 and dot sigma. And let's square this. So Sn vector squared. This matrix squared would be
h bar over 2 squared times n dot sigma squared, which is 1. And sigma squared is 1. Therefore, this spin operator
along the n direction squares to h bar r squared
over 2 times 1. Now, the trace of this
Sn operator is also 0. Why? Because the trace
means that you're going to sum the
elements in the diagonal. Well, you have a sum
of matrices here. And therefore, you will have
to sum the diagonals of each. But each of the
sigmas has 0 trace. We wrote it there. Trace of sigma 1 is 0. All the Pauli matrices have
0 trace, so this has 0 trace. So you have these two relations. And again, this tells you that
the eigenvalues of this matrix can be plus minus h bar over 2. Because the eigenvalues
satisfy the same equation as the matrix. Therefor,e plus
minus h bar over 2. And this one says that the
eigenvalues add up to 0. So the eigenvalues of S hat n
vector are plus h bar over 2 and minus h bar over 2. We did that last time, but we do
that by just taking that matrix and finding the eigenvalues. But this shows that its
property is almost manifest. And this is fundamental
for the interpretation of this operator. Why? Well, we saw that if n
points along the z-direction, it becomes the operator sz. If it points about
the x-direction, it becomes the operator sx. If it points along
y, it becomes sy. But in an arbitrary
direction, it's a funny thing. But it still has
the key property. If you measured the spin
along an arbitrary direction, you should find only plus h bar
over 2 or minus h bar over 2. Because after all, the
universe is isotopic. It doesn't depend on direction. So a spin one-half particle. If you find out that whenever
you measure the z component, it's either plus
minus h bar over 2. Well, when you
measure any direction, it should be plus
minus h bar over 2. And this shows that this
operator has those eigenvalues. And therefore, it makes sense
that this is the operator that measures spins in
an arbitrary direction. There's a little more
of an aside in there, in the notes about
something that will be useful and fun to do. And it corresponds to
the case in which you have two triplets of
operators-- x1, x2, x3. These are operators now. And y equal y1, y2, y3. Two triplets of operators. So you define the dot
product of these two triplets as xi yi summed. That's the definition. Now, the dot product of
two triplets of operators defined that way
may not commute. Because the operators x
and y may not commute. So this new dot product
of both phase operators is not commutative-- probably. It may happen that
these operators commute, in which case x dot y
is equal to y dot x. Similarly, you can
define the cross product of these two things. And the k-th component is
epsilon i j k xi yj like this. Just like you would define
it for two number vectors. Now, what do you know about
the cross product in general? It's anti-symmetric. A cross B is equal
to minus B cross A. But this one won't be
because the operators x and y may not commute. Even x cross x may be nonzero. So one thing I will ask you
to compute in the homework is not a long calculation. It's three lines. But what is S cross S equal to? Question there? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes, it's
the sum [INAUDIBLE]. Just in the same
way that here you're summing over i's and j's to
produce the cross product. So whenever an
index is repeated, we'll assume it's summed. And when it is not summed,
I will put to the right, not summed explicitly--
the words-. Because in some
occasions, it matters. So how much is this? It will involve i, h
bar, and something. And you will try to
find out what this is. It's a cute thing. All right, any other questions? More questions? Nope. OK. So now, finally,
we get to that part of the course that has to
do with linear algebra. And I'm going to
do an experiment. I'm going to do it
differently than I did it in the previous years. There is this nice book. It's here. I don't know if you
can read from that far, but it has a pretty-- you might
almost say an arrogant title. It says, Linear Algebra
Done Right by Sheldon Axler. This is the book, actually,
MIT's course 18.700 of linear algebra uses. And when you first get the
book that looks like that, you read it and open--
I'm going to show you that this is not that well done. But actually, I think
it's actually true. The title is not a lie. It's really done right. I actually wish I had learned
linear algebra this way. It may be a little
difficult if you've never done any linear algebra. You don't know what
the matrix is-- I don't think that's
the case anybody here. A determinant, or eigenvalue. If you never heard
any of those words, this might be a little hard. But if you've heard
those words and you've had a little linear
algebra, this is quite nice. Now, this book has
also a small problem. Unless you study
it seriously, it's not all that easy to grab
results that you need from it. You have to study it. So I don't know if
it might help you or not during this semester. It may. It's not necessary to get it. Absolutely not. But it is quite lovely. And the emphasis is
quite interesting. It really begins from
very basic things and logically
develops everything and asks at every point
the right questions. It's quite nice. So what I'm going to do
is-- inspired by that, I want to introduce some of
the linear algebra little by little. And I don't know very
well how this will go. Maybe there's too much detail. Maybe it's a lot of
detail, but not enough so it's not all that great. I don't know, you
will have to tell me. But we'll try to get
some ideas clear. And the reason I want
to get some ideas clear is that good books
on this subject allow you to understand
how much structure you have to put in a vector space
to define certain things. And unless you do
this carefully, you probably miss some
of the basic things. Like many physicists
don't quite realize that talking about the
matrix representation, you don't need brass
and [INAUDIBLE] to talk about the matrix
representation of an operator. At first sight, it seems
like you'd need it, but you actually don't. Then, the differences between
a complex and a vector space-- complex and a real vector
space become much clearer if you take your time
to understand it. They are very different. And in a sense,
complex vector spaces are more powerful, more
elegant, have stronger results. So anyway, it's enough
of an introduction. Let's see how we do. And let's just begin
there for our story. So we begin with vector
spaces and dimensionality. Yes. AUDIENCE: Quick question. The length between
the trace of matrix equals 0 and [INAUDIBLE] is
proportional to the identity. One is the product of
the eigenvalues is 1 and the other one was
the sum is equal to 0. Are those two statements
related causally, or are they just separate
statements [INAUDIBLE]? PROFESSOR: OK, the
question is, what is the relation between
these two statements? Those are separate observations. One does not imply the other. You can have matrices that
square to the identity, like the identity itself,
and don't have 0 trace. So these are
separate properties. This tells us that
the eigenvalue squared are h bar over 2. And this one tells me that
lambda 1 plus lambda 2-- there are two
eigenvalues-- are 0. So from here, you deduce
that the eigenvalues could be plus
minus h bar over 2. And in fact, have to be
plus minus h bar over 2. All right, so let's
talk about vector spaces and dimensionality. Spaces and dimensionality. So why do we care about this? Because the end result
of our discussion is that the states
of a physical system are vectors in a
complex vector space. That's, in a sense, the
result we're going to get. Observables, moreover,
are linear operators on those vector spaces. So we need to understand what
are complex vector spaces, what linear operators on them mean. So as I said,
complex vector spaces have subtle properties that make
them different from real vector spaces and we want
to appreciate that. In a vector space,
what do you have? You have vectors and
you have numbers. So the two things must exist. The numbers could be the
real numbers, in which case we're talking about
the real vector space. And the numbers could be
complex numbers, in which case we're talking about the
complex vector space. We don't say the vectors are
real, or complex, or imaginary. We just say there are vectors
and there are numbers. Now, the vectors can be
added and the numbers can be multiplied by
vectors to give vectors. That's basically
what is happening. Now, these numbers can
be real or complex. And the numbers-- so there
are vectors and numbers. And we will focus on
just either real numbers or complex numbers,
but either one. So these sets of
numbers form what is called in
mathematics a field. So I will not define the field. But a field-- use the
letter F for field. And our results. I will state results whenever--
it doesn't matter whether it's real or complex, I
may use the letter F to say the numbers are in F.
And you say real or complex. What is a vector space? So the vector space,
V. Vector space, V, is a set of vectors with an
operation called addition-- and we represent it as plus--
that assigns a vector u plus v in the vector space when u and
v belong to the vector space. So for any u and v
in the vector space, there's a rule called addition
that assigns another vector. This also means that this
space is closed under addition. That is, you cannot get out
of the vector space by adding vectors. The vector space must
contain a set that is consistent in that
you can add vectors and you're always there. And there's a multiplication. And a scalar
multiplication by elements of the numbers of F such
that a, which is a number, times v belongs to
the vector space when a belongs to the numbers
and v belongs to the vectors. So every time you
have a vector, you can multiply by those
numbers and the result of that multiplication
is another vector. So we say the space is also
closed under multiplication. Now, these properties
exist, but they must-- these operations
exist, but they must satisfy the
following properties. So the definition
is not really over. These operations satisfy-- 1. u plus v is equal to v plus u. The order doesn't matter
how you sum vectors. And here, u and v in V. 2. Associative. So u plus v plus w is
equal to u plus v plus w. Moreover, two numbers a times b
times v is the same as a times bv. You can add with the
first number on the vector and you add with the second. 3. There is an additive identity. And that is what? It's a vector 0 belonging
to the vector space. I could write an arrow. But actually, for
some reason they just don't like to write
it because they say it's always ambiguous
whether you're talking about the 0
number or the 0 vector. We do have that problem
also in the notation in quantum mechanics. But here it is, here is
a 0 vector such that 0 plus any vector v is equal to v. 4. Well, in the field,
in the set of numbers, there's the number 1, which
multiplied by any other number keeps that number. So the number 1 that
belongs to the field satisfies that 1 times any
vector is equal to the vector. So we declare that that number
multiplied by other numbers is an identity. [INAUDIBLE] identity
also multiplying vectors. Yes, there was a question. AUDIENCE: [INAUDIBLE]. PROFESSOR: There is
an additive identity. Additive identity, the 0 vector. Finally, distributive laws. No. One second. One, two, three--
the zero vector. Oh, actually in my list I
put them in different orders in the notes, but never mind. 5. There's an additive inverse
in the vector space. So for each v belonging
to the vector space, there is a u belonging
to the vector space such that v plus u is equal to 0. So additive identity
you can find for every element
its opposite vector. It always can be found. And last is this
[INAUDIBLE] which says that a times u plus
v is equal to au plus av, and a plus b on v is
equal to av plus bv. And a's and b's
belong to the numbers. a and b's belong to the field. And u and v belong
to the vector space. OK. It's a little disconcerting. There's a lot of things. But actually, they
are quite minimal. It's well done, this definition. They're all kind
of things that you know that follow quite
immediately by little proofs. You will see more in
the notes, but let me just say briefly
a few of them. So here is the additive
identity, the vector 0. It's easy to prove that
this vector 0 is unique. If you find another 0 prime that
also satisfies this property, 0 is equal to 0 prime. So it's unique. You can also show that 0 times
any vector is equal to 0. And here, this 0
belongs to the field and this 0 belongs
to the vector space. So the 0-- you had to postulate
that the 1 in the field does the right thing, but
you don't need to postulate that 0, the number 0,
multiplied by a vector is 0. You can prove that. And these are not
difficult to prove. All of them are
one-line exercises. They're done in that book. You can look at them. Moreover, another one. a any number times the 0 vector
is equal to the 0 vector. So in this case, those
both are vectors. That's also another property. So the 0 vector and the 0 number
really do the right thing. Then, another property,
the additive inverse. This is sort of interesting. So the additive inverse,
you can prove it's unique. So the additive
inverse is unique. And it's called-- for v, it's
called minus v, just a name. And actually, you can prove
it's equal to the number minus 1 times the vector. Might sound totally trivial
but try to prove them. They're all simple, but they're
not trivial, all these things. So you call it minus v, but
it's actually-- this is a proof. OK. So examples. Let's do a few examples. I'll have five examples
that we're going to use. So I think the main thing for
a physicist that I remember being confused about
is the statement that there's no characterization
that the vectors are real or complex. The vectors are
the vectors and you multiply by a real
or complex numbers. So I will have one example
that makes that very dramatic. As dramatic as it can be. So one example of vector spaces,
the set of N component vectors. So here it is,
a1, a2, up to a n. For example, with capital N.
With a i belongs to the real and i going from 1 up
to N is a vector space over r, the real numbers. So people use that
terminology, a vector space over the kind of numbers. You could call it
also a real vector space, that would be the same. You see, these
components are real. And you have to
think for a second if you believe all of them are
true or how would you do it. Well, if I would
be really precise, I would have to tell
you a lot of things that you would find boring. That, for example, you have
this vector and you add a set of b's. Well, you add the components. That's the definition of plus. And what's the definition
of multiplying by a number? Well, if a number is
multiplied by this vector, it goes in and
multiplies everybody. Those are implicit, or you
can fill-in the details. But if you define
them that way, it will satisfy all the properties. What is the 0 vector? It must be the one
with all entries 0. What is the additive inverse? Well, change the sign
of all these things. So it's kind of obvious that
this satisfies everything, if you understand how the sum
and the multiplication goes. Another one, it's
kind of similar. 2. The set of M cross N matrices
with complex entries. Complex entries. So here you have it,
a1 1, a1 2, a1 N. And here it goes up
to aM1, aM2, aMN. With all the a i j's belonging
to the complex numbers, then-- I'll erase here. Then you have that this
is a complex vector space. Is a complex vector space. How do you multiply by a number? You multiply a number times
every entry of the matrices. How do sum two matrices? They have the same size, so
you sum each element the way it should be. And that should
be a vector space. Here is an example
that is, perhaps, a little more surprising. So the space of 2 by
2 Hermitian matrices is a real vector space. You see, this can be easily
thought [INAUDIBLE] naturally thought as a real vector space. This is a little surprising
because Hermitian matrices have i's. You remember the most
general Hermitian matrix was of the form--
well, a plus-- no, c plus d, c minus d,
a plus ib, a minus ib, with all these numbers
c, d, b in real. But they're complex numbers. Why is this naturally
a real vector space? The problem is that if
you multiply by a number, it should still be a
Hermitian matrix in order for it to be a vector space. It should be in the vector. But if you multiply by a real
number, there's no problem. The matrix remains Hermitian. You multiplied by
a complex number, you use the Hermiticity. But an i somewhere here
for all the factors and it will not be Hermitian. So this is why it's
a real vector space. Multiplication by real
numbers preserves Hermiticity. So that's surprising. So again, illustrates
that nobody would say this is a real vector. But it really should be thought
as a vector over real numbers. Vector space over real numbers. Two more examples. And they are kind
of interesting. So the next example is the set
of polynomials as vector space. So that, again, is sort of
a very imaginative thing. The set of polynomials p of z. Here, z belongs to some
field and p of z, which is a function of z, also
belongs to the same field. And each polynomial
has coefficient. So any p of z is a0
plus a1 z plus a2 z squared plus-- up to some an zn. A polynomial is
supposed to end That's pretty important
about polynomials. So the dots don't go up forever. So here it is, the a i's
also belong to the field. So looked at this polynomials. We have the letter z and
they have these coefficients which are numbers. So a real polynomial-- you
know 2 plus x plus x squared. So you have your real numbers
times this general variable that it's also
supposed to be real. So you could have it real. You could have it complex. So that's a polynomial. How is that a vector space? Well, it's a vector
space-- the space p of F of those polynomials--
of all polynomials is a vector space over
F. And why is that? Well, you can take--
again, there's some implicit definitions. How do you sum polynomials? Well, you sum the
independent coefficients. You just sum them
and factor out. So there's an obvious
definition of sum. How do you multiply a
polynomial by a number? Obvious definition, you
multiply everything by a number. If you sum polynomials,
you get polynomials. Given a polynomial, there
is a negative polynomial that adds up to 0. There's a 0 when all
the coefficients is 0. And it has all the
nice properties. Now, this example
is more nontrivial because you would
think, as opposed to the previous examples,
that this is probably infinite dimensional because
it has the linear polynomial, the quadratic, the cubic,
the quartic, the quintic, all of them together. And yes, we'll see
that in a second. So set of polynomials. 5. Another example, 5. The set F infinity of
infinite sequences. Sequences x1, x2, infinite
sequences where the x i's are in the field. So you've got an
infinite sequence and you want to add
another infinite sequence. Well, you add the first
element, the second elements. It's like an infinite
column vector. Sometimes mathematicians like to
write column vectors like that because it's practical. It saves space on a page. The vertical one, you
start writing and the pages grow very fast. So here's an infinite sequence. And think of it as a
vertical one if you wish. And all elements
are here, but there are infinitely many
in every sequence. And of course, the set of all
infinite sequences is infinite. So this is a vector
space over F. Again, because all
the numbers are here, so it's a vector space over F. And last example. Our last example is a
familiar one in physics, is the set of complex
functions in an interval. Set of complex functions on
an interval x from 0 to L. So a set of complex
functions f of x I could put here on an
interval [INAUDIBLE]. So this is a complex
vector space. Vector space. The last three
examples, probably you would agree that there
are infinite dimensional, even though I've not defined
what that means very precisely. But that's what we're going
to try to understand now. We're supposed to understand
the concept of dimensionality. So let's get to
that concept now. So in terms of dimensionality,
to build this idea you need a definition. You need to know the term
subspace of a vector space. What is a subspace
of a vector space? A subspace of a vector space
is a subset of the vector space that is still a vector space. So that's why it's
called subspace. It's different from subset. So a subspace of V is a subset
of V that is a vector space. So in particular, it
must contain the vector 0 because any vector space
contains the vector 0. One of the ways you sometimes
want to understand the vector space is by representing it as
a sum of smaller vector spaces. And we will do that when
we consider, for example, angular momentum in detail. So you want to write a vector
space as a sum of subspaces. So what is that called? It's called a direct sum. So if you can write--
here is the equation. You say V is equal to u1
direct sum with u2 direct sum with u3 direct sum with u m. When we say this, we
mean the following. That the ui's are
subspaces of V. And any V in the
vector space can be written uniquely as a1
u1 plus a2 u2 plus a n u n with ui [INAUDIBLE]
capital Ui. So let me review
what we just said. So you have a
vector space and you want to decompose it in
sort of basic ingredients. This is called a direct sum. V is a direct sum of subspaces. Direct sum. And the Ui's are
subspaces of V. But what must happen for
this to be true is that once you take
any vector here, you can write it as
a sum of a vector here, a vector here, a vector
here, a vector everywhere. And it must be done uniquely. If you can do this
in more than one way, this is not a direct sum. These subspaces kind of overlap. They're not doing the
decomposition in a minimal way. Yes. AUDIENCE: Does the expression
of V have to be a linear combination of the
vectors of the U, or just sums of the U sub i's? PROFESSOR: It's some
linear combination. Look, the interpretation,
for example, R2. The normal vector space R2. You have an intuition quite
clearly that any vector here is a unique sum of this
component along this subspace and this component
along this subspace. So it's a trivial example,
but the vector space R2 has a vector subspace R1 here
and a vector subspace R1. Any vector in R2
is uniquely written as a sum of these two vectors. That means that R2
is really R1 plus R1. Yes. AUDIENCE: [INAUDIBLE]. Is it redundant to say that
that-- because a1 u1 is also in big U sub 1. PROFESSOR: Oh. Oh, yes. You're right. No, I'm sorry. I shouldn't write those. I'm sorry. That's absolutely right. If I had that in my
notes, it was a mistake. Thank you. That was very good. Did I have that in my notes? No, I had it as you said it. True. So can be written
uniquely as a vector in first, a vector in the second. And the a's are
absolutely not necessary. OK. So let's go ahead then and
say the following things. So here we're
going to try to get to the concept of
dimensionality in a precise way. Yes. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right,
the last one is m. Thank you. All right. The concept of dimensionality
of a vector space is something that you
intuitively understand. It's sort of how many
linearly independent vectors you need to describe the
whole set of vectors. So that is the number
you're trying to get to. I'll follow it up in a
slightly rigorous way to be able to do infinite
dimensional space as well. So we will consider something
called a list of vectors. List of vectors. And that will be something
like v1, v2 vectors in a vector space up to vn. Any list of vectors
has finite length. So we don't accept infinite
lists by definition. You can ask, once you
have a list of vectors, what is the vector subspace
spanned by this list? How much do you
reach with that list? So we call it the
span of the list. The span of the list, vn. And it's the set of all linear
combinations a1 v1 plus a2 v2 plus a n vn for ai in the field. So the span of the list
is all possible products of your vectors on the list
are-- and put like that. So if we say that the
list spans a vector space, if the span of the list
is the vector space. So that's natural language. We say, OK, this list
spans the vector space. Why? Because if you produce
the span of the list, it fills a vector space. OK, so I could say it that way. So here is the definition,
V is finite dimensional if it's spanned by some list. If V is spanned by some list. So why is that? Because if the list is-- a
definition, finite dimensional. If it's spanned by some list. If you got your list, by
definition it's finite length. And with some set of
vectors, you span everything. And moreover, it's
infinite dimensional if it's not finite dimensional. It's kind of silly,
but infinite-- a space V is infinite dimensional if
it is not finite dimensional. Which is to say that there is
no list that spans the space. So for example, this definition
is tailored in a nice way. Like let's think
of the polynomials. And we want to see if it's
finite dimensional or infinite dimensional. So you claim it's
finite dimensional. Let's see if it's
finite dimensional. So we make a list
of polynomials. The list must have some length,
at least, that spans it. You put all these
730 polynomials that you think span the list,
span the space, in this list. Now, if you look at
the list, it's 720. You can check one
by one until you find what is the one
of highest order, the polynomial of
highest degree. But if the highest degree
is say, z to the 1 million, then any polynomial that has
a z to the 2 million cannot be spanned by this one. So there's no finite
list that can span this, so this set-- the
example in 4 is infinite dimensional for sure. Example 4 is
infinite dimensional. Well, example one is
finite dimensional. You can see that
because we can produce a list that spans the space. So look at the example 1. It's there. Well, what would be the list? The list would be-- list. You would put a vector
e1, e2, up to en. And the vector e1
would be 1, 0, 0, 0, 0. The vector e2 would
be 0, 1, 0, 0, 0. And go on like that. So you put 1's and 0's. And you have n of them. And certainly, the most general
one is a1 times e1 a2 times e2. And you got the list. So example 1 is
finite dimensional. A list of vectors is
linearly independent. A list is linearly independent
if a list v1 up to vn is linearly independent, If
a1 v1 plus a2 v2 plus a n vn is equal to 0 has the unique
solution a1 equal a2 equal all of them equal 0. So that is to mean that
whenever this list satisfies this property-- if you want
to represent the vector 0 with this list, you
must set all of them equal to 0, all
the coefficients. That's clear as well
in this example. If you want to
represent the 0 vector, you must have 0 component
against the basis vector x and basis vector y. So the list of this
vector and this vector is linearly independent
because the 0 vector must have 0 numbers
multiplying each of them. So finally, we define
what is a basis. A basis of V is a
list of vectors in V that spans V and is
linearly independent. So what is a basis? Well, you should
have enough vectors to represent every vector. So it must span V. And
what else should it have? It shouldn't have extra
vectors that you don't need. It should be minimal. It should be all
linearly independent. You shouldn't have
added more stuff to it. So any finite dimensional
vector space has a basis. It's easy to do it. There's another thing
that one can prove. It may look kind of obvious,
but it requires a small proof that if you have-- the
bases are not unique. It's something we're going
to exploit all the time. One basis, another
basis, a third basis. We're going to change
basis all the time. Well, the bases are not
unique, but the length of the bases of a vector
space is always the same. So the length of the
list is-- a number is the same whatever
base you choose. And that length
is what is called the dimension of
the vector space. So the dimension
of a vector space is the length of any
bases of V. And therefore, it's a well-defined concept. Any base of a finite vector
space has the same length, and the dimension
is that number. So there was a question. Yes? AUDIENCE: Is there
any difference between bases [INAUDIBLE]? PROFESSOR: No, absolutely not. You could have a
basis, for example, of R2, which is this vector. The first and the
second is this vector. And any vector is a
linear superposition of these two vectors with some
coefficients and it's unique. You can find the coefficients. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes. But you see, here is exactly
what I wanted to make clear. We're putting the
vector space and we're putting the least
possible structure. I didn't say how to take the
inner product of two vectors. It's not a definition
of a vector space. It's something we'll put later. And then, we will be able
to ask whether the basis is orthonormal or not. But the basis exists. Even though you have no
definition of an inner product, you can talk about basis
without any confusion. You can also talk about
the matrix representation of an operator. And you don't need
an inner product, which is sometimes very unclear. You can talk about the
trace of an operator and you don't need
an inner product. You can talk about
eigenvectors and eigenvalues and you don't need
an inner product. The only thing you
need the inner product is to get numbers. And we'll use them to use
[INAUDIBLE] to get numbers. But it can wait. It's better than
you see all that you can do without
introducing more things, and then introduce them. So let me explain a
little more this concept. We were talking about this
base, this vector space 1, for example. And we produced a list that
spans e1, e2, up to en. And those were these vectors. Now, this list not
only spans, but they are linearly independent. If you put a1 times
this plus a2 times this and you set
it all equal to 0. Well, each entry will be
0, and all the a's are 0. So these e's that you put
here on that list is actually a basis. Therefore, the length of that
basis is the dimensionality. And this space has
dimension N. You should be able to prove that
this space has been dimension m times N. Now, let me do the
Hermitian-- these matrices. And try to figure out
the dimensionality of the space of
Hermitian matrices. So here they are. This is the most general
Hermitian matrix. And I'm going to produce for
you a list of four vectors. Vectors-- yes, they're matrices,
but we call them vectors. So here is the list. The unit matrix, the first
Pauli matrix, the second Pauli matrix, and the
third Pauli matrix. All right, let's see how
far do we get from there. OK, this is a list of
vectors in the vector space because all of
them are Hermitian. Good. Do they span? Well, you calculate the most
general Hermitian matrix of this form. You just put arbitrary
complex numbers and require that the matrix
be equal to its matrix complex conjugate and transpose. So this is the most general one. Do I obtain this
matrix from this one's? Yes I just have to put 1 times
c plus a times sigma 1 plus b times sigma 2 plus
d times sigma 3. So any Hermitian
matrix can be obtained as the span of this list. Is this list
linearly independent? So I have to go here
and set this equal to 0 and see if it sets to 0
all these coefficients. Well, it's the same thing as
setting to 0 all this matrix. Well, if c plus d and c minus
d are 0, then c and d are 0. If this is 0, it must be a 0
and b 0, so all of them are 0. So yes, it's
linearly independent. It spans. Therefore, you've proven
completely rigorously that this vector
space is dimension 4. This vector space--
I will actually leave it as an exercise for
you to show that this vector space is infinite dimensional. You say, of course, it's
infinite dimensional. It has infinite sequences. Well, you have to
show that if you have a finite list of
those infinite sequences, like 300 sequences,
they span that. They cannot span that. So it takes a little work. It's interesting
to think about it. I think you will enjoy trying
to think about this stuff. So that's our discussion
of dimensionality. So this one is a little
harder to make sure it's infinite dimensional. And this one is, yet, a
bit harder than that one but it can also be done. This is infinite dimensional. And this is infinite
dimensional. In the last two minute, I want
to tell you a little bit-- one definition and let
you go with that, is the definition of
a linear operator. So here is one thing. So you can be more general,
and we won't be that general. But when you talk
about linear maps, you have one vector space and
another vector space, v and w. This is a vector space and
this is a vector space. And in general, a map from
here is sometimes called, if it satisfies the
property, a linear map. And the key thing is
that in all generality, these two vector spaces may
not have the same dimension. It might be one vector space and
another very different vector space. You go from one to the other. Now, when you have
a vector space v and you map to the
same vector space, this is also a
linear map, but this is called an operator
or a linear operator. And what is a linear
operator therefore? A linear operator is
a function T. Let's call the linear operator T.
It takes v to v. In which way? Well, T acting u plus v,
on the sum of vectors, is Tu plus T v. And T
acting on a times a vector is a times T of the vector. These two things make
it into something we call a linear operator. It acts on the sum
of vectors linearly and on a number times a vector. The number goes out and
you act on the vector. So all you need to know for
what a linear operator is, is how it acts on basis vectors. Because any vector
on the vector space is a superposition
of basis vectors. So if you tell me how it
acts on the basis vectors, you know everything. So we will figure out how
the matrix representation of the operators arises from how
it acts on the basis vectors. And you don't need
an inner product. The reason people think
of this is they say, oh, the T i j
matrix element of T is the inner product of the
operator between i and j. And this is true. But for that you
need [? brass ?] and inner product,
all these things. And they're not necessary. We'll define this without that. We don't need it. So see you next time,
and we'll continue that. [APPLAUSE] Thank you.
