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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.

thanks

Any fix for the one ear sound?

amazing amazing professor 🙂

my god the cameraman dosen't stop zooming

Aula maravilhosa!!!

Very nice lecture

I highly recommend watching this introduction to linear algebra https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

Thank me later 🙂

I suspect the cameraman is overdoing it. Just a sugestion.

Great lecture by the way!

Thank you for making this free and online, it really helps!

You know that the class is awesome when you are compelled to applaud ate the end of it! Very few teachers can achieve that.

42K views on a quantum mechanics lecture gives me hope for the world

1h17 Mistake, it's "-c" instead of c, he confused sigma2.

For another great book on linear algebra and vector spaces I would highly recommended Linear Algebra and Matrix Theory by Jimmie Gilbert and Linda Gilbert! It’s a great book for both mathematicians and physicists. The road paved by the authors to reach hermitian operators and unitary spaces is second to none.