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visit MIT OpenCourseWare ocw.mit.edu. PROFESSOR: OK, so
let's get started. I just wanted to
make one announcement before we start the lecture. So Prof. Zwiebach is a way again today,
which is why I'm lecturing. And his office hours he's
obviously not going to have, but Prof. Harrow has kindly agreed
to take them over. So today I'll have office hours
four to five, and then Prof. Harrow will have office
hours afterwards five to six. So feel free to
come and talk to us. So today we're going to
try and cover a few things. So we're going to spend a
little bit of time talking about eigenvalues and vectors,
which we've-- finishing this discussion from last time. Then we'll talk about inner
products and inner product spaces. And then we'll
talk about-- we'll introduce Dirac's
notation, some of which we've already been using. And then, depending
on time, we'll also talk a little bit more
about linear operators. OK? So let's start with
where we were last time. So we were talking about
T-invariant subspaces. So we had U is a
T-invariant subspace if the following is satisfied. If T of U is equal to--
if this thing, which is all vectors that are
generated by T from vectors that live in U. So if
T is inside U itself. OK? And we can define this in
general for any U. However, one class of these invariant
subspaces are very useful. So if we take U to
be one dimensional. OK/ and so that really means
that U I can write as some whatever field I'm defining
my vector space over. Every element of this subspace
U is just some scalar multiple of a single vector U. So this
is a one dimensional thing. Now if we have a T-invariant
subspace of this one-- if this is going to be
a T-invariant objective, then we get a very
simple equation that you've seen before. So we're taking all vectors
in U acting on them with T, and if it stays
within U, then it has to be able to be
written like this. So we have some operator
acting on our vector space producing something in
the same vector space, just rescaling it. OK? For sum lambda, which
we haven't specified. And you've seen
this equation before in terms of matrices
and vectors, right? This is an eigenvalue equation. So these are eigenvalues
and these are eigenvectors. But now they're just
an abstract version of what you've discussed before. And we'll come back
to this in a moment. One thing that we just
defined at the end is the spectrum of an operator. The spectrum of T is
equal to all eigenvalues of that operator. And so later on
these will become-- this object will
become important. Let's just concentrate on this
and ask what does it mean. So if we have lambda
being an eigenvalues, what does that tell us? What does this equation tell us? Well, it tells us that
on U. So all I'm doing is taking this term over to
the other side of the equation and inserting the
identity operator I. So this is in itself
an operator now, right? And so this tells us also that
this operator, because it maps something that's non-zero
to the null vector, this is not injective, OK? And you can even write the
null space of T-- of T minus I lambda is equal to
all eigenvectors with eigenvalue lambda. OK? So every eigenvector
with eigenvalue lambda, T acting on it is just going
to give me lambda times the eigenvector again,
and so this will vanish. So for all eigenvectors
with that eigenvalue. And we've previously seen that,
if something is not injective, it's also not invertible, right? So this lets us write
something quite nice down. So there's a theorem. Let me write it out. So if we let T is in the space
of linear operators acting on this vector
space v, and we have a set of eigenvalues, lambda
1, lambda 2, lambda n, distinct eigenvalues,
eigenvalues of T, and the corresponding
eigenvectors, which we will call U. OK, so the
sum set U1, U2, up to Un with the correspondence
by their label. So then we know that
this list is actually a linearly independent set. So we can prove this
one very quickly. So let's do that. So let's assume it's false. So the proof is y contradiction,
so assume it's false. And what does that mean? Well, that means that there
is a non-trivial relation. I could write down some relation
C1U1 plus C2U2 plus CkUk equals 0 without
all the C's being 0. And what we'll do
is we'll actually say OK, let's do let--
so we'll let there be a k, a value of k that's
less than or equal to n, such that this holds for
Ci not equal to 0. So we're postulating that
there is some linear dependence on some of these things. So what we can do is
then act on this vector here with T minus lambda k times
the identity acting on this. So this is C1U1 plus
dot dot dot plus CkUk. OK? And what do we get here? So we're going to get, if
act on this piece of it, this is an eigenvector,
so T acting on this one would just give us
lambda 1, right? And so we're going to get
products of lambda 1 minus lambda k for this
piece, et cetera. So this will give us C1 lambda
1 minus lambda k U1 plus dot dot dot up to the Ck minus
1 lambda k minus 1 minus lambda k Uk minus 1. And then when we act
on this one here, so this one has an eigenvalue--
the eigenvalue corresponding to the eigenvector is lambda k,
so that last term gets killed. So we get plus 0 lots of Uk. And we know this is still 0. And now we've
established, in fact, these things here
are just numbers. All of these things. So we've actually
written down a relation that involves less than k. Actually, I should
have said this. Let there be a least k
less than or equal to, and such that we have
linear dependence. But what we've just shown
is that, in fact, there's a smaller space that's also
linear independent, right? So we've contradicted what
we assumed to start with. And you can just repeat
this procedure, OK? And so this is a contradiction. And so, in fact, there must
be no non-trivial relation even for k equals n
between these vectors, OK? Another brief
theorem that we won't prove, although we'll sort of
see why it works in a moment, is, again, for T in linear
operators on v with v being a finite dimensional
complex vector space. OK? There is at least
one eigenvalue. So I guess for this-- so T
has at least one eigenvalue. Now remember, in
the last lecture, we looked at a matrix,
two by two matrix, that was rotations in the
xy-plane and found there were, in fact,
no eigenvalues. But that's because we were
looking at a real vector space. So we were looking at
rotations of the real plane. So this is something
that you can prove. We will see why it's true,
but we won't prove it. And so one way of saying
this is to go to a basis. And so everything we've said
so far about eigenvalues and eigenvectors has
not been referring to any particular basis. And in fact, eigenvalues
are basis independent. But we can use a basis. And then we have matrix
representations of operators that we've talked about. And sort of this
operator equation, or the operator statement
that T minus lambda I-- so as operator statement
T minus lambda I U equals 0 is equivalent to saying that--
well, we've said it here. We said that it's
equivalent to saying that it's not invertible. This operator is not invertible. But that's also
equivalent to saying that the matrix representation
of it in any basis is not invertible. And by here we just mean
inverses as in the inverses that you've taken the many
matrices in your lives. And so what that means then,
if-- I'm sure you remember. If a matrix is not
invertible, that means it has a
vanishing determinant. So it has debt of this. Now you can think
of this as a matrix. This determinant has to be 0. And remembering we can
write this thing out. And so it has lambdas
on the diagonal, and then whatever entries
T has wherever it wants. This just gives us a
polynomial in lambda, right? So this gives us some f of
lambda, which is a polynomial. And if you remember,
this is called the characteristic polynomial. Characteristic. Right? And so we can write it, if we
want, as just some f of lambda is equal to just,
in this case, it's going to be just lambda
minus some lambda 1. I have to be able to
write it like this. I can just break it
up into these terms here, where the lambda I's,
the 0's of this polynomial are, in general, complex
and can be repeated. Now what can happen is that
you have, in the worst case -- I don't know if it's the
worst case, but in one case, you could have all of
the singularities-- all of the the 0's
being at the same place. And you could have a eigenvalue
that is in full degenerate here. Right? So if we, say, have
lambda 1 occurring twice in this sequence, then we set
out to a degenerate eigenvalue. And in principle, you could
have just a single eigenvalue that's in full degenerate. But you can always write this. There has to be one
lambda there at least, one lambda I there at least. And so you can see why
this is true, right? Now if you're in a
real vector space, you don't get to say that,
because this polynomial may only have complex
roots, and they're not part of the space
you're talking about. OK? So it can be repeated, and
this is called degeneracy. OK, so are there any questions? AUDIENCE: Can you turn it? It should be lambda
I minus T, just so that it matches the next line. PROFESSOR: Thank you, OK. Thank you. I could have flipped the sign
on the next line as well. So any other questions? No? OK, so let's move on and we
can talk about inner products. And so first, what
is an inner product? So an inner product is a map,
but it's a very specific map. So an inner product
on a vector space V is a map from V cross
V to the field, F. And that's really
what it's going to be. Now who has seen an
inner product somewhere? OK, what do we call it? AUDIENCE: Dot product. PROFESSOR: Dot product, right. So we can learn a
lot from thinking about this simple case, so
the motivation for thinking about this is really
the dot product. So we have a vector space Rn. OK? And on that vector space, we
might have two vectors, a, which I'm going to write as a1. a2 dot dot dot. a2 dot dot dot. an, and b. So we have two vectors, and
these are in vector space V. Then we can define
the dot product, which is an example of one
of these in a product. So a dot b. We can even put little
vectors over these. And so our definitions that
we've used for many years is that this is a1 b1 plus
a2 b2 plus dot dot dot an bn. And you see that this
does what we want it. So it takes two vectors which
live in our vector space. And from that, you
get a number, right? So this lives in R. So this is a nice
example in a product. And we can look at what
properties it gives us. So what do we know
about this dot product? Well, we know some properties
that it has is that a dot b. So it doesn't care which order
you give the arguments in, all right? Also, if I take
the same vector, I know that this is got to be
greater than or equal to 0, right? Because this is going
to be our length. And the only case where it's
0 is when the vector is 0. Well we can write this.
a dotted in to, say, b to 1 b1 plus b to 1 b2. So this b2's are real numbers,
and these b's are vectors, right? So this thing we can just
write is equal to b to one a dot b1 plus b to 2 a do b2. And make them
vectors everywhere. OK, so we've got
three nice properties. And you can write
down more if you want, but this will be enough for us. And the other thing that
we can do with this is we can define the length
of a vector, right? So we can say this is for
this defines a length. And more generally, we only call
this the norm of the vector. And that, of course, you
know is that mod a squared is just equal to a
dot a, all right? So this is our
definition of the norm. OK so this definition over
here is really by no means unique in satisfying
these properties. So if I wrote down something
where, instead of just say a1 b1 plus a1 b2 et cetera, I
wrote down some positive number times a1 b1 times some
other positive number a2 b2, et cetera, that would also
satisfy all of these properties up here. So it's not unique. And so you could
consider another the dot product,
which we would write as just c1 a1 b1 plus c2 a2
b2 plus some cn an bn, where the c's are just
positive real numbers. That would satisfy
all of the things that we know about our
standard dot product. But for obvious reasons,
we don't choose to do this, because it's not a
very natural definition to put these random
positive numbers along here. But we could. And I guess one
other thing that we have is the Schwarz inequality. And so this is the a dot b. So the absolute value of
the dot product of a dot b is less than or
equal to the product of the norms of
the vectors, right? And so one of the
problems in the piece is to consider this in
the more abstract sense, but this is very easy to
show for real vectors, right? So this is all very nice. So we've talked about Rn. What we really are
going to worry about is complex vector spaces. And so there we have
a little problem. And the problem
comes in defining what we mean by a normal, right? Because if I say now that this
vector has complex components and write this
thing here, I'm not guaranteed that this is
a real number, right? And so I need to be
a little bit careful. So let's just talk
about complex spaces. And we really want to have a
useful definition of a length. So let's let z be in this thing,
in interdimensional complex space. So really my z is equal
to z1 z2 zn, where the zI line as
being in c, right? So how can define a
link for this object? Well, we have to do it
sort of in two steps. So already know how
to define the length for a complex number, right? It's just the absolute
value, the distance from the origin in
the complex plane. But now we need to
do this in terms of a more complicated
vector space. And so we can really
think of this as equal to the sum of the squares
of z1, of the absolute values of these complex numbers. OK? Which if we write it out,
looks like z1 star z1 plus. OK? And so we should now, thinking
about the inner product, we should be thinking
that the appearance of complex conjugation is
not entirely unnatural. So if we ask about the
length of a vector here, then that's going to arise
from an inner product. OK? This object we want to arise
from our inner product. So we can now define our general
in a product with the following axioms. So firstly, we want to basically
maintain the properties that we've written down
here, because we don't want to make our dot product not
being in an inner product anymore. That'd be kind of silly. So let's define our inner
product in the following way. I'm going to write it
in a particular way. So the inner product is
going to be, again, a map. And it's going to take our
vector space, two elements of the vector
space to the field. And I'm in a complex
vector space. So it's a map that I'm
going to right like this that takes v cross v to c. OK? And what I mean here is you put
the two elements of your vector space in these positions
in this thing, OK? And so really a b is
what I mean by this. Where a and b-- so let
me write it this way. So this thing is in c for a
and b are in the v, right? So these things
dots are where I'm going to plug-in my vectors. And so this inner product
should satisfy some axioms. And they look very much like
what we've written here. So the first one is a
slight modification. We want that a b is
equal not to b a, but to its complex
conjugate, OK? And this is related to
what I was discussing here. But from this, we can see that
the product of a with itself is always real, because it
and its complex conjugate are the same. So we know that a a is real. And we're also going
to demand a definition of this inner a
product that this is greater than or equal to 0. And it's only 0 if a equals 0. Right? So that's pretty much unchanged. And then we want the same
sort of distributivity. We do want to have that a
inner producted with B to 1 b plus B to 2 b2 should be equal
to B to 1 a b1 plus B to 2 a b2 where the [INAUDIBLE] are
just complex numbers, right? And that's what we
need to ask of this. And then we can make a
sensible definitions of it that will give us a
useful norm as well. Now I'll just make one remark. This notation here,
this is due to Dirac. And so it's very
prevalent in physics. You will see in most purely
mathematical literature you will see this
written just like this. So let me write it as a b and
put these things in explicitly. And sometimes you'll even see
a combination of these written like this, OK/ But they
all mean the same thing. Compared to what
we've written up here, this seems a little asymmetric
between the two items, right? Well firstly, these
are isometric. And then down here
we've shown something about that we demand something
about the second argument, but we don't demand the same
thing about the first argument. So why not? Can anyone see? I guess what we would demand
is exactly the same thing the other way around. So we would demand
another thing that would be sort of alpha
1 a plus alpha 2 a2 b is equal to-- well,
something like this. Well, we would
actually demand this. a1 b. But I don't actually
need to demand that, because that follows
from number one, right? I take axiom one,
apply it to this, and I automatically
get this thing here. OK? And notice what's
arisen is-- actually let's just go through
that, because you really do want to see these complex
conjugates appearing here, because they are important. So this follows. So 1 plus 3, imply this. Let's just do this. So let's start with
this expression and start with this piece. And we know that
this will then be given by axiom one by b alpha
1 a 1 plus alpha 2 a2 complex conjugate, all right? And then by this linearity
of the second argument we can now distribute
this piece, right? We can write this is alpha
1 b a1 plus alpha 2 b a2, all complex conjugated. Which let's put
all the steps in. Is alpha 1 star
and this one star. And then, again, by the first
argument, by the first axiom, we can flip these and get rid
of the complex conjugation. And that gives us this
one up here, right? So we only need to define
this linearity, distributive property on one
side of this thing. We could have chosen to define
it here and wouldn't have needed that one, but we didn't. OK, so let's look at
a couple of examples. And the first one is a
finite dimensional example. And we're going to
take v is equal to cn. And our definition
is going to be a pretty natural
generalization of what we've written down before. So a and b are elements of cn. And this is just
going to be a1 star b1 plus a2 star b2
plus an star bn. Another piece of chalk. So the only difference from dot
product in real vector space is that we've put this
complex conjugates here. And that you can check
satisfies all of these axioms. Another example is
actually an example of an infinite
dimensional vector space. Let's take v is the set
of all complex functions, all f of x that are in c with x
living in some finite interval. OK? And so a natural norm to
define on this space-- and this is something that we
can certainly talk about in recitations-- is that if
I have f and g in this vector space v, then f g
I'm going to define-- this is my definition of
what the dot product is-- is the integral from 0 to
l f star of x g of x dx. OK? If you think of this as
arising from evaluating f at a set of discrete
points where you've got a finite dimensional
vector space, and then letting the space
between those points go to 0, this is kind of the
natural thing to arise. It's really an integral
as a limit of a sum. And over here, of course,
I could write this one as just the sum over
i of ai star bi. i equals 1 to n. And so this is the integral
is infinite dimensional generalization of the
sum, and so we have this. And that might be something
to talk about in recitations. OK? So we've gone from
having just a vector space to having a
vector space where we've added this new operation on it,
this inner product operation. And that lets us do things
that we couldn't do before. So firstly, it lets us
talk about orthogonality. Previously we couldn't
ask any question about two objects
within our vector space. This let's us ask a
question about two objects. So if we have the inner product
a b in some vector space V, then if this is 0, we
say they're orthogonal. We say that the vectors
a and b are orthogonal. And I'm sure you know
what orthogonal means in terms of Rn, but this
is just the statement of what it means in a
abstract vector space. This is the definition
of orthogonality. So this is one thing if
we have a set of vectors. e1 e2 en, such that ei ej is
equal to delta ij, chronic to delta ij, this
set is orthonormal. Again, a word you've
seen many times. OK, so we can also define
the components of vectors now in basis dependent way. We're going to choose ei to be
a set of vectors in our vector space V. We previously
had things that form a basis, a basis
of V. And if we also demand that they're
orthonormal, then we can-- well, we can always decompose
any vector in V in terms of its basis, right? But if it's also
orthonormal, then we can write a, which is
a is some vector in V. a is equal to the sum over i
equals 1 to n of some ai ei. So we can do that for any basis. But then we can take this vector
and form its inner product with the basis vectors. So we can look at
what ek a is, right? So we have our basis vectors
ek, and we take one of them and we dot product it
into this vector here. And this is
straightforward to c. This is going to be equal to
the sum over i equals 1 to n ai. And then it's going to be the
inner product of ek with ei, right? Because of this
distributive property here. OK? But we also know that, because
this is an orthonormal basis, this thing here is a delta
function, delta ik, right? And so I can, in fact,
do this sum, and I get and this is equal to ak. And so we've
defined what we mean by a component of this
vector in this basis ei. They're defined by
this inner product. So we can also talk
about the norm, which I think, unsurprisingly, we
are going to take the norm to be, again, equal to
this, just as we did in Rn, but now it's the more general
definition of my inner product that defines our norm. And because of our
axiom-- so because of number two in particular,
this is a sensible norm. It's always going to be
greater than or equal to 0. OK? And conveniently we can also
change this Schwarz inequality. So instead of the one
that's specific to Rn, that becomes a b. All right, so let's
cross that one out. This is what it becomes. And in the current
p set, you've got to prove this is true, right? We can also write down
a triangle inequality, which is really something
that norms should satisfy. So the norm of a plus
b should be less than or equal to the norm of
a plus the norm of b. And the R3 version of
this is the longest side of a triangle is shorter than
the two shorter sides, right? So this is fine. OK, so you might
ask why we're doing all of this seemingly
abstract mathematics. Well, so now we're in a
place where we can actually talk about the space where
all of our quantum states are going to live. And so these inner
product space-- these vector spaces that
we've given an inner product. We can call them
inner product spaces. So we have a vector space
with an inner product is actually we call
a Hilbert space. And so this needs
a little qualifier. So if this is a finite
dimensional vector space, then this is straight. It is just a Hilbert space. Let me write it here. So let's write it as a
finite dimensional vector space with an inner
product is a Hilbert space. But if we have an infinite
dimensional vector space, we need to be a
little bit careful. For an infinite
dimensional vector space, we again need an inner product. We need to make sure that
this space is complete. OK? And this is a kind
of technical point that I don't want to
spend too much time on, but if you think about-- well,
let me just write it down. Vector space. Let me write it here. And I haven't defined what this
complete vector space means. But if we have an infinite
dimensional vector space that is complete
or we make it complete, then we have an inner product. We also get a Hilbert space. And all of quantum mechanical
states live in a Hilbert space. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes, that's true. So how's that? So we need to define what we
mean by complete though, right? So I don't want to
spend much time on this. But we can just do an example. If we take the space of-- let V
equal the space of polynomials on an interval 0 to L, say. So this means I've got all pn's. P0 plus p1x plus pn xn. There are things that will
live in the completed vector space that are not
of this form here. So for example, if I
take n larger and larger, I could write down
this polynomial. I could write pn of x is the sum
over i equals 1 up to n of i-- x to the i over i
factorial, right? And all of pn's live in
this space of polynomials. But their limit,
as n becomes large, there's a sequence of these
call it a cushy sequence that, as n goes to infinity,
I generate something that's actually
not a polynomial. So I generate e
of x, which lives in the completion of this,
but itself not a polynomial. Don't worry about this
too much, but in order to really define
a Hilbert space, we have to be a
little bit careful for infinite dimensional cases. OK, so a few more things
that we can do to talk about. Well how do we make
a orthonormal basis? So I presume you've all
heard of Gram-Schmidt? The Gram-Schmidt procedure? Yep. OK, so that's how we
make a orthonormal basis. And just the way
you do it in R3, you do it the same way in
your arbitrary vector space. So we have the
Gram-Schmidt procedure. So you can define this-- so we
have a list v1, v2, vn are just vectors in our vector space
that are linearly independent. So we can construct
another list. There's also orthonormal, so
it's a very useful thing for us to have. And so you could define
this recursively. You can just write
that ej is equal to vj minus the sum over
i less than j of ei. So this thing divided
by its length. And so by the sum,
you're orthogonalizing ej versus all of the previous
ei's that you've already defined. And then you normalize it by
dividing by its length, right? So that's something
that's very useful. And the last thing I want to
say about these inner product spaces is that we can use them--
these inner products at least, is that we can use them to
find the orthogonal complement of something, of
anything really. So let's let u-- so we
have a vector space V, and I can just choose some
things in that and make a set. So u is the set of
v that are in V. So it doesn't need
to be a subspace. It's just a set. For example, if v
Rn, I could just choose vectors pointing
along two directions, and that would give me my set. But that's not a subspace,
because it doesn't contain some multiple of this
vector times some multiple of this vector, which would
be pointing over here. So this is just a set so far. We can define u
perpendicular, which we'll call the orthogonal
complement of u. And this is defined
as u perpendicular is equal to the set
of v's in V such that v u is equal to 0 for
all u in U. All of the things that live in this space are
orthogonal to everything that lives in U. OK And in fact, this one is
a subspace automatically. So it is a vector space. So if I took my example of
choosing the x direction and y direction for my set here,
then everything perpendicular to the x direction
and y direction is actually everything
perpendicular to the xy-plane, and so that is actually
a subspace of R3. And so there's a nice theorem
that you can think about, but it's actually
kind of obvious. So if u is a subspace,
then I can actually write that V is equal
to the direct sum of U plus its orthogonal
complement, OK? So that one's kind of fairly
straightforward to prove, but we won't do it now. OK, so in the last
little bit, I want to talk more about this
notation that I've introduced, that Dirac introduced. What can we say? If I can find a
[INAUDIBLE] here. Are there any
questions about this? Yep. AUDIENCE: So when we find space
and the idea of basis balance, why is that [INAUDIBLE]
decompose things into plane waves when we're
not actually [INAUDIBLE]? PROFESSOR: So it's because
it's-- basically it works. Mathematically,
we're doing things that are not quite legitimate. And so we can generalize the
Hilbert space a little bit, such that these non
normalizable things can live in this generalized space. But really the answer
is that it works, but no physical system is going
to correspond to something like that. So if I take plane
waves, that's not a physically realizable thing. It gives us an easy
way to, instead of talking about
some wave packet that some superposition
of plane waves, we can talk about the
plane waves by themselves and then form the wave packet
afterwards, for example. Does that answer the question
a little bit at least? Yep. AUDIENCE: If p could be written
as a sum of U [INAUDIBLE], why is U not [INAUDIBLE]? PROFESSOR: Well, just
think about the case that I was talking about. So if we're looking
at R3 and we take U to be the set the unit
vector in the x direction, the unit vector in
the y direction, that's not a
subspace, as I said, because I can take the unit
vector in the x direction plus the unit vector
in the y direction. It goes in the 45
degree direction. And it's not in the things
I've written down originally. So then if I talk
about the subspace, the things spanned by x hat and
y hat, then I have a subspace. It's the whole xy-plane. And the things are
orthogonal to it in R3 are just the things
proportion to z hat. And so then I've got the
things in this x hat and y hat, and the thing that's
in here is z hat. And so that really is the basis
for my R3 that I started with. That contains everything. And more generally, the reason
I need to make this a subspace is just because-- so I define
U by some set of vectors that I'm putting into it. The things that are
orthogonal to that are automatically
already everything that's orthogonal
to it, so there's no combination of the things
in the orthogonal complement that's not already
in that complement. Because I'm saying that
this is everything in V that's orthogonal to these
things in this subspace. So I could write down
some arbitrary vector v, and I could aways write
it as a projection onto things that live
in here and things that don't live in
this one, right? And what I'm doing by
defining this complement is I'm getting rid of the bits
that are proportional to things in this, OK? All right any-- yep? AUDIENCE: So an
orthogonal complement is automatically a subspace? PROFESSOR: Yes. AUDIENCE: But that
doesn't necessarily mean that any random collection
of vectors is a subspace. PROFESSOR: No. All right, so let's
move on and talk about the Dirac's notation. And let's do it here. So three or four
lectures ago, we started talking
about these objects, and we were calling
them kets, right? And they were things that
live in our vector space V. So these are just a way of
writing down our vectors, OK? So when I write down
the inner product, which we have on the wall
above, one of the bits of it looks lot like this. So we can really think of
a b, the b being a ket. We know that b is a
vector, and here we're writing in a particular
way of writing things in terms of a ket. And what we can do
is actually think about breaking this
object, this inner product up into two pieces. So remember the dot product is
taking two vectors, a and b. One of them, we already have
written it like a vector, because a ket is a vector. What Dirac did in breaking
this up is he said, OK, well this
thing is a bracket, and so he's going to call this
one a ket, and this is a bra. So this object
with something it. The things inside
these you should think of as just
labeling these things. OK? Now we already know
this thing here. So these kets are
things that live in-- I should say this
is direct notation. OK, so we already
know these kets are things that live
in the vector space. But what are the bras? Well, they're not vectors
in V. So b is a vector, so maybe I should've
called this one b to be a little less confusing. So b is a ket, and
this is something that lives in our
vector space V. This inner product we're writing
in terms of bra and a ket. The bra, what does
it actually do? So I'm going to use it to
make this inner product. And so what it's doing
is it's taking a vector and returning a complex number. The inner product takes
v cross v goes to c. But if I think of it as the
action of this bra on this ket, then the action is that
this bra eats a vector and spits back a
complex number, OK? So a is actually a map. OK? So these bras live in
a very different place than the kets do. Although they are going to
be very closely related. So firstly, it's not in V. You
should be careful if you ever say that, because
it's not right. We actually say that it
belongs to a dual space, which we label as V star, because it
is very dependent on V, right? It's mapped from V to c. And I shouldn't even say
this is a linear map. Now what is V star? Well, at the moment
it's just the space of all linear maps from V to c. Me But it itself
is a vector space. So we can define
addition of these maps. We can define addition on V star
and also a scalar modification of these maps. And so what that means is
that I can define some bra w That's equal to alpha lots
of another one plus B to b. And all of these live
in this V star space. Let me write that explicitly. So a b and w live in V star, OK? And the way we define
this is actually through the inner product. We define it such
that-- so I take all vectors v in the
vector space big V, and the definition of
w is that this holds. And then basically
from the properties of the inner product, you
inherit the vector structure, the vector space structure. So this tells us V
star is a vector space. Let's go over here. And there's actually
a correspondence between the objects in the
original vector space V and those that live in V star. So we can say for
any v in V, there's a unique-- I should
write it like this. Any ket v in the
vector space, there is a unique bra, which I'm
also going to label by v, and this lives in V star. And so we can show uniqueness
by assuming it doesn't work. So let's assume that
there exists a v and a v prime in here such
that v-- so we'll assume that this
one is not unique, but there are two
things, v and v prime. And then we can
construct-- from this, I can take this over
to this side here, and I just get that
v w minus v prime w is equal to 0, which I
can then use the skew symmetry of these
objects to write as w v minus w v prime star. So I've just changed the
order of both of them. And then I can use
the property of kets. I can combine them linearly. So I know this is equal
to w v minus v prime star. And essentially,
that's it, because I know this has to be true for
every w in the vector space V. So this thing is equal to 0. And so the only thing that can
annihilate every other vector is going to be 0 for
my definition, in fact, of the inner product. So this implies
that v minus v prime equals 0, the null vector, which
implies that v equals v prime. So our assumption was wrong,
and so this is unique. OK, let's see. And so we actually
have really a one to one correspondence between
things in the vector space and things in the
joule space, OK? And so we can actually
label the bras by the same thing that's
labeling the kets. So I can really do what I've
done in the top line up there and have something--
everything is labeled by the same little v. Both
the thing in the big vector space, big V, and
the thing in V star are label by the same thing. And more generally,
I could say that v-- so there's a correspondence
between this thing and this thing. And notice the stars
appearing here. They came out of how we
define the inner product. OK, so really, in fact, any
linear map you write down, any linear map like this
defines one of these bras, because every linear
map that takes to V to c lives in V star. So there has to be an element
that corresponds to it. And just if you want
to think about kind of a concrete way of
talking about these, I can think of-- if I think
of this as a column vector, v1 to vn, the way I should
think about the bras is that they are really what you
want to write as row vectors. And they may have to have
the conjugates of the thing. The components are conjugated. OK, and now you can ask what
the dot product looks like. Alpha v is then just this
matrix multiplication. But it's matrix multiplication
of an n by 1 thing by 1 by n thing. Alpha 1 star alpha 2 star alpha
n star times this one here. So v1 vn. And this is now just
matrix multiplication. I guess I can
write it like this. So they're really,
really quite concrete. They're as concrete
as the kets are. So you can construct
them as vectors like strings of
numbers in this way. So I guess we should finish. So I didn't get to talk
about linear operators, but we will resume
there next week. Are there any questions about
this last stuff or anything? No? OK. So you next week, or see
you tomorrow, some of you. Thanks. [APPLAUSE]
