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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: It's good to be back. I really want to thank
both Aaron and Will who took my teaching
duties over last week. You've been receiving
updates of the lecture notes, and, in particular,
as I don't want to go back over
some things, I would like you to read some of
the material you have there. In particular, the
part on projectors has been developed further. We will meet projectors
a lot in this space, because in quantum mechanics,
whenever you do a measurement, the effect of a
measurement is to act on a stage with a projector. So projectors are
absolutely important. And orthogonal
projectors are the ones that we're going to
use-- are the ones that are relevant in
quantum mechanics. There's a property, for
example, of projectors that is quite neat that is used
in maximization and fitting problems. And you will see
that in the PSET. In the PSET, the
last problem has to do with using a
projector to find best approximations to some
functions using polynomials. So there's lots of things
to say about projectors, and we'll find them along
when we go and do later stuff in the course. So please read that
part on projectors. The other thing is that much
of what we're going to do uses the notation that we have
for describing inner products-- for example, u, v. And
then, as we've mentioned, and this is in the notes--
this, in the bracket notation, becomes something like this. And the bracket notation
of quantum mechanics is fairly nice for
many things, and it's used sometimes for
some applications. Everybody uses the bracket
notation for some applications. I hope to get to
one of those today. So much of the theory
we've developed is done with this as
the inner product. Nevertheless, the translation
to the language of bras and kets is very quick. So the way the notes are
going to be structured-- and we're still
working on the notes, and they're going
to chang a bit-- is that everything
regarding the math is being developed
more in this notation, but then we turn
into bras and kets and just go quickly
over all you've seen, just how it looks
with bras and kets so that you're familiar. Then, in the later
part of the course, we'll use sometimes bras and
kets, and sometimes this. And sometimes some
physicisists use this notation with parentheses. So for example,
Weinberg's recent book on quantum mechanics
uses this notation. It doesn't use bras and
kets I think at all. So you have to be ready
to work with any notation. The bra and ket notation
has some nice properties that make it very fast
to do things with it. It is very efficient. Nevertheless, in some ways this
notation is a little clearer. So many of the things we'll
develop is with this notation. So today I'm going
to develop the idea of the Hermitian
conjugator for an operator, or the adjoint of an operator. And this idea is generally a
little subtle, a little hard to understand. But we'll just go at it slowly
and try to make it very clear. So adjoints or Hermitian
operators, or Hermitian conjugates-- adjoints
or Hermitian conjugates. So the idea of Adjoints,
or Hermition conjugates, really begins with some
necessary background on what they're called--
linear functionals. It sounds complicated,
but it's not. What is a linear functional? A linear functional
on V-- on a vector space V-- is a linear
map from V to the numbers F. We've always been
calling F the numbers. So it's just that, something
that, once you have a vector, you get a number
and it's linear. So a linear function of Phi,
if it's a linear functional, Phi on v belongs to F.
Phi acts on a vector, v, that belongs to the vector
space and gives you a number. So "linear" means Phi of v1 plus
v2 is Phi of v1 plus Phi of v2. And Phi of av, for a is
number, is a Phi of v. So seems simple,
and indeed it is. And we can construct examples
of linear functionals, some trivial ones, for example. Let Phi be a map that takes the
vector space, reals in three dimensions, to the real numbers. So how does it act? Phi acts on a vector, which
is x1, x2, and x3-- three components. And it must give
a numbers, so it could be 3x1 minus x2 plus
7x3, as simple as that. It's linear. x1, x2, and x3 are the
coordinates of a single vector. And whenever you have
this vector, that is, this triplet-- now, I
could have written it like this-- Phi of x1,
x2, and x3, as a vector. It looks like that. But it's easier to use
horizontal notation, so we'll write it like that. And, if you have
an inner product on this space-- on this three
dimensional vector space-- there's something you can say. Actually this Phi is
equal-- and this we call the vector V-- is
actually equal to u, inner product with v, where u is
the vector that has components 3, minus 1, and
7, because if you take the inner product of
this vector with this vector, in three dimensions
real vector spaces-- inner product is a dot product. And then we make the dot
product of u with the vector V. Maybe I should have
called it v1, v2, v3. I'll change that--
v1, v2, v3 here are components of the
vector-- v1, v2, and v3, not to be confused
with three vectors. This whole thing is a vector
V. So this linear functional, that, given a vector
gives me a number. The clever thing is
that the inner product is this thing that gives
you numbers out of vectors. So you've reconstructed
this linear functional as the inner product of some
vector with the vector you're acting on, so, where
u is given by that. The most important result
about linear functionals is that this is not an accident. This kind be that
very generally. So any time you give
me a linear functional, I can find a vector that,
using the inner product, acts on the vector you're
acting on the same way as the linear function of thus. The most general
linear functional is just some most general
vector acting this way. So let's state
that and prove it. So this is a theorem, it's
not a definition or anything like that. Let Phi be a linear
functional on v. Then there is a unique vector u
belonging to the vector space such that Phi acting
on v is equal to u, v. Since this is such
a canonical thing, you could even
invent a notation. Call this the linear
functional created by u, acting on v. Everybody
doesn't use this, but you could call it like that. This is a linear
functional acting on v, but it's labeled by u, which
is the vector that you've use there. This is important enough that we
better understand why it works. So I'll prove it. We're going to use
an orthonormal basis, say e1 up to en is an
orthonormal, O-N, basis. AUDIENCE: That
means we're assuming v is finite dimensional here? PROFESSOR: Sorry? AUDIENCE: We're assuming V is
finite dimensional, correct? PROFESSOR: Yeah, it's
finite dimensional I'm going to prove it using
a finite basis like that. Is true finite dimensional? I presume yes. AUDIENCE: If it's
not [INAUDIBLE]. PROFESSOR: What hypothesis? AUDIENCE: You say continuous
when you're talking [INAUDIBLE]. PROFESSOR: OK, I'll check. But let's just prove this one
finite dimensional like this. Let's take that. And now write the vector as a
superposition of these vectors. Now we know how to do that. We just have the components
of v along each basis vector. For example, the
component of v along e1 is precisely e1, v.
So then you go on like that until
you go en, v, en. I think you've derived this
a couple of times already, but this is a statement
you can review, and let's take it to be correct. Now let's consider what is
Phi acting on a v like that. Well, it's a linear map, so
it takes on a sum of vectors by acting on the
vectors, each one. So it should act on from
this plus that, plus that, plus that. Now, it acts on this vector. Well, this is a number. The number goes out. It's a linear function. So this is e1, v, Phi of e1, all
the way up to en, v Phi of en. Now this is a number,
so let's bring it into the inner product. Now, if you brought it in
on the side of V as a number it would go in just
like the number. If you bring it
into the left side, remember it's
conjugate homogeneous, so this enters as
a complex number. So this would be e1, Phi of
e1 star times V plus en, Phi of en star, v. And then we have
our result that this Phi of v has been written now. The left input is different
on each of these terms, but the right input is the same. So at this moment linearity
on the first input says that you can put here e1,
Phi of e1 star plus up to en, Phi of en star, v. And this
is the vector you were looking for, the vector U. Kind of
simple, at the end of the day you just used the basis
and made it clearer. It can always be constructed. Basically, the vector you want
is e1 times Phi of u1 star plus en up to Phi of en star. So if you know what the linear
map does to the basis vectors, you construct the
vector this way. Vector is done. The only thing to be
proven is that it's unique. Uniqueness is rather easy
to prove at this stage. Suppose you know that u
with v works and gives you the right answer. Well, you ask, is
there a u prime that also gives the
right answer for all v? Well, pass it to the
other side, and you would have u minus u prime,
would have zero inner product with v for all v. Pass
to the other side, take the difference,
and it's that. So u minus u prime
is a vector that has zero inner product
with any vector. And any such thing
as always zero. And perhaps the easiest way
to show that, in case you haven't seen that
before, if x with v equals 0 for all for all v.
What can you say about x? Well, take v is the value
for any v. So take v equal x. So you take x, x is equal to 0. And by the axioms of
the inner product, if a vector has 0 inner
product with itself, it's 0. So at this stage, you go
u minus u prime equals 0, and u is equal to u prime. So it's definitely
unique, you can't find another one that works. So we have this thing. This theorem is proven. And now let's use to
define this the adjoint, which is a very
interesting thing. So the adjoing, or
Hermitian conjugate, sometimes called
adjoint-- physicists use the name Hermitian
conjugate, which is more appropriate. Well, I don't know if
it's more appropriate. It's more pictorial if you
have a complex vector space. And if you're accustomed
with linear algebra about Hermition matrices,
and what they are, and that will show
up a little later, although with a
very curious twist. So given an operator T belonging
to the set of linear operators on a vector space, you
can define T dagger, also belonging to
l of v. So this is the aim-- constructing an
operator called the Hermitian conjugate. Now the way we're
going to do it is going to be defining
something that is a T star. Well, I said "T star" because
mathematicians in fact call it star. And most mathematicians,
they complex conjugate if a number is not
z star but z bar. So that's why we
call it T star and I may make this mistake
a few times today. We're going to use dagger. And so I will make
a definition that will tell you what T
dagger is supposed to be, acting on things. But it might not be obvious,
at least at first sight, that it's a linear operator. So let's see how does this go. Here is the claim. Consider the
following thing-- u, T, v-- this inner product of
u with T, v. And think of it as a linear functional. Well, it's certainly
a linear functional of v. It's a linear
functional because if you put a times v the a goes out. And if you put v1 plus
v2 you get it's linear. So it's linear, but
it's not the usual one's that we've been building, in
which the linear functional looks like u with v. I
just put an operator there. So by this theorem,
there must be some vector that this can be
represented as this acting with that vector inside here,
because any linear operator is some vector acting on the
vector-- on the vector v. Any linear functional, I'm
sorry-- not linear operator. Any linear functional--
this is a linear functional. And every linear function can
be written as some vector acting on v. So there must
be a vector here. Now this vector surely
will depend on what u is. So we'll give it a name. It's a vector that depends on
U. I'll write it as T dagger u. At this moment, T dagger
is just a map from v to v. We said that this
thing that we must put here depends on u, and
it must be a vector. So it's some thing that
takes u and produces another vector
called T dagger on u. But we don't know
what T dagger is, and we don't even
know that it's linear. So at this moment it's just
a map, and it's a definition. This defines what T dagger
u is, because some vector-- it could be calculated exactly
the same way we calculated the other ones. So let's try to see
why it is linear. Claim T dagger belongs to
the linear operators in v. So how do we do that? Well, we can say the following. Consider u1 plus
u1 acting on Tv. Well, by definition, this would
be the T dagger of u1 plus u2, some function on u1 plus
u2, because whatever is here gets acted by T dagger
times v. On the other hand, this thing is equal
to u1, Tv plus u2, Tv, which is equal to T dagger
u1, v plus T dagger u2, v. And, by linearity, here you
get equal to T dagger u1 plus T dagger on u2. And then comparing this too--
and this is true for arbitrary v-- you find that T dagger,
acting on this sum of vectors, is the same as this thing. And similarly, how about au, Tv? Well, this is equal to T dagger
on au, v. Now, T dagger on au, do you think the a goes
out as a or as a bar? Sorry? a or a-bar? What do you think T dagger
and au is supposed to be? a, because it's supposed to be
a linear operator, so no dagger here. You see-- well, I
didn't show it here. Any linear operator, T on av,
is supposed to be a T of v. And we're saying
T dagger is also a linear operator
in the vector space. So this should be with an a. We'll see what we get. Well, the a can go out here,
and it becomes a star u1, Tv, which is equal. I'm going through the left side. By definition, a
bar T dagger of u, v. And now the
constant can go in, and it goes back as
a, T dagger u, v. So this must be equal
to that, and you get what we're claiming here,
which is T dagger on au, is equal to a T dagger of u. So the operator is linear. So we've defined something
this way, and it's linear, and it's doing all
the right things. Now, you really feel
proud at this stage. This is still not
all that intuitive. What does this all do? So we're going to do
an example, and we're going to do one more property. Let me do one more property
and then stop for a second. So here is one property--
ST dagger is supposed to be T dagger S dagger. So how do you get that? Not hard-- u, STv. Well, STv is really the
same as S acting on Tv. Now the first S can be
brought to the other side by the definition that
you can bring something to the other side. Put in a dagger. So the S is brought
there, and you get S dagger on u, T on v. And
then the T can be brought here and act on this one, and
you get T dagger S dagger u, v. So this thing is the
dagger of this thing, and that's the statement here. There's yet one more
simple property, that the dagger of S dagger
is S. You take dagger twice and you're back to
the same operator. Nothing has changed. So how do you do that? Take, for example, this-- take
u, put S dagger here, and put v. Now, by definition,
this is equal to-- you put the operator on the
other side, adding a dagger. So that's why we put
that one like this. The operator gets
daggers, so now you've got the double dagger. So at this moment,
however, you have to do something
to simplify this. The easiest thing to do is
probably the following-- to just flip these
two, which you can do the order
by putting a star. So this is equal. The left hand side
is equal to this. And now this S dagger can be
moved here and becomes an S. So this is u, Sv, and
you still have the star. And now reverse this by
eliminating the star, so you have S--
I'm sorry, I have this notation completely wrong. Sv-- this is u. The u's v's are easily confused. So this is v, and this is u. I move the S, and then finally
I have Su, v without a star. I flipped it again. So then you compare these two,
and you get the desired result. OK, so we've gone
through this thing, which is the main result
of daggers, and I would like to see if
there are questions. Anything that has been unclear
as we've gone along here? And question? OK. No questions. So let's do a simple
example, and it's good because it's useful to
practice with explicit things. So here's an example. There's a vector space V,
which is three complex numbers, three component vectors--
complex vectors. So a v is equal to v1, v2, v3--
three numbers are all the vi. Each one belongs to
the complex number. So three complex numbers makes
a vector space like this. So somebody comes
along and gives you the following linear map--
T on a vector, v1, v1, v3, gives you another vector. It's a linear map. So what is it? It's 0 times v1 plus 2v2 plus
iv3 for the first component. The first component
of the new vector-- I put the 0v1 just so you see that
it just depends on v2 and v3. The second component is
v1 minus iv2 plus 0v3. Those are not vectors. These are components. These are numbers. So this is just
a complex number. This is another complex
number, as it should be. Acting on three
complex numbers gives you, linearly, three other ones. And then the third component--
they don't have space there, so I'll put it here--
3iv1 plus v2 plus 7v3. And the question
is two questions. Find T dagger, and write
the matrix representations of T and T dagger. Write the matrices
T and T dagger using the standard basis
in which the three basis vectors are 1, 0, 0, 0, 1,
1, 0, 0, 1, 0, and 0, 0, 1. These are the three basis
vectors-- e1, e2, and e3. You know, to write the matrix
you need the basis vectors. So that's a problem. It's a good problem
in order to practice, to see that you
understand how to turn an operator into a matrix. And you don't get confused. Is it a row? Is it a column? How does it go? So let's do this. So first we're going to try to
find the rules for T dagger. So we have the following. You see, you use
the basic property. u on Tv is equal
to T dagger u on v. So let's try to compute the left
hand side, and then look at it and try to see if we could
derive the right hand side. So what is u supposed to be
a three component vector? So for that use, u
equals u1, u2, u3. OK, now implicit in all that is
that when somebody tells you-- OK, you've got a three
dimensional complex vector space what is the inner product? The inner product
is complex conjugate of the first component. That's first component of the
second, plus complex conjugate of the second times
star, times star. So it's just a generalization
of the dot product, but you complex conjugate
the first entries. So what is this? I should take the complex
conjugate of the first term here-- u1-- times the first one. So I have 2v2 plus iv3. This is the left hand side,
plus the complex conjugate of the second
component-- there's the second component--
so u2 times v1 minus iv2 plus-- well, 0v3--
his time I won't write it-- plus u3 bar
times the last vector, which is 3iv1 plus v2 plus 7v3. OK, that's the left hand side. I think I'm going to use
this blackboard here, because otherwise
the numbers are going to be hard to see from
one side to the other. So this information,
those two little proofs, are to be deleted. And now we have
this left hand side. Now, somehow when
you say, OK, now I'm going to try to figure out this
right hand side your head goes and looks in there
and says well, in the left hand side the
u's are sort of the ones that are alone, and the
v's are acted upon. Here the v's must be alone. So what I should do
is collect along v. So let's collect along v. So
let's put "something" times v1 plus "something" like v2
plus "something" like v3. And then I will know what
is the vector T star this. So let's do that. So v1, let's collect. So you get u2 bar for
this v1, and 3iu3 bar. v2 will have 2u1 bar
minus iu2 bar plus u3 bar. I think I got them right. OK. And then v3, let's collect--
iu1 bar, nothing here, and v3 7u3 bar. OK, and now I must say, OK,
this is the inner product of T dagger u times v3. So actually, T dagger on
u, which is u1, u2, u3, must be this vector with three
components for which this thing is the inner product of this
vector with the vector V. So I look at this
I say, well, what was the formula for
the inner product? Well, you complex conjugate
the first entry of this and multiply by the
first entry of that. Complex conjugate
the second entry. So here I should
put u2 minus 3iu3, because the complex
conjugate of that is that as multiplied by v1. So here I continue--
2u1 plus iu2 plus u3. And, finally,
minus iu1 plus 7u3. And that's the answer
for this operator. So the operator
is there for you. The only thing we haven't
done is the matrices. Let me do a little
piece of one, and you try to compute the rest. Make sure you understand it. So suppose you get T
on the basis vector e1. It's easier than what it looks. I'm going to have to
write some things in order to give you a few
components, but then once you get a little practice,
or you look what it means, it will become clear. So what is T on e1? Well, it's T on
the vector 1, 0, 0. T on the vector 1, 0, 0-- look
at the top formula ther3-- is equal to 0, 1, and 3i. Top formula-- the v1 is
1, and all others are 0. And this is e2 plus 3ie3. So how do you read,
now, matrix elements? You remember the
formula that T on ei is supposed to be
Tkiek-- sum over k. So this thing is supposed
to be equal to T11e1 plus T21e2 plus T31e3. Your sum over the first index,
T of e1, is there for that. So then I read this, and I
see that T21 is equal to 1. This is equal to 3i. And this is equal to 0. So you've got a
piece of the matrix, and the rest I will just
tell you how you see it. But you should check it. You don't have to write
that much after you have a little
practice with this. But, the matrix T--
what you've learned is that you have 0, 1, and 3i. So 0, 1, and 3i
are these numbers, in fact-- 0, 1, and 3i. And they go vertical. So 2, minus i, and 1
is the next column. 2, minus i, and 1
is the next column, and the third one would be
i-- look at the v3 there. It has an i for the first entry,
a 0 for the second, and a 7. So this is the matrix. How about the matrix T dagger? Same thing-- once you've
done one, don't worry. Don't do the one. So this you look for
the first column. It's going to be a 0-- no u1
here-- a 2, and a minus i. 0, 2, and a minus i, then 1,
i, and 0, minus 3i, 1, and 7. And those are it. And look how nice. The second one is in fact
the Hermitian conjugate of the other. Transpose and complex
conjugate gives it to you. So that example suggests
that that, of course, is not an accident. So what do you need
for that to happen? Nobody said that what you're
supposed to do to find T dagger is transpose some
complex conjugate, but somehow that's what you
do once you have the matrix, or at least what
it seems that you do when you have the matrix. So let's see if we can
get that more generally. So end of example. Look at T dagger u,
v is equal to u, Tv. We know this is
the key equation. Everything comes from this. Now take u and v to be
orthonormal vectors, so u equal ei, and v equal ej. And these are orthonormal. The e's are going to be
orthonormal each time we say basis vectors--
e, orthonormal. So put them here, so you
get T dagger on ei times ej is equal to ei, Tej. Now use the matrix action
on these operators. So T dagger on ei is
supposed to be T dagger kiek. The equation is something
worth knowing by heart. What is the matrix
representation? If the index of the
vector goes here, the sum index goes like that. So then you have ej here,
and here you have ei, and you have Tkjek. So now this basis orthonormal. This is a number, and
this is the basis. The number goes out. T dagger ki-- remember,
it's on the left side, so it should go out with a star. And then you have ekej. That's orthonormal,
so it's delta kej. The number here
goes out as well, and the inner product
gives delta ik. So what do we get? T dagger ji star
is equal to Tij. First, change i for j, so
it looks more familiar. So then you have T dagger
ij star is equal to Tji. And then take complex
conjugate, so that finally you have T dagger ij is
equal to Tji star. And that shows that, as long as
you have an orthonormal basis you can see the Hermitian
conjugate of the operator by taking the matrix,
and then what you usually call the Hermitian
conjugate of the matrix. But I want to
emphasize that, if you didn't have an
orthonormal basis-- if you have your operator, and you want
to calculate the dagger of it, and you find its
matrix representation. You take the Hermitian
conjugate of the matrix. It would be wrong if your basis
vectors are not orthonormal. It just fails. So what would happen if
the basis vectors are not orthonormal? Instead of having ei with
ej giving you delta iej, you have that ei with
ej is some number. And you can call it aij, or
alpha iej, or gij, I think, is maybe a better name. So if the basis is not
orthonormal, then ei with ej is some sort of gij. And then you go back here. And, instead of having deltas
here, you would have g's. So you would have the T
dagger star ki with gkj is equal to Tkj, gik. And there's no such
simple thing as saying, oh, well you just take
the matrix and complex conjugate and transpose. That's not the dagger. It's more complicated than that. If this matrix
should be invertible, you could pass this
to the other side using the inverse
of this matrix. And you can find a
formula for the dagger in terms of the g matrix, its
inverses and multiplications. So what do you learn from here? You learn a fundamental
fact, that the statement that an operator-- for
example, you have T. And you can find T
dagger as the adjoint. The adjoint operator, or the
Hermitian conjugate operator, has a basis
independent definition. It just needs that
statement that we've written many times
now, that T dagger u, v is defined via this relation. And it has nothing
to do with a basis. It's true for arbitrary vectors. Nevertheless, how you
construct T dagger, if you have a basis--
well, sometimes it's a Hermitian
conjugate matrix, if your basis is orthonormal. But that statement, that
the dagger is the Hermitian conjugate basis, is a
little basis dependent, is not a universal
fact about the adjoint. It's not always
constructed that way. And there will be examples
where you will see that. Questions? No questions? Well, let's do brackets
for a few minutes so that you see a few
properties of them. With the same language, I'll
write formulas that we've-- OK, I wrote a formula here, in fact. So for example,
this formula-- if I want to write it with bras
and kets, I would write u Tv. And I could also
write it as u T v, because remember this
means-- the bra and the ket-- just says a way to make clear
that this object is a vector. But this vector is obtained
by acting T on the vector v. So it's T on the vector
v, because a vector v is just something, and when you put
it like that that's still the vector v. The kit
doesn't do much to it. It's almost like
putting an arrow, so that's why this thing is
really this thing as well. Now, on the other
hand, this thing-- let's say that this is
equal to v, T dagger u star. So then you would put here
that this is v T dagger u star. So this formula is something
that most people remember in physics, written perhaps
a little differently. Change v and u so that
this left hand side now reads u T dagger v.
And it has a star, and the right hand side
would become v T u. And just complex conjugated it. So u T dagger v is equal to v
T u star-- a nice formula that says how do you get to
understand what T dagger is. Well, if you know
T dagger's value in between any set
of states, then you know-- well, if you know T
between any set of states u and v, then you
can figure out what T dagger is between any same two
states by using this formula. What you have to do
is that this thing is equal to the reverse thing. So you go from right to
left and reverse it here. So you go v, then T, then
u, and you put a star, and that gives you that object. Another thing that we've
been doing all the time when we calculate, for
example, ei, T on ej. What is this? Well, you know what this is. Let's write it like that-- ei. Now T on ej is
the matrix T kjek. If this is an orthonormal
basis, here is a delta iek. So this is nothing else but Tij. So another way of writing that
matrix element, ij, of a matrix is to put an ei, an
ej here, and a T here. So people write it like
that-- Tij is ei comma Tej. Or, in bracket language,
they put ei T ej. So I need it to be
flexible and just be able to pass from one
notation to the other, because it helps you. One of the most helpful
things in this object is to understand, for example,
in bra and ket notation, what is the following object? What is ei ei? This seems like the
wrong kind of thing, because you were supposed to
have bras acting on vectors. So this would be on
the left of that, but otherwise it
would be too trivial. If it would be on the left of
it, it would give you a number. But think of this thing as
a object that stands there. And it's repeated
endlessly, so it's summed. So what is this object? Well, this object is a
sum of things like that, so this is really e1 e1 plus e2
e2, and it goes on like that. Well, let it act on a vector. This kind of object
is an operator. Whenever you have
the bra and the ket sort of in this wrong
position-- the ket first, and the bra afterwards--
this is, in Dirac's notation, an operator, a
particular operator. And you will see in general
how it is the general operator very soon. So look at this. You have something like
that, and why do we call it an operator? We call it an operator
business if it acts on a vector--
you put a vector here, a bra-- this becomes a
number, and there's still a vector left. So this kind of structure,
acting on something like that, gives a vector, because
this thing goes in here, produces a number, and
the vector is left there. So for example, if you act with
this thing on the vector a-- an arbitrary vector
a-- what do you get? Whatever this operator
is is acted on a. Well, you remember that these
thing are the components of a, and these are the basis vectors. So this is nothing else
but the vector a again. You see, you can start with
a equals some alpha i's with ei's, and then you calculate
what are the alpha i's. You put an ej a, and this ej
on that gives you alpha j. So alpha j-- these
numbers are nothing else but these things, these numbers. So here you have the
number times the vector. The only difference is that
this is like ei alpha i. The number has
been to the right. So this thing acting on any
vector is the vector itself. So this is perhaps the
most fundamental relation in bracket notation, is that
the identity operator is this. Yes. AUDIENCE: Is that just 1 e
of i, or sum over all e of i? PROFESSOR: It's sum of over all. So here implicit sum is
the sum of all up to en en. You will see, if you
take just one of them, you will get what is an
orthogonal projector. Now this allows you
to do another piece of very nice Dirac notation. So let's do that. Suppose you have an
operator T. You put a 1 in front of it-- a T
and a 1 in front of it. And then you say, OK,
this 1, I'll put ei ei. Then comes the T, and then
comes the ej ej-- another 1. And then you look at
that and you suddenly see a number lying there. Why? Because this thing
is some number. So this is the magic
of the Dirac notation. You write all this
thing, and suddenly you see numbers have been
created in between. This number is nothing else
but this matrix representation of the operator. T, between this, is Tij. So this is ei Tij ej. So this formula is
very fundamental. It shows that the most general
operator that you can ever invent is some sort
of ket before a bra, and then you superimpose
them with these numbers which actually happen to be the matrix
representation of the operator. So the operator can
be written as a sum of, if this is an n by n matrix
n squared thinks of this form-- 1 with 1, 1 with 2, 1
with 3, and all of them. Bu then, you know this
formula is so important that people make
sure that you realize that you're summing
over i and j. So just put it there. Given an operator, these
are its matrix elements. And this is the operator written
back in abstract notation. The whole operator is
back there for you. I want to use the last
part of the lecture to discuss a theorem that
is pretty interesting, that allows you to
understand things about all these Hermitian
operators and unitary operators much more clearly. And it's a little
mysterious, this theorem, and let's see how it goes. So any questions about this
Dirac notation at this moment, anything that I wrote there? It takes a while
to get accustomed to the Dirac notation. But once you get
the hang of it, it's sort of fun and
easy to manipulate. No questions? Can't be. You can prove all
kinds of things with this matrix
representation of the identity. For example, you
can prove easily something you proved already,
that when you multiply two operators the
matrices multiply. You can prove all
kinds of things. Pretty much
everything we've done can also be proven this way. OK, so here comes the theorem
I want to ask you about. Suppose somebody comes
along, and they tell you, well, you know,
here's a vector v, and I'm going to have a linear
operator acting on this space. So the operator's
going to be T, and I'm going act with the vector v. And moreover, I
find that this is 0 for all vectors v belonging
to the vector space. And the question is-- what can
we say about this operator? From all vectors it's just 0. So is this operator 0, maybe? Does it have to be 0? Can it be something else? OK, we've been talking about
real and complex vector spaces. And we've seen that
it's different. The inner product is
a little different. But let's think about this. Take two dimensions,
real vector space. The operator that
takes any vector and rotates it by 90 degrees,
that's a linear operator. And that is a non-trivial linear
operator, and it gives you 0. So case settled-- there's
no theorem here, nothing you can say about this operator. It may be non-zero. But here comes the catch. If you're talking complex
vector spaces, T is 0. It just is 0, can't
be anything else. Complex vector
spaces are different. You can't quite do that thing--
rotate all vectors by something and do things. So that's a theorem
we want to understand. Theorem-- let v be a
complex inner product space. By that is a complex vector
space with an inner product. Then v, Tv equals
0 for all v implies that the operator is just 0. I traced a lot of my
confusions in quantum mechanics to not knowing
about this theorem, that somehow it must be true. I don't know why
it should be true, but somehow it's not, because
it really has exceptions. So here it is. We tried to prove that. It's so important, I think,
that it should be proven. And how could you prove that? And at first sight it seems
it's going to be difficult, because, if I do
just a formal proof, how is it going to
know that I'm not talking real or
complex vector spaces. So it must make a crucial
difference in the proof whether it's real or complex. So this property really sets
the complex vector spaces quite apart from the real ones. So let's see what
you would need to do. Well, here's a strategy--
if I could prove that u, Tv is equal to 0 for all
u and all v. You see, the problem here is that
these two are the same vector. They're all vectors, but
they're the same vector. If I could prove that this
is 0 for all u and v, then what would I say? I would say, oh, if this
is 0 for all u and v, then pick u equal to Tv. And then you find
that Tv, Tv is 0, therefore Tv is the 0 vector. By the axiom of
the inner product, for all v is a 0 vector,
so T kills all vectors, therefore T is 0. So if I could prove this
is true, I would be done. Now, of course,
that's the difficulty. Well, I wouldn't say of course. This takes a leap
of faith to believe that this is the way
you're going to prove that. You could try to prove this,
and then it would follow. But maybe that's
difficult to prove. But actually that's
possible to prove. But how could you ever
prove that this is true? You could prove it
if you could somehow rewrite u and Tv as some
sort of something with a T and something plus some
other thing with a T, and that other thing plus some--
all kinds of things like that. Because the things in which
this is the same as that are 0. So if you can do that-- if you
could re-express this left hand side as a sum of things of
that kind-- that would be 0. So let's try. So what can you try? You can put u plus v here,
and T of u plus v. That would be 0, because that's
a vector, same vector here. But that's not equal to this,
because it has the u, Tu, and it has the v Tv. And it has this in
a different order. So maybe we can subtract u
minus v, T of u minus v. Well, we're getting
there, but all this is question marks-- u, Tu, v,
Tv-- these cancel-- u, Tu, v, Tv. But, the cross-products,
what are they? Well here you have a u, Tv. And here you have a v, Tu. And do they cancel? No. Let's see. u, Tv, and up here
is u minus Tv about. But there's another minus,
so there's another one there. And v, Tu has a minus,
minus is a plus. So actually this gives me
two of this plus two of that. OK, it shouldn't have
been so easy anyway. So here is where you have to
have the small inspiration. Somehow it shouldn't
have worked, you know. If this had worked, the
theorem would read different. You could use a
real vector space. Nothing is imaginary there. So the fact that you have
a complex vector space might help. So somehow you have
to put i's there. So let's try i's here. So you put u plus iv
and T of u plus iv. Well, then you probably have
to subtract things as well, so u minus iv, T of u minus iv. These things will be 0 because
of the general structure-- the same operator here as here. And let's see what they are. Well, there's u, Tu,
and here's minus u, Tu, so the diagonal things go away--
the minus iv, minus iv, iv, and a T. You have minus
iv, minus iv subtracted, so that also cancels. So there's the cross-products. Now you will say, well,
just like the minus signs, you're not going to get
anything because you're going to get 2 and 2. Let's see. Let's see what we
get with this one. You get u with Tiv,
so you get i u, Tv. But look, this i on the left,
however, when you take it out, becomes a minus i, so
you get minus i v, Tu. And the other
products [INAUDIBLE]. So let's look what you get
here-- a u with a minus iv and a minus here
gives you a 2 here. And the other
term, v, Tu-- well, this goes out as a plus i. But with a minus, it becomes
a minus i, so v, Tu is this. So there's a 2 here. So that's what these
terms give you. And now you've succeeded. Why? Because the relative
sign is negative. So who cares? You can divide by i,
and divide this by i. You are constructing something. So let me put here what you get. I can erase this blackboard. So what do we get? I claim that if you put
one quarter of u plus v, T u plus v minus u minus
v, T of u minus v, then, let's see, what do
we need to keep? We need to keep u and Tv. So divide this by
i plus 1 over i u plus iv, T of
u plus iv minus 1 over i, u minus iv,
T of u minus iv. And close it. You've divided by i. You get here four of these
ones, zero of these ones, and you got the
answer you wanted. So this whole thing
is written like that, and now, since this
is equal to u with Tv, by the conditions
of the theorem, any vector-- any vector
here-- these are all 0. You've shown that this is 0,
and therefore the operator is 0. And you should be
very satisfied, because the proof
made use of the fact that it was a
complex vector space. Otherwise you could
not add vectors with an imaginary number. And the imaginary
number made it all work. So the theorem is there. It's a pretty useful
theorem, so let's use it for the most
obvious application. People say that,
whenever you find that v, Tv is real for all
v, then this operator is Hermitian, or self-adjoint. That is, then, it implies
T dagger equals T. So let's show that. So let's take v, Tv. Proof. You take v, Tv, and
now this thing is real. So since this is real, you can
say it's equal to v, Tv star. Now, because it's real--
that's the assumption. The number is real. Now, the star off an
inner product is Tv, v. But on the other
hand, this operator, by the definition of
adjoint, can be moved here. And this is equal
to T dagger v, v. So now you have done
this is equal to this. So if you put it
to one side, you get that T dagger minus T
on v times v is equal to 0. Or, since any inner
product that is 00-- it's complex conjugate is
0-- you can write it as v, T dagger minus v is 0 for all v. And so this is an actually
well known statement, that any operator that gives you
real things must be Hermitian. But it's not obvious, because
that theorem is not obvious. And now you can use a
theorem and say, well, since this is true for all
v, T dagger minus T is 0, and T dagger is equal to T. Then
you can also show, of course, if T dagger is equal to
T, this thing is real. So in fact, this
arrow is both ways. And this way is very easy, but
this way uses this theorem. There's another
kind of operators that are called
unitary operators. We'll talk a little more
about them next time. And they preserve
the norm of vectors. People define them
from you, and you see that they preserve
the norm of vectors. On the other hand, you
sometimes find an operator that preserves every norm. Is it unitary? You will say, yes, must be. How do you prove it? You need again that theorem. So this theorem is
really quite fundamental to understand the
properties of operators. And we'll continue
that next time. All right.
