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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Now, a theorem that
was quite powerful and applied to complex vector
spaces for old V longing to V complex vector space. This implied that the
operator was zero, and it's not true for
real vector spaces. And we gave a proof
that made it clear that indeed, a proof wouldn't
work for a real vector space. I will ask you in the homework
to do something that presumably would be the first thing
you would do if you had to try to understand why this
is true-- take two by two matrices, and just see why
it has to be 0 in one case, and it doesn't have to
be 0 in the other case. And I think that will give you
a better perspective on why this happens. And then once you do
it for a two by two, and you see how it works, you
can do it for n by n matrices, and it will be clear as well. So we'll use this theorem. Our immediate application of
this theorem was a well-known result that we could
prove now rigorously-- that t is equal
to t dagger, which is to say the operator
is equal to the adjoined. And I think I will call
this-- and in the notes you will see always the
adjoint, as opposed to Hermitian congregate. And I will say
whenever the operator is equal to the adjoint,
that it is Hermitian. So a Hermitian operator
is equivalent to saying that v Tv is a real number
for all v. And we proved that. In other words, physically,
Hermitian operators have real expectation values. This is an expectation
value, because-- as you remember in
the bracket notation-- v Tv, you can write it v T v-- the same thing. So it is an expectation value,
so it's an important thing, because we usually deal
with Hermitian operators, and we want expectation
values of Hermitian operators to be real. Now that we're talking
about Hermitian operators, I delay a complete discussion
of diagonalization, and diagonalization
of several operators simultaneously, for
the next lecture. Today, I want to move forward
a bit and do some other things. And this way we spread out
a little more the math, and we can begin to look
more at physical issues, and how they apply here. But at any rate, we're
just here already, and we can prove two
basic things that are the kinds of things
that an 805 student should be able to prove at any time. They're really very simple. And it's a kind of proof
that is very short-- couple of lines--
something you should be able to reproduce at any time. So the first theorem
says the eigenvalues of Hermitian operators-- H is for Hermitian-- are real. And I will do a little bit of
notation, in which I will start with an expression, and
evaluate it sort of to the left and to the right. So when you have an
equation, you start here, and you start evaluating there. So I will start with this-- consider v Tv. And I will box it
as being the origin, and I will start evaluating it. Now if v is an eigenvector-- so let v be an eigenvector so
that Tv is equal to lambda v. And now we say consider
that expression in the box. And you try to evaluate it. So one way to evaluate it-- I evaluate it to the left, and
then evaluate to the right. Is the naive evaluation-- t on v is lambda v, so
substitute it there. v, lambda v-- we know it. And then by homogeneity,
this lambda goes out, and therefore it's lambda v v. On the other hand, we have
that we can go to the right. And what is the way you move an
operator to the first position? By putting a dagger. That's a definition. So this is by definition. Now, we use that the
operator is Hermitian. So this is equal to Tv v.
And this is by T Hermitian. Then you can apply again the
equation of the eigenvalues, so this is lambda v v. And
by conjugate homogeneity of the first input,
this is lambda star v v. So at the end of
the day, you have something on the extreme
left, and something on the extreme right. v-- if there is
an eigenvector, v can be assumed to be non-zero. The way we are saying
things in a sense, 0-- we also think of
it as an eigenvector, but it's a trivial one. But the fact that
there's an eigenvalue means there's a non-zero v
that solves this equation. So we're using that non-zero
v. And therefore, this is a number that is non-zero. You bring one to the
other side, and you have lambda minus lambda
star times v v equals 0. This is different from 0. This is different from 0. And therefore, lambda
is equal to lambda star. So it's a classic proof-- relatively straightforward. The second theorem is
as simple to prove. And it's already interesting. And it states that
different eigenvectors of a Hermitian operator-- well, different eigenvalues
of Hermitian operators correspond to orthogonal
eigenfunctions, or eigenvectors. So different eigenvalues
of Hermitian ops correspond to orthogonal eigenfunctions-- eigenvectors, I'm sorry. So what are we saying here? We're saying that
suppose you have a v1 that [INAUDIBLE] T
gives you a lambda1 v1. That's one eigenvalue. You have another one-- v2
is equal to lambda 2 v2, and lambda 1 is
different from lambda 2. Now, just focusing on a fact
that is going to show up later, is going to make
life interesting, is that some
eigenvalues may have a multiplicity of eigenvectors. In other words, If a
vector v is an eigenvector, minus v is an eigenvector,
square root of three v is an eigenvector, but that's
a one-dimensional subspace. But sometimes for
a given eigenvalue, there may be a
higher dimensional subspace of eigenvectors. That's a problem of degeneracy,
and it's very interesting-- makes life really interesting
in quantum mechanics. So if you have degeneracy,
and that set of eigenvectors form a subspace, and
you can choose a basis, and you could have
several vectors here. Now what do you do in that case? The theorem doesn't say much,
so it means choose any one. If you had the bases
there, choose any one. The fact remains that if these
two eigenvalues are different, then you will be able to show
that the eigenvectors are orthogonal. So if you have some
space of eigenvectors-- a degenerate higher-dimensional
space of eigenvectors, one eigenvalue, and another
space with another-- any vector here is orthogonal
to any vector there. So how do you show this? How do you show this property? Well you have to
involve v1 and v2, so you're never
going to be using the property that gives
Hermitian, unless you have an inner product. So if you don't have any
idea how to prove that, you presumably at some stage
realize that you probably have to use an inner product. And we should mix
the vectors, so maybe a V2 inner product with this. So we'll take a v2
inner product with T v1. And this is interesting,
because we can use it, that T v1 is lambda 1 v1 to
show that this is just lambda 1 v2 v1. And that already brings
all kinds of good things. You're interested in
this inner product. You want to show it's
0, so it shows up. So it's a good idea. So we have evaluated
this, and now you have to think of evaluating
it in a different way. Again, the operator
is Hermitian, so it's asking you to
move it to the other side and exploit to that. So we'll move it to the
other side a little quicker this time. It goes as T dagger, but
T dagger is equal to T, because it's Hermitian. So this is the center
of the equation. We go one way. We go the other
way-- this time down. So we'll put T v2 v1, and
this is equal to lambda-- let me go a little slow here-- lambda 2 v2 v1. Your impulse should be it
goes out as lambda 2 star, but the eigenvalues
are already real, so it goes out as
lambda 2 v2 v1, because the operator
is Hermitian. So at this moment, you
have these two equations. You bring, say, this
to the right-hand side, and you get lambda 1 minus
lambda 2 v1 v2 is equal to 0. And since the eigenvalues
are supposed to be different, you conclude that v1 inner
product with v2 is 0. So that's the end of the proof. And those are the two properties
that are very quickly proven with rather little effort. So where do we go from now? Well there's one more
class of operators that are crucial in the physics. They are perhaps as important
as the Hermitian operators, if not more. They are some operators that
are called unitary operators, and the way I will introduce
them is as follows-- so I will say-- it's an economical way to
introduce them-- so we'll talk about unitary operators. If S is unitary,
and mathematicians call it anisometry-- if you find that S
acting on any vector-- if you take the norm, it's
equal to the norm of the vector for all u in the vector space. So let's follow this, and
make a couple of comments. An example-- a trivial example-- this operator lambda
times the identity. Lambda times the
identity acts on vectors. What does it do,
lambda times identity? The identity does
nothing on the vector, and lambda stretches it. So lambda, in order not to
change the length of any vector should be kind of 1. Well, in fact, it suffices-- it's unitary-- if the absolute
value of lambda is equal to 1. Because then lambda is a phase,
and it just rotates the vector. Or in other words, you
know that the norm of av is equal to the absolute
value of a times the norm of v, where
this is a number. And remember these two
norms are different. This is the norm of a vector. This is the normal
of a complex number. And therefore, if you
take lambda i u-- norm-- is lambda u is equal
absolute value of lambda u, and absolute value of lambda
is equal to 1 is the answer. So that's a simple unitary
operator, but an important one. Another observation--
what are the vectors annihilated by this operator u? Zero-- it's the only vector,
because any other vector that's nonzero has some length,
so it's not killed. So it kills only zero. So the null space of S
is equal to the 0 vector. So this operator has no
kernel, nothing nontrivial is put to zero. It's an invertible operator. So s is invertible. So that's a few things
that you get very cheaply. Now from this
equation, S u equals u-- if you square that
equation, you would have S u S u is equal to u u. Maybe I should
probably call it v. I don't know why I called
it u, but let's stick to u. Now, remember that
we can move operators from one side to the other. So I'll move this
one to that side. If you move an S here,
you would put an S dagger. But since the dagger
of an S dagger is S, you can move also the S
to that side as S dagger. So u S dagger, S u-- you see that. If you want to
move this one, you can move it by putting another
dagger, and you get that one. And this is u u,
and therefore you get u S dagger, S minus
the identity acting on u is equal to 0 for all u. So for every vector, this is
true, because this is true. We just squared it. And now you have our
favorite theorem, that says if this is true
in a complex vector space, this is 0, and
therefore, you've shown that S dagger S is equal to 1. So that's another property
of unitary operators. In fact that's the way
it's many times defined. Unitary operators
sometimes are said to be operators whose
inverse is S dagger. I will not go into
the subtleties of what steps in all
these things I'm saying are true or not true
for infinite dimensional operators-- infinite dimensional
vector spaces. So I will assume, and it
will be true in our examples, that if S dagger is an
inverse from the left, it's also an inverse
from the right. And perhaps everything is true
for infinite dimensional vector spaces, but I'm
not 100% positive. So S dagger is the
inverse of S. And that's a pretty important thing. So one last comment on unitary
operators has to do with basis. So suppose you have an
orthonormal basis, e1 up to en. Now you can define
another basis. fi equal-- I'll
change to a letter U-- U e i where U is
unitary, so it's like the S. In fact,
most books in physics call it U for unitary. So maybe I should have changed
that letter in there, too, as well. So suppose you change basis. You put-- oh, there
was something else I wanted to say before. Thanks to this
equation, consider now the following thing-- S U Sv. SUSv-- you can move this S, for
example, to the other side-- S dagger S U v, and S dagger
S is equal to 1, and it's Uv. So this is a pretty
nice property. We started from the
fact that it preserved the norm of a single
vector, of all vectors, and now you see that in fact,
it preserved the inner product. So if you have two vectors, to
compare their inner product, compute them after action with
U or before action with U, and it doesn't
make a difference. So suppose you define
a second basis here. You have one orthonormal basis. You define basis
vectors like this. Then the claim is that the
f1 up to fn is orthonormal. And for that you simply
do the following-- you just check f i f j
is equal to U e i, U e j. By this property, you can
delete both U's, rules, and therefore this is e i, e j. And that's delta i j. So the new basis is orthonormal. If you play with
these things, it's easy to get some extra
curious fact here. Let's think of the matrix
representation of the operator U. Well, we know
how these things are, and let's think of
this in the basis e basis. So U k i is equal to ek U e i. That's the definition
of U in the basis e-- the matrix elements
of U. You can try to figure out what
is Uki in the f basis. How does operator U
look in the f basis? Well, let's just do
it without thinking. So in the f basis,
I would put fk U fi. Well, but fk is U ek,
so I'll put Uek Ufi. Now we can delete both
U's, and it's ek fi. And I can remember what
fi was, which is ek U ei. And it's just the same
as the one we had there. So the operator,
unitary operator, looks the same in both bases. That might seem strange or
a coincidence, but it's not. So I leave it to
you to think about, and visualize why
did that happen. What's the reason? So the bracket notation-- we've been using
it here and there-- and I will ask you to
please read the notes. The notes will be posted this
afternoon, and they will have-- not maybe all we've
done today, but they will have some of
what we'll do today, and all of what
we've been doing. And the way it's
done-- it's first done in this sort of
inner product language, and then things are done
in the bracket language. And it's a little
repetitious, and I'm trying to take out some
things here and there, so it's less repetitious. But at this moment it's
probably worth reading it, and reading it again. Yes. AUDIENCE: [INAUDIBLE] if you
have two orthonormal bases, is the transformation between
them necessarily unitary? PROFESSOR: Yes, yes. All right. So as I was saying we're going
to go into the Dirac notation again. And here's an example of a
place where everybody, I think, tends to use Dirac notation. And the reason is
a little curious, and you will appreciate
it quite fast. So this will be
the case of where we return to x and p operators,
on a non-denumerable basis. So we're going to
try to do x and p. now this is the classic
of Dirac notation. It's probably-- as I said--
the place where everybody likes to use Dirac notation. And the reason it's efficient
is because it prevents you from confusing two things. So I've written in the notes,
and we have all these v's that belong to the vector space. And then we put
this, and we still say it belongs to
the vector space. And this is just a decoration
that doesn't do much. And we can play with this. Now, in the non-denumerable
basis, the catch-- and the possible confusion-- is that the label is not quite
a vector in the vector space. So that is the reason why
the notation is helpful, because it helps you
distinguish two things that you could confuse. So here we go. We're going to talk
about coordinate x, and the x operator,
and the states. Well, this is a state space. So what kind of states
do we have here? Well, we've talked
about wave functions, and we could give
the value of the wave function of different places. We're going to go for a
more intrinsic definition. We're going to try to
introduce position states. And position states
will be called this-- x. Now, what is the meaning
of this position state? We should think of this
intuitively as a particle at x. Now here's how you can
go wrong with this thing, if you stop thinking
for a second. What is, then, ax? Is it ax, a being a number. Is it the same thing? No, not at all. This is a particle
at the coordinate ax, and this is a particle at x
with some different amplitude-- very different. So this is not true-- typical mistake. This is not minus x. That's totally different. So there's no such
thing as this, either. It doesn't mean anything. And the reason is that these
things are not our vectors. Our vector is this whole thing
that says a particle at x. Maybe to make a
clearer impression, imagine you're in
three dimensions, and you have an x vector. So then you have to ket this. This is the ket particle at x. x is now a vector. It's a three-dimensional vector. This is a vector,
but it's a vector in an infinite
dimensional space, because the particle
can be anywhere. So this is a vector
in quantum mechanics. This is a complex vector space. This is a real vector space,
and it's the label here. So again, minus x is
not minus x vector. It's not the vector. The addition of the bra
has moved you from vectors that you're familiar
with, to states that are a little more abstract. So the reason this notation is
quite good is because this is the number, but this i---
or this is a coordinate, and this is a vector already. So these are going to
be our basis states, and they are non-denumerable. And here you can
have that all x must belong to the real numbers,
because we have particles in a line, while this thing
can be changed by real numbers. The states can be multiplied
by complex numbers, because we're doing
quantum mechanics. So if you want to define
a vector space-- now, this is all infinite dimension. It's a little
worse in this sense the basis is non-denumerable. If I use this basis, I cannot
make a list of all the basis vectors. So for an inner
product, we will take the following-- we
will take x with y to be delta of x minus y. That will be our inner product. And it has all the properties
of the inner product that we may want. And what else? Well at this moment,
we can try to-- this is physically
sensible, let me say, because if you have a
particle at one point and a particle at another
point, the amplitude that this particle at one point
is at this other point is 0. And these states are
not normalizable. They correspond to a
particle at the point, so once you try to normalize
them, you get infinity, and you can't do much. But what you can do here is
state more of the properties, and learn how to
manipulate this. So remember we had one was
the sum of all e i e i. The unit operator was that. Well, let's try to
write a similar one. The unit operator will
be the sum over all x's. And you could say,
well, looks reasonable, but maybe there's a 1/2
in here, or some factor. Well, no factor is needed. You can check that-- that you've defined
this thing properly. So let me do it. So act to on this
so-called resolution of the identity with the vector
y, so 1 on y is equal to y. And now let's add
on the right xxy. This is delta of x minus y. And then when you
integrate, you get y. So we're fine. So this looks a
little too abstract, but it's not the abstract if you
now introduce wave functions. So let's do wave functions. So you have a particle, a
state of the particle psi. Time would be
irrelevant, so I will put just this psi like that
without the bottom line. And let's look at it. Oh, I want to say
one more thing. The x operator acts on
the x states to give x x. So these are eigenstates
of the x operator. We declare them to be
eigenstates of the x operator with eigenvalue x. That's their physical
interpretation. I probably should
have said before. Now, if we have a psi as a
state or a vector, how do we get the wave function? Well, in this language
the wave function, which we call psi
of x, is defined to be the overlap of x with psi. And that makes sense,
because this overlap is a function of this label
here, where the particle is. And therefore, the result
is a complex number that is dependent on x. So this belongs to
the complex numbers, because inner products
can have complex numbers. Now, I didn't put any
complex number here, but when you form states,
you can superpose states with complex numbers. So this psi of x will
come out this way. And now that you
are armed with that, you can even think of
this in a nicer way. The state psi is
equal to 1 times psi. And then use the rest of this
formula, so this is integral-- dx x x psi. And again, the bracket
notation is quite nice, because the bra
already meets the ket. This is a number, and
this is dx x psi of x. This equation has a
nice interpretation. It says that the state is a
superposition of the basis states, the position
states, and the component of your original state
along the basis state x is precisely the value
of the wave function at x. So the wave function
at x is giving you the weight of the state x
as it enters into the sum. So one can compute more things. You will get practice in
this type of computations. There are just a limited type
of variations that you can do, so it's not that complicated. Basically, you can introduce
resolutions of the identity wherever you need them. And if you introduce too
many, you waste time, but you typically get
the answer anyway. So it's not too serious. So suppose you
want to understand what is the inner
product of two states. Put the resolution of
the identity in between. So put phi, and then put
the integral dx x x psi. Well, the integral goes out,
and you get phi x x psi. And remember, if
x psi is psi of x, phi x is the complex conjugate,
so it's phi star of x. And you knew that. If you have two wave
functions, and you want to compute the overlap, you
integrate the complex conjugate of one against the other. So this notation is doing
all what you want from this. You want to compute a
matrix element of x. Well, put another resolution
of the identity here. So this would be
integral dx phi-- the x hat is here. And then you put x x psi. The x hat on x is x. That's what this operator does,
so you get integral dx of-- I'll put x phi x x psi, which
is what you expect it to be-- integral of x phi
star of x, psi of x. Now we can do exactly the same
thing with momentum states. So I don't want to bore you,
so I just list the properties-- basis states are momenta
where the momenta is real. p prime p is equal delta
of p minus p prime. One is the integral dp of p p. And p hat p is equal to p p. So these are the momentum bases. They're exactly analogous. So all what we've
done for x is true. The completeness
and normalization work well together,
like we checked there, and everything is true. The only thing that you need
to make this more interesting is a relation between the
x basis and the p basis. And that's where
physics comes in. Anybody can define these two,
but then a physical assumption as to what you really mean
by momentum is necessary. And what we've said is
that the wave function of a particle with
momentum p is e to the i px over h bar over
square root of 2 pi h-- convenient normalization,
but that was it. That was our physical
interpretation of the wave function of a
particle with some momentum. And therefore, if this is
a wave function, that's xp. A state of momentum p
has this wave function. So we write this. OK, there are tricks you can
do, and please read the notes. But let's do a
little computation. Suppose you want to
compute what is p on psi. You could say, well, I
don't know why would I want to do something with that? Looks simple enough. Well, it's simple
enough, but you could say I want to see that
in terms of wave functions, coordinate space wave functions. Well, if you want to see them in
terms of coordinate space wave functions, you have to introduce
a complete set of states. So introduce p x x psi. Then you have this
wave function, and oh, this is sort
of known, because it's the complex conjugate
of this, so it's integral dx px over h bar,
square root of 2 pi h bar times psi of x. And this was the
Fourier transform-- what we call the Fourier
transform of the wave function. So we can call it psi tilde
of p, just to distinguish it, because we called
psi with x, psi of x. So if I didn't put
a tilde, you might think it's the same
functional form, but it's the momentum
space wave function. So here is the wave
function in the p basis. It's the Fourier
transform of the wave function in the x basis. One last computation, and
then we change subjects again. It's the classic
computation that you have now a mixed
situation, in which you have the momentum operator
states and the coordinate bra. So what is the
following expression-- X p hat psi? OK. What is your temptation? Your temptation is
to say, look, this is like the momentum
operator acting on the wave function in the x basis. It can only be h bar
over i d dx of psi of x. That's probably what it means. But the notation
is clear enough, so we can check if that
is exactly what it is. We can manipulate
things already. So let's do it. So for that, I first have to
try to get rid of this operator. Now the only way I know how
to get rid of this operator p is because it has eigenstates. So it suggests very strongly
that we should introduce momentum states, complete them. So I'll put v p x p hat p p psi. And now I can
evaluate the little-- because p hat and p is little
p, or p without the hat. So this is p xp p psi. Now you can look
at that, and think carefully what should you do. And there's one
thing that you can do is look at the equation on top. And this is a way to
avoid working very hard. So look at the equation on
top-- x p is equal to that. How do I get a p
to multiply this? I can get a p to
multiply this xp by doing h bar over i d dx of x p. Because if I see it there, I
see that differentiating by d dx brings down an ip over h bar. So if I multiply by h
bar over i, I get that. So let's do this. Now I claim we can
take the h over i d dx out of this integral. And the reason is that first,
it's not an x integral. It's a p integral, and nothing
else except this factor depends on x. So I take it out and I
want to bring it back, it will only act on
this, because this is not x dependent. So you should think of psi, psi
doesn't have an x dependence. Psi is a state, and here is p-- doesn't have an x dependence? You say no, it
does, it looks here. No, but it doesn't have it,
because it's been integrated. It really doesn't
have x dependence. So we can take this out. We'll have h over i d dx. And now we have vp x p p psi. And now by completeness,
this is just 1. So this becomes x psi. So h bar over i d dx of x psi,
which is what we claimed it would be. So this is rigorous-- a rigorous derivation. There's no guessing. We've introduced
complete states until you can see how things act. But the moral is here
that you shouldn't have to go through this
more than once in your life, or practice it. But once you see something
like that, you think. You're using x
representation, and you're talking about the operator p. It cannot be anything like that. If you want to practice
something different, show that the analogue p x
hat psi is equal i h bar d dp of psi tilde. So it's the opposite relation. All right. Questions? Yes. AUDIENCE: So how's
one supposed to-- so what it appears
is happening is you're basically taking
some state like psi, and you're basically writing
in terms of some basis. And then you're basically using
the [INAUDIBLE] coordinates of this thing. But the question is, what does
this basis actually look like? Like, what do these vectors-- because if you put them
in their own coordinates, they're just infinite. PROFESSOR: Yup. AUDIENCE: They're
not even delta-- I mean-- PROFESSOR: They are
delta functions. AUDIENCE: [INAUDIBLE] PROFESSOR: These vectors
are delta functions because if you have a state
that has this as the position state of a particle,
you find the wave function by doing x on it. That's our definition
of a wave function. And its infinite. So there's is not too much
one can say about this. If people want to work
more mathematically, the more comfortable
way, what you do is, instead of taking
infinite things, you put everything
on a big circle. And then you have
a Fourier series and they transform as sums,
and everything goes into sums. But there's no real need. These operations are safe. And we managed to do them,
and we're OK with them. Other questions? Yes. AUDIENCE: [INAUDIBLE] PROFESSOR: Probably not. You know, infinite
bases are delicate. Hilbert spaces are infinite
dimensional vector spaces, and they-- not every infinite dimensional
space is a Hilbert space. The most important
thing of a Hilbert space is this norm, this
inner product. But the other important thing
is some convergence facts about sequences of vectors
that converge to points that are on the space. So it's delicate. Infinite dimensional
spaces can be pretty bad. A Banach space is
not a Hilbert space. It's more com-- AUDIENCE: [INAUDIBLE] PROFESSOR: Only
for Hilbert spaces, and basically, this problem
of a particle in a line, or a particle in three space
is sufficiently well known that we're totally
comfortable with this somewhat singular operation. So the operator x
or the operator p may not be what mathematicians
like them to be-- bounded operators in Hilbert spaces. But we know how not to
make mistakes with them. And if you have a very
subtle problem, one day you probably have
to be more careful. But for the problems we're
interested in now, we don't. So our last topic today is
uncertainties and uncertainty relations. I probably won't get through all
of it, but we'll get started. And so we'll have uncertainties. And we will talk
about operators, and Hermitian operators. So here is the
question, basically-- if you have a state,
we know the result of a measurement
of an observable is the eigenvalue of
a Hermitian operator. Now, if the state is an
eigenstate of the Hermitian operator, you measure
the observable, and out comes eigenvalue. And there's no uncertainty
in the measured observable, because the measured
observable is an eigenvalue and its state is an eigenstate. The problem arises when
the state that you're trying to measure this
property is not an eigenstate of the observable. So you know that
the interpretation of quantum mechanics is a
probabilistic distribution. You sometimes get one thing,
sometimes get another thing, depending on the
amplitudes of the states to be in those
particular eigenstates. But there's an uncertainty. At this time, you
don't know what the measured value will be. So we'll define the
uncertainty associated to a Hermitian
operator, and we want to define this uncertainty. So A will be a
Hermitian operator. And you were talking
about the uncertainty. Now the uncertainty
of that operator-- the first thing that
you should remember is you can't talk about the
uncertainty of the operator unless you give me a state. So all the formulas we're going
to write for uncertainties are uncertainties of
operators in some state. So let's call the state psi. And time will not be relevant,
so maybe I should delete the-- well, I'll leave that
bar there, just in case. So we're going to try
to define uncertainty. But before we do that, let's
try to define another thing-- the expectation value. Well, the expectation value-- you know it. The expectation value of A,
and you could put a psi here if you wish, to remind you
that it depends on the state-- is, well, psi A psi. That's what we call
expectation value. In the inner product
notation would be psi A psi. And one thing you know-- that this thing is real,
because the expectation values of Hermitian operators is real. That's something we reviewed
at the beginning of the lecture today. So now comes the
question, what can I do to define an
uncertainty of an operator? And an uncertainty-- now
we've said already something. I wish to define an
uncertainty that is such that the uncertainty is 0 if
the state is an eigenstate, and the uncertainty
is different from 0 if it's not an eigenstate. In fact, I wish that
the uncertainty is 0 if and only if the state
is an eigenstate. So actually, we
can achieve that. And in some sense, I think,
the most intuitive definition is the one that
I will show here. It's that we define the
uncertainty, delta A, and I'll put the psi here. So this is called
the uncertainty of A in the state psi. So we'll define it a simple way. What else do we want? We said this should
be 0 if and only if the state is an eigenstate. Second, I want this thing
to be a real number-- in fact, a positive number. What function do we know
in quantum mechanics that can do that magic? Well, it's the norm. The norm function is
always real and positive. So this-- we'll try to
set it equal to a norm. So it's the norm of the state
A minus the expectation value of A times 1 acting on psi. This will be our definition
of the uncertainty. So it's the norm of this vector. Now let's look at this. Suppose the norm
uncertainty is 0. And if the uncertainty is
0, this vector must be 0. So A minus expectation
value of A on psi is 0. Or A psi is equal to
expectation value of A on psi. The 1 doesn't do much. Many people don't write the 1. I could get tired
and stop writing it. You should-- probably it's
good manners to write the i, but it's not all that necessary. You don't get that confused. If there's an operator
and a number here, it must be an identity matrix. So the uncertainty is 0, the
vector is 0, then this is true. Now, you say, well, this
equation looks kind of funny, but it says that psi
is an eigenstate of A, because this is a number. It looks a little
funny, because we're accustomed to A psi lambda
psi, but this is a number. And in fact, let me
show you one thing. If you have A psi
equal lambda psi-- oh, I should say here
that psi is normalized. If psi would not be normalized,
you change the normalization. You change the uncertainty. So it should be normalized. And look at this-- if you
have a psi equal lambda psi, do the inner product with psi. Psi comma A psi would
be equal to lambda, because psi inner
product with psi is 1. But what is this? This is the
expectation value of A. So actually, given
our definition, the eigenvalue of some
operator on this state is the expectation value of
the operator in the state. So back to the argument-- if the uncertainty is 0,
the state is an eigenstate. And the eigenvalue happens
to be the expectation value-- that is, if the
uncertainty is 0. On the other hand, if you are
in an eigenstate, you're here. Then lambda is A,
and this equation shows that this vector is
0, and therefore you get 0. So you've shown that this norm
or this uncertainty is 0, if and only if the state
is an eigenstate. And that's a very
powerful statement. The statement that's
always known by everybody is that if you have
an eigenstate-- yes-- no uncertainty. But if there's no uncertainty,
you must have an eigenstate. That's the second
part, and uses the fact that the only vector with
0 norm is the zero vector-- a thing that we use
over and over again. So let me make a
couple more comments on how you compute this. So that's the
uncertainty so far. So the uncertainty
vanishes in that case. Now, we can square this
equation to find a formula that is perhaps more familiar-- not necessarily more
useful, but also good. For computations,
it's pretty good-- delta A of psi, which
is real-- we square it. Well, the norm square
is the inner product of this A minus A
psi A minus A psi. Norm squared is the inner
product of these two vectors. Now, the thing
that we like to do is to move this
factor to that side. How do you move a factor on the
first input to the other input? You take the adjoint. So I should move
it with an adjoint. So what do I get? Psi, and then I get the
adjoint and this factor again. Now, I should put a dagger
here, but let me not put it, because A is Hermitian. And moreover, expectation
value of A is real. Remember-- so no
need for the dagger, so you can put the dagger,
and then explain that this is Hermitian and this is real-- or just not put it. And now look at this. This is a typical calculation. You'll do it many, many times. You just spread out the things. So let me just do it once. Here you get A squared minus
A expectation value of A minus expectation value of
A A plus expectation value of A squared psi. So I multiplied everything,
but you shouldn't be all that-- I should put a 1
here, probably-- shouldn't worry about this much. This is just a number and
an A, a number and an A. The order doesn't matter. These two terms are
really the same. Well, let me go
slowly on this once. What is the first term? It's psi A squared psi, so
it's the expectation value of A squared. Now, what is this term? Well, you have a number
here, which is real. It goes out of whatever you're
doing, and you have psi A psi. So this is expectation
value of A. And from the leftover psi A
psi, you get another expectation value of A. So this is A A. Here the same thing--
the number goes out, and you're left with
a psi A psi, which is another expectation value
of A, so you get minus A A. And you have a plus
expectation value of A squared. And I don't need the i anymore,
because the expectation values have been taken. And this always happens. It's a minus here, a
minus here, and a plus here, so there's just one
minus at the end of the day. One minus at the end of the
day, and a familiar, or famous formula comes out that
delta of A on psi squared is equal to the expectation
value of A squared minus expectation value of A squared. Which shows something
quite powerful. This has connections, of course,
with statistical mechanics and standard deviations. It's a probabilistic
interpretation of this formula, but one fact that this
has allowed us to prove is that the expectation
value of A squared is always greater
or equal than that, because this number is
positive, because it is the square of a
real positive number. So that's a slightly
non-trivial thing, and it's good to know it. And this formula, of
course, is very well known. Now, I'm going to leave a funny
geometrical interpretation of the uncertainty. Maybe you will find
it illuminating, in some ways turning into
pictures all these calculations we've done. I think it actually
adds value to it, and I don't think
it's very well known, or it's kind of funny, because
it must not be very well known. But maybe people don't
find it that suggestive. I kind of find it suggestive. So here's what I want
to say geometrically. You have this vector space,
and you have a vector psi. Then you come along, and
you add with the operator A. Now the fact that
this thing is not and eigenstate means that
after you add with A, you don't keep in
the same direction. You go in different directions. So here is A psi. So what can we say here? Well, actually
here is this thing. Think of this vector
space spanned by psi. Let's call it U psi. So it's that line there. You can project this
in here, orthogonally. Here is the first claim-- the vector that you
get up to here-- this vector-- is nothing
else but expectation value of A times psi. And that makes sense, because
it's a number times psi. But precisely the orthogonal
projection is this. And here, you get an
orthogonal vector. We'll call it psi perp. And the funny thing
about this psi perp is that its length is
precisely the uncertainty. So all this, but
you could prove-- I'm going to do it. I'm going to show you all
these things are true, but it gives you a
bit of an insight. you have a vector. A moves you out. What is the uncertainty is
this vertical projection-- vertical thing is
the uncertainty. If you're down there,
you get nothing. So how do we prove that? Well, let's construct a
projector down to the space U psi, which is psi psi. This is a projector,
just like any e1. e1 is a projection into
the direction of 1. Well, take your first
basis vector to be psi, and that's a projection to psi. So let's see what it-- so the projection to psi. So now let's see what it gives
you when it acts on A psi-- this project acting on A psi
is equal to psi psi A psi. And again, the usefulness
of bracket notation is kind of nice here. So what is this? The expectation
value of A. So indeed psi expectation value
of A is what you get when you project this down. So then, the rest
is sort of simple. If you take psi, and
subtract from psi-- well, I'll subtract
from psi, psi times expectation value of A. I'm
sorry, I was saying it wrong. If you think the
original vector-- A psi, and subtract from
it what we took out, which is psi times expectation
value of A, the projected thing-- this is some vector. But the main thing is that this
vector is orthogonal to psi. Why? If you take a psi on the left,
this is orthogonal to psi. And how do you see it? Put the psi from the left. And what do you get here? Psi A psi, which is
expectation value of A, psi psi, which is 1, and
expectation value A is 0. So this is a vector psi perp. And this is, of course,
A minus expectation value of A acting on the state psi. Well, precisely the norm of
psi perp is the norm of this, but that's what we defined
to be the uncertainty. So indeed, the norm of psi
perp is delta A of psi. So our ideas of projectors
and orthogonal projectors allow you to understand better
what is the uncertainty-- more pictorially. You have pictures of vectors,
and orthogonal projections, and you want to make
the uncertainty 0, you have to push
the A psi into psi. You have to be an
eigenstate, and you're there. Now, the last thing of-- I'll use the last five minutes
to motivate the uncertainty, the famous uncertainty theorem. And typically, the uncertainly
theorem is useful for A and B-- two Hermitian operators. And it relates the uncertainty
in A on the state psi to the uncertainty
in B of psi, saying it must be greater than
or equal than some number. Now, if you look
at that, and you think of all the math
we've been talking about, you maybe know
exactly how you're supposed to prove the
uncertainty theorem. Well, what does
this remind you of? Cauchy-Schwarz--
Schwarz inequality, I'm sorry-- not Cauchy-Schwarz. Why? Because for Schwarz inequality,
you have norm of u, norm of v is greater than or equal than
the norm of the inner product of u and v-- absolute value of the
inner product of u and v. Remember, in this thing,
this is norm of a vector, this is norm of a vector, and
this is value of a scalar. And our uncertainties are norms. So it better be that. That inequality is the only
inequality that can possibly give you the answer. So how would you set this up? You would say define-- as we'll say f equal A
minus A acting on psi, and g is equal to B
minus B acting on psi. And then f f, or f f
is delta A squared. f g g is delta B squared. And you just need to compute
the inner product of f g, because you need the mixed one. So if you want to
have fun, try it. We'll do it next time anyway. All right that's it for today.
