The following
content is provided under a Creative
Commons license. Your support will help MIT
OpenCourseWare continue to offer high quality
educational resources for free. To make a donation or to
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Going to
get started right away. I want to make a comment
about energy time uncertainty relations. And we talked last
time about the fact that the energy time uncertainty
relation really tells you something about how
fast a state can change. So an interesting way
to try to evaluate that is to consider
a system that has to state that time
equals 0 and compute the overlap with a
state at time equals t. Now this overlap is a
very schematic thing. It's a bracket. It's a number. And you know what that means. You can keep in
mind what that is. It's an integral over space
possibly of psi star at t equals 0 x psi t x. It's a complete integral,
its inner product. So you may want to understand
this, because, in a sense, this is telling you how
quickly a state can change. At time equals 0, this
overlap is just 1. A little later, this
overlap is going to change and perhaps after some time
the overlap is going to be 0. And we're going to say that we
actually have changed a lot. So this number is very
interesting to compute. And in fact, we might
as well square it, because it's a complex number. So to understand better
what it is we'll square it. And we'll have to evaluate this. Now how could you evaluate this? Well, we'll assume that this
system that governs this time evolution has a time
independent Hamiltonian. Once this evolution is done by
a time independent Hamiltonian, you can wonder what it is. Now it's quite
interesting, and you will have to discuss
that in the homework because it will
actually help you prove that version of the
time energy uncertainty relationship that says that the
quickest time state can turn orthogonal to itself is
bounded by some amount. Cannot do it infinitely fast. So you want to know how
fast this can change. Now it's very surprising what
it depends on, this thing. Because suppose you had
an energy eigenstate, suppose psi at time equals
0 is an energy eigenstate. What would happen later on? Well, you know that
energy eigenstates evolve with a phase,
an exponential of e to the minus iht over h bar. So actually if you had
an energy eigenstate, this thing would remain
equal to 1 for all times. So if this is going
to be non-zero, it's because it's
going to have-- you have to have a state is
not an energy eigenstate. That you're going to have
an uncertainty in the energy and energy uncertainty. So the curious thing is that you
can evaluate this, and expand it as a power in t, and go,
say, to quadratic ordering in t evaluating what this is. And this only depends
on the uncertainty of h, and t, and things like that. So only the uncertainty of
h matters at this moment. So this would be quite
interested, I think, for you to figure out
and to explore in detail. That kind of analysis
has been the center of attention recently, having
to do with quantum computation. Because in a sense,
in a quantum computer, you want to change states
quickly and do operations. So how quickly you can
change a state is crucial. So in fact, the people that
proved these inequalities that you're going to find
say that this eventually will limit the speed
of a quantum computer, and more slow that computers
become twice as fast every year or so, and double the speed. So this limit
apparently, it's claimed, will allow 80
duplications of the speed until you hit this limit
in a quantum computer due to quantum mechanics. So you will be investigating
this in next week's homework. Today we want to begin
quickly with an application of the uncertainty principal to
find exact bounds of energies for quantum systems, for ground
states of quantum systems. So this will be a very precise
application of the uncertainty principle. Then we'll turn to a
completion of these things that we've been
talking about having to do with linear algebra. We'll get to the main
theorems of the subject. In a sense, the most important
theorems of the subject. These are the spectral
theorem that tells you about what operators
can be diagonalized. And then a theorem that
leads to the concept of a complete set of
commuting observables. So really pretty key
mathematical ideas. And the way we're good to
do it, I think you will see, that we have gained a lot by
learning the linear algebra concepts in a
slightly abstract way. I do remember doing this
proof that we're going today in previous years,
and it was always considered the most complicated
lecture of the course. Just taking the
[? indices ?] went crazy, and lots of formulas, and
the notation was funny. And now we will do
the proof, and we'll write a few little things. And we'll just try to
imagine what's going on. And will be, I think, easier. I hope you will agree. So let's begin with an example
of a use of the uncertainty principle. So example. So this will be
maybe for 20 minutes. Consider this Hamiltonian for
a one dimensional particle that, in fact, you've
considered before, alpha x to the fourth, for which
you did some approximations. You know that the
expectation value of the energy in
the ground state. You've done it numerically. You've done it variationally. And variationally you knew
that the energy at every stage was smaller at a given bound. The uncertainty
principle is not going to give us an upper bound. It's going to give
us a lower bound. So it's a really
nice thing, because between the variational
principal and the uncertainty principle, we can narrow the
energy of this ground state to a window. In one of the problems that
you're doing for this week-- and I'm sorry only
today you really have all the tools after you
hear this discussion-- you do the same for the
harmonic oscillator. You do a variational estimate
for the ground state energy. You do the uncertainty principle
bound for ground state energy. And you will see
these two bounds meet. And therefore after you've
done this to bounds, you found the
ground state energy of the harmonic oscillator,
so it's kind of a neat thing. So we want to estimate
the ground state energy. So we first write some words. We just say H in
the ground state will be given by the
expectation value of p squared in the ground state plus
alpha times the expectation value of x to the fourth
in the ground state. Haven't made much
progress, but you have, because you're starting to
talk about the right variables. Now this thing to
that you have to know is that you have
a potential that is like this, sort
of a little flatter than x squared potential. And what can we say about
the expectation value of the momentum on
the ground state and the expectation value
of x in the ground state? Well, the expectation value
of x should be no big problem. This is a symmetric
potential, therefore wave functions in a one
dimensional quantum mechanics problems are either
symmetric or anti symmetric. It could not be anti
symmetric because it's a ground state and
kind of have a 0. So it's asymmetric. Has no nodes. So there's the wave
function of the ground state, the symmetric,
and the expectation value of x in the ground state is 0. Similarly, the expectation value
of the momentum in the ground state, what is it? Is? 0 too. And you can imagine
just computing it. It would be the integral of psi. Psi is going to be a real
d dx h bar over i psi. This is a total derivative. If it's a bound state,
it's 0 at the ends. This is 0. So actually, we have a
little advantage here. We have some control
over what p squared is, because the uncertainty in p
in the ground state-- well, the uncertainty in p squared
is the expectation value of p squared minus the
expectation value of p squared. So in the ground
state, this is 0. So delta p squared
in the ground state is just p squared
on the ground state. Similarly, because
the expectation value of x is equal to 0, delta x
squared in the ground state is equal to expectation value of
x squared in the ground state. So actually, this expectation
of p squared is delta p. And we want to use the
uncertainty principle, so that's progress. We've related something we want
to estimate to an uncertainty. Small complication is that
we have an expectation value of x to the fourth. Now we learned-- maybe
I can continue here. We learned that the expectations
value for an operator squared is bigger than or equal
to the expectation value of the operator squared. So the expectation
value of x to the fourth is definitely bigger than the
expectation value of x squared squared. And this is true on any state. This was derived when
we did uncertainty. We proved that the
uncertainty squared is positive, because
the norm of a vector, and that gave you this thing. So here you think of the
operator as x squared. So the operator squared
is x to the fourth. And here's the operator
expectation value squared. So this is true for
the ground state. It's also here true for any
state, so is the ground state. And this x squared
now is delta x. So this is delta x on the
ground state to the fourth. So look what we have. We have that the expectation
value of H on the ground state is strictly equal to delta p
on the ground state squared over 2m plus alpha. And we cannot do a Priorean
equality here, so we have this. This is so far an equality. But because of this, this
thing is bigger than that. Well, alpha is supposed
to be positive. So this is bigger than delta
p ground state squared over 2m plus alpha delta x on the
ground state to the fourth. OK, so far so good. We have a strict thing, this. And the order of the inequality
is already showing up. We're going to get, if
anything, a lower bound. You're going to be bigger
than or equal to something. So what is next? Next is the
uncertainty principle. We know that delta p delta
x is greater than or equal to h bar over 2 in any state. So the delta p ground state
and delta x on the ground state still should be equal to that. Therefore delta p ground
state is bigger than or equal than h over 2 delta x
in the ground state like that. So this inequality still
the right direction. So we can replace
this by something that is bigger than
this quantity day without disturbing the logic. So we have H ground
state now is greater than or equal to replace
the delta p by this thing here, h squared over 8,
because this is squared and there's another 2m delta x
ground state squared plus alpha delta x ground
state to the fourth. And that's it. We've obtained this inequality. So here you say, well, this
is good but how can I use it? I don't know what delta
x is in the ground state, so what have I gained? Well, let me do
a way of thinking about this that can help you. Plot the right hand side
as a function of delta x on the ground. So you don't know
how much it is, delta x on the ground
state, so just plot it. So if you plot this
function, there will be a divergence as
this delta x goes to 0, then it will be a minimum. It will be a positive minimum,
because this is all positive. And then it will go up again. So the right hand side as a
function of delta x is this. So here it comes. You see I don't know
what delta x is. Suppose delta x
happens to be this. Well, then I know
that the ground state energy is bigger
than that value. But maybe that's not delta x. Delta x may be is this
on the ground state. And then if it's that, well,
the ground state energy is bigger than this
value over here. Well, since I just
don't know what it is, the worst situation
is if delta x is here, and therefore definitely H must
be bigger than the lowest value that this can take. So the claim is that
H of gs, therefore is greater than or equal than
the minimum of this function h squared over 8m, and
I'll just write here delta x squared plus alpha delta
x to the fourth over delta x. The minimum of this function
over that space is the bound. So I just have to do a
calculus problem here. This is the minimum. I should take the derivative
with respect to delta x. Find delta x and substitute. Of course, I'm not
going to do that here, but I'll tell you a formula
that will do that for you. A over x squared
plus Bx to the fourth is minimized for x squared
is equal to 1 over 2-- it's pretty awful numbers. 2 to the 1/3 A
over B to the 1/3. And its value at that point is
2 to the 1/3 times 3/2 times A to the 2/3 times B to the 1/3. A little bit of arithmetic. So for this function,
it turns out that A is whatever
coefficient is here. B is whatever
coefficient is there, so this is supposed
to be the answer. And you get H on
the ground state is greater than or equal to
2 to the 1/3 3/8 h squared square root of alpha over m to
the 2/3, which is about 0.4724 times h squared square root
of alpha over m to the 2/3. And that's our bound. How good or how
bad is the bound? It's OK. It's not fabulous. The real answer is done
numerically is 0.668. I think I remember
variational principal gave you something like 0.68 or 0.69. And this one says
it's bigger than 0.47. It gives you something. So the important thing is
that it's completely rigorous. Many times people use
the uncertainty principle to estimate ground
state energies. Those estimates
are very hand wavy. You might as well just
do dimensional analysis. You don't gain anything. You don't know the factors. But this is completely rigorous. I never made an approximation
or anything here. Every step was logical. Every inequality was exact. And therefore, this
is a solid result. This is definitely true. It doesn't tell you an
estimate of the answer. If you dimensional analysis, you
say the answer is this times 1, and that's as good as you can
do with dimensional analysis. It's not that bad. The answer turns out to be 0.7. But the uncertainty
principle really, if you're careful, sometimes,
not for every problem, you can do a rigorous thing
and find the rigorous answer. OK, so are there any questions? Your problem in
the homework will be to do this for the
harmonic oscillator and find the two bounds. Yes? AUDIENCE: How does
the answer change if we don't look at
the ground state? PROFESSOR: How do they what? PROFESSOR: How does
the answer change if we look at a state different
from the ground state? PROFESSOR: Different
from the ground state? So the question was
how would this change if I would try to do something
different from the ground state. I think for any
state, you would still be able to say that
the expectation value of the momentum is 0. Now the expectation value
of x still would be 0. So you can go through
some steps here. The problem here being
that I don't have a way to treat any other
state differently. So I would go ahead,
and I would have said for any stationary state,
or for any energy eigenstate, all of what I said is true. So I don't get a new one. These things people actually
keep working and writing papers on this stuff. People sometimes find bounds
that are a little original. Yes? AUDIENCE: How do you know
the momentum expectation is 0 again? PROFESSOR: The momentum
expectation for a bound state goes like this. So you want to figure
out what is psi p psi. And you do the following. That's integral. Now psi in these
problems can be chosen to be real, so I won't bother. It's psi of x h bar
over i d dx of psi. So this is equal to h bar over
2i the integral dx of d dx of psi squared. So at this moment, you say
well that's just h bar over 2i, the value of psi squared at
infinity and at minus infinity. And since it's a bound
state, it's 0 here, 0 there, and it's equal to 0. A state that would have
expectation value of momentum you would expect
it to be moving. So this state is
there is static. It's stationary It doesn't have
expectation value of momentum. Yes? AUDIENCE: Is the reason that
you can't get a better estimate for the things that are
on the ground state, because if you consider
the harmonic oscillator, the uncertainty delta x
[? delta p ?] from the ground state [INAUDIBLE] you
go up to higher states. PROFESSOR: Right, I
think that's another way. AUDIENCE: [INAUDIBLE] higher
state using the absolute. PROFESSOR: Yeah. That's a [INAUDIBLE]. So the ground state of the
harmonic oscillator saturates the uncertainty principal
and the others don't. So this argument, I think,
is just good for ground state energies. One more question. AUDIENCE: It appears that
this method really works. Doesn't particularly work well
if we have a potential that has an odd power, because we
can't use [? the packet ?], like x [INAUDIBLE] expectation
value x to the fourth is something, some
power, expectation. PROFESSOR: Right, if
it's an odd power, the method doesn't work well. But actually for an
odd power, the physics doesn't work well either,
because the system doesn't have ground states. And so let's say
that if you had x to the fourth plus some x
cubed, the physics could still make sense. But then it's not clear
I can do the same. Actually you can
do the same for x to the eighth and any sort
of powers of this type. But I don't think it
works for x to the sixth. You can try a few things. OK, so we leave the
uncertainty principle and begin to understand
more formerly the operators for which there's no uncertainty
and you can simultaneously diagonalize them. So we're going to find
operators like A and B, that they commute. And then sometimes
you can simultaneously diagonalize them. Yes? You have a question. AUDIENCE: So part of [INAUDIBLE]
we use here is [INAUDIBLE], right? PROFESSOR: Right. AUDIENCE: If we get an asset--
is there any way that we can better our [INAUDIBLE]
principle based on the wave function with non saturated? Can we get an upper
bound for just [INAUDIBLE] principle
with an h bar over it? PROFESSOR: Can I
get an upper bound? I'm not sure I
understand your question. AUDIENCE: [INAUDIBLE] the fact
that the [INAUDIBLE] principle will not be saturated. Can you put the bound for
just taking [INAUDIBLE]? PROFESSOR: Yeah, certainly. You might have some
systems in which you know that this
uncertainty might be bigger than the one warranted
by the uncertainty principles. And you use that information. But on general grounds,
it's hard to know that a system might come
very close to satisfy the uncertainty principle
in its ground state. We don't know. There are systems that come
very close in the ground state to satisfy this and
some that are far. If they are far, you
must have some reason to understand that to use it. So I don't know. So let me turn now to
this issue of operators and diagonalization
of operators. Now you might be irritated
a little even by the title. Diagonalization of operation. You'll be talking about
diagonalization of matrices. Well, there's a
way to state what we mean by diagonalizing
an operator in such a way that we can talk later
about the matrix. So what is the point here? You have an operator,
and it's presumably an important operator
in your theory. You want to understand
this operator better. So you really are
faced with a dilemma. How do I get some insight
into this operator? Perhaps the simplest thing
you could do is to say, OK let me choose some ideal
basis of the vector space, such as that operator
is as simple as possible in that basis. So that's the origin
of this thing. Find a basis in the state
space, so the operator looks as simple as possible. So you say that you can
diagonalize an operator if you can find the basis
such that the operator has just diagonal entries. So let me just
write it like this. So if you can find a basis in
V where the matrix representing the operator is
diagonal, the operator is said to be diagonalizable. So to be diagonalizable
is just a statement that there is some basis
where you look at the matrix representation
operator, and you find that it takes form
as a diagonal. So let's try to understand
this conceptually and see what actually
it's telling us. It tells us actually a lot. Suppose t is diagonal in
some basis u1 up to un. So what does it mean
for it to be diagonal? Well, you may remember
all these definitions we had about matrix action. If T acting on a ui
is supposed to be Tki uk in some basis sum over k. You act on ui, and
you get a lot of u's. And these are the matrix
elements of the operator. Now the fact that
it's diagonalizable means that in some basis, the
u basis, this is diagonal. So ki in this sum only
happens to work out when k is equal to i. And that's one number and you
get back to the vector ui. So if it's diagonal
in this basis, you have the T on u1 is lambda
a number times u1 T on u2 is lambda 2 in u2. And Tun equal lambda n un. So what you learn is
that this basis vector-- so you learn something
that maybe you thought it's tautological. It's not tautological. You learn that if you have
a set of basis vectors in which the
operator is diagonal, these basis vectors are
eigenvectors of the operator. And then you learn something
that is quite important, that an operator is
diagonalizable if, and only if, it has
a set of eigenvectors that span the space. So the statement
is very important. An operator T is
that diagonalizable if it has a set of eigenvectors
that span the space. Span V. If and only if. If this double f. If and only if. So here it's diagonalizable,
and we have a basis, and it has a set of
these are eigenvectors. So diagonalizable
realizable really means that it has a
set of eigenvectors that span the space. On the other hand, if you have
the set of eigenvectors that span the space, you have a
set of u's that satisfy this, and then you read that, oh
yeah, this matrix is diagonal, so it's diagonalizable. So a simple statement,
but an important one, because there are examples of
matrices that immediately you know you're never going
to succeed to diagonalize. So here is one matrix, 0 0 1 0. This matrix has
eigenvalues, so you do the characteristic equation
lambda squared equals 0. So the only eigenvalue
is lambda equals 0. And let's see how
many eigenvectors you would have for
lambda equals 0. Well, you would have if
this is T, T on some vector a b must be equal to 0. So this is 0 1 0 0 on a b,
which is b and 0, must be zero. So b is equal to 0. So the only
eigenvector here-- I'll just write it here and then
move to the other side. The only eigenvector
for lambda equals 0, the only eigenvector
is with b equals 0. So it's 1 0. One eigenvector only. No more eigenvectors. By the theorem,
or by this claim, you know it's a two dimensional
vector space you just can't diagonalize this matrix. It's impossible. Can't be done. OK, a couple more
things that I wish to say about this process
of diagonalization. Well, the statement that
an operator is diagonal is a statement about the
existence of some basis. Now you can try to figure
out what that basis is, so typically what is the
problem that you face? Typically you have a
vector spaces V. Sorry? AUDIENCE: I have a question. PROFESSOR: Yes? If you had an infinite
dimensional space and you had an operator
whose eigenvectors do not span the space, can it
still have eigenvectors, or does it not have any then? PROFESSOR: No. You said it has
some eigenvectors, but they don't span the space. So it does have
some eigenvectors. AUDIENCE: So my question
is was what I just said a logical contradiction in
an infinite dimensional space? PROFESSOR: To have
just some eigenvectors? I think-- PROFESSOR: I'm looking
more specifically at a dagger for instance. PROFESSOR: Yes. AUDIENCE: In the
harmonic oscillator, you I think mentioned at some
point that it does not have-- PROFESSOR: So the
fact that you can' diagonalize this
thing already implies that it's even worse
in higher dimensions. So some operator
may be pretty nice, and you might still be
able to diagonalize it, so you're going to lack
eigenvectors in general. You're going to
lack lots of them. And there are going to
be blocks of Jordan. Blocks are called things that
are above the diagonal, things that you can't do much about. Let me then think concretely now
that you have a vector space, and you've chosen
some basis v1 vn. And then you look
at this operator T, and of course, you chose
an arbitrary basis. There's no reason why
its matrix representation would be diagonal. So T on the basis v-- Tij. Sometimes to be very explicit
we write Tij like that-- is not diagonal. Now if it's not
diagonal, the question is whether you can find a
basis where it is diagonal. And then you try, of
course, changing basis. And you change basis--
you've discussed that in the homework--
with a linear operator. So you use a linear
operator to produce another basis, an
invertible in your operator. So that you get these vectors
uk being equal to some operator A times vk. So this is going to be
the u1's up to un's are going to be another basis. The n vector here
is the operator acting with the n
vector on this thing. And then you prove, in the
homework, a relationship between these matrix elements
of T in the new basis, in the u basis. And the matrix elements
of T in the v basis. You have a
relationship like this, or you have more explicitly
Tij in the basis u is equal to A minus
1 ik Tkp of v Apj. So this is what happens. This is the new
operator in this basis. And typically what
you're trying to do is find this matrix A
that makes this thing into a diagonal matrix. Because we say in the u basis
the operator is diagonal. I want to emphasize that there's
a couple of ways in which you can think of diagonalization. Sort of a passive
and an active way. You can imagine the
operator, and you say look, this
operator I just need to find some basis in
which it is diagonal. So I'm looking for a basis. The other way of
thinking of this operator is to think that A minus
1 TA is another operator, and it's diagonal
in original basis. So it might have seem funny
to you, but let's stop again and say this again. You have an operator, and the
question of diagonalization is whether there is some basis
in which it looks diagonal, its matrix is diagonal. But the equivalent
question is whether there is an operator A
such that this is diagonal in the original basis. To make sure that you see
that, consider the following. So this is diagonal
in the original basis. So in order to see that, think
of Tui is equal to lambda i ui. We know that the
u's are supposed to be this basis of eigenvectors
where the matrix is diagonal, so here you got it. Here the i not summed. It's pretty important. There's a problem with
this eigenvalue notation. I don't know how
to do it better. If you have several eigenvalues,
you want to write this, but you don't want this to
think that you're acting on u1 and you get lambda 1 u1. Not the sum right here. OK, but they ui is
equal to A on vi. So therefore this
is lambda i A on vi. And then you'll
act with A minus 1. Act with A minus 1 from
the left with the operator. So you get A minus 1 TA vi
is equal to lambda i vi. So what do you see? You see an operator
that is actually diagonal in the v basis. So this operator is diagonal
in the original basis. That's another way of
thinking of the process of diagonalization. There's one last remark,
which is that the columns of A are the eigenvectors, in fact. Columns of A are
the eigenvectors. Well, how do you see that? It's really very simple. You can convince
yourself in many ways, but the uk are the eigenvectors. But what are uk's? I have it somewhere. There it is. A on vk. And A on vk is this
matrix representation is sum over i Aik vi. So now if this is
the original basis, the vi's are your
original basis, then you have the
following, that the vi's can be thought as the basis
vectors and represented by columns with a
1 in the ith entry. So this equation is saying
nothing more, or nothing less than uk, in terms of
matrices or columns, is equal to A1k v1 plus Ank
vn, which is just A1k A2k Ank. Because vi is the
ith basis vector. So 1 0's only in
the ith position. So these are the eigenvectors. And they're thought as linear
combinations of the vi's. The vi's are the
original basis vectors. So the eigenvectors
are these numbers. OK, we've talked
about diagonlization, but then there's a term that is
more crucial for our operators that we're interested in. We're talking about
Hermitian operators. So the term that is going
to be important for us is unitarily diagonalizable. What is a unitarily
diagonalizable operator? Two ways again of
thinking about this. And perhaps the first
way is the best. And I will say it. A matrix is set to be
unitarily diagonalizable if you have an orthonormal
basis of eigenvectors. Remember diagonalizable meant
a basis of eigenvectors. That's all it means. Unitarily diagonalizable
means orthonormal basis of eigenvectors. So T has an orthonormal
basis of eigenvectors. Now that's a very
clear statement. And it's a fabulous
thing if you can achieve, because you basically
have broken down the space into basis spaces, each one
of them with a simple thing before your operators. And they're orthonormal,
so it's the simplest possible calculational tool. So it's ideal if
you can have this. Now the way we think
of this is that you start with--
concretely, you start with a T of some basis v
that is an orthonormal basis. Start with an orthonormal
basis, and then pass to another
orthonormal basis u. So you're going to pass to
another orthonormal basis u with some operator. But what you have
learned is that if you want to pass from v
orthonormal to another basis u, a vector that is
also orthonormal, the way to go from
one to the other is through a unitary operator. Only unitary operators pass you
from orthonormal to orthonormal basis. Therefore really, when
you start with your matrix in an orthonormal basis
that is not diagonal, the only thing you can
hope is that T of u will be equal to sum u dagger,
or u minus 1, T of v u. Remember, for a unitary
operator, where u is unitary, the inverse is the dagger. So you're doing a
unitary transformation, and you find the matrix that
is presumably then diagonal. So basically, unitarily
diagonalizable is the statement
that if you start with the operator in an
arbitrary orthonormal basis, then there's some unitary
operator that takes you to the privilege basis in which
your operator is diagonal, is still orthonormal. But maybe in a more simple
way, unitarily diagonalizable is just a statement
that you can find an orthonormal basis
of eigenvectors. Now the main theorem
of this subject, perhaps one of the
most important theorems of linear algebra, is
the characterization of which operators have such
a wonderful representation. What is the most
general operator T that will have an orthonormal
basis of eigenvectors? Now we probably have heard that
Hermitian operators do the job. Hermitian operators have that. But that's not the
most general ones. And given that you want
the complete result, let's give you the
complete result. The operators that have
this wonderful properties are called normal
operators, and they satisfy the following property. M is normal if M dagger, the
adjoint of it, commutes with M. So Hermitian
operators are normal, because M dagger is equal
to M, and they commute. Anti Hermitian mission
operators are also normal, because anti Hermitian
means that dagger is equal to minus M, and
it still commutes with M. Unitary operators have
U dagger U equal to U U dagger equals to 1. So U and U dagger
actually commute as well. So Hermitian, anti
Hermitian, unitary, they're all normal operators. What do we know about
normal operators? There's one important result
about normal operators, a lemma. If M is normal and
W is an eigenvector, such that MW is
equal to lambda W. Now normal operators need
not have real eigenvalues, because they include
unitary operators. So here I should write
Hermitian, anti Hermitian, and unitary are normal. So here is what a normal
operator is doing. You have a normal operator. It has an eigenvector
with some eigenvalue. Lambda is a complex
number in principle. Then the following
result is true, then M dagger omega is also
an eigenvector of M dagger. And it has eigenvalue
lambda star. Now this is not all
that easy to show. It's a few lines, and
it's done in the notes. I ask you to see. It's actually very elegant. What is the usual
strategy to prove things like that Is to say oh, I want
to show this is equal to that, so I want to show that
this binds that is 0. So I have a vector
that is zero what is the easiest way
to show that it's 0? If I can show it's norm is 0. So that's a typical
strategy that you use to prove equalities. You say, oh, it's a vector
that must be 0 as my equality. Let's see if it's 0. Let's find its norm,
and you get it. So that's a result. So with this stated, we
finally have the main result that we wanted to get to. And I will be very
sketchy on this. The notes are complete,
but I will be sketchy here. It's called the
spectral theorem. Let M be an operator in
a complex vector space. The vector space has
an orthonormal basis of eigenvectors of M if
and only if M is normal. So the normal operators are it. You want to have a complete set
of orthonormal eigenvectors. Well, this will only
happen if your operator is normal, end of story. Now there's two things
about this theorem is to show that if it's
diagonalizable, it is normal, and the other thing is to
show, that if it's normal, it can be diagonalized. Of course, you can
imagine the second one is harder than the first. Let me do the first
one for a few minutes. And then say a couple of
words about the second. And you may discuss
this in recitation. It's a little
mathematical, but it's all within the kind of
things that we do. And really it's fairly
physical in a sense. We're accustomed to do
such kinds of arguments. So suppose it's
unitarily diagonalizable, which means that M-- so
if you have U dagger, MU is equal to a
diagonal matrix, DM. I'm talking now matrices. So these are all matrices,
a diagonal matrix. There's no basis to the
notion of a diagonal operator, because if you have
a diagonal operator, it may not look diagonal
in another basis. Only the identity operator is
diagonal in all basis, but not the typical diagonal operator. So unitarily
diagonalizable, as we said, you make it-- it's
gone somewhere. Here. You act with an
inverse matrices, and you get the diagonal matrix. So from this, you find that
M is equal to U DM U dagger by acting with U on the left and
with U dagger from the right, you solve for M, and it's this. And then M dagger is the
dagger of these things. So it's U to DM dagger U dagger. The U's sort of
remain the same way, but the diagonal matrix
is not necessarily real, so you must put the
dagger in there. And now M dagger M. To check
that the matrix is normal that commutator should be 0. So M dagger M. You
do this times that. You get U DM dagger. U dagger U. That's one. DM U dagger. And M M dagger you multiply
the other direction you get U DM DM dagger U dagger. So the commutator of M dagger
M is equal to U DM dagger DM minus DM Dm dagger U dagger. But any two diagonal
matrices commute. They may not be that simple. Diagonal matrices are not
the identity matrices, but for sure they commute. You multiply elements along
with diagonal so this is 0. So certainly any unitarily
diagonalizable matrix is normal. Now the other part
of the proof, which I'm not going to speak about,
it's actually quite simple. And it's based on the fact that
any matrix in a complex vector space has at least
one eigenvalue. So what you do is you pick
out that eigenvalue and it's eigenvector, and
change the basis to use that eigenvector
instead of your other vectors. And then you look at the matrix. And after you use
that eigenvector, the matrix has a lot of 0's
here and a lot of 0's here. And then the matrix has been
reduced in dimension mansion, and then you go step by step. So basically, it's the
fact that any operator has at least one eigenvalue
and at least one eigenvector. It allows you to go down. And normality is analogous
to Hermiticity in some sense. And the statement that
you have an eigenvector generally tells you that this
thing is full of 0's, but then you don't know that
there are 0's here. And either normality
or Hermiticity shows that there are
0's here, and then you can proceed at lower dimensions. So you should look at
the proof because it will make clear to you that
you understand what's going on. OK but let's take it for granted
now you have these operators and can be diagonalized. Then we have the
next thing, which is simultaneous diagonalization. What is simultaneous
diagonalization? It's an awfully important thing. So we will now focus on
simultaneous diagonalization of Hermitian operators. So simultaneous diagonalization
of Hermitian ops. Now as we will emphasize
towards the end, this is perhaps one of
the most important ideas in quantum mechanics. It's this stuff that allows
you to label and understand your state system. Basically you need
to diagonalize more than one operator
most of the time. You can say OK, you found
the energy eigenstates. You're done. But if you find your
energy eigenstates and you think you're
done, maybe you are if you have all these
energy eigenstates tabulated. But if you have
a degeneracy, you have a lot of states that
have the same energy. And what's different about them? They're certainly
different because you've got several states, but
what's different about them? You may not know,
unless you figure out that they have different
physical properties. If they're different, something
must be different about them. So you need more
than one operator, and your facing the problem of
simultaneously diagonalizing things, because states cannot
be characterized just by one property, one observable. Would be simple if you could,
but life is not that simple. So you need more
than one observable, and then you ask
when can they be simultaneously diagonalizable. Well, the statement is clear. If you have two
operators, S and T that belong to the linear
operators in a vector space, they can be simultaneously
diagonalized if there is a basis for
which every basis vector is eigenstate of this and
an eigenstate of that. Common set of eigenstates. So they can be
simultaneously diagonalized. Diagonalizable is that there
is a basis where this basis is comprised of the
eigenvectors of the operator. So this time you require
more, that that basis be at the same time a basis
set of eigenvectors of this and a set of eigenvectors
of the second one. So a necessary condition for
simultaneous diagonalization is that they commute. Why is that? The fact that two operators
commute or they don't commute is an issue that is
basis independent. If they don't commute,
the order gives something different, and that you
can see in every basis. So if they don't commute
and they're simultaneously diagonalizable, there would
be a basis in which both are diagonal and they
still wouldn't commute. But you know that
diagonal matrices commute. So if two operators
don't commute, they must not
commute in any base, therefore there can't
be a basis in which both are at the same time diagonal. So you need, for
simultaneous diagonalizable, you need that S and P commute. Now that may not be enough,
because not all operators can be diagonalized. So the fact that they commute
is necessary, but not everything can be diagonalizable. Well, you've learned that
every normal operator, every Hermitian operator
is diagonalizable. And then you got now
a claim of something that could possibly be true. Is the fact that whenever you
have two Hermitian operators, each one can be
diagonalized by themselves. And they commute. There is a simultaneous
set of eigenvectors of 1 that are eigenvectors of
the first and eigenvectors of the second. So the statement is that
if S and T are commuting Hermitian operators, they can
be simultaneously diagonalized. So this theorem would
be quite easy to show if there would be
no degeneracies, and that's what we're
going to do first. But then we'll consider
the case of degeneracies. So I'm going to consider
the following possibilities. Perhaps neither one has
a degenerate spectrum. What does it mean a
degenerate spectrum? Same eigenvalue
repeated many times. But that is a wishful
thinking situation. So either both are
non degenerate, either one is non degenerate
and the other is degenerate, or both are degenerate. And that causes a very
interesting complication. So let's say there's
going to be two cases. It will suffice. In fact, it seems
that there are three, but two is enough to consider. There is no degeneracy
in T. So suppose one operator has no
degeneracy, and let's call it T. So that's one possibility. And then S may be degenerate,
or it may not be degenerate. And the second possibility
is that both S and T are degenerate. So I'll take care
of case one first. And then we'll discuss
case two, and that will complete our discussion. So suppose there's
no the degeneracy in the spectrum
of T. So case one. So what does that mean? It means that T
is non degenerate. There's a basis U1
can be diagonalized to UM, orthonormal by
the spectral theorem. And there's eigenvectors T
U-- these are eigenvectors. Lambda I Ui. And lambda I is
different to lambda j for i different from j. So all the eigenvalues,
again, it's not summed here. All the eigenvalues
are different. So what do we have? Well, each of
those eigenvectors, each of the Ui's that
are eigenvectors, generate invariant subspaces. There are T invariant subspaces. So each one, each
vector U1 you can imagine multiplying by
all possible numbers, positive and negative. And that's an invariant
one dimensional subspace, because if you act with T,
it's a T invariant space, you get the number
times a vector there. So the question that
you must ask now is you want to know if these
are simultaneous eigenvectors. So you want to figure
out what about S. How does S work with this thing? So you can act with
S from the left. So you get STUi is
equal to lambda i SUi. So here each Ui generates
an invariant subspace Ui T invariant. But S and T commute, so you have
T SUi is equal to lambda i SUi. And look at that equation again. This says that
this vector belongs to the invariant subspace Ui,
because it satisfies exactly the property that T acting on
it is equal to lambda i Ui. And it couldn't belong to
any other of the subspaces, because all the
eigenvalues are different. So spaces that are in Ui are the
spaces-- vectors that are in Ui are precisely all those
vectors that are left invariant by the action of T.
They're scaled only. So this vector is also in Ui. If this vector is
in Ui, SUi must be some number Wi times Ui. And therefore you've shown that
Ui is also an eigenvector of S, possibly with a different
eigenvalue of course. Because the only
thing that you know is that SUi is in this space. You don't know how big it is. So then you've shown
that, indeed, these Ui's that were eigenstates of T
are also eigenstates of S. And therefore
that's the statement of simultaneously
diagonalizable. They have the common
set of eigenvectors. So that's this part. And it's relatively
straight forward. Now we have to do case two. Case two is the interesting one. This time you're going
to have degeneracy. We have to have a notation
that is good for degeneracy. So if S is degeneracies,
has degeneracies, what happens with this operator? It will have-- remember,
a degenerate operator has eigenstates that form higher
than one dimensional spaces. If you have different
eigenvalues, each one generates a one
dimensional operator invariant subspace. But if you have
degeneracies, there are operators-- there are
spaces of higher dimensions that are left invariant. So for example, let Uk
denote the S invariant subspace of some
dimension Dk, which is greater or equal than 1. I will go here first. We're going to define Uk to
be the set of all vectors so that SU is equal
to lambda k U. And this will have dimension
of Uk is going to be Dk. So look what's happening. Basically the fact is
that for some eigenvalues, say the kth eigenvalue, you
just get several eigenvectors. So if you get several
eigenvectors not just scaled off each other,
these eigenvectors correspond to that
eigenvalue span of space. It's a degenerate subspace. So you must imagine
that as having a subspace of some
dimensionality with some basis vectors that span this thing. And they all have
the same eigenvector. Now you should really
have visualized this in a simple way. You have this subspace like a
cone or something like that, in which every vector
is an eigenvector. So every vector, when it's
acted by S is just scaled up. And all of them are
scaled by the same amount. That is what this
statement says. And corresponding to this thing,
you have a basis of vectors, of these eigenvectors,
and we'll call them Uk1, the first one,
the second, up to U Dk1, because it's a
subspace that we say it has the dimensionality Dk. So look at this thing. Somebody tells you
there's an operator. It has degenerate spectrum. You should start imagining all
kind of invariant subspaces of some dimensionality. If it has degeneracy, it's
a degeneracy each time the Dk is greater than
1, because if it's one dimensional, it's
just one basis vector one eigenvector, end of the story. Now this thing, by
the spectral theorem, this is an orthonormal basis. There's no problem, when you
have a degenerate subspace, to find an orthonormal basis. The theorem guarantees it,
so these are all orthonormal. So at the end of the day,
you have a decomposition of the vector space, V as U1
plus U2 plus maybe up to UM. And all of these vector
spaces like U's here, they may have some with just
no degeneracy, and some with degeneracy 2,
degeneracy 3, degeneracy 4. I don't know how
much degeneracy, but they might have
different degeneracy. Now what do we say next? Well, the fact that S
is a Hermitian operator says it just can
be diagonalized, and we're can find
all these spaces, and the basis for
the whole thing. So the basis would look U1
of the first base up to U d1 of the first base. These are the basis
vectors of the first plus the basis
vectors of the second. All the basis vectors U1
up to Udm of the mth space. All this is the list. This is the basis
of V. So I've listed the basis of V, which a basis
for U1, all these vectors. U2, all of this. So you see, we're not
calculating anything. We're just trying to
understand the picture. And why is this operator,
S, diagonal in this basis? It's clear. Because every vector here, every
vector is an eigenvector of S. So when you act with
S on any vector, you get that vector
times a number. But that vector is
orthogonal to all the rest. So when you have some
U and S and another U, this gives you a vector
proportional to U. And this is another vector. The matrix element is 0,
because they're all orthogonal. So it should be obvious why
this list produces something that is completely
orthogonal-- a diagonal matrix. So S, in this basis, looks
like the diagonal matrix in which you have lambda 1 d1
times up to lambda m dm times. Now I'll have to go until
2:00 to get the punchline. I apologize, but we
can't stop right now. We're almost there,
believe it or not. Two more things. This basis is good, but
actually another basis would also be good. I'll write this other basis
would be a V1 acting on the U1 up to V1 acting on that U1 up to
a Vm acting on this U1 up to Vm acting on that U1. This is m. m. This is dm. And here it's not U1. It's Ud1. You see, in the first
collection of vectors, I act with an operator V1 up
to here with an operator Vm. All of them with Vk being
a unitary operator in Uk. In every subspace, there
are unitary operators. So you can have
these bases and act with a unitary operator
of the space U1 here. A unitary operator
with a space U2 here. A unitary operator
of the space Un here. Hope you're following. And what happens if this
operator is unitary, this is still an
orthonormal basis in U1. These are still
orthonormal basis in Um. And therefore this is
an orthonormal basis for the whole thing, because
anyway those different spaces are orthogonal to each other. It's an orthogonal
decomposition. Everything is orthogonal
to everything. So this basis would be equally
good to represent the operator. Yes? AUDIENCE: [INAUDIBLE]
arbitrary unitary operators? PROFESSOR: Arbitrary unitary
operators at this moment. Arbitrary. So here comes the catch
as to the main property that now you want to establish
is that the spaces Uk are also T invariant. You see, the spaces
Uk were defined to be S invariant subspaces. And now the main important
thing is that they are also T invariant because they
commute with that. So let's see why
that is the case. Suppose U belongs to Uk. And then let's look at the
vector-- examine the vector Tu. What happens to Tu? Well, you want to act on
S on Tu to understand it. But S and T commute,
so this is T SU. But since U belongs
to Uk, that's the space with
eigenvalue lambda k. So this is lambda k times
u, so you have Tu here. So Tu acted with S
gives you lambda k Tu. So Tu is in the
invariant subspace Uk. What's happening here is now
something very straightforward. You try to imagine how does
the matrix T look in the basis that we have here. Here is this basis. how does this matrix T look? Well, this matrix keeps
the invariant subspaces. So you have to think of
it blocked diagonally. If it acts on it-- here
are the first vectors that you're considering, the U1. Well if you act on it
with T of the U1 subspace, you stay in the U1 subspace. So you don't get anything else. So you must have
0's all over here. And you can have a matrix here. And if you act on the second
Uk U2, you get a vector in U2, so it's orthogonal to
all the other vectors. So you get a matrix here. And you get a matrix here. So actually you get a
blocked diagonal matrix in which the blocks
correspond to the degeneracy. So if there's a degeneracy
d1 here, it's a d1 times d1. And d2 times d2. So actually you
haven't simultaneously diagonalized them. That's the problem
of degeneracy. You haven't, but you
now have the tools, because this operator is
Hermitian, therefore it's Hermitian here, and
here, and here, and here. So you can diagonalize here. But what do you need
for diagonalizing here? You need a unitary matrix. Call it V1. For here you need
another unitary matrix. Call it V2. Vn. And then this matrix
becomes diagonal. But then what about
the old matrix? Well, we just explained here
that if you change the basis by unitary matrices, you
don't change the first matrix. So actually you succeeded. You now can diagonalize
this without destroying your earlier result. And you managed to
diagonalize the whole thing. So this is for two
operators in the notes. You'll see why it simply extends
for three, four, and five, or arbitrary number
of operators. See you next time.
