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PROFESSOR: Going to
get started right away.
00:00:24.130 --> 00:00:29.570
I want to make a comment
about energy time uncertainty
00:00:29.570 --> 00:00:31.460
relations.
00:00:31.460 --> 00:00:35.530
And we talked last
time about the fact
00:00:35.530 --> 00:00:39.260
that the energy time uncertainty
relation really tells you
00:00:39.260 --> 00:00:43.760
something about how
fast a state can change.
00:00:43.760 --> 00:00:48.580
So an interesting way
to try to evaluate
00:00:48.580 --> 00:00:51.950
that is to consider
a system that
00:00:51.950 --> 00:00:55.950
has to state that time
equals 0 and compute
00:00:55.950 --> 00:01:00.550
the overlap with a
state at time equals t.
00:01:00.550 --> 00:01:06.880
Now this overlap is a
very schematic thing.
00:01:06.880 --> 00:01:08.210
It's a bracket.
00:01:08.210 --> 00:01:09.710
It's a number.
00:01:09.710 --> 00:01:11.960
And you know what that means.
00:01:11.960 --> 00:01:14.240
You can keep in
mind what that is.
00:01:14.240 --> 00:01:21.780
It's an integral over space
possibly of psi star at t
00:01:21.780 --> 00:01:29.600
equals 0 x psi t x.
00:01:29.600 --> 00:01:33.820
It's a complete integral,
its inner product.
00:01:33.820 --> 00:01:39.650
So you may want to understand
this, because, in a sense,
00:01:39.650 --> 00:01:44.310
this is telling you how
quickly a state can change.
00:01:44.310 --> 00:01:48.820
At time equals 0, this
overlap is just 1.
00:01:48.820 --> 00:01:52.020
A little later, this
overlap is going to change
00:01:52.020 --> 00:01:56.880
and perhaps after some time
the overlap is going to be 0.
00:01:56.880 --> 00:02:02.120
And we're going to say that we
actually have changed a lot.
00:02:02.120 --> 00:02:06.300
So this number is very
interesting to compute.
00:02:06.300 --> 00:02:10.530
And in fact, we might
as well square it,
00:02:10.530 --> 00:02:12.740
because it's a complex number.
00:02:12.740 --> 00:02:16.760
So to understand better
what it is we'll square it.
00:02:16.760 --> 00:02:21.340
And we'll have to evaluate this.
00:02:21.340 --> 00:02:24.310
Now how could you evaluate this?
00:02:24.310 --> 00:02:29.670
Well, we'll assume that this
system that governs this time
00:02:29.670 --> 00:02:32.425
evolution has a time
independent Hamiltonian.
00:02:39.740 --> 00:02:44.630
Once this evolution is done by
a time independent Hamiltonian,
00:02:44.630 --> 00:02:46.120
you can wonder what it is.
00:02:49.480 --> 00:02:51.640
Now it's quite
interesting, and you
00:02:51.640 --> 00:02:53.660
will have to discuss
that in the homework
00:02:53.660 --> 00:02:55.190
because it will
actually help you
00:02:55.190 --> 00:02:59.810
prove that version of the
time energy uncertainty
00:02:59.810 --> 00:03:04.650
relationship that says that the
quickest time state can turn
00:03:04.650 --> 00:03:09.490
orthogonal to itself is
bounded by some amount.
00:03:09.490 --> 00:03:11.880
Cannot do it infinitely fast.
00:03:11.880 --> 00:03:16.250
So you want to know how
fast this can change.
00:03:16.250 --> 00:03:22.010
Now it's very surprising what
it depends on, this thing.
00:03:22.010 --> 00:03:28.740
Because suppose you had
an energy eigenstate,
00:03:28.740 --> 00:03:33.070
suppose psi at time equals
0 is an energy eigenstate.
00:03:33.070 --> 00:03:35.740
What would happen later on?
00:03:35.740 --> 00:03:38.160
Well, you know that
energy eigenstates
00:03:38.160 --> 00:03:41.560
evolve with a phase,
an exponential of e
00:03:41.560 --> 00:03:45.820
to the minus iht over h bar.
00:03:45.820 --> 00:03:51.320
So actually if you had
an energy eigenstate,
00:03:51.320 --> 00:03:55.235
this thing would remain
equal to 1 for all times.
00:03:57.740 --> 00:04:00.770
So if this is going
to be non-zero,
00:04:00.770 --> 00:04:04.375
it's because it's
going to have--
00:04:04.375 --> 00:04:07.360
you have to have a state is
not an energy eigenstate.
00:04:07.360 --> 00:04:10.720
That you're going to have
an uncertainty in the energy
00:04:10.720 --> 00:04:12.340
and energy uncertainty.
00:04:12.340 --> 00:04:17.745
So the curious thing is that you
can evaluate this, and expand
00:04:17.745 --> 00:04:23.735
it as a power in t, and go,
say, to quadratic ordering
00:04:23.735 --> 00:04:26.456
in t evaluating what this is.
00:04:26.456 --> 00:04:30.670
And this only depends
on the uncertainty of h,
00:04:30.670 --> 00:04:32.840
and t, and things like that.
00:04:32.840 --> 00:04:38.450
So only the uncertainty of
h matters at this moment.
00:04:38.450 --> 00:04:40.700
So this would be quite
interested, I think,
00:04:40.700 --> 00:04:45.690
for you to figure out
and to explore in detail.
00:04:45.690 --> 00:04:50.650
That kind of analysis
has been the center
00:04:50.650 --> 00:04:56.150
of attention recently, having
to do with quantum computation.
00:04:56.150 --> 00:04:59.700
Because in a sense,
in a quantum computer,
00:04:59.700 --> 00:05:03.640
you want to change states
quickly and do operations.
00:05:03.640 --> 00:05:07.270
So how quickly you can
change a state is crucial.
00:05:07.270 --> 00:05:10.970
So in fact, the people that
proved these inequalities
00:05:10.970 --> 00:05:14.760
that you're going to find
say that this eventually
00:05:14.760 --> 00:05:19.110
will limit the speed
of a quantum computer,
00:05:19.110 --> 00:05:25.440
and more slow that computers
become twice as fast every year
00:05:25.440 --> 00:05:29.210
or so, and double the speed.
00:05:29.210 --> 00:05:32.440
So this limit
apparently, it's claimed,
00:05:32.440 --> 00:05:37.870
will allow 80
duplications of the speed
00:05:37.870 --> 00:05:43.430
until you hit this limit
in a quantum computer
00:05:43.430 --> 00:05:44.760
due to quantum mechanics.
00:05:44.760 --> 00:05:52.940
So you will be investigating
this in next week's homework.
00:05:52.940 --> 00:05:56.460
Today we want to begin
quickly with an application
00:05:56.460 --> 00:06:00.980
of the uncertainty principal to
find exact bounds of energies
00:06:00.980 --> 00:06:04.830
for quantum systems, for ground
states of quantum systems.
00:06:04.830 --> 00:06:09.340
So this will be a very precise
application of the uncertainty
00:06:09.340 --> 00:06:10.970
principle.
00:06:10.970 --> 00:06:14.150
Then we'll turn to a
completion of these things
00:06:14.150 --> 00:06:16.310
that we've been
talking about having
00:06:16.310 --> 00:06:18.830
to do with linear algebra.
00:06:18.830 --> 00:06:24.350
We'll get to the main
theorems of the subject.
00:06:24.350 --> 00:06:27.070
In a sense, the most important
theorems of the subject.
00:06:27.070 --> 00:06:30.240
These are the spectral
theorem that tells you
00:06:30.240 --> 00:06:35.140
about what operators
can be diagonalized.
00:06:35.140 --> 00:06:38.520
And then a theorem that
leads to the concept
00:06:38.520 --> 00:06:41.610
of a complete set of
commuting observables.
00:06:41.610 --> 00:06:46.030
So really pretty key
mathematical ideas.
00:06:46.030 --> 00:06:51.500
And the way we're good to
do it, I think you will see,
00:06:51.500 --> 00:06:56.080
that we have gained a lot by
learning the linear algebra
00:06:56.080 --> 00:06:59.270
concepts in a
slightly abstract way.
00:06:59.270 --> 00:07:03.390
I do remember doing this
proof that we're going today
00:07:03.390 --> 00:07:06.060
in previous years,
and it was always
00:07:06.060 --> 00:07:10.310
considered the most complicated
lecture of the course.
00:07:10.310 --> 00:07:12.830
Just taking the
[? indices ?] went crazy,
00:07:12.830 --> 00:07:16.540
and lots of formulas, and
the notation was funny.
00:07:16.540 --> 00:07:19.620
And now we will do
the proof, and we'll
00:07:19.620 --> 00:07:22.170
write a few little things.
00:07:22.170 --> 00:07:24.810
And we'll just try to
imagine what's going on.
00:07:24.810 --> 00:07:29.150
And will be, I think, easier.
00:07:29.150 --> 00:07:30.460
I hope you will agree.
00:07:30.460 --> 00:07:35.036
So let's begin with an example
of a use of the uncertainty
00:07:35.036 --> 00:07:35.535
principle.
00:07:35.535 --> 00:07:38.800
So example.
00:07:38.800 --> 00:07:42.980
So this will be
maybe for 20 minutes.
00:07:42.980 --> 00:07:48.220
Consider this Hamiltonian for
a one dimensional particle
00:07:48.220 --> 00:07:50.240
that, in fact, you've
considered before,
00:07:50.240 --> 00:07:54.510
alpha x to the fourth, for which
you did some approximations.
00:07:58.600 --> 00:08:02.290
You know that the
expectation value
00:08:02.290 --> 00:08:06.460
of the energy in
the ground state.
00:08:06.460 --> 00:08:08.220
You've done it numerically.
00:08:08.220 --> 00:08:10.950
You've done it variationally.
00:08:10.950 --> 00:08:15.710
And variationally you knew
that the energy at every stage
00:08:15.710 --> 00:08:18.940
was smaller at a given bound.
00:08:18.940 --> 00:08:22.570
The uncertainty
principle is not going
00:08:22.570 --> 00:08:25.780
to give us an upper bound.
00:08:25.780 --> 00:08:27.390
It's going to give
us a lower bound.
00:08:27.390 --> 00:08:29.540
So it's a really
nice thing, because
00:08:29.540 --> 00:08:32.610
between the variational
principal and the uncertainty
00:08:32.610 --> 00:08:37.039
principle, we can narrow the
energy of this ground state
00:08:37.039 --> 00:08:39.210
to a window.
00:08:39.210 --> 00:08:42.140
In one of the problems that
you're doing for this week--
00:08:42.140 --> 00:08:44.460
and I'm sorry only
today you really
00:08:44.460 --> 00:08:48.590
have all the tools after you
hear this discussion-- you
00:08:48.590 --> 00:08:50.505
do the same for the
harmonic oscillator.
00:08:50.505 --> 00:08:54.740
You do a variational estimate
for the ground state energy.
00:08:54.740 --> 00:08:59.420
You do the uncertainty principle
bound for ground state energy.
00:08:59.420 --> 00:09:02.820
And you will see
these two bounds meet.
00:09:02.820 --> 00:09:05.320
And therefore after you've
done this to bounds,
00:09:05.320 --> 00:09:07.480
you found the
ground state energy
00:09:07.480 --> 00:09:12.100
of the harmonic oscillator,
so it's kind of a neat thing.
00:09:12.100 --> 00:09:15.570
So we want to estimate
the ground state energy.
00:09:15.570 --> 00:09:17.930
So we first write some words.
00:09:17.930 --> 00:09:21.180
We just say H in
the ground state
00:09:21.180 --> 00:09:25.060
will be given by the
expectation value of p squared
00:09:25.060 --> 00:09:30.290
in the ground state plus
alpha times the expectation
00:09:30.290 --> 00:09:35.210
value of x to the fourth
in the ground state.
00:09:35.210 --> 00:09:38.850
Haven't made much
progress, but you
00:09:38.850 --> 00:09:43.310
have, because you're starting to
talk about the right variables.
00:09:43.310 --> 00:09:46.920
Now this thing to
that you have to know
00:09:46.920 --> 00:09:48.950
is that you have
a potential that
00:09:48.950 --> 00:09:53.530
is like this, sort
of a little flatter
00:09:53.530 --> 00:09:57.900
than x squared potential.
00:09:57.900 --> 00:10:02.280
And what can we say about
the expectation value
00:10:02.280 --> 00:10:04.420
of the momentum on
the ground state
00:10:04.420 --> 00:10:09.380
and the expectation value
of x in the ground state?
00:10:09.380 --> 00:10:14.530
Well, the expectation value
of x should be no big problem.
00:10:14.530 --> 00:10:17.940
This is a symmetric
potential, therefore
00:10:17.940 --> 00:10:21.990
wave functions in a one
dimensional quantum mechanics
00:10:21.990 --> 00:10:25.180
problems are either
symmetric or anti symmetric.
00:10:25.180 --> 00:10:27.430
It could not be anti
symmetric because it's
00:10:27.430 --> 00:10:29.790
a ground state and
kind of have a 0.
00:10:29.790 --> 00:10:31.620
So it's asymmetric.
00:10:31.620 --> 00:10:33.180
Has no nodes.
00:10:33.180 --> 00:10:35.800
So there's the wave
function of the ground
00:10:35.800 --> 00:10:38.850
state, the symmetric,
and the expectation value
00:10:38.850 --> 00:10:40.580
of x in the ground state is 0.
00:10:43.310 --> 00:10:47.930
Similarly, the expectation value
of the momentum in the ground
00:10:47.930 --> 00:10:49.275
state, what is it?
00:10:49.275 --> 00:10:49.775
Is?
00:10:52.550 --> 00:10:54.120
0 too.
00:10:54.120 --> 00:10:58.740
And you can imagine
just computing it.
00:10:58.740 --> 00:11:02.030
It would be the integral of psi.
00:11:02.030 --> 00:11:08.190
Psi is going to be a real
d dx h bar over i psi.
00:11:08.190 --> 00:11:09.830
This is a total derivative.
00:11:09.830 --> 00:11:13.810
If it's a bound state,
it's 0 at the ends.
00:11:13.810 --> 00:11:14.645
This is 0.
00:11:19.400 --> 00:11:23.280
So actually, we have a
little advantage here.
00:11:23.280 --> 00:11:29.530
We have some control
over what p squared is,
00:11:29.530 --> 00:11:39.630
because the uncertainty in p
in the ground state-- well,
00:11:39.630 --> 00:11:45.670
the uncertainty in p squared
is the expectation value
00:11:45.670 --> 00:11:51.490
of p squared minus the
expectation value of p squared.
00:11:51.490 --> 00:11:57.120
So in the ground
state, this is 0.
00:11:57.120 --> 00:12:01.990
So delta p squared
in the ground state
00:12:01.990 --> 00:12:05.010
is just p squared
on the ground state.
00:12:07.810 --> 00:12:10.570
Similarly, because
the expectation value
00:12:10.570 --> 00:12:16.730
of x is equal to 0, delta x
squared in the ground state
00:12:16.730 --> 00:12:23.130
is equal to expectation value of
x squared in the ground state.
00:12:23.130 --> 00:12:27.950
So actually, this expectation
of p squared is delta p.
00:12:27.950 --> 00:12:30.950
And we want to use the
uncertainty principle,
00:12:30.950 --> 00:12:32.670
so that's progress.
00:12:32.670 --> 00:12:36.266
We've related something we want
to estimate to an uncertainty.
00:12:39.580 --> 00:12:42.600
Small complication is that
we have an expectation
00:12:42.600 --> 00:12:46.850
value of x to the fourth.
00:12:46.850 --> 00:12:53.540
Now we learned-- maybe
I can continue here.
00:12:53.540 --> 00:12:57.510
We learned that the expectations
value for an operator squared
00:12:57.510 --> 00:13:01.060
is bigger than or equal
to the expectation
00:13:01.060 --> 00:13:04.300
value of the operator squared.
00:13:04.300 --> 00:13:10.300
So the expectation
value of x to the fourth
00:13:10.300 --> 00:13:16.930
is definitely bigger than the
expectation value of x squared
00:13:16.930 --> 00:13:19.590
squared.
00:13:19.590 --> 00:13:22.270
And this is true on any state.
00:13:24.800 --> 00:13:27.900
This was derived when
we did uncertainty.
00:13:27.900 --> 00:13:30.530
We proved that the
uncertainty squared
00:13:30.530 --> 00:13:33.150
is positive, because
the norm of a vector,
00:13:33.150 --> 00:13:35.820
and that gave you this thing.
00:13:35.820 --> 00:13:39.340
So here you think of the
operator as x squared.
00:13:39.340 --> 00:13:42.050
So the operator squared
is x to the fourth.
00:13:42.050 --> 00:13:45.630
And here's the operator
expectation value squared.
00:13:45.630 --> 00:13:50.120
So this is true for
the ground state.
00:13:50.120 --> 00:13:55.070
It's also here true for any
state, so is the ground state.
00:13:55.070 --> 00:13:58.580
And this x squared
now is delta x.
00:13:58.580 --> 00:14:05.120
So this is delta x on the
ground state to the fourth.
00:14:05.120 --> 00:14:07.425
So look what we have.
00:14:07.425 --> 00:14:12.460
We have that the expectation
value of H on the ground state
00:14:12.460 --> 00:14:17.890
is strictly equal to delta p
on the ground state squared
00:14:17.890 --> 00:14:23.120
over 2m plus alpha.
00:14:23.120 --> 00:14:28.660
And we cannot do a Priorean
equality here, so we have this.
00:14:28.660 --> 00:14:30.700
This is so far an equality.
00:14:30.700 --> 00:14:37.320
But because of this, this
thing is bigger than that.
00:14:37.320 --> 00:14:39.850
Well, alpha is supposed
to be positive.
00:14:39.850 --> 00:14:46.990
So this is bigger than delta
p ground state squared over
00:14:46.990 --> 00:14:53.555
2m plus alpha delta x on the
ground state to the fourth.
00:14:58.400 --> 00:15:00.040
OK, so far so good.
00:15:00.040 --> 00:15:02.850
We have a strict thing, this.
00:15:02.850 --> 00:15:06.270
And the order of the inequality
is already showing up.
00:15:06.270 --> 00:15:09.830
We're going to get, if
anything, a lower bound.
00:15:09.830 --> 00:15:12.575
You're going to be bigger
than or equal to something.
00:15:18.080 --> 00:15:19.610
So what is next?
00:15:19.610 --> 00:15:22.000
Next is the
uncertainty principle.
00:15:22.000 --> 00:15:26.780
We know that delta p delta
x is greater than or equal
00:15:26.780 --> 00:15:30.540
to h bar over 2 in any state.
00:15:30.540 --> 00:15:36.090
So the delta p ground state
and delta x on the ground state
00:15:36.090 --> 00:15:38.260
still should be equal to that.
00:15:38.260 --> 00:15:43.660
Therefore delta p ground
state is bigger than
00:15:43.660 --> 00:15:51.170
or equal than h over 2 delta x
in the ground state like that.
00:15:55.020 --> 00:15:58.480
So this inequality still
the right direction.
00:15:58.480 --> 00:16:05.140
So we can replace
this by something
00:16:05.140 --> 00:16:08.050
that is bigger than
this quantity day
00:16:08.050 --> 00:16:10.620
without disturbing the logic.
00:16:10.620 --> 00:16:14.800
So we have H ground
state now is greater
00:16:14.800 --> 00:16:20.490
than or equal to replace
the delta p by this thing
00:16:20.490 --> 00:16:27.860
here, h squared over 8,
because this is squared
00:16:27.860 --> 00:16:37.230
and there's another 2m delta x
ground state squared plus alpha
00:16:37.230 --> 00:16:42.060
delta x ground
state to the fourth.
00:16:42.060 --> 00:16:43.010
And that's it.
00:16:43.010 --> 00:16:46.970
We've obtained this inequality.
00:16:52.340 --> 00:17:00.830
So here you say, well, this
is good but how can I use it?
00:17:00.830 --> 00:17:04.270
I don't know what delta
x is in the ground state,
00:17:04.270 --> 00:17:07.099
so what have I gained?
00:17:07.099 --> 00:17:11.540
Well, let me do
a way of thinking
00:17:11.540 --> 00:17:14.290
about this that can help you.
00:17:14.290 --> 00:17:21.599
Plot the right hand side
as a function of delta
00:17:21.599 --> 00:17:23.380
x on the ground.
00:17:23.380 --> 00:17:25.270
So you don't know
how much it is,
00:17:25.270 --> 00:17:29.840
delta x on the ground
state, so just plot it.
00:17:29.840 --> 00:17:32.400
So if you plot this
function, there
00:17:32.400 --> 00:17:37.910
will be a divergence as
this delta x goes to 0,
00:17:37.910 --> 00:17:39.250
then it will be a minimum.
00:17:39.250 --> 00:17:43.420
It will be a positive minimum,
because this is all positive.
00:17:43.420 --> 00:17:45.125
And then it will go up again.
00:17:47.880 --> 00:17:52.100
So the right hand side as a
function of delta x is this.
00:17:52.100 --> 00:17:53.950
So here it comes.
00:17:53.950 --> 00:17:57.490
You see I don't know
what delta x is.
00:17:57.490 --> 00:18:00.640
Suppose delta x
happens to be this.
00:18:00.640 --> 00:18:02.760
Well, then I know
that the ground state
00:18:02.760 --> 00:18:06.460
energy is bigger
than that value.
00:18:06.460 --> 00:18:09.070
But maybe that's not delta x.
00:18:09.070 --> 00:18:14.280
Delta x may be is this
on the ground state.
00:18:14.280 --> 00:18:18.280
And then if it's that, well,
the ground state energy
00:18:18.280 --> 00:18:21.830
is bigger than this
value over here.
00:18:21.830 --> 00:18:25.570
Well, since I just
don't know what it is,
00:18:25.570 --> 00:18:30.570
the worst situation
is if delta x is here,
00:18:30.570 --> 00:18:38.150
and therefore definitely H must
be bigger than the lowest value
00:18:38.150 --> 00:18:41.130
that this can take.
00:18:41.130 --> 00:18:47.000
So the claim is that
H of gs, therefore
00:18:47.000 --> 00:18:53.070
is greater than or equal than
the minimum of this function
00:18:53.070 --> 00:18:57.620
h squared over 8m, and
I'll just write here
00:18:57.620 --> 00:19:06.900
delta x squared plus alpha delta
x to the fourth over delta x.
00:19:06.900 --> 00:19:12.995
The minimum of this function
over that space is the bound.
00:19:15.560 --> 00:19:22.770
So I just have to do a
calculus problem here.
00:19:22.770 --> 00:19:25.760
This is the minimum.
00:19:25.760 --> 00:19:29.260
I should take the derivative
with respect to delta x.
00:19:29.260 --> 00:19:32.060
Find delta x and substitute.
00:19:32.060 --> 00:19:34.140
Of course, I'm not
going to do that here,
00:19:34.140 --> 00:19:38.170
but I'll tell you a formula
that will do that for you.
00:19:38.170 --> 00:19:43.800
A over x squared
plus Bx to the fourth
00:19:43.800 --> 00:19:50.660
is minimized for x squared
is equal to 1 over 2--
00:19:50.660 --> 00:19:52.250
it's pretty awful numbers.
00:19:52.250 --> 00:19:58.550
2 to the 1/3 A
over B to the 1/3.
00:19:58.550 --> 00:20:12.390
And its value at that point is
2 to the 1/3 times 3/2 times
00:20:12.390 --> 00:20:16.760
A to the 2/3 times B to the 1/3.
00:20:19.470 --> 00:20:22.335
A little bit of arithmetic.
00:20:30.420 --> 00:20:33.210
So for this function,
it turns out
00:20:33.210 --> 00:20:37.390
that A is whatever
coefficient is here.
00:20:37.390 --> 00:20:39.280
B is whatever
coefficient is there,
00:20:39.280 --> 00:20:42.130
so this is supposed
to be the answer.
00:20:42.130 --> 00:20:46.130
And you get H on
the ground state
00:20:46.130 --> 00:20:52.830
is greater than or equal to
2 to the 1/3 3/8 h squared
00:20:52.830 --> 00:21:02.430
square root of alpha over m to
the 2/3, which is about 0.4724
00:21:02.430 --> 00:21:09.030
times h squared square root
of alpha over m to the 2/3.
00:21:09.030 --> 00:21:11.410
And that's our bound.
00:21:11.410 --> 00:21:13.740
How good or how
bad is the bound?
00:21:13.740 --> 00:21:14.540
It's OK.
00:21:14.540 --> 00:21:17.190
It's not fabulous.
00:21:17.190 --> 00:21:30.710
The real answer is done
numerically is 0.668.
00:21:30.710 --> 00:21:33.580
I think I remember
variational principal gave you
00:21:33.580 --> 00:21:38.170
something like 0.68 or 0.69.
00:21:38.170 --> 00:21:43.790
And this one says
it's bigger than 0.47.
00:21:43.790 --> 00:21:45.190
It gives you something.
00:21:45.190 --> 00:21:50.700
So the important thing is
that it's completely rigorous.
00:21:50.700 --> 00:21:54.310
Many times people use
the uncertainty principle
00:21:54.310 --> 00:21:57.190
to estimate ground
state energies.
00:21:57.190 --> 00:21:59.560
Those estimates
are very hand wavy.
00:21:59.560 --> 00:22:02.800
You might as well just
do dimensional analysis.
00:22:02.800 --> 00:22:04.570
You don't gain anything.
00:22:04.570 --> 00:22:06.500
You don't know the factors.
00:22:06.500 --> 00:22:08.185
But this is completely rigorous.
00:22:08.185 --> 00:22:12.240
I never made an approximation
or anything here.
00:22:12.240 --> 00:22:14.160
Every step was logical.
00:22:14.160 --> 00:22:16.580
Every inequality was exact.
00:22:16.580 --> 00:22:20.250
And therefore, this
is a solid result.
00:22:20.250 --> 00:22:22.980
This is definitely true.
00:22:22.980 --> 00:22:25.970
It doesn't tell you an
estimate of the answer.
00:22:25.970 --> 00:22:31.400
If you dimensional analysis, you
say the answer is this times 1,
00:22:31.400 --> 00:22:36.350
and that's as good as you can
do with dimensional analysis.
00:22:36.350 --> 00:22:37.530
It's not that bad.
00:22:37.530 --> 00:22:41.480
The answer turns out to be 0.7.
00:22:41.480 --> 00:22:44.110
But the uncertainty
principle really,
00:22:44.110 --> 00:22:47.660
if you're careful, sometimes,
not for every problem,
00:22:47.660 --> 00:22:52.190
you can do a rigorous thing
and find the rigorous answer.
00:22:52.190 --> 00:22:54.400
OK, so are there any questions?
00:23:00.280 --> 00:23:01.980
Your problem in
the homework will
00:23:01.980 --> 00:23:04.900
be to do this for the
harmonic oscillator
00:23:04.900 --> 00:23:06.911
and find the two bounds.
00:23:06.911 --> 00:23:07.410
Yes?
00:23:07.410 --> 00:23:08.910
AUDIENCE: How does
the answer change
00:23:08.910 --> 00:23:10.492
if we don't look at
the ground state?
00:23:10.492 --> 00:23:11.900
PROFESSOR: How do they what?
00:23:11.900 --> 00:23:12.890
PROFESSOR: How does
the answer change
00:23:12.890 --> 00:23:15.320
if we look at a state different
from the ground state?
00:23:15.320 --> 00:23:19.190
PROFESSOR: Different
from the ground state?
00:23:19.190 --> 00:23:21.970
So the question was
how would this change
00:23:21.970 --> 00:23:24.650
if I would try to do something
different from the ground
00:23:24.650 --> 00:23:25.720
state.
00:23:25.720 --> 00:23:30.640
I think for any
state, you would still
00:23:30.640 --> 00:23:35.470
be able to say that
the expectation
00:23:35.470 --> 00:23:37.700
value of the momentum is 0.
00:23:37.700 --> 00:23:42.430
Now the expectation value
of x still would be 0.
00:23:42.430 --> 00:23:48.460
So you can go through
some steps here.
00:23:48.460 --> 00:23:52.760
The problem here being
that I don't have a way
00:23:52.760 --> 00:23:55.260
to treat any other
state differently.
00:23:55.260 --> 00:23:57.690
So I would go ahead,
and I would have
00:23:57.690 --> 00:24:03.830
said for any stationary state,
or for any energy eigenstate,
00:24:03.830 --> 00:24:07.660
all of what I said is true.
00:24:07.660 --> 00:24:09.110
So I don't get a new one.
00:24:12.210 --> 00:24:16.140
These things people actually
keep working and writing papers
00:24:16.140 --> 00:24:17.880
on this stuff.
00:24:17.880 --> 00:24:21.810
People sometimes find bounds
that are a little original.
00:24:21.810 --> 00:24:23.124
Yes?
00:24:23.124 --> 00:24:25.415
AUDIENCE: How do you know
the momentum expectation is 0
00:24:25.415 --> 00:24:25.914
again?
00:24:25.914 --> 00:24:30.170
PROFESSOR: The momentum
expectation for a bound state
00:24:30.170 --> 00:24:30.940
goes like this.
00:24:30.940 --> 00:24:35.379
So you want to figure
out what is psi p psi.
00:24:35.379 --> 00:24:36.420
And you do the following.
00:24:36.420 --> 00:24:39.350
That's integral.
00:24:39.350 --> 00:24:42.230
Now psi in these
problems can be chosen
00:24:42.230 --> 00:24:44.710
to be real, so I won't bother.
00:24:44.710 --> 00:24:52.240
It's psi of x h bar
over i d dx of psi.
00:24:52.240 --> 00:25:01.525
So this is equal to h bar over
2i the integral dx of d dx
00:25:01.525 --> 00:25:02.765
of psi squared.
00:25:07.340 --> 00:25:13.960
So at this moment, you say
well that's just h bar over 2i,
00:25:13.960 --> 00:25:19.800
the value of psi squared at
infinity and at minus infinity.
00:25:19.800 --> 00:25:23.360
And since it's a bound
state, it's 0 here, 0 there,
00:25:23.360 --> 00:25:27.020
and it's equal to 0.
00:25:27.020 --> 00:25:30.800
A state that would have
expectation value of momentum
00:25:30.800 --> 00:25:34.520
you would expect
it to be moving.
00:25:34.520 --> 00:25:37.610
So this state is
there is static.
00:25:37.610 --> 00:25:41.630
It's stationary It doesn't have
expectation value of momentum.
00:25:41.630 --> 00:25:42.250
Yes?
00:25:42.250 --> 00:25:45.172
AUDIENCE: Is the reason that
you can't get a better estimate
00:25:45.172 --> 00:25:47.607
for the things that are
on the ground state,
00:25:47.607 --> 00:25:50.042
because if you consider
the harmonic oscillator,
00:25:50.042 --> 00:25:53.451
the uncertainty delta x
[? delta p ?] from the ground
00:25:53.451 --> 00:25:57.370
state [INAUDIBLE] you
go up to higher states.
00:25:57.370 --> 00:25:59.770
PROFESSOR: Right, I
think that's another way.
00:25:59.770 --> 00:26:05.580
AUDIENCE: [INAUDIBLE] higher
state using the absolute.
00:26:05.580 --> 00:26:06.920
PROFESSOR: Yeah.
00:26:06.920 --> 00:26:08.150
That's a [INAUDIBLE].
00:26:08.150 --> 00:26:11.920
So the ground state of the
harmonic oscillator saturates
00:26:11.920 --> 00:26:15.280
the uncertainty principal
and the others don't.
00:26:15.280 --> 00:26:19.630
So this argument, I think,
is just good for ground state
00:26:19.630 --> 00:26:21.750
energies.
00:26:21.750 --> 00:26:23.902
One more question.
00:26:23.902 --> 00:26:26.510
AUDIENCE: It appears that
this method really works.
00:26:26.510 --> 00:26:29.638
Doesn't particularly work well
if we have a potential that
00:26:29.638 --> 00:26:32.932
has an odd power, because we
can't use [? the packet ?],
00:26:32.932 --> 00:26:35.430
like x [INAUDIBLE] expectation
value x to the fourth
00:26:35.430 --> 00:26:37.670
is something, some
power, expectation.
00:26:37.670 --> 00:26:40.820
PROFESSOR: Right, if
it's an odd power,
00:26:40.820 --> 00:26:42.450
the method doesn't work well.
00:26:42.450 --> 00:26:45.000
But actually for an
odd power, the physics
00:26:45.000 --> 00:26:50.010
doesn't work well either,
because the system doesn't
00:26:50.010 --> 00:26:51.410
have ground states.
00:26:51.410 --> 00:26:56.020
And so let's say
that if you had x
00:26:56.020 --> 00:26:59.470
to the fourth plus some x
cubed, the physics could still
00:26:59.470 --> 00:27:00.740
make sense.
00:27:00.740 --> 00:27:04.590
But then it's not clear
I can do the same.
00:27:04.590 --> 00:27:06.080
Actually you can
do the same for x
00:27:06.080 --> 00:27:09.570
to the eighth and any sort
of powers of this type.
00:27:09.570 --> 00:27:12.470
But I don't think it
works for x to the sixth.
00:27:12.470 --> 00:27:14.780
You can try a few things.
00:27:14.780 --> 00:27:19.390
OK, so we leave the
uncertainty principle
00:27:19.390 --> 00:27:28.480
and begin to understand
more formerly the operators
00:27:28.480 --> 00:27:33.120
for which there's no uncertainty
and you can simultaneously
00:27:33.120 --> 00:27:34.860
diagonalize them.
00:27:34.860 --> 00:27:39.150
So we're going to find
operators like A and B,
00:27:39.150 --> 00:27:41.050
that they commute.
00:27:41.050 --> 00:27:43.070
And then sometimes
you can simultaneously
00:27:43.070 --> 00:27:43.810
diagonalize them.
00:27:43.810 --> 00:27:44.310
Yes?
00:27:44.310 --> 00:27:45.645
You have a question.
00:27:45.645 --> 00:27:50.082
AUDIENCE: So part of [INAUDIBLE]
we use here is [INAUDIBLE],
00:27:50.082 --> 00:27:50.582
right?
00:27:50.582 --> 00:27:51.290
PROFESSOR: Right.
00:27:54.558 --> 00:27:59.031
AUDIENCE: If we get an asset--
is there any way that we
00:27:59.031 --> 00:28:03.272
can better our [INAUDIBLE]
principle based on the wave
00:28:03.272 --> 00:28:05.008
function with non saturated?
00:28:05.008 --> 00:28:07.988
Can we get an upper
bound for just
00:28:07.988 --> 00:28:10.521
[INAUDIBLE] principle
with an h bar over it?
00:28:10.521 --> 00:28:12.020
PROFESSOR: Can I
get an upper bound?
00:28:12.020 --> 00:28:15.260
I'm not sure I
understand your question.
00:28:15.260 --> 00:28:18.966
AUDIENCE: [INAUDIBLE] the fact
that the [INAUDIBLE] principle
00:28:18.966 --> 00:28:20.250
will not be saturated.
00:28:20.250 --> 00:28:26.542
Can you put the bound for
just taking [INAUDIBLE]?
00:28:33.240 --> 00:28:36.070
PROFESSOR: Yeah, certainly.
00:28:36.070 --> 00:28:38.460
You might have some
systems in which
00:28:38.460 --> 00:28:40.270
you know that this
uncertainty might
00:28:40.270 --> 00:28:43.730
be bigger than the one warranted
by the uncertainty principles.
00:28:43.730 --> 00:28:45.620
And you use that information.
00:28:45.620 --> 00:28:49.420
But on general grounds,
it's hard to know
00:28:49.420 --> 00:28:52.730
that a system might come
very close to satisfy
00:28:52.730 --> 00:28:55.120
the uncertainty principle
in its ground state.
00:28:55.120 --> 00:28:56.050
We don't know.
00:28:56.050 --> 00:28:59.220
There are systems that come
very close in the ground state
00:28:59.220 --> 00:29:01.800
to satisfy this and
some that are far.
00:29:01.800 --> 00:29:04.790
If they are far, you
must have some reason
00:29:04.790 --> 00:29:06.710
to understand that to use it.
00:29:06.710 --> 00:29:10.320
So I don't know.
00:29:10.320 --> 00:29:15.080
So let me turn now to
this issue of operators
00:29:15.080 --> 00:29:25.156
and diagonalization
of operators.
00:29:30.090 --> 00:29:36.350
Now you might be irritated
a little even by the title.
00:29:36.350 --> 00:29:37.800
Diagonalization of operation.
00:29:37.800 --> 00:29:42.970
You'll be talking about
diagonalization of matrices.
00:29:42.970 --> 00:29:46.160
Well, there's a
way to state what
00:29:46.160 --> 00:29:51.660
we mean by diagonalizing
an operator in such a way
00:29:51.660 --> 00:29:55.990
that we can talk later
about the matrix.
00:29:55.990 --> 00:29:58.670
So what is the point here?
00:29:58.670 --> 00:30:01.360
You have an operator,
and it's presumably
00:30:01.360 --> 00:30:05.090
an important operator
in your theory.
00:30:05.090 --> 00:30:09.250
You want to understand
this operator better.
00:30:09.250 --> 00:30:12.280
So you really are
faced with a dilemma.
00:30:12.280 --> 00:30:17.480
How do I get some insight
into this operator?
00:30:17.480 --> 00:30:20.610
Perhaps the simplest thing
you could do is to say,
00:30:20.610 --> 00:30:26.530
OK let me choose some ideal
basis of the vector space,
00:30:26.530 --> 00:30:29.370
such as that operator
is as simple as possible
00:30:29.370 --> 00:30:30.530
in that basis.
00:30:30.530 --> 00:30:32.870
So that's the origin
of this thing.
00:30:32.870 --> 00:30:37.090
Find a basis in the state
space, so the operator
00:30:37.090 --> 00:30:39.310
looks as simple as possible.
00:30:39.310 --> 00:30:43.800
So you say that you can
diagonalize an operator
00:30:43.800 --> 00:30:51.950
if you can find the basis
such that the operator has
00:30:51.950 --> 00:30:53.300
just diagonal entries.
00:30:56.420 --> 00:30:58.770
So let me just
write it like this.
00:31:03.160 --> 00:31:20.990
So if you can find a basis in
V where the matrix representing
00:31:20.990 --> 00:31:34.950
the operator is
diagonal, the operator
00:31:34.950 --> 00:31:39.150
is said to be diagonalizable.
00:31:46.900 --> 00:31:49.870
So to be diagonalizable
is just a statement
00:31:49.870 --> 00:31:54.370
that there is some basis
where you look at the matrix
00:31:54.370 --> 00:31:56.410
representation
operator, and you find
00:31:56.410 --> 00:31:59.300
that it takes form
as a diagonal.
00:31:59.300 --> 00:32:04.770
So let's try to understand
this conceptually
00:32:04.770 --> 00:32:07.220
and see what actually
it's telling us.
00:32:07.220 --> 00:32:09.440
It tells us actually a lot.
00:32:09.440 --> 00:32:24.590
Suppose t is diagonal in
some basis u1 up to un.
00:32:27.620 --> 00:32:31.460
So what does it mean
for it to be diagonal?
00:32:31.460 --> 00:32:34.790
Well, you may remember
all these definitions
00:32:34.790 --> 00:32:37.580
we had about matrix action.
00:32:37.580 --> 00:32:42.700
If T acting on a ui
is supposed to be
00:32:42.700 --> 00:32:49.970
Tki uk in some basis sum over k.
00:32:49.970 --> 00:32:54.070
You act on ui, and
you get a lot of u's.
00:32:54.070 --> 00:32:57.610
And these are the matrix
elements of the operator.
00:32:57.610 --> 00:33:01.150
Now the fact that
it's diagonalizable
00:33:01.150 --> 00:33:05.010
means that in some basis, the
u basis, this is diagonal.
00:33:05.010 --> 00:33:10.465
So ki in this sum only
happens to work out
00:33:10.465 --> 00:33:12.640
when k is equal to i.
00:33:12.640 --> 00:33:16.440
And that's one number and you
get back to the vector ui.
00:33:16.440 --> 00:33:19.860
So if it's diagonal
in this basis,
00:33:19.860 --> 00:33:28.060
you have the T on u1 is lambda
a number times u1 T on u2
00:33:28.060 --> 00:33:31.140
is lambda 2 in u2.
00:33:31.140 --> 00:33:35.883
And Tun equal lambda n un.
00:33:38.990 --> 00:33:43.740
So what you learn is
that this basis vector--
00:33:43.740 --> 00:33:46.010
so you learn something
that maybe you
00:33:46.010 --> 00:33:47.910
thought it's tautological.
00:33:47.910 --> 00:33:49.350
It's not tautological.
00:33:49.350 --> 00:33:53.210
You learn that if you have
a set of basis vectors
00:33:53.210 --> 00:33:56.140
in which the
operator is diagonal,
00:33:56.140 --> 00:34:02.450
these basis vectors are
eigenvectors of the operator.
00:34:02.450 --> 00:34:06.510
And then you learn something
that is quite important,
00:34:06.510 --> 00:34:12.100
that an operator is
diagonalizable if,
00:34:12.100 --> 00:34:16.179
and only if, it has
a set of eigenvectors
00:34:16.179 --> 00:34:19.230
that span the space.
00:34:19.230 --> 00:34:21.830
So the statement
is very important.
00:34:21.830 --> 00:34:29.830
An operator T is
that diagonalizable
00:34:29.830 --> 00:34:43.717
if it has a set of eigenvectors
that span the space.
00:34:43.717 --> 00:34:50.980
Span V. If and only if.
00:34:50.980 --> 00:34:53.210
If this double f.
00:34:53.210 --> 00:34:56.770
If and only if.
00:34:56.770 --> 00:35:00.490
So here it's diagonalizable,
and we have a basis,
00:35:00.490 --> 00:35:03.720
and it has a set of
these are eigenvectors.
00:35:03.720 --> 00:35:06.000
So diagonalizable
realizable really
00:35:06.000 --> 00:35:08.560
means that it has a
set of eigenvectors
00:35:08.560 --> 00:35:11.370
that span the space.
00:35:11.370 --> 00:35:14.590
On the other hand, if you have
the set of eigenvectors that
00:35:14.590 --> 00:35:19.520
span the space, you have a
set of u's that satisfy this,
00:35:19.520 --> 00:35:23.250
and then you read that, oh
yeah, this matrix is diagonal,
00:35:23.250 --> 00:35:26.220
so it's diagonalizable.
00:35:26.220 --> 00:35:30.050
So a simple statement,
but an important one,
00:35:30.050 --> 00:35:35.600
because there are examples of
matrices that immediately you
00:35:35.600 --> 00:35:39.330
know you're never going
to succeed to diagonalize.
00:35:39.330 --> 00:35:42.648
So here is one matrix, 0 0 1 0.
00:35:46.060 --> 00:35:50.450
This matrix has
eigenvalues, so you
00:35:50.450 --> 00:35:54.560
do the characteristic equation
lambda squared equals 0.
00:35:54.560 --> 00:35:58.790
So the only eigenvalue
is lambda equals 0.
00:35:58.790 --> 00:36:01.350
And let's see how
many eigenvectors
00:36:01.350 --> 00:36:04.050
you would have for
lambda equals 0.
00:36:04.050 --> 00:36:09.320
Well, you would have if
this is T, T on some vector
00:36:09.320 --> 00:36:13.560
a b must be equal to 0.
00:36:13.560 --> 00:36:25.750
So this is 0 1 0 0 on a b,
which is b and 0, must be zero.
00:36:25.750 --> 00:36:28.990
So b is equal to 0.
00:36:28.990 --> 00:36:33.090
So the only
eigenvector here-- I'll
00:36:33.090 --> 00:36:36.190
just write it here and then
move to the other side.
00:36:36.190 --> 00:36:38.770
The only eigenvector
for lambda equals 0,
00:36:38.770 --> 00:36:41.510
the only eigenvector
is with b equals 0.
00:36:41.510 --> 00:36:43.610
So it's 1 0.
00:36:43.610 --> 00:36:45.610
One eigenvector only.
00:36:45.610 --> 00:36:47.620
No more eigenvectors.
00:36:47.620 --> 00:36:51.780
By the theorem,
or by this claim,
00:36:51.780 --> 00:36:55.305
you know it's a two dimensional
vector space you just
00:36:55.305 --> 00:36:57.310
can't diagonalize this matrix.
00:36:57.310 --> 00:36:58.560
It's impossible.
00:36:58.560 --> 00:37:01.700
Can't be done.
00:37:01.700 --> 00:37:05.760
OK, a couple more
things that I wish to
00:37:05.760 --> 00:37:08.550
say about this process
of diagonalization.
00:37:12.300 --> 00:37:17.820
Well, the statement that
an operator is diagonal
00:37:17.820 --> 00:37:23.370
is a statement about the
existence of some basis.
00:37:23.370 --> 00:37:26.820
Now you can try to figure
out what that basis is,
00:37:26.820 --> 00:37:30.950
so typically what is the
problem that you face?
00:37:30.950 --> 00:37:35.264
Typically you have a
vector spaces V. Sorry?
00:37:35.264 --> 00:37:36.430
AUDIENCE: I have a question.
00:37:36.430 --> 00:37:37.055
PROFESSOR: Yes?
00:37:39.555 --> 00:37:43.190
If you had an infinite
dimensional space
00:37:43.190 --> 00:37:49.340
and you had an operator
whose eigenvectors do not
00:37:49.340 --> 00:37:53.175
span the space, can it
still have eigenvectors,
00:37:53.175 --> 00:37:56.090
or does it not have any then?
00:37:56.090 --> 00:37:56.860
PROFESSOR: No.
00:37:56.860 --> 00:37:58.930
You said it has
some eigenvectors,
00:37:58.930 --> 00:38:00.394
but they don't span the space.
00:38:00.394 --> 00:38:01.810
So it does have
some eigenvectors.
00:38:01.810 --> 00:38:03.899
AUDIENCE: So my question
is was what I just
00:38:03.899 --> 00:38:08.060
said a logical contradiction in
an infinite dimensional space?
00:38:08.060 --> 00:38:10.450
PROFESSOR: To have
just some eigenvectors?
00:38:10.450 --> 00:38:10.950
I think--
00:38:10.950 --> 00:38:12.680
PROFESSOR: I'm looking
more specifically
00:38:12.680 --> 00:38:14.825
at a dagger for instance.
00:38:14.825 --> 00:38:15.450
PROFESSOR: Yes.
00:38:15.450 --> 00:38:17.160
AUDIENCE: In the
harmonic oscillator,
00:38:17.160 --> 00:38:21.660
you I think mentioned at some
point that it does not have--
00:38:21.660 --> 00:38:23.980
PROFESSOR: So the
fact that you can'
00:38:23.980 --> 00:38:27.060
diagonalize this
thing already implies
00:38:27.060 --> 00:38:29.450
that it's even worse
in higher dimensions.
00:38:29.450 --> 00:38:32.720
So some operator
may be pretty nice,
00:38:32.720 --> 00:38:36.370
and you might still be
able to diagonalize it,
00:38:36.370 --> 00:38:39.154
so you're going to lack
eigenvectors in general.
00:38:39.154 --> 00:38:40.570
You're going to
lack lots of them.
00:38:40.570 --> 00:38:42.730
And there are going to
be blocks of Jordan.
00:38:42.730 --> 00:38:46.420
Blocks are called things that
are above the diagonal, things
00:38:46.420 --> 00:38:47.750
that you can't do much about.
00:38:51.610 --> 00:38:56.530
Let me then think concretely now
that you have a vector space,
00:38:56.530 --> 00:39:03.380
and you've chosen
some basis v1 vn.
00:39:03.380 --> 00:39:08.920
And then you look
at this operator T,
00:39:08.920 --> 00:39:12.540
and of course, you chose
an arbitrary basis.
00:39:12.540 --> 00:39:16.000
There's no reason why
its matrix representation
00:39:16.000 --> 00:39:17.210
would be diagonal.
00:39:17.210 --> 00:39:23.540
So T on the basis v-- Tij.
00:39:23.540 --> 00:39:27.440
Sometimes to be very explicit
we write Tij like that--
00:39:27.440 --> 00:39:30.700
is not diagonal.
00:39:30.700 --> 00:39:34.310
Now if it's not
diagonal, the question
00:39:34.310 --> 00:39:38.390
is whether you can find a
basis where it is diagonal.
00:39:38.390 --> 00:39:42.050
And then you try, of
course, changing basis.
00:39:42.050 --> 00:39:44.090
And you change basis--
you've discussed
00:39:44.090 --> 00:39:46.710
that in the homework--
with a linear operator.
00:39:46.710 --> 00:39:49.800
So you use a linear
operator to produce
00:39:49.800 --> 00:39:53.490
another basis, an
invertible in your operator.
00:39:53.490 --> 00:39:58.670
So that you get these vectors
uk being equal to some operator
00:39:58.670 --> 00:40:01.280
A times vk.
00:40:01.280 --> 00:40:06.560
So this is going to be
the u1's up to un's are
00:40:06.560 --> 00:40:09.690
going to be another basis.
00:40:09.690 --> 00:40:12.350
The n vector here
is the operator
00:40:12.350 --> 00:40:15.680
acting with the n
vector on this thing.
00:40:15.680 --> 00:40:19.340
And then you prove, in the
homework, a relationship
00:40:19.340 --> 00:40:27.010
between these matrix elements
of T in the new basis,
00:40:27.010 --> 00:40:28.310
in the u basis.
00:40:31.300 --> 00:40:35.500
And the matrix elements
of T in the v basis.
00:40:39.830 --> 00:40:42.650
You have a
relationship like this,
00:40:42.650 --> 00:40:50.720
or you have more explicitly
Tij in the basis u
00:40:50.720 --> 00:41:03.602
is equal to A minus
1 ik Tkp of v Apj.
00:41:09.370 --> 00:41:10.955
So this is what happens.
00:41:14.040 --> 00:41:16.550
This is the new
operator in this basis.
00:41:16.550 --> 00:41:19.020
And typically what
you're trying to do
00:41:19.020 --> 00:41:24.020
is find this matrix A
that makes this thing
00:41:24.020 --> 00:41:26.090
into a diagonal matrix.
00:41:26.090 --> 00:41:30.020
Because we say in the u basis
the operator is diagonal.
00:41:36.560 --> 00:41:41.740
I want to emphasize that there's
a couple of ways in which you
00:41:41.740 --> 00:41:44.230
can think of diagonalization.
00:41:44.230 --> 00:41:47.490
Sort of a passive
and an active way.
00:41:47.490 --> 00:41:51.050
You can imagine the
operator, and you
00:41:51.050 --> 00:41:54.610
say look, this
operator I just need
00:41:54.610 --> 00:41:58.860
to find some basis in
which it is diagonal.
00:41:58.860 --> 00:42:02.280
So I'm looking for a basis.
00:42:02.280 --> 00:42:06.070
The other way of
thinking of this operator
00:42:06.070 --> 00:42:17.940
is to think that A minus
1 TA is another operator,
00:42:17.940 --> 00:42:22.930
and it's diagonal
in original basis.
00:42:22.930 --> 00:42:29.020
So it might have seem funny
to you, but let's stop again
00:42:29.020 --> 00:42:31.170
and say this again.
00:42:31.170 --> 00:42:34.620
You have an operator, and the
question of diagonalization
00:42:34.620 --> 00:42:39.540
is whether there is some basis
in which it looks diagonal,
00:42:39.540 --> 00:42:41.060
its matrix is diagonal.
00:42:41.060 --> 00:42:44.650
But the equivalent
question is whether there
00:42:44.650 --> 00:42:48.690
is an operator A
such that this is
00:42:48.690 --> 00:42:52.800
diagonal in the original basis.
00:42:52.800 --> 00:42:57.680
To make sure that you see
that, consider the following.
00:42:57.680 --> 00:43:03.170
So this is diagonal
in the original basis.
00:43:08.120 --> 00:43:16.240
So in order to see that, think
of Tui is equal to lambda i ui.
00:43:16.240 --> 00:43:19.010
We know that the
u's are supposed
00:43:19.010 --> 00:43:23.350
to be this basis of eigenvectors
where the matrix is diagonal,
00:43:23.350 --> 00:43:25.210
so here you got it.
00:43:25.210 --> 00:43:27.200
Here the i not summed.
00:43:30.160 --> 00:43:31.820
It's pretty important.
00:43:31.820 --> 00:43:34.300
There's a problem with
this eigenvalue notation.
00:43:34.300 --> 00:43:36.580
I don't know how
to do it better.
00:43:36.580 --> 00:43:40.340
If you have several eigenvalues,
you want to write this,
00:43:40.340 --> 00:43:44.950
but you don't want this to
think that you're acting on u1
00:43:44.950 --> 00:43:46.910
and you get lambda 1 u1.
00:43:46.910 --> 00:43:49.320
Not the sum right here.
00:43:49.320 --> 00:43:55.640
OK, but they ui is
equal to A on vi.
00:44:00.130 --> 00:44:05.560
So therefore this
is lambda i A on vi.
00:44:09.200 --> 00:44:12.080
And then you'll
act with A minus 1.
00:44:12.080 --> 00:44:16.000
Act with A minus 1 from
the left with the operator.
00:44:16.000 --> 00:44:25.610
So you get A minus 1 TA vi
is equal to lambda i vi.
00:44:28.540 --> 00:44:30.090
So what do you see?
00:44:30.090 --> 00:44:33.270
You see an operator
that is actually
00:44:33.270 --> 00:44:37.130
diagonal in the v basis.
00:44:37.130 --> 00:44:41.730
So this operator is diagonal
in the original basis.
00:44:41.730 --> 00:44:43.950
That's another way of
thinking of the process
00:44:43.950 --> 00:44:46.160
of diagonalization.
00:44:46.160 --> 00:44:53.700
There's one last remark,
which is that the columns of A
00:44:53.700 --> 00:44:57.770
are the eigenvectors, in fact.
00:44:57.770 --> 00:45:02.343
Columns of A are
the eigenvectors.
00:45:06.580 --> 00:45:09.670
Well, how do you see that?
00:45:09.670 --> 00:45:11.700
It's really very simple.
00:45:11.700 --> 00:45:14.030
You can convince
yourself in many ways,
00:45:14.030 --> 00:45:19.760
but the uk are the eigenvectors.
00:45:19.760 --> 00:45:22.480
But what are uk's?
00:45:22.480 --> 00:45:23.470
I have it somewhere.
00:45:23.470 --> 00:45:24.290
There it is.
00:45:24.290 --> 00:45:25.510
A on vk.
00:45:31.800 --> 00:45:36.740
And A on vk is this
matrix representation
00:45:36.740 --> 00:45:45.528
is sum over i Aik vi.
00:45:50.800 --> 00:45:54.320
So now if this is
the original basis,
00:45:54.320 --> 00:45:57.560
the vi's are your
original basis,
00:45:57.560 --> 00:46:04.240
then you have the
following, that the vi's
00:46:04.240 --> 00:46:09.530
can be thought as the basis
vectors and represented
00:46:09.530 --> 00:46:16.740
by columns with a
1 in the ith entry.
00:46:16.740 --> 00:46:24.590
So this equation is saying
nothing more, or nothing less
00:46:24.590 --> 00:46:32.300
than uk, in terms of
matrices or columns,
00:46:32.300 --> 00:46:49.230
is equal to A1k v1 plus Ank
vn, which is just A1k A2k Ank.
00:46:51.910 --> 00:46:56.590
Because vi is the
ith basis vector.
00:46:56.590 --> 00:47:01.680
So 1 0's only in
the ith position.
00:47:07.110 --> 00:47:12.860
So these are the eigenvectors.
00:47:12.860 --> 00:47:16.070
And they're thought as linear
combinations of the vi's.
00:47:16.070 --> 00:47:19.180
The vi's are the
original basis vectors.
00:47:19.180 --> 00:47:23.950
So the eigenvectors
are these numbers.
00:47:23.950 --> 00:47:33.130
OK, we've talked
about diagonlization,
00:47:33.130 --> 00:47:38.890
but then there's a term that is
more crucial for our operators
00:47:38.890 --> 00:47:40.870
that we're interested in.
00:47:40.870 --> 00:47:43.170
We're talking about
Hermitian operators.
00:47:43.170 --> 00:47:46.420
So the term that is going
to be important for us
00:47:46.420 --> 00:47:53.020
is unitarily diagonalizable.
00:47:58.240 --> 00:48:04.180
What is a unitarily
diagonalizable operator?
00:48:04.180 --> 00:48:06.800
Two ways again of
thinking about this.
00:48:06.800 --> 00:48:10.950
And perhaps the first
way is the best.
00:48:10.950 --> 00:48:13.030
And I will say it.
00:48:13.030 --> 00:48:19.740
A matrix is set to be
unitarily diagonalizable
00:48:19.740 --> 00:48:24.720
if you have an orthonormal
basis of eigenvectors.
00:48:27.840 --> 00:48:34.390
Remember diagonalizable meant
a basis of eigenvectors.
00:48:34.390 --> 00:48:35.470
That's all it means.
00:48:35.470 --> 00:48:40.420
Unitarily diagonalizable
means orthonormal basis
00:48:40.420 --> 00:48:42.190
of eigenvectors.
00:48:42.190 --> 00:48:53.845
So T has an orthonormal
basis of eigenvectors.
00:48:59.750 --> 00:49:03.826
Now that's a very
clear statement.
00:49:06.860 --> 00:49:09.870
And it's a fabulous
thing if you can achieve,
00:49:09.870 --> 00:49:14.820
because you basically
have broken down the space
00:49:14.820 --> 00:49:20.270
into basis spaces, each one
of them with a simple thing
00:49:20.270 --> 00:49:21.420
before your operators.
00:49:21.420 --> 00:49:23.860
And they're orthonormal,
so it's the simplest
00:49:23.860 --> 00:49:26.510
possible calculational tool.
00:49:26.510 --> 00:49:29.710
So it's ideal if
you can have this.
00:49:29.710 --> 00:49:33.730
Now the way we think
of this is that you
00:49:33.730 --> 00:49:36.370
start with--
concretely, you start
00:49:36.370 --> 00:49:42.735
with a T of some basis v
that is an orthonormal basis.
00:49:48.340 --> 00:49:52.700
Start with an orthonormal
basis, and then
00:49:52.700 --> 00:49:57.940
pass to another
orthonormal basis u.
00:49:57.940 --> 00:50:01.110
So you're going to pass to
another orthonormal basis
00:50:01.110 --> 00:50:04.390
u with some operator.
00:50:04.390 --> 00:50:06.570
But what you have
learned is that if you
00:50:06.570 --> 00:50:12.600
want to pass from v
orthonormal to another basis
00:50:12.600 --> 00:50:17.040
u, a vector that is
also orthonormal,
00:50:17.040 --> 00:50:19.330
the way to go from
one to the other
00:50:19.330 --> 00:50:21.520
is through a unitary operator.
00:50:21.520 --> 00:50:25.560
Only unitary operators pass you
from orthonormal to orthonormal
00:50:25.560 --> 00:50:27.530
basis.
00:50:27.530 --> 00:50:31.860
Therefore really, when
you start with your matrix
00:50:31.860 --> 00:50:35.320
in an orthonormal basis
that is not diagonal,
00:50:35.320 --> 00:50:41.600
the only thing you can
hope is that T of u
00:50:41.600 --> 00:50:52.900
will be equal to sum u dagger,
or u minus 1, T of v u.
00:50:52.900 --> 00:51:02.500
Remember, for a unitary
operator, where u is unitary,
00:51:02.500 --> 00:51:05.420
the inverse is the dagger.
00:51:05.420 --> 00:51:08.040
So you're doing a
unitary transformation,
00:51:08.040 --> 00:51:12.580
and you find the matrix that
is presumably then diagonal.
00:51:12.580 --> 00:51:16.720
So basically, unitarily
diagonalizable
00:51:16.720 --> 00:51:18.390
is the statement
that if you start
00:51:18.390 --> 00:51:23.820
with the operator in an
arbitrary orthonormal basis,
00:51:23.820 --> 00:51:26.840
then there's some unitary
operator that takes you
00:51:26.840 --> 00:51:31.080
to the privilege basis in which
your operator is diagonal,
00:51:31.080 --> 00:51:32.900
is still orthonormal.
00:51:32.900 --> 00:51:39.430
But maybe in a more simple
way, unitarily diagonalizable
00:51:39.430 --> 00:51:41.830
is just a statement
that you can find
00:51:41.830 --> 00:51:46.050
an orthonormal basis
of eigenvectors.
00:51:46.050 --> 00:51:52.010
Now the main theorem
of this subject,
00:51:52.010 --> 00:51:54.110
perhaps one of the
most important theorems
00:51:54.110 --> 00:51:57.540
of linear algebra, is
the characterization
00:51:57.540 --> 00:52:02.020
of which operators have such
a wonderful representation.
00:52:07.500 --> 00:52:09.870
What is the most
general operator T
00:52:09.870 --> 00:52:14.320
that will have an orthonormal
basis of eigenvectors?
00:52:14.320 --> 00:52:20.390
Now we probably have heard that
Hermitian operators do the job.
00:52:20.390 --> 00:52:22.720
Hermitian operators have that.
00:52:22.720 --> 00:52:24.830
But that's not the
most general ones.
00:52:24.830 --> 00:52:29.280
And given that you want
the complete result,
00:52:29.280 --> 00:52:33.400
let's give you the
complete result.
00:52:33.400 --> 00:52:36.740
The operators that have
this wonderful properties
00:52:36.740 --> 00:52:40.030
are called normal
operators, and they
00:52:40.030 --> 00:52:41.930
satisfy the following property.
00:52:41.930 --> 00:52:56.400
M is normal if M dagger, the
adjoint of it, commutes with M.
00:52:56.400 --> 00:52:59.520
So Hermitian
operators are normal,
00:52:59.520 --> 00:53:04.750
because M dagger is equal
to M, and they commute.
00:53:04.750 --> 00:53:07.940
Anti Hermitian mission
operators are also normal,
00:53:07.940 --> 00:53:10.360
because anti Hermitian
means that dagger
00:53:10.360 --> 00:53:14.920
is equal to minus M, and
it still commutes with M.
00:53:14.920 --> 00:53:21.180
Unitary operators have
U dagger U equal to U U
00:53:21.180 --> 00:53:23.260
dagger equals to 1.
00:53:23.260 --> 00:53:27.750
So U and U dagger
actually commute as well.
00:53:27.750 --> 00:53:32.370
So Hermitian, anti
Hermitian, unitary, they're
00:53:32.370 --> 00:53:33.706
all normal operators.
00:53:37.470 --> 00:53:39.590
What do we know about
normal operators?
00:53:42.190 --> 00:53:45.390
There's one important result
about normal operators,
00:53:45.390 --> 00:53:46.060
a lemma.
00:53:48.970 --> 00:54:07.970
If M is normal and
W is an eigenvector,
00:54:07.970 --> 00:54:17.110
such that MW is
equal to lambda W.
00:54:17.110 --> 00:54:22.090
Now normal operators need
not have real eigenvalues,
00:54:22.090 --> 00:54:25.200
because they include
unitary operators.
00:54:25.200 --> 00:54:32.042
So here I should write
Hermitian, anti Hermitian,
00:54:32.042 --> 00:54:35.715
and unitary are normal.
00:54:41.050 --> 00:54:45.460
So here is what a normal
operator is doing.
00:54:45.460 --> 00:54:46.810
You have a normal operator.
00:54:46.810 --> 00:54:49.760
It has an eigenvector
with some eigenvalue.
00:54:49.760 --> 00:54:54.640
Lambda is a complex
number in principle.
00:54:54.640 --> 00:54:57.190
Then the following
result is true,
00:54:57.190 --> 00:55:06.300
then M dagger omega is also
an eigenvector of M dagger.
00:55:06.300 --> 00:55:08.920
And it has eigenvalue
lambda star.
00:55:17.380 --> 00:55:20.830
Now this is not all
that easy to show.
00:55:20.830 --> 00:55:23.320
It's a few lines, and
it's done in the notes.
00:55:23.320 --> 00:55:24.170
I ask you to see.
00:55:24.170 --> 00:55:25.585
It's actually very elegant.
00:55:28.702 --> 00:55:31.740
What is the usual
strategy to prove things
00:55:31.740 --> 00:55:37.210
like that Is to say oh, I want
to show this is equal to that,
00:55:37.210 --> 00:55:40.540
so I want to show that
this binds that is 0.
00:55:40.540 --> 00:55:42.760
So I have a vector
that is zero what
00:55:42.760 --> 00:55:45.400
is the easiest way
to show that it's 0?
00:55:45.400 --> 00:55:47.780
If I can show it's norm is 0.
00:55:47.780 --> 00:55:49.900
So that's a typical
strategy that you
00:55:49.900 --> 00:55:52.350
use to prove equalities.
00:55:52.350 --> 00:55:56.430
You say, oh, it's a vector
that must be 0 as my equality.
00:55:56.430 --> 00:55:57.740
Let's see if it's 0.
00:55:57.740 --> 00:56:00.130
Let's find its norm,
and you get it.
00:56:00.130 --> 00:56:02.280
So that's a result.
00:56:02.280 --> 00:56:11.695
So with this stated, we
finally have the main result
00:56:11.695 --> 00:56:13.740
that we wanted to get to.
00:56:16.680 --> 00:56:20.660
And I will be very
sketchy on this.
00:56:24.700 --> 00:56:28.200
The notes are complete,
but I will be sketchy here.
00:56:28.200 --> 00:56:30.005
It's called the
spectral theorem.
00:56:35.770 --> 00:56:48.105
Let M be an operator in
a complex vector space.
00:56:55.820 --> 00:57:12.420
The vector space has
an orthonormal basis
00:57:12.420 --> 00:57:31.780
of eigenvectors of M if
and only if M is normal.
00:57:31.780 --> 00:57:37.370
So the normal operators are it.
00:57:37.370 --> 00:57:43.270
You want to have a complete set
of orthonormal eigenvectors.
00:57:43.270 --> 00:57:46.900
Well, this will only
happen if your operator
00:57:46.900 --> 00:57:50.580
is normal, end of story.
00:57:50.580 --> 00:57:59.200
Now there's two things
about this theorem
00:57:59.200 --> 00:58:06.650
is to show that if it's
diagonalizable, it is normal,
00:58:06.650 --> 00:58:11.110
and the other thing is to
show, that if it's normal,
00:58:11.110 --> 00:58:12.780
it can be diagonalized.
00:58:12.780 --> 00:58:17.270
Of course, you can
imagine the second one
00:58:17.270 --> 00:58:19.620
is harder than the first.
00:58:19.620 --> 00:58:23.050
Let me do the first
one for a few minutes.
00:58:23.050 --> 00:58:28.010
And then say a couple of
words about the second.
00:58:28.010 --> 00:58:31.660
And you may discuss
this in recitation.
00:58:31.660 --> 00:58:35.150
It's a little
mathematical, but it's all
00:58:35.150 --> 00:58:37.350
within the kind of
things that we do.
00:58:37.350 --> 00:58:40.970
And really it's fairly
physical in a sense.
00:58:40.970 --> 00:58:43.860
We're accustomed to do
such kinds of arguments.
00:58:43.860 --> 00:58:54.270
So suppose it's
unitarily diagonalizable,
00:58:54.270 --> 00:59:02.880
which means that M-- so
if you have U dagger,
00:59:02.880 --> 00:59:09.490
MU is equal to a
diagonal matrix, DM.
00:59:09.490 --> 00:59:11.140
I'm talking now matrices.
00:59:11.140 --> 00:59:14.550
So these are all matrices,
a diagonal matrix.
00:59:17.370 --> 00:59:23.130
There's no basis to the
notion of a diagonal operator,
00:59:23.130 --> 00:59:26.380
because if you have
a diagonal operator,
00:59:26.380 --> 00:59:29.120
it may not look diagonal
in another basis.
00:59:29.120 --> 00:59:34.570
Only the identity operator is
diagonal in all basis, but not
00:59:34.570 --> 00:59:36.330
the typical diagonal operator.
00:59:36.330 --> 00:59:40.440
So unitarily
diagonalizable, as we said,
00:59:40.440 --> 00:59:43.850
you make it-- it's
gone somewhere.
00:59:43.850 --> 00:59:44.540
Here.
00:59:44.540 --> 00:59:46.900
You act with an
inverse matrices,
00:59:46.900 --> 00:59:48.640
and you get the diagonal matrix.
00:59:48.640 --> 01:00:01.190
So from this, you find that
M is equal to U DM U dagger
01:00:01.190 --> 01:00:05.610
by acting with U on the left and
with U dagger from the right,
01:00:05.610 --> 01:00:09.830
you solve for M, and it's this.
01:00:09.830 --> 01:00:15.670
And then M dagger is the
dagger of these things.
01:00:15.670 --> 01:00:22.100
So it's U to DM dagger U dagger.
01:00:22.100 --> 01:00:26.490
The U's sort of
remain the same way,
01:00:26.490 --> 01:00:30.100
but the diagonal matrix
is not necessarily real,
01:00:30.100 --> 01:00:34.200
so you must put the
dagger in there.
01:00:34.200 --> 01:00:40.540
And now M dagger M. To check
that the matrix is normal
01:00:40.540 --> 01:00:42.360
that commutator should be 0.
01:00:42.360 --> 01:00:46.380
So M dagger M. You
do this times that.
01:00:46.380 --> 01:00:51.030
You get U DM dagger.
01:00:51.030 --> 01:00:53.790
U dagger U. That's one.
01:00:53.790 --> 01:00:57.416
DM U dagger.
01:00:57.416 --> 01:01:02.110
And M M dagger you multiply
the other direction
01:01:02.110 --> 01:01:08.640
you get U DM DM dagger U dagger.
01:01:08.640 --> 01:01:22.750
So the commutator of M dagger
M is equal to U DM dagger DM
01:01:22.750 --> 01:01:29.880
minus DM Dm dagger U dagger.
01:01:29.880 --> 01:01:34.290
But any two diagonal
matrices commute.
01:01:34.290 --> 01:01:36.430
They may not be that simple.
01:01:36.430 --> 01:01:38.970
Diagonal matrices are not
the identity matrices,
01:01:38.970 --> 01:01:40.770
but for sure they commute.
01:01:40.770 --> 01:01:45.570
You multiply elements along
with diagonal so this is 0.
01:01:45.570 --> 01:01:53.260
So certainly any unitarily
diagonalizable matrix
01:01:53.260 --> 01:01:54.970
is normal.
01:01:54.970 --> 01:01:56.910
Now the other part
of the proof, which
01:01:56.910 --> 01:02:03.170
I'm not going to speak about,
it's actually quite simple.
01:02:03.170 --> 01:02:08.440
And it's based on the fact that
any matrix in a complex vector
01:02:08.440 --> 01:02:11.500
space has at least
one eigenvalue.
01:02:11.500 --> 01:02:16.610
So what you do is you pick
out that eigenvalue and it's
01:02:16.610 --> 01:02:19.110
eigenvector, and
change the basis
01:02:19.110 --> 01:02:22.510
to use that eigenvector
instead of your other vectors.
01:02:22.510 --> 01:02:24.730
And then you look at the matrix.
01:02:24.730 --> 01:02:27.012
And after you use
that eigenvector,
01:02:27.012 --> 01:02:31.490
the matrix has a lot of 0's
here and a lot of 0's here.
01:02:31.490 --> 01:02:35.920
And then the matrix has been
reduced in dimension mansion,
01:02:35.920 --> 01:02:38.130
and then you go step by step.
01:02:38.130 --> 01:02:41.590
So basically, it's the
fact that any operator
01:02:41.590 --> 01:02:45.190
has at least one eigenvalue
and at least one eigenvector.
01:02:45.190 --> 01:02:47.150
It allows you to go down.
01:02:47.150 --> 01:02:55.530
And normality is analogous
to Hermiticity in some sense.
01:02:55.530 --> 01:02:59.130
And the statement that
you have an eigenvector
01:02:59.130 --> 01:03:03.800
generally tells you that this
thing is full of 0's, but then
01:03:03.800 --> 01:03:06.170
you don't know that
there are 0's here.
01:03:06.170 --> 01:03:09.990
And either normality
or Hermiticity
01:03:09.990 --> 01:03:11.830
shows that there are
0's here, and then you
01:03:11.830 --> 01:03:13.490
can proceed at lower dimensions.
01:03:13.490 --> 01:03:16.580
So you should look at
the proof because it
01:03:16.580 --> 01:03:22.160
will make clear to you that
you understand what's going on.
01:03:22.160 --> 01:03:26.480
OK but let's take it for granted
now you have these operators
01:03:26.480 --> 01:03:32.750
and can be diagonalized.
01:03:32.750 --> 01:03:37.787
Then we have the
next thing, which
01:03:37.787 --> 01:03:39.120
is simultaneous diagonalization.
01:03:42.700 --> 01:03:45.550
What is simultaneous
diagonalization?
01:03:45.550 --> 01:03:47.670
It's an awfully important thing.
01:03:47.670 --> 01:03:51.690
So we will now focus on
simultaneous diagonalization
01:03:51.690 --> 01:03:53.380
of Hermitian operators.
01:03:53.380 --> 01:04:02.150
So simultaneous diagonalization
of Hermitian ops.
01:04:06.210 --> 01:04:11.990
Now as we will emphasize
towards the end,
01:04:11.990 --> 01:04:16.250
this is perhaps one of
the most important ideas
01:04:16.250 --> 01:04:18.210
in quantum mechanics.
01:04:18.210 --> 01:04:22.540
It's this stuff that allows
you to label and understand
01:04:22.540 --> 01:04:24.240
your state system.
01:04:24.240 --> 01:04:28.000
Basically you need
to diagonalize
01:04:28.000 --> 01:04:30.340
more than one operator
most of the time.
01:04:30.340 --> 01:04:33.340
You can say OK, you found
the energy eigenstates.
01:04:33.340 --> 01:04:34.930
You're done.
01:04:34.930 --> 01:04:37.270
But if you find your
energy eigenstates
01:04:37.270 --> 01:04:39.980
and you think you're
done, maybe you
01:04:39.980 --> 01:04:44.250
are if you have all these
energy eigenstates tabulated.
01:04:44.250 --> 01:04:46.390
But if you have
a degeneracy, you
01:04:46.390 --> 01:04:51.640
have a lot of states that
have the same energy.
01:04:51.640 --> 01:04:53.241
And what's different about them?
01:04:53.241 --> 01:04:54.990
They're certainly
different because you've
01:04:54.990 --> 01:04:57.770
got several states, but
what's different about them?
01:04:57.770 --> 01:05:00.600
You may not know,
unless you figure out
01:05:00.600 --> 01:05:04.010
that they have different
physical properties.
01:05:04.010 --> 01:05:07.250
If they're different, something
must be different about them.
01:05:07.250 --> 01:05:09.550
So you need more
than one operator,
01:05:09.550 --> 01:05:13.220
and your facing the problem of
simultaneously diagonalizing
01:05:13.220 --> 01:05:17.500
things, because states cannot
be characterized just by one
01:05:17.500 --> 01:05:19.440
property, one observable.
01:05:19.440 --> 01:05:24.160
Would be simple if you could,
but life is not that simple.
01:05:24.160 --> 01:05:27.520
So you need more
than one observable,
01:05:27.520 --> 01:05:31.250
and then you ask
when can they be
01:05:31.250 --> 01:05:34.090
simultaneously diagonalizable.
01:05:34.090 --> 01:05:35.710
Well, the statement is clear.
01:05:35.710 --> 01:05:40.720
If you have two
operators, S and T that
01:05:40.720 --> 01:05:46.050
belong to the linear
operators in a vector space,
01:05:46.050 --> 01:05:49.650
they can be simultaneously
diagonalized
01:05:49.650 --> 01:05:58.330
if there is a basis for
which every basis vector is
01:05:58.330 --> 01:06:02.210
eigenstate of this and
an eigenstate of that.
01:06:02.210 --> 01:06:04.420
Common set of eigenstates.
01:06:04.420 --> 01:06:07.330
So they can be
simultaneously diagonalized.
01:06:07.330 --> 01:06:12.920
Diagonalizable is that there
is a basis where this basis is
01:06:12.920 --> 01:06:16.410
comprised of the
eigenvectors of the operator.
01:06:16.410 --> 01:06:19.760
So this time you require
more, that that basis
01:06:19.760 --> 01:06:24.690
be at the same time a basis
set of eigenvectors of this
01:06:24.690 --> 01:06:28.320
and a set of eigenvectors
of the second one.
01:06:28.320 --> 01:06:36.450
So a necessary condition for
simultaneous diagonalization
01:06:36.450 --> 01:06:38.570
is that they commute.
01:06:38.570 --> 01:06:40.760
Why is that?
01:06:40.760 --> 01:06:44.240
The fact that two operators
commute or they don't commute
01:06:44.240 --> 01:06:48.130
is an issue that is
basis independent.
01:06:48.130 --> 01:06:51.010
If they don't commute,
the order gives something
01:06:51.010 --> 01:06:54.730
different, and that you
can see in every basis.
01:06:54.730 --> 01:06:59.850
So if they don't commute
and they're simultaneously
01:06:59.850 --> 01:07:04.680
diagonalizable, there would
be a basis in which both
01:07:04.680 --> 01:07:07.300
are diagonal and they
still wouldn't commute.
01:07:07.300 --> 01:07:09.960
But you know that
diagonal matrices commute.
01:07:09.960 --> 01:07:12.890
So if two operators
don't commute,
01:07:12.890 --> 01:07:15.130
they must not
commute in any base,
01:07:15.130 --> 01:07:18.260
therefore there can't
be a basis in which both
01:07:18.260 --> 01:07:21.350
are at the same time diagonal.
01:07:21.350 --> 01:07:25.500
So you need, for
simultaneous diagonalizable,
01:07:25.500 --> 01:07:29.420
you need that S and P commute.
01:07:29.420 --> 01:07:34.660
Now that may not be enough,
because not all operators
01:07:34.660 --> 01:07:35.820
can be diagonalized.
01:07:35.820 --> 01:07:38.900
So the fact that they commute
is necessary, but not everything
01:07:38.900 --> 01:07:40.560
can be diagonalizable.
01:07:40.560 --> 01:07:44.530
Well, you've learned that
every normal operator,
01:07:44.530 --> 01:07:47.800
every Hermitian operator
is diagonalizable.
01:07:47.800 --> 01:07:51.820
And then you got now
a claim of something
01:07:51.820 --> 01:07:54.160
that could possibly be true.
01:07:54.160 --> 01:07:58.590
Is the fact that whenever you
have two Hermitian operators,
01:07:58.590 --> 01:08:01.330
each one can be
diagonalized by themselves.
01:08:01.330 --> 01:08:03.230
And they commute.
01:08:03.230 --> 01:08:08.030
There is a simultaneous
set of eigenvectors of 1
01:08:08.030 --> 01:08:11.040
that are eigenvectors of
the first and eigenvectors
01:08:11.040 --> 01:08:12.240
of the second.
01:08:12.240 --> 01:08:25.710
So the statement is that
if S and T are commuting
01:08:25.710 --> 01:08:34.990
Hermitian operators, they can
be simultaneously diagonalized.
01:08:49.569 --> 01:08:57.420
So this theorem would
be quite easy to show
01:08:57.420 --> 01:08:59.294
if there would be
no degeneracies,
01:08:59.294 --> 01:09:01.680
and that's what we're
going to do first.
01:09:01.680 --> 01:09:04.510
But then we'll consider
the case of degeneracies.
01:09:08.430 --> 01:09:13.100
So I'm going to consider
the following possibilities.
01:09:13.100 --> 01:09:16.700
Perhaps neither one has
a degenerate spectrum.
01:09:16.700 --> 01:09:18.790
What does it mean a
degenerate spectrum?
01:09:18.790 --> 01:09:22.229
Same eigenvalue
repeated many times.
01:09:22.229 --> 01:09:26.120
But that is a wishful
thinking situation.
01:09:26.120 --> 01:09:28.960
So either both are
non degenerate,
01:09:28.960 --> 01:09:32.890
either one is non degenerate
and the other is degenerate,
01:09:32.890 --> 01:09:35.939
or both are degenerate.
01:09:35.939 --> 01:09:40.130
And that causes a very
interesting complication.
01:09:40.130 --> 01:09:43.439
So let's say there's
going to be two cases.
01:09:43.439 --> 01:09:44.670
It will suffice.
01:09:44.670 --> 01:09:46.630
In fact, it seems
that there are three,
01:09:46.630 --> 01:09:49.510
but two is enough to consider.
01:09:49.510 --> 01:10:01.020
There is no degeneracy
in T. So suppose
01:10:01.020 --> 01:10:05.600
one operator has no
degeneracy, and let's call
01:10:05.600 --> 01:10:07.990
it T. So that's one possibility.
01:10:07.990 --> 01:10:12.300
And then S may be degenerate,
or it may not be degenerate.
01:10:12.300 --> 01:10:19.500
And the second possibility
is that both S and T
01:10:19.500 --> 01:10:20.125
are degenerate.
01:10:25.780 --> 01:10:30.150
So I'll take care
of case one first.
01:10:30.150 --> 01:10:33.530
And then we'll discuss
case two, and that
01:10:33.530 --> 01:10:36.730
will complete our discussion.
01:10:36.730 --> 01:10:42.380
So suppose there's
no the degeneracy
01:10:42.380 --> 01:10:46.860
in the spectrum
of T. So case one.
01:10:49.450 --> 01:10:52.200
So what does that mean?
01:10:52.200 --> 01:10:56.290
It means that T
is non degenerate.
01:10:56.290 --> 01:11:00.280
There's a basis U1
can be diagonalized
01:11:00.280 --> 01:11:10.690
to UM, orthonormal by
the spectral theorem.
01:11:10.690 --> 01:11:15.170
And there's eigenvectors T
U-- these are eigenvectors.
01:11:15.170 --> 01:11:17.480
Lambda I Ui.
01:11:17.480 --> 01:11:22.010
And lambda I is
different to lambda j
01:11:22.010 --> 01:11:25.606
for i different from j.
01:11:25.606 --> 01:11:30.540
So all the eigenvalues,
again, it's not summed here.
01:11:30.540 --> 01:11:36.300
All the eigenvalues
are different.
01:11:36.300 --> 01:11:39.640
So what do we have?
01:11:39.640 --> 01:11:42.490
Well, each of
those eigenvectors,
01:11:42.490 --> 01:11:46.630
each of the Ui's that
are eigenvectors,
01:11:46.630 --> 01:11:49.905
generate invariant subspaces.
01:11:49.905 --> 01:11:52.570
There are T invariant subspaces.
01:11:52.570 --> 01:11:57.580
So each one, each
vector U1 you can
01:11:57.580 --> 01:12:00.020
imagine multiplying by
all possible numbers,
01:12:00.020 --> 01:12:01.330
positive and negative.
01:12:01.330 --> 01:12:04.910
And that's an invariant
one dimensional subspace,
01:12:04.910 --> 01:12:08.700
because if you act with T,
it's a T invariant space,
01:12:08.700 --> 01:12:12.760
you get the number
times a vector there.
01:12:12.760 --> 01:12:15.930
So the question that
you must ask now
01:12:15.930 --> 01:12:20.770
is you want to know if these
are simultaneous eigenvectors.
01:12:20.770 --> 01:12:26.350
So you want to figure
out what about S.
01:12:26.350 --> 01:12:30.010
How does S work with this thing?
01:12:30.010 --> 01:12:34.640
So you can act with
S from the left.
01:12:34.640 --> 01:12:43.340
So you get STUi is
equal to lambda i SUi.
01:12:47.081 --> 01:12:57.960
So here each Ui generates
an invariant subspace Ui T
01:12:57.960 --> 01:12:58.460
invariant.
01:13:02.540 --> 01:13:13.796
But S and T commute, so you have
T SUi is equal to lambda i SUi.
01:13:13.796 --> 01:13:16.170
And look at that equation again.
01:13:16.170 --> 01:13:20.160
This says that
this vector belongs
01:13:20.160 --> 01:13:31.370
to the invariant subspace Ui,
because it satisfies exactly
01:13:31.370 --> 01:13:36.610
the property that T acting on
it is equal to lambda i Ui.
01:13:36.610 --> 01:13:40.560
And it couldn't belong to
any other of the subspaces,
01:13:40.560 --> 01:13:44.700
because all the
eigenvalues are different.
01:13:44.700 --> 01:13:53.180
So spaces that are in Ui are the
spaces-- vectors that are in Ui
01:13:53.180 --> 01:13:56.700
are precisely all those
vectors that are left invariant
01:13:56.700 --> 01:13:59.800
by the action of T.
They're scaled only.
01:13:59.800 --> 01:14:03.590
So this vector is also in Ui.
01:14:03.590 --> 01:14:07.480
If this vector is
in Ui, SUi must
01:14:07.480 --> 01:14:12.640
be some number Wi times Ui.
01:14:12.640 --> 01:14:18.940
And therefore you've shown that
Ui is also an eigenvector of S,
01:14:18.940 --> 01:14:22.836
possibly with a different
eigenvalue of course.
01:14:22.836 --> 01:14:24.335
Because the only
thing that you know
01:14:24.335 --> 01:14:26.340
is that SUi is in this space.
01:14:26.340 --> 01:14:29.250
You don't know how big it is.
01:14:29.250 --> 01:14:34.850
So then you've shown
that, indeed, these Ui's
01:14:34.850 --> 01:14:45.280
that were eigenstates of T
are also eigenstates of S.
01:14:45.280 --> 01:14:47.910
And therefore
that's the statement
01:14:47.910 --> 01:14:51.290
of simultaneously
diagonalizable.
01:14:51.290 --> 01:14:55.340
They have the common
set of eigenvectors.
01:14:55.340 --> 01:14:58.770
So that's this part.
01:14:58.770 --> 01:15:04.020
And it's relatively
straight forward.
01:15:04.020 --> 01:15:07.380
Now we have to do case two.
01:15:07.380 --> 01:15:09.280
Case two is the interesting one.
01:15:12.540 --> 01:15:16.240
This time you're going
to have degeneracy.
01:15:16.240 --> 01:15:19.410
We have to have a notation
that is good for degeneracy.
01:15:19.410 --> 01:15:28.610
So if S is degeneracies,
has degeneracies,
01:15:28.610 --> 01:15:31.820
what happens with this operator?
01:15:31.820 --> 01:15:38.050
It will have-- remember,
a degenerate operator
01:15:38.050 --> 01:15:43.440
has eigenstates that form higher
than one dimensional spaces.
01:15:43.440 --> 01:15:45.840
If you have different
eigenvalues,
01:15:45.840 --> 01:15:52.360
each one generates a one
dimensional operator invariant
01:15:52.360 --> 01:15:53.670
subspace.
01:15:53.670 --> 01:15:55.480
But if you have
degeneracies, there
01:15:55.480 --> 01:15:58.960
are operators-- there are
spaces of higher dimensions that
01:15:58.960 --> 01:16:00.330
are left invariant.
01:16:00.330 --> 01:16:09.330
So for example, let Uk
denote the S invariant
01:16:09.330 --> 01:16:16.680
subspace of some
dimension Dk, which
01:16:16.680 --> 01:16:18.790
is greater or equal than 1.
01:16:22.190 --> 01:16:29.160
I will go here first.
01:16:29.160 --> 01:16:39.380
We're going to define Uk to
be the set of all vectors
01:16:39.380 --> 01:16:51.510
so that SU is equal
to lambda k U.
01:16:51.510 --> 01:16:59.150
And this will have dimension
of Uk is going to be Dk.
01:16:59.150 --> 01:17:01.570
So look what's happening.
01:17:01.570 --> 01:17:05.240
Basically the fact is
that for some eigenvalues,
01:17:05.240 --> 01:17:09.870
say the kth eigenvalue, you
just get several eigenvectors.
01:17:09.870 --> 01:17:13.510
So if you get several
eigenvectors not just scaled
01:17:13.510 --> 01:17:15.690
off each other,
these eigenvectors
01:17:15.690 --> 01:17:19.200
correspond to that
eigenvalue span of space.
01:17:19.200 --> 01:17:21.750
It's a degenerate subspace.
01:17:21.750 --> 01:17:25.140
So you must imagine
that as having
01:17:25.140 --> 01:17:29.280
a subspace of some
dimensionality with some basis
01:17:29.280 --> 01:17:32.840
vectors that span this thing.
01:17:32.840 --> 01:17:35.460
And they all have
the same eigenvector.
01:17:35.460 --> 01:17:38.940
Now you should really
have visualized
01:17:38.940 --> 01:17:40.800
this in a simple way.
01:17:40.800 --> 01:17:45.520
You have this subspace like a
cone or something like that,
01:17:45.520 --> 01:17:50.250
in which every vector
is an eigenvector.
01:17:50.250 --> 01:17:54.410
So every vector, when it's
acted by S is just scaled up.
01:17:54.410 --> 01:17:57.720
And all of them are
scaled by the same amount.
01:17:57.720 --> 01:18:01.320
That is what this
statement says.
01:18:01.320 --> 01:18:12.310
And corresponding to this thing,
you have a basis of vectors,
01:18:12.310 --> 01:18:14.790
of these eigenvectors,
and we'll call
01:18:14.790 --> 01:18:28.700
them Uk1, the first one,
the second, up to U Dk1,
01:18:28.700 --> 01:18:31.280
because it's a
subspace that we say
01:18:31.280 --> 01:18:35.270
it has the dimensionality Dk.
01:18:35.270 --> 01:18:37.540
So look at this thing.
01:18:37.540 --> 01:18:39.240
Somebody tells you
there's an operator.
01:18:39.240 --> 01:18:40.810
It has degenerate spectrum.
01:18:40.810 --> 01:18:46.120
You should start imagining all
kind of invariant subspaces
01:18:46.120 --> 01:18:48.540
of some dimensionality.
01:18:48.540 --> 01:18:51.970
If it has degeneracy, it's
a degeneracy each time
01:18:51.970 --> 01:18:54.870
the Dk is greater than
1, because if it's
01:18:54.870 --> 01:18:58.740
one dimensional, it's
just one basis vector one
01:18:58.740 --> 01:19:01.250
eigenvector, end of the story.
01:19:01.250 --> 01:19:04.250
Now this thing, by
the spectral theorem,
01:19:04.250 --> 01:19:06.490
this is an orthonormal basis.
01:19:06.490 --> 01:19:10.010
There's no problem, when you
have a degenerate subspace,
01:19:10.010 --> 01:19:11.810
to find an orthonormal basis.
01:19:11.810 --> 01:19:17.080
The theorem guarantees it,
so these are all orthonormal.
01:19:17.080 --> 01:19:20.300
So at the end of the day,
you have a decomposition
01:19:20.300 --> 01:19:28.470
of the vector space, V as U1
plus U2 plus maybe up to UM.
01:19:31.660 --> 01:19:36.330
And all of these vector
spaces like U's here, they
01:19:36.330 --> 01:19:39.340
may have some with just
no degeneracy, and some
01:19:39.340 --> 01:19:42.990
with degeneracy 2,
degeneracy 3, degeneracy 4.
01:19:42.990 --> 01:19:44.700
I don't know how
much degeneracy,
01:19:44.700 --> 01:19:48.420
but they might have
different degeneracy.
01:19:48.420 --> 01:19:53.350
Now what do we say next?
01:19:53.350 --> 01:19:58.870
Well, the fact that S
is a Hermitian operator
01:19:58.870 --> 01:20:01.210
says it just can
be diagonalized,
01:20:01.210 --> 01:20:03.550
and we're can find
all these spaces,
01:20:03.550 --> 01:20:06.020
and the basis for
the whole thing.
01:20:06.020 --> 01:20:15.110
So the basis would look U1
of the first base up to U d1
01:20:15.110 --> 01:20:17.430
of the first base.
01:20:17.430 --> 01:20:19.659
These are the basis
vectors of the first
01:20:19.659 --> 01:20:21.200
plus the basis
vectors of the second.
01:20:21.200 --> 01:20:28.155
All the basis vectors U1
up to Udm of the mth space.
01:20:31.460 --> 01:20:32.970
All this is the list.
01:20:32.970 --> 01:20:43.650
This is the basis
of V. So I've listed
01:20:43.650 --> 01:20:48.130
the basis of V, which a basis
for U1, all these vectors.
01:20:48.130 --> 01:20:50.540
U2, all of this.
01:20:50.540 --> 01:20:52.720
So you see, we're not
calculating anything.
01:20:52.720 --> 01:20:55.740
We're just trying to
understand the picture.
01:20:55.740 --> 01:21:03.940
And why is this operator,
S, diagonal in this basis?
01:21:03.940 --> 01:21:04.720
It's clear.
01:21:04.720 --> 01:21:11.940
Because every vector here, every
vector is an eigenvector of S.
01:21:11.940 --> 01:21:14.940
So when you act with
S on any vector,
01:21:14.940 --> 01:21:17.270
you get that vector
times a number.
01:21:17.270 --> 01:21:20.190
But that vector is
orthogonal to all the rest.
01:21:20.190 --> 01:21:25.270
So when you have some
U and S and another U,
01:21:25.270 --> 01:21:29.430
this gives you a vector
proportional to U.
01:21:29.430 --> 01:21:31.030
And this is another vector.
01:21:31.030 --> 01:21:34.900
The matrix element is 0,
because they're all orthogonal.
01:21:34.900 --> 01:21:39.190
So it should be obvious why
this list produces something
01:21:39.190 --> 01:21:43.130
that is completely
orthogonal-- a diagonal matrix.
01:21:43.130 --> 01:21:48.890
So S, in this basis, looks
like the diagonal matrix
01:21:48.890 --> 01:22:04.700
in which you have lambda 1 d1
times up to lambda m dm times.
01:22:07.260 --> 01:22:11.900
Now I'll have to go until
2:00 to get the punchline.
01:22:11.900 --> 01:22:15.340
I apologize, but we
can't stop right now.
01:22:15.340 --> 01:22:21.010
We're almost there,
believe it or not.
01:22:21.010 --> 01:22:22.705
Two more things.
01:22:26.720 --> 01:22:33.890
This basis is good, but
actually another basis
01:22:33.890 --> 01:22:35.570
would also be good.
01:22:35.570 --> 01:22:41.450
I'll write this other basis
would be a V1 acting on the U1
01:22:41.450 --> 01:22:50.960
up to V1 acting on that U1 up to
a Vm acting on this U1 up to Vm
01:22:50.960 --> 01:22:53.040
acting on that U1.
01:22:53.040 --> 01:22:55.240
This is m.
01:22:55.240 --> 01:22:55.920
m.
01:22:55.920 --> 01:22:58.420
This is dm.
01:22:58.420 --> 01:23:00.210
And here it's not U1.
01:23:00.210 --> 01:23:00.830
It's Ud1.
01:23:04.280 --> 01:23:08.660
You see, in the first
collection of vectors,
01:23:08.660 --> 01:23:15.420
I act with an operator V1 up
to here with an operator Vm.
01:23:15.420 --> 01:23:33.730
All of them with Vk being
a unitary operator in Uk.
01:23:33.730 --> 01:23:38.660
In every subspace, there
are unitary operators.
01:23:38.660 --> 01:23:41.930
So you can have
these bases and act
01:23:41.930 --> 01:23:46.140
with a unitary operator
of the space U1 here.
01:23:46.140 --> 01:23:48.350
A unitary operator
with a space U2 here.
01:23:48.350 --> 01:23:52.980
A unitary operator
of the space Un here.
01:23:52.980 --> 01:23:55.720
Hope you're following.
01:23:55.720 --> 01:23:59.230
And what happens if this
operator is unitary,
01:23:59.230 --> 01:24:02.320
this is still an
orthonormal basis in U1.
01:24:02.320 --> 01:24:07.680
These are still
orthonormal basis in Um.
01:24:07.680 --> 01:24:10.870
And therefore this is
an orthonormal basis
01:24:10.870 --> 01:24:14.810
for the whole thing, because
anyway those different spaces
01:24:14.810 --> 01:24:17.320
are orthogonal to each other.
01:24:17.320 --> 01:24:19.270
It's an orthogonal
decomposition.
01:24:19.270 --> 01:24:21.700
Everything is orthogonal
to everything.
01:24:21.700 --> 01:24:26.817
So this basis would be equally
good to represent the operator.
01:24:26.817 --> 01:24:27.316
Yes?
01:24:27.316 --> 01:24:31.150
AUDIENCE: [INAUDIBLE]
arbitrary unitary operators?
01:24:31.150 --> 01:24:34.710
PROFESSOR: Arbitrary unitary
operators at this moment.
01:24:34.710 --> 01:24:37.170
Arbitrary.
01:24:37.170 --> 01:24:42.300
So here comes the catch
as to the main property
01:24:42.300 --> 01:24:53.220
that now you want to establish
is that the spaces Uk are also
01:24:53.220 --> 01:24:54.043
T invariant.
01:24:56.990 --> 01:24:59.870
You see, the spaces
Uk were defined
01:24:59.870 --> 01:25:03.370
to be S invariant subspaces.
01:25:03.370 --> 01:25:07.950
And now the main important
thing is that they are also T
01:25:07.950 --> 01:25:12.660
invariant because they
commute with that.
01:25:12.660 --> 01:25:15.310
So let's see why
that is the case.
01:25:15.310 --> 01:25:18.950
Suppose U belongs to Uk.
01:25:21.710 --> 01:25:26.970
And then let's look at the
vector-- examine the vector Tu.
01:25:26.970 --> 01:25:29.270
What happens to Tu?
01:25:29.270 --> 01:25:35.740
Well, you want to act on
S on Tu to understand it.
01:25:35.740 --> 01:25:40.880
But S and T commute,
so this is T SU.
01:25:40.880 --> 01:25:44.260
But since U belongs
to Uk, that's
01:25:44.260 --> 01:25:46.890
the space with
eigenvalue lambda k.
01:25:46.890 --> 01:25:55.110
So this is lambda k times
u, so you have Tu here.
01:25:55.110 --> 01:26:00.260
So Tu acted with S
gives you lambda k Tu.
01:26:00.260 --> 01:26:05.785
So Tu is in the
invariant subspace Uk.
01:26:14.320 --> 01:26:20.570
What's happening here is now
something very straightforward.
01:26:20.570 --> 01:26:29.680
You try to imagine how does
the matrix T look in the basis
01:26:29.680 --> 01:26:31.660
that we have here.
01:26:31.660 --> 01:26:34.240
Here is this basis.
01:26:34.240 --> 01:26:38.340
how does this matrix T look?
01:26:38.340 --> 01:26:46.910
Well, this matrix keeps
the invariant subspaces.
01:26:46.910 --> 01:26:50.070
So you have to think of
it blocked diagonally.
01:26:53.850 --> 01:26:56.990
If it acts on it-- here
are the first vectors
01:26:56.990 --> 01:26:59.100
that you're considering, the U1.
01:26:59.100 --> 01:27:03.190
Well if you act on it
with T of the U1 subspace,
01:27:03.190 --> 01:27:05.140
you stay in the U1 subspace.
01:27:05.140 --> 01:27:06.580
So you don't get anything else.
01:27:06.580 --> 01:27:09.490
So you must have
0's all over here.
01:27:09.490 --> 01:27:12.320
And you can have a matrix here.
01:27:12.320 --> 01:27:17.720
And if you act on the second
Uk U2, you get a vector in U2,
01:27:17.720 --> 01:27:20.150
so it's orthogonal to
all the other vectors.
01:27:20.150 --> 01:27:22.330
So you get a matrix here.
01:27:22.330 --> 01:27:24.070
And you get a matrix here.
01:27:24.070 --> 01:27:30.150
So actually you get a
blocked diagonal matrix
01:27:30.150 --> 01:27:35.350
in which the blocks
correspond to the degeneracy.
01:27:35.350 --> 01:27:41.540
So if there's a degeneracy
d1 here, it's a d1 times d1.
01:27:41.540 --> 01:27:45.690
And d2 times d2.
01:27:45.690 --> 01:27:48.710
So actually you
haven't simultaneously
01:27:48.710 --> 01:27:50.300
diagonalized them.
01:27:50.300 --> 01:27:52.160
That's the problem
of degeneracy.
01:27:52.160 --> 01:27:55.560
You haven't, but you
now have the tools,
01:27:55.560 --> 01:27:59.020
because this operator is
Hermitian, therefore it's
01:27:59.020 --> 01:28:03.060
Hermitian here, and
here, and here, and here.
01:28:03.060 --> 01:28:05.840
So you can diagonalize here.
01:28:05.840 --> 01:28:08.880
But what do you need
for diagonalizing here?
01:28:08.880 --> 01:28:10.950
You need a unitary matrix.
01:28:10.950 --> 01:28:13.400
Call it V1.
01:28:13.400 --> 01:28:16.140
For here you need
another unitary matrix.
01:28:16.140 --> 01:28:18.430
Call it V2.
01:28:18.430 --> 01:28:19.910
Vn.
01:28:19.910 --> 01:28:24.140
And then this matrix
becomes diagonal.
01:28:24.140 --> 01:28:26.990
But then what about
the old matrix?
01:28:26.990 --> 01:28:31.400
Well, we just explained here
that if you change the basis
01:28:31.400 --> 01:28:37.330
by unitary matrices, you
don't change the first matrix.
01:28:37.330 --> 01:28:39.900
So actually you succeeded.
01:28:39.900 --> 01:28:45.030
You now can diagonalize
this without destroying
01:28:45.030 --> 01:28:46.130
your earlier result.
01:28:46.130 --> 01:28:49.430
And you managed to
diagonalize the whole thing.
01:28:49.430 --> 01:28:52.180
So this is for two
operators in the notes.
01:28:52.180 --> 01:28:56.060
You'll see why it simply extends
for three, four, and five,
01:28:56.060 --> 01:28:58.410
or arbitrary number
of operators.
01:28:58.410 --> 01:29:00.860
See you next time.