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ROBERT FIELD: OK.
00:00:23.780 --> 00:00:30.540
So, today is the first of a
pair of lectures taking us
00:00:30.540 --> 00:00:33.690
from the Schrodinger picture
to the Heisenberg picture.
00:00:33.690 --> 00:00:37.580
Or the wave function picture
to the matrix picture.
00:00:37.580 --> 00:00:41.280
Now almost everyone who
does quantum mechanics
00:00:41.280 --> 00:00:45.240
is rooted in the
Heisenberg picture.
00:00:45.240 --> 00:00:49.980
We use the Heisenberg
picture because the structure
00:00:49.980 --> 00:00:54.930
of the problem is immediately
evident when you write down
00:00:54.930 --> 00:00:57.140
what you know.
00:00:57.140 --> 00:01:03.680
And it's also mostly the way
you program your computers
00:01:03.680 --> 00:01:06.630
to solve any problem.
00:01:06.630 --> 00:01:14.240
So I'm going to get you to the
Heisenberg picture by dealing
00:01:14.240 --> 00:01:17.150
with the two-level
problem, which is--
00:01:17.150 --> 00:01:21.240
it should be called one of
the exactly solved problems,
00:01:21.240 --> 00:01:25.900
except it's abstract as
opposed to harmonic oscillator,
00:01:25.900 --> 00:01:29.270
particle in an infinite
box, rigid rotor.
00:01:29.270 --> 00:01:33.590
It's a exactly-solved
problem, and it leads us to--
00:01:33.590 --> 00:01:35.660
or guides us to--
00:01:35.660 --> 00:01:36.500
the new approach.
00:01:39.480 --> 00:01:42.630
I'm going to
approach this problem
00:01:42.630 --> 00:01:45.440
the algebraic or
Schrodinger way.
00:01:45.440 --> 00:01:50.340
And then I'm going to
describe it in a matrix way
00:01:50.340 --> 00:01:52.995
and introduce the new
language and the new notation.
00:01:57.750 --> 00:02:00.750
So I can say, at least at
the end of this lecture,
00:02:00.750 --> 00:02:03.770
I'll be able to say
the word matrix element
00:02:03.770 --> 00:02:05.780
and everybody will
know what it is,
00:02:05.780 --> 00:02:09.389
and I'll stop talking
about integrals.
00:02:09.389 --> 00:02:13.470
And one of the scary
things is, when
00:02:13.470 --> 00:02:17.630
we go to the matrix
picture, we stop
00:02:17.630 --> 00:02:19.520
looking at the wave function.
00:02:19.520 --> 00:02:22.250
We never think about
the wave functions.
00:02:22.250 --> 00:02:26.380
And so there are all sorts
of things like phases
00:02:26.380 --> 00:02:31.630
that we have to make sure
we're not screwing up.
00:02:31.630 --> 00:02:36.250
Because we're playing fast
and loose with symbols,
00:02:36.250 --> 00:02:39.340
and sometimes they can bite
you if you don't know what
00:02:39.340 --> 00:02:40.670
you're dealing with.
00:02:40.670 --> 00:02:44.740
But mostly, we're going to
get rid of wave functions,
00:02:44.740 --> 00:02:50.470
because we know all the
solutions to standard problems,
00:02:50.470 --> 00:02:54.010
and we know a lot of integrals
involving those solutions
00:02:54.010 --> 00:02:56.020
to standard problems.
00:02:56.020 --> 00:02:58.510
And so basically, all
we need is an index
00:02:58.510 --> 00:03:02.760
saying this wave function or
this state, this integral,
00:03:02.760 --> 00:03:03.980
and you have--
00:03:03.980 --> 00:03:07.200
and then you can
write down everything
00:03:07.200 --> 00:03:11.460
you need in matrix
notation, and then you
00:03:11.460 --> 00:03:14.910
can tell your friendly computer,
OK, solve the problem for me.
00:03:18.420 --> 00:03:18.920
OK.
00:03:18.920 --> 00:03:24.770
So once I give you
the matrix picture,
00:03:24.770 --> 00:03:29.930
then I will talk about how
you find the eigenvalues
00:03:29.930 --> 00:03:33.890
and eigenvectors of the
matrix picture using
00:03:33.890 --> 00:03:36.020
a unitary transformation.
00:03:36.020 --> 00:03:38.930
And then I'll generalize
from the two-level problem,
00:03:38.930 --> 00:03:41.420
which is exactly soluble.
00:03:41.420 --> 00:03:44.202
You don't need a computer
for it, but it's convenient--
00:03:44.202 --> 00:03:46.160
because you don't have
to write anything down--
00:03:46.160 --> 00:03:51.420
to N levels where
N can be infinite.
00:03:51.420 --> 00:03:58.980
And the N-level problem
is in principle difficult
00:03:58.980 --> 00:04:04.140
because even a computer can't
diagonalize an infinite matrix.
00:04:04.140 --> 00:04:07.920
And all of your basis sets,
all of your standard problems
00:04:07.920 --> 00:04:10.950
involve an infinite
number of functions.
00:04:10.950 --> 00:04:15.120
And so I've introduced
a notation and a concept
00:04:15.120 --> 00:04:21.180
of solving a matrix equation,
but you can't do it unless you,
00:04:21.180 --> 00:04:23.190
say, let's make
an approximation,
00:04:23.190 --> 00:04:26.160
and it's called nondegenerate
perturbation theory.
00:04:26.160 --> 00:04:28.290
And this is the
tool for learning
00:04:28.290 --> 00:04:30.270
almost everything
you want to know
00:04:30.270 --> 00:04:33.080
about complicated problems.
00:04:33.080 --> 00:04:37.120
And it's not
complicated to apply
00:04:37.120 --> 00:04:38.870
nondegenerate
perturbation theory.
00:04:38.870 --> 00:04:41.870
It's just ugly.
00:04:41.870 --> 00:04:43.700
But it's really
valuable because you
00:04:43.700 --> 00:04:46.040
don't have to
remember how to solve
00:04:46.040 --> 00:04:48.710
a particular complicated
differential equation,
00:04:48.710 --> 00:04:50.840
you just write
down what you know
00:04:50.840 --> 00:04:53.660
and you do some simple
stuff, and bang, you've
00:04:53.660 --> 00:04:55.460
got a solution to the problem.
00:04:55.460 --> 00:04:57.560
And this is really what
I want you to come away
00:04:57.560 --> 00:04:58.850
from this course with--
00:04:58.850 --> 00:05:03.830
the concept that anything that
requires quantum mechanics
00:05:03.830 --> 00:05:09.390
you can solve using some
form of perturbation theory.
00:05:09.390 --> 00:05:12.125
So hold onto your seats.
00:05:25.740 --> 00:05:29.420
So the Schrodinger picture
is differential equations.
00:05:38.360 --> 00:05:41.270
And they're often coupled
differential equations.
00:05:41.270 --> 00:05:44.000
And you don't want
to go there, usually.
00:05:44.000 --> 00:05:46.960
And we're going to replace
that with linear algebra.
00:05:52.490 --> 00:05:55.880
Now many of you have not had
a course in linear algebra.
00:05:55.880 --> 00:06:00.834
And so that should make
you say, can I do this?
00:06:00.834 --> 00:06:02.500
And the answer is,
yeah, you can do this
00:06:02.500 --> 00:06:04.990
because the linear
algebra you are going
00:06:04.990 --> 00:06:09.010
to need in this course
is based on one concept,
00:06:09.010 --> 00:06:14.200
and that is that the solution
of coupled linear homogeneous
00:06:14.200 --> 00:06:19.495
equations involves diagonalizing
or solving a determinant.
00:06:22.090 --> 00:06:26.210
And when you solve this
determinant of the equation,
00:06:26.210 --> 00:06:30.090
it's equivalent to
diagonalizing a matrix.
00:06:30.090 --> 00:06:33.890
And that's the language we're
going to be using all the time.
00:06:33.890 --> 00:06:35.660
And so the only thing
I'm not going to do
00:06:35.660 --> 00:06:38.520
is prove this fundamental
theorem of linear algebra,
00:06:38.520 --> 00:06:44.480
but the notation is easy to use
and the tricks are very simple.
00:06:47.800 --> 00:06:52.290
So, we have with
exactly-solved problems
00:06:52.290 --> 00:06:58.030
complete sets of the energy
levels of the wave functions.
00:06:58.030 --> 00:07:00.420
And we know a lot of integrals.
00:07:00.420 --> 00:07:06.630
Psi i operator psi j d tau.
00:07:06.630 --> 00:07:10.150
We know these not
by evaluating them,
00:07:10.150 --> 00:07:14.130
but because the functions are so
simple that we can write these
00:07:14.130 --> 00:07:17.470
as simply a function of the
initial and final quantum
00:07:17.470 --> 00:07:17.970
numbers.
00:07:21.170 --> 00:07:24.660
And so all of a
sudden, we forget
00:07:24.660 --> 00:07:26.580
about the wave functions,
because all the work
00:07:26.580 --> 00:07:29.390
has been done for us.
00:07:29.390 --> 00:07:30.850
We could do it too, but why?
00:07:33.600 --> 00:07:36.860
So we start with the
two-level problem.
00:07:36.860 --> 00:07:39.140
And the two-level
problem says we have
00:07:39.140 --> 00:07:44.270
two states, psi 1 and psi 2.
00:07:44.270 --> 00:07:47.020
Still in the
Schrodinger picture.
00:07:47.020 --> 00:07:54.270
And this has an energy which
we could we could call E1,
00:07:54.270 --> 00:08:03.550
and that would be H11,
the diagonal integral
00:08:03.550 --> 00:08:08.750
of the Hamiltonian between the
1 function and the 1 function,
00:08:08.750 --> 00:08:09.520
and this is E2.
00:08:13.660 --> 00:08:18.430
Now these two states
have an interaction.
00:08:18.430 --> 00:08:21.650
They're connected by
an interaction term
00:08:21.650 --> 00:08:25.570
which causes them to repel
each other equal and opposite
00:08:25.570 --> 00:08:26.830
amounts.
00:08:26.830 --> 00:08:31.880
And so this is H12.
00:08:31.880 --> 00:08:36.289
Again, an integral, which you
usually don't have to evaluate,
00:08:36.289 --> 00:08:39.890
because it's basically
done for you.
00:08:39.890 --> 00:08:44.760
And then we get E
plus and E minus,
00:08:44.760 --> 00:08:47.640
and the corresponding
eigenfunction.
00:08:50.680 --> 00:08:53.200
So this is the
two-level problem,
00:08:53.200 --> 00:08:56.200
and it's expressed in
terms of three parameters--
00:08:56.200 --> 00:09:02.670
H11, H22, and H12.
00:09:02.670 --> 00:09:08.540
And we get the two energy levels
and the two eigenfunctions.
00:09:08.540 --> 00:09:11.290
And we know how to solve this
problem in the Schrodinger
00:09:11.290 --> 00:09:15.700
picture, and I'll do that first.
00:09:15.700 --> 00:09:24.380
And so, H11 is just this
integral, psi 1 star H.
00:09:24.380 --> 00:09:27.410
And we put a hat on H
for the time being--
00:09:27.410 --> 00:09:30.380
1 d tau.
00:09:30.380 --> 00:09:41.950
And H22 is a different integral
of psi 2 star H psi 2 d tau.
00:09:41.950 --> 00:09:50.345
And H12 is psi 1
star H psi 2 tau.
00:09:52.990 --> 00:10:00.670
And we know that the Hamiltonian
operator is Hermitian,
00:10:00.670 --> 00:10:05.440
and so we can write
also that this is
00:10:05.440 --> 00:10:14.500
psi 2 star H star psi I d tau.
00:10:14.500 --> 00:10:19.480
Anyway, we call this thing V.
And there's some subtleties.
00:10:19.480 --> 00:10:25.550
If the integral between these
two things is imaginary,
00:10:25.550 --> 00:10:34.190
then the 1-2 and 2-1
integrals have opposite signs.
00:10:34.190 --> 00:10:36.710
Because the Hamiltonian
is Hermitian.
00:10:36.710 --> 00:10:39.380
But let's just think of
this as just one number.
00:10:39.380 --> 00:10:44.960
So we have E1, E2, and delta.
00:10:49.620 --> 00:10:54.330
So the two-level problem
is going to let us find
00:10:54.330 --> 00:10:56.250
the eigenfunctions--
00:10:56.250 --> 00:10:58.290
the plus and minus
eigenfunctions.
00:10:58.290 --> 00:10:59.290
C plus 1.
00:11:06.330 --> 00:11:11.040
So we have eigenfunctions
corresponding to-- which
00:11:11.040 --> 00:11:15.960
are eigenfunctions belonging to
the eigenvalues, E plus and E
00:11:15.960 --> 00:11:17.160
minus.
00:11:17.160 --> 00:11:20.690
And there are a linear
combination of the two states.
00:11:20.690 --> 00:11:23.040
Now, it's a two-level problem.
00:11:23.040 --> 00:11:27.730
This is a state space that
contains only two states.
00:11:27.730 --> 00:11:29.160
It's not an approximation.
00:11:31.770 --> 00:11:35.100
And so completeness says
we can write any function
00:11:35.100 --> 00:11:38.520
we want as a linear combination
of the functions and the basis
00:11:38.520 --> 00:11:40.440
set.
00:11:40.440 --> 00:11:45.180
And so our job is going to
be to find these coefficients
00:11:45.180 --> 00:11:48.441
and also to find the
energy eigenvalues.
00:11:55.990 --> 00:11:56.500
OK.
00:11:56.500 --> 00:11:59.620
So let's start to do some work.
00:11:59.620 --> 00:12:01.750
We know this thing.
00:12:01.750 --> 00:12:03.550
And let's start some--
00:12:03.550 --> 00:12:08.740
and we know that
H psi plus-minus
00:12:08.740 --> 00:12:15.010
has to be equal to the
eigenenergies psi plus-minus.
00:12:15.010 --> 00:12:19.840
Now, in order to get
something useful out of this,
00:12:19.840 --> 00:12:22.830
we left multiply by psi 1.
00:12:27.040 --> 00:12:27.650
OK.
00:12:27.650 --> 00:12:30.290
And when you left
multiply by psi 1-- well,
00:12:30.290 --> 00:12:31.497
let's just write it out.
00:12:35.240 --> 00:12:42.590
We're going to get
psi 1 HC 1 plus.
00:12:42.590 --> 00:12:45.070
So let's just write this out.
00:12:45.070 --> 00:12:53.610
We have H11 C1 plus-minus.
00:12:53.610 --> 00:13:00.120
Because this is C
plus-minus, and we
00:13:00.120 --> 00:13:06.360
have the integral of psi 1
with itself or the Hamiltonian
00:13:06.360 --> 00:13:07.690
is H11.
00:13:07.690 --> 00:13:14.200
And then we get another term.
00:13:14.200 --> 00:13:16.080
I'm doing this
differently from my notes,
00:13:16.080 --> 00:13:28.360
so the other term is C
plus-minus psi 2 V. OK.
00:13:28.360 --> 00:13:36.420
So plugging in what I have over
here and knowing that H11--
00:13:36.420 --> 00:13:42.390
we know that V is psi 1 H psi
2, and so this is what we get.
00:13:45.060 --> 00:13:50.530
We do the same thing, and we
have the same left-hand side,
00:13:50.530 --> 00:13:59.790
but we can also now
express this as that H--
00:14:07.220 --> 00:14:10.172
that psi plus-minus
is an eigenfunction.
00:14:13.130 --> 00:14:15.710
And so we have the
same left-hand side,
00:14:15.710 --> 00:14:17.390
but on the right-hand
side, we now
00:14:17.390 --> 00:14:28.060
get integral psi 1 star
H psi plus-minus d tau.
00:14:28.060 --> 00:14:41.540
But that's equal to psi star E
plus-minus C1 plus-minus psi 1
00:14:41.540 --> 00:14:48.640
plus C2 plus-minus psi 2 d tau.
00:14:48.640 --> 00:14:49.190
OK.
00:14:49.190 --> 00:14:52.700
And so now we're talking about
orthonormality of the wave
00:14:52.700 --> 00:14:55.880
function, and because we don't
have any operators in here,
00:14:55.880 --> 00:15:04.980
then this thing becomes simply
E plus-minus times C1 plus-minus
00:15:04.980 --> 00:15:10.550
plus 0 C2 plus-minus.
00:15:14.210 --> 00:15:16.330
There should be a 1 here.
00:15:16.330 --> 00:15:16.830
OK.
00:15:16.830 --> 00:15:21.040
Because psi 1, psi
2, that's orthogonal.
00:15:21.040 --> 00:15:23.320
So now we have an equation
that's quite useful.
00:15:29.470 --> 00:15:31.980
So we're going to combine
these two equations,
00:15:31.980 --> 00:15:41.122
and we get C1 plus
H11 plus C2 plus--
00:15:41.122 --> 00:15:44.220
I'm sorry, this is
plus-minus, plus-minus.
00:15:44.220 --> 00:15:51.080
V is equal to E
plus-minus C1 plus-minus.
00:15:54.870 --> 00:15:57.010
Now we need another equation.
00:15:57.010 --> 00:16:00.980
And so we do the same thing and
we left multiply by psi 2 star
00:16:00.980 --> 00:16:03.170
and integrate.
00:16:03.170 --> 00:16:05.870
And we get another
equation, and that
00:16:05.870 --> 00:16:19.770
is C1 plus-minus the plus
C2 plus-minus H22 minus E
00:16:19.770 --> 00:16:25.630
plus-minus is equal to 0.
00:16:25.630 --> 00:16:33.020
So we combine the two equations
that we have derived, solving--
00:16:33.020 --> 00:16:39.920
because both equations can be
solved for C1 plus-minus or C2
00:16:39.920 --> 00:16:41.210
plus-minus.
00:16:41.210 --> 00:16:44.810
So we equate the C1
plus-minus over C2 plus-minus
00:16:44.810 --> 00:16:46.430
from the two equations.
00:16:46.430 --> 00:16:51.710
And we get this wonderful
result, V over H11 minus E
00:16:51.710 --> 00:17:04.450
plus-minus is equal to H22
minus E plus-minus over V.
00:17:04.450 --> 00:17:07.050
Now you're not going
to ever do this.
00:17:07.050 --> 00:17:13.800
So yes, you can attempt
to reconstruct this
00:17:13.800 --> 00:17:16.890
from what I read on the board
or what's in your notes,
00:17:16.890 --> 00:17:21.359
but the important point is that
we're just using what we know.
00:17:21.359 --> 00:17:24.450
We say we have
eigenfunctions, and we
00:17:24.450 --> 00:17:29.790
want to find something about
the coefficients of the basis
00:17:29.790 --> 00:17:31.900
functions in each of
the eigenfunctions.
00:17:31.900 --> 00:17:33.850
And we want to find
the eigenvalues,
00:17:33.850 --> 00:17:36.890
and we get this equation.
00:17:36.890 --> 00:17:41.560
So, this is easy.
00:17:41.560 --> 00:17:44.230
We relate this to--
00:17:44.230 --> 00:17:47.860
we just multiply
through and we get V2,
00:17:47.860 --> 00:17:57.989
V squared is equal to H11
minus E plus-minus times H22
00:17:57.989 --> 00:17:58.780
minus E plus-minus.
00:17:58.780 --> 00:18:07.100
Or we solve, and we know that
E plus-minus is equal-- this
00:18:07.100 --> 00:18:08.900
is a quadratic equation.
00:18:08.900 --> 00:18:14.620
A quadratic and E plus-minus,
and so we have this result,
00:18:14.620 --> 00:18:26.380
H11 plus H22 plus or
minus this H11 squared--
00:18:26.380 --> 00:18:48.680
H11 plus H22 squared minus 4
H11 H22 minus V squared over 2.
00:18:48.680 --> 00:18:51.660
This is just a
quadratic equation.
00:18:51.660 --> 00:18:56.450
So we have the eigenenergies
expressed in terms
00:18:56.450 --> 00:18:57.680
of the quantities we know.
00:19:08.230 --> 00:19:10.210
We simplify the notation.
00:19:10.210 --> 00:19:19.410
E bar is H11 plus H22 over 2.
00:19:19.410 --> 00:19:26.630
And delta is H11
minus H22 over 2.
00:19:29.690 --> 00:19:33.600
So when we do that, we
simplify the algebra
00:19:33.600 --> 00:19:37.590
and we end up discovering that
the eigenvalue equation is
00:19:37.590 --> 00:19:46.350
E plus-minus is equal to E bar
plus or minus delta squared
00:19:46.350 --> 00:19:50.130
plus V squared square root.
00:19:50.130 --> 00:19:54.580
That's a simple result. This is
something you should remember.
00:19:57.810 --> 00:19:59.780
So if you have a
two-level problem,
00:19:59.780 --> 00:20:03.547
the energy eigenvalues are the
average energy plus or minus
00:20:03.547 --> 00:20:04.130
this quantity.
00:20:08.390 --> 00:20:11.130
This is an exact solution
for a two-level problem.
00:20:11.130 --> 00:20:12.944
For all two-level problems.
00:20:16.430 --> 00:20:16.970
OK.
00:20:16.970 --> 00:20:19.280
And we even simplify
the notation more.
00:20:19.280 --> 00:20:25.080
We call X delta
squared plus V squared.
00:20:25.080 --> 00:20:35.560
And so this becomes E
bar plus-minus X squared.
00:20:35.560 --> 00:20:37.600
Pretty compact.
00:20:37.600 --> 00:20:40.600
Now, the reason for
simplifying the notation
00:20:40.600 --> 00:20:43.090
is that we're going
to derive the values
00:20:43.090 --> 00:20:47.820
for the eigenfunctions, and
they involve a lot of symbols.
00:20:47.820 --> 00:20:51.020
And we want to make this
as compact as possible,
00:20:51.020 --> 00:20:54.630
and so we'll do that.
00:20:54.630 --> 00:20:55.130
OK.
00:20:55.130 --> 00:20:57.020
So when we started
out, it looked
00:20:57.020 --> 00:21:02.060
like we were going to
solve for these quantities,
00:21:02.060 --> 00:21:06.830
but we took a detour and solved
for the eigenenergies first.
00:21:06.830 --> 00:21:10.100
And this is one of the things
that happens in linear algebra.
00:21:10.100 --> 00:21:12.830
You get something
you can get easily--
00:21:12.830 --> 00:21:15.430
quickly before you get the
other stuff that you want.
00:21:18.510 --> 00:21:23.410
OK, but the second
part of the job
00:21:23.410 --> 00:21:26.660
is to find these
mixing coefficients.
00:21:26.660 --> 00:21:31.240
And so one thing
you do is you say,
00:21:31.240 --> 00:21:45.330
well, we insist that the
wave functions be orthog--
00:21:45.330 --> 00:21:47.980
normalized.
00:21:47.980 --> 00:21:51.820
After that a lot
of algebra ensues.
00:21:51.820 --> 00:21:58.780
And I'm not going to even
attempt to work through it.
00:21:58.780 --> 00:22:02.020
You may want to work through
it, but what I recommend doing
00:22:02.020 --> 00:22:06.400
is looking at the
solution and then
00:22:06.400 --> 00:22:08.530
checking to see
whether it does things
00:22:08.530 --> 00:22:11.270
that you expect it has to do.
00:22:11.270 --> 00:22:15.250
Because one of the things
that kind of inspection
00:22:15.250 --> 00:22:22.520
leads you to is factors of
two errors and sign errors.
00:22:22.520 --> 00:22:23.030
OK.
00:22:23.030 --> 00:22:26.840
But, C plus-minus--
C1 plus-minus
00:22:26.840 --> 00:22:33.960
is equal to 1/2
times 1 plus or minus
00:22:33.960 --> 00:22:42.880
delta over the square root
of X. Two square roots.
00:22:42.880 --> 00:22:43.380
Whoops.
00:22:49.970 --> 00:22:51.970
OK, no square root
in here because we
00:22:51.970 --> 00:22:53.850
get square root there.
00:22:53.850 --> 00:23:02.440
OK and C2 plus-minus
is 1/2 1 minus or plus
00:23:02.440 --> 00:23:07.720
delta square root,
x square root.
00:23:07.720 --> 00:23:09.640
OK?
00:23:09.640 --> 00:23:13.810
So these two things are
kind of simple-looking,
00:23:13.810 --> 00:23:16.570
but there's an awful
lot of compression
00:23:16.570 --> 00:23:19.900
and select and
clever manipulation
00:23:19.900 --> 00:23:21.250
in order to get this.
00:23:21.250 --> 00:23:23.320
But the important
thing to notice
00:23:23.320 --> 00:23:27.850
is that we have the
energy difference divided
00:23:27.850 --> 00:23:33.640
by this thing, this
delta squared plus V
00:23:33.640 --> 00:23:39.130
squared that expresses
the importance
00:23:39.130 --> 00:23:40.810
of the two-level interaction.
00:23:40.810 --> 00:23:43.855
And these two things
enter with opposite signs.
00:23:50.130 --> 00:23:50.770
OK.
00:23:50.770 --> 00:23:53.350
And so one of the checks
that's really easy to do
00:23:53.350 --> 00:24:00.770
is let's let V go to 0, and
let's let V go to infinity.
00:24:00.770 --> 00:24:03.940
OK, remember, our Hamiltonian--
00:24:03.940 --> 00:24:04.900
the two-level problem--
00:24:08.290 --> 00:24:10.130
I can't write it
as a matrix yet,
00:24:10.130 --> 00:24:18.670
so remember that we
had E1, E2, and V.
00:24:18.670 --> 00:24:22.210
And this was the higher energy,
this was the lower energy,
00:24:22.210 --> 00:24:28.760
and so delta is E1 minus E2--
00:24:28.760 --> 00:24:30.190
I better not write that.
00:24:34.040 --> 00:24:35.980
OK.
00:24:35.980 --> 00:24:46.540
So we normally write E1 over E2,
and we have this V interaction
00:24:46.540 --> 00:24:48.220
between them.
00:24:48.220 --> 00:24:49.120
OK.
00:24:49.120 --> 00:24:58.120
And so if the interaction
integral between the two
00:24:58.120 --> 00:25:06.100
functions is 0, then E1
is the higher energy,
00:25:06.100 --> 00:25:08.500
and it corresponds to--
00:25:08.500 --> 00:25:12.550
it should correspond
to psi 1 alone.
00:25:12.550 --> 00:25:20.560
Well, so the higher energy we're
taking the plus combinations
00:25:20.560 --> 00:25:28.640
here, and if V is equal to 0,
then this is delta over delta.
00:25:28.640 --> 00:25:32.110
And this is either 2 or 0.
00:25:32.110 --> 00:25:33.580
And when it's the higher--
00:25:33.580 --> 00:25:37.870
the upper sine, it's
2 divided by 2 or 1,
00:25:37.870 --> 00:25:40.030
exactly what you expect.
00:25:40.030 --> 00:25:44.200
And this one is 1 minus 1.
00:25:44.200 --> 00:25:46.930
So these two are
behaving, right?
00:25:46.930 --> 00:25:50.380
And the V to 0 rate.
00:25:50.380 --> 00:25:57.720
Now, in the V going to
infinity, then this is infinite.
00:25:57.720 --> 00:26:16.120
So that just means that
this term goes away, right?
00:26:16.120 --> 00:26:24.230
And so we have C1 plus and minus
is equal to square root of 2--
00:26:24.230 --> 00:26:27.260
1 over the square root of 2.
00:26:27.260 --> 00:26:29.710
And same here.
00:26:29.710 --> 00:26:31.920
So what we get is what
we call 50/50 mixing.
00:26:37.410 --> 00:26:39.740
OK, so this also works.
00:26:42.250 --> 00:26:45.640
So I don't say that this can--
00:26:45.640 --> 00:26:49.720
this confirms that I have not
made an algebraic mistake,
00:26:49.720 --> 00:26:56.120
I haven't, but it's a good test
because it's really easy to do.
00:26:56.120 --> 00:26:58.220
And if you're not
getting what you expect,
00:26:58.220 --> 00:27:01.100
you know you've either
blown a sign or a factor
00:27:01.100 --> 00:27:07.260
or two, which are the two things
that you fear most in a wave
00:27:07.260 --> 00:27:10.630
function free approach.
00:27:10.630 --> 00:27:14.290
Because you've got nothing to
do except check your algebra,
00:27:14.290 --> 00:27:18.320
and usually, you made the
mistake because it was subtle,
00:27:18.320 --> 00:27:20.710
and you'll make it again
when you're checking it.
00:27:20.710 --> 00:27:23.050
So these kinds of checks
are really valuable.
00:27:28.365 --> 00:27:28.865
OK.
00:27:33.310 --> 00:27:37.750
So once you know that this
is likely to be correct,
00:27:37.750 --> 00:27:39.850
then you do some
other things, and you
00:27:39.850 --> 00:27:45.760
check the wave functions you've
derived with your C plus-minus
00:27:45.760 --> 00:27:55.460
1 and C plus-minus two are
both normalized and orthogonal.
00:28:01.340 --> 00:28:10.160
And that the energy you get is
such that psi plus is E plus.
00:28:10.160 --> 00:28:12.560
And you know what E plus was.
00:28:12.560 --> 00:28:16.550
And so you just go
in and you calculate
00:28:16.550 --> 00:28:20.930
what the energy should be
giving using the values for PSI
00:28:20.930 --> 00:28:28.370
plus-minus And so we plug those
into the original equations.
00:28:28.370 --> 00:28:34.160
And you can also, again,
show that psi plus-minus
00:28:34.160 --> 00:28:50.590
a star H psi plus-minus
is equal to 0
00:28:50.590 --> 00:28:56.240
because the eigenfunctions
are orthogonal.
00:28:56.240 --> 00:28:59.450
We already did that, but we
plugged it into an equation
00:28:59.450 --> 00:29:01.190
here, and we got
it a second time.
00:29:07.240 --> 00:29:14.770
So in the lecture notes, there
is a lengthy algebraic proof
00:29:14.770 --> 00:29:17.380
or demonstration
that E plus-minus
00:29:17.380 --> 00:29:20.540
is equal to plus or
minus square of x,
00:29:20.540 --> 00:29:22.570
which are already
derived, but then I
00:29:22.570 --> 00:29:23.830
just did it the long way.
00:29:23.830 --> 00:29:24.730
OK .
00:29:24.730 --> 00:29:26.710
So this is it--
00:29:26.710 --> 00:29:31.090
we are about to move from
the Schrodinger picture
00:29:31.090 --> 00:29:34.700
to the matrix picture.
00:29:34.700 --> 00:29:46.970
So the trick now is to
go to linear algebra,
00:29:46.970 --> 00:29:49.520
go to the matrix
picture and learn
00:29:49.520 --> 00:29:53.600
how to just write the equations
and what the language is
00:29:53.600 --> 00:29:54.575
and show that it works.
00:29:57.460 --> 00:30:04.480
So suppose you have two
square matrices, A and B,
00:30:04.480 --> 00:30:09.490
they're n by n squared
matrices, you know the rules
00:30:09.490 --> 00:30:11.307
for matrix multiplication?
00:30:15.290 --> 00:30:23.000
So if you wanted, the mn
element of this product or two
00:30:23.000 --> 00:30:28.550
matrices, you would
go, n equals--
00:30:28.550 --> 00:30:29.640
i equals 1.
00:30:32.320 --> 00:30:38.850
Let's use the same notation.
j equals 1 to n of Am--
00:30:38.850 --> 00:30:40.544
because I want the mn--
00:30:40.544 --> 00:30:42.270
j Bjn.
00:30:47.220 --> 00:30:50.040
And so when you multiply
two square matrices,
00:30:50.040 --> 00:30:51.600
you get back a square matrix.
00:30:51.600 --> 00:30:53.950
And this picture
is not a bad one.
00:30:53.950 --> 00:30:57.020
So you can say, all right.
00:31:03.410 --> 00:31:08.510
So if you need a little
q to remind yourself,
00:31:08.510 --> 00:31:14.390
you take this row and
multiply term by term
00:31:14.390 --> 00:31:16.940
and add the results--
00:31:16.940 --> 00:31:20.630
this row in that column,
and you get a number here.
00:31:20.630 --> 00:31:24.260
And you just repeat that,
and it's really easy
00:31:24.260 --> 00:31:25.820
to tell your
computer to do this,
00:31:25.820 --> 00:31:28.130
and it's rather tedious
to do it yourself
00:31:28.130 --> 00:31:30.080
if it's more than a
two-by-two matrix.
00:31:33.060 --> 00:31:35.420
OK.
00:31:35.420 --> 00:31:40.290
So what about this thing C?
00:31:40.290 --> 00:31:51.940
Well, C is a N
row column matrix.
00:31:51.940 --> 00:31:54.120
So it's N rows, 1 column.
00:31:56.800 --> 00:32:00.280
It looks like this--
00:32:00.280 --> 00:32:04.090
C1, C2, Cn.
00:32:06.910 --> 00:32:13.290
And so if you want to multiply
a square matrix by a vector,
00:32:13.290 --> 00:32:15.880
we know the rules too, OK?
00:32:15.880 --> 00:32:19.930
And again, this picture
is a useful one.
00:32:19.930 --> 00:32:22.900
So let's just draw
something like this.
00:32:22.900 --> 00:32:29.320
We do this and this,
and that gives you
00:32:29.320 --> 00:32:30.800
one element in the column.
00:32:34.970 --> 00:32:36.490
OK.
00:32:36.490 --> 00:32:40.420
Now, the last thing that
I want to remind you of--
00:32:40.420 --> 00:32:41.860
I better use this board.
00:32:46.680 --> 00:32:59.570
If we have a two-state problem,
then the vector c1 is 1, 0;
00:32:59.570 --> 00:33:02.960
the vector c2--
00:33:02.960 --> 00:33:05.840
this should be a lower
case c, because we
00:33:05.840 --> 00:33:10.100
tend to use lowercase letters
for vectors and uppercase
00:33:10.100 --> 00:33:11.135
letters for matrices.
00:33:16.040 --> 00:33:32.340
And so if we do c1 dagger
c2, or c1 dagger c1.
00:33:35.410 --> 00:33:41.570
Now this dagger means conjugate
transpose, except, well,
00:33:41.570 --> 00:33:47.450
there isn't anything to do
except convert a matrix--
00:33:47.450 --> 00:33:58.250
a vector into-- so that
becomes a row and this
00:33:58.250 --> 00:33:59.390
becomes a column.
00:34:03.110 --> 00:34:05.740
And so a row times the
column gives a number.
00:34:08.360 --> 00:34:11.770
And that number is going
to be-- well, let's do it.
00:34:11.770 --> 00:34:16.719
We have 1, 0; 0, 1.
00:34:16.719 --> 00:34:21.790
1 times 0 is 0, 0 times
1 is 0, and so this is 0.
00:34:21.790 --> 00:34:25.000
Well that's orthogonality.
00:34:25.000 --> 00:34:32.730
This, we have 1 times 1 is 1.
00:34:32.730 --> 00:34:34.030
0 times 0 is 0.
00:34:34.030 --> 00:34:37.760
So it's 1 plus 0.
00:34:37.760 --> 00:34:38.770
Right.
00:34:38.770 --> 00:34:40.429
OK, all right.
00:34:46.960 --> 00:34:47.460
OK.
00:34:50.400 --> 00:34:58.530
So the Schrodinger equation
becomes in matrix language--
00:34:58.530 --> 00:35:01.900
and now, one notation is--
00:35:01.900 --> 00:35:04.020
I'm going to stop using
the double underline.
00:35:04.020 --> 00:35:06.360
That means boldface.
00:35:06.360 --> 00:35:09.500
And we don't use hats anymore.
00:35:09.500 --> 00:35:12.340
Or at least if we were
really consistent,
00:35:12.340 --> 00:35:15.400
when we go away from
the Schrodinger picture,
00:35:15.400 --> 00:35:20.320
we don't put hats on operators,
we make them boldface letters.
00:35:20.320 --> 00:35:21.000
OK.
00:35:21.000 --> 00:35:26.580
Now, the thing is we're so
comfortable in matrix land,
00:35:26.580 --> 00:35:29.740
that we don't use either.
00:35:29.740 --> 00:35:30.260
OK.
00:35:30.260 --> 00:35:33.780
But remember, we're talking
about different things.
00:35:33.780 --> 00:35:36.950
So the Schrodinger
equation looks like this.
00:35:45.600 --> 00:35:53.070
And so we have a matrix,
delta v, v delta times 1, 0.
00:35:56.810 --> 00:36:06.350
And that's delta, v; or delta
times 1, 0 plus v times 0, 1.
00:36:06.350 --> 00:36:07.760
Isn't that interesting?
00:36:07.760 --> 00:36:12.410
Remember, the Hamiltonian
or any operator operating
00:36:12.410 --> 00:36:16.640
on a function gives rise
to a linear combination
00:36:16.640 --> 00:36:19.280
of the functions
and the basis set.
00:36:19.280 --> 00:36:22.921
And so here's one of the
functions, here's the other.
00:36:22.921 --> 00:36:23.420
OK.
00:36:23.420 --> 00:36:27.990
Nothing very mysterious
has happened here.
00:36:27.990 --> 00:36:36.990
So this very equation is
going to be HC is equal to EC.
00:36:43.180 --> 00:36:47.340
OK, so how do we approach this?
00:37:01.550 --> 00:37:11.220
Well I have to
introduce a new symbol,
00:37:11.220 --> 00:37:15.840
and that's going to be this
symbol T. It's a matrix.
00:37:18.390 --> 00:37:20.670
It's a unitary matrix.
00:37:20.670 --> 00:37:24.470
And we want it to have
a special properties.
00:37:24.470 --> 00:37:27.620
Those special properties
will be shown here.
00:37:27.620 --> 00:37:31.230
OK, so first of all,
we have this matrix--
00:37:31.230 --> 00:37:34.860
T11, T--
00:37:34.860 --> 00:37:37.420
AUDIENCE: Your Hamiltonian
matrix is incorrect.
00:37:37.420 --> 00:37:41.320
So I think you mean to
say E bar plus-minus?
00:37:41.320 --> 00:37:43.240
ROBERT FIELD: I'm sorry?
00:37:43.240 --> 00:37:44.470
AUDIENCE: I think the diagonal
elements of the Hamiltonian
00:37:44.470 --> 00:37:45.430
matrix should be--
00:37:45.430 --> 00:37:46.388
ROBERT FIELD: Yeah, OK.
00:37:46.388 --> 00:37:54.420
If we wanted the
eigenvalues, OK?
00:37:54.420 --> 00:37:55.470
This is--
00:37:55.470 --> 00:37:59.470
AUDIENCE: So the infinite
matrix in this case should be--
00:37:59.470 --> 00:38:02.560
the diagonal elements
should be E1 and E2.
00:38:02.560 --> 00:38:05.120
So I think you meant
E bar plus delta,
00:38:05.120 --> 00:38:09.130
and then E bar minus delta.
00:38:09.130 --> 00:38:10.080
ROBERT FIELD: Yeah.
00:38:10.080 --> 00:38:11.100
OK.
00:38:11.100 --> 00:38:13.530
There is something in
the notes and something
00:38:13.530 --> 00:38:15.840
in my notes which--
00:38:15.840 --> 00:38:18.240
we can always write
the Hamiltonian
00:38:18.240 --> 00:38:30.780
as E bar 0, 0, E bar plus
delta v v minus delta.
00:38:30.780 --> 00:38:35.360
We can always take out
this constant term.
00:38:35.360 --> 00:38:39.020
And this is the thing
we're always working on.
00:38:39.020 --> 00:38:42.730
And so rather than--
00:38:42.730 --> 00:38:46.720
and so we could
call this H prime.
00:38:46.720 --> 00:38:49.570
Or we can simply say, oh,
we have these two things--
00:38:49.570 --> 00:38:53.260
this always gets added
in, and it's not affected.
00:38:53.260 --> 00:38:59.050
If we take this
diagonal constant matrix
00:38:59.050 --> 00:39:03.830
and apply a transformation to
it, a unitary transformation,
00:39:03.830 --> 00:39:07.000
you get that matrix again,
you get-- it's nothing.
00:39:07.000 --> 00:39:23.060
OK, so we wanted to
describe some special matrix
00:39:23.060 --> 00:39:27.710
where the transpose
of that matrix
00:39:27.710 --> 00:39:30.790
is equal to the
inverse of that matrix.
00:39:30.790 --> 00:39:34.060
Now-- yes?
00:39:34.060 --> 00:39:37.110
AUDIENCE: So shouldn't
there be a minus on there?
00:39:37.110 --> 00:39:39.660
On the prior equation?
00:39:39.660 --> 00:39:42.230
For the bottom right entry.
00:39:42.230 --> 00:39:44.470
The primary delta and the--
00:39:44.470 --> 00:39:46.330
first line.
00:39:46.330 --> 00:39:48.036
First line.
00:39:48.036 --> 00:39:49.660
ROBERT FIELD: This
is delta minus the--
00:39:49.660 --> 00:39:50.243
AUDIENCE: Yes.
00:39:50.243 --> 00:39:52.621
But up there, you
just had delta delta.
00:39:52.621 --> 00:39:53.371
ROBERT FIELD: Yep.
00:39:56.710 --> 00:39:57.664
Thank you.
00:40:01.020 --> 00:40:01.750
OK.
00:40:01.750 --> 00:40:04.950
Now, when you have
a matrix, you really
00:40:04.950 --> 00:40:11.940
like to have T minus 1T is
equal to the unit matrix.
00:40:11.940 --> 00:40:16.820
Getting the inverse of a
matrix in a general problem
00:40:16.820 --> 00:40:19.700
is really awful.
00:40:19.700 --> 00:40:25.420
But for a unitary matrices,
all you do to get the inverse
00:40:25.420 --> 00:40:26.950
is to take--
00:40:26.950 --> 00:40:28.465
you flip it on the diagonal.
00:40:31.270 --> 00:40:40.980
Now strictly, the conjugate
transpose of a unitary matrix
00:40:40.980 --> 00:40:42.660
is the inverse.
00:40:42.660 --> 00:40:45.480
But often we have real matrices.
00:40:45.480 --> 00:40:50.190
But the important thing is
always this flipping along--
00:40:50.190 --> 00:40:52.500
flipping the matrix
on the diagonal.
00:40:52.500 --> 00:40:56.180
And that gives you the
inverse unless there
00:40:56.180 --> 00:40:58.940
is stuff here that is complex.
00:40:58.940 --> 00:40:59.440
OK.
00:41:02.690 --> 00:41:06.650
So this conjugate transpose,
it would look like--
00:41:18.482 --> 00:41:20.870
OK?
00:41:20.870 --> 00:41:21.370
OK.
00:41:21.370 --> 00:41:25.630
Now, what we want to do
is derive the matrix form
00:41:25.630 --> 00:41:28.060
of the Schrodinger
equation using
00:41:28.060 --> 00:41:33.920
this unitary transformation.
00:41:33.920 --> 00:41:38.010
So we start out again
with HC is equal to EC.
00:41:44.260 --> 00:41:52.190
And now, we insert TT dagger.
00:41:55.010 --> 00:41:59.030
And this is one of the
things where you screw up.
00:41:59.030 --> 00:42:02.180
Whether you insert TT
dagger or T dagger T,
00:42:02.180 --> 00:42:03.680
because they're both 1.
00:42:03.680 --> 00:42:06.830
And if you use the wrong
one, all of your phases,
00:42:06.830 --> 00:42:08.870
everything is wrong.
00:42:08.870 --> 00:42:11.330
But it's still correct,
the equations are correct,
00:42:11.330 --> 00:42:15.050
but the things you've memorized
are no longer correct.
00:42:15.050 --> 00:42:18.620
OK, so we're going
to insert this unit
00:42:18.620 --> 00:42:32.950
matrix between H and C.
And of course, we have to--
00:42:32.950 --> 00:42:37.050
we would have to insert
this on the other side.
00:42:37.050 --> 00:42:47.850
T T dagger C and
E. But that's one.
00:42:47.850 --> 00:42:50.570
OK, so we don't do anything
on the left-hand side.
00:42:53.310 --> 00:42:55.790
And now we left
multiply by T dagger.
00:43:05.340 --> 00:43:06.550
OK.
00:43:06.550 --> 00:43:12.740
And we call this H twiddle,
and we call this C twiddle.
00:43:15.430 --> 00:43:24.160
So we have H twiddle, C
twiddle is equal to EC twiddle.
00:43:24.160 --> 00:43:26.850
Now we're cooking.
00:43:26.850 --> 00:43:32.160
OK, because we construct
this unitary matrix
00:43:32.160 --> 00:43:37.350
to cause this Hamiltonian,
the transformed Hamiltonian
00:43:37.350 --> 00:43:38.460
to be in diagonal form.
00:43:48.850 --> 00:43:58.450
So we say that H twiddle,
which is T dagger T,
00:43:58.450 --> 00:44:06.800
is equal to E plus, E minus,
0, 0 for the 2 by 2 problem.
00:44:11.870 --> 00:44:14.730
OK, that leads to
some requirements.
00:44:14.730 --> 00:44:16.450
What are the elements
of the T matrix?
00:44:19.620 --> 00:44:30.930
But the important thing is
we say that T diagonalizes H.
00:44:30.930 --> 00:44:35.624
So this magic matrix gives
you the energy eigenvalues.
00:44:41.060 --> 00:44:49.450
But we also have T dagger
C is equal to C twiddle.
00:44:49.450 --> 00:44:54.510
And so this gives you
the linear combination
00:44:54.510 --> 00:44:59.752
of the column vectors that
correspond to the eigenvector.
00:45:06.840 --> 00:45:08.210
So what is it?
00:45:08.210 --> 00:45:31.520
We have here T dagger, which
is TT, T12 T, T21 T, T22 T
00:45:31.520 --> 00:45:46.970
times C. And that's supposed
to be equal to C twiddle or C1
00:45:46.970 --> 00:45:50.060
twiddle, C2 twiddle.
00:45:50.060 --> 00:45:56.070
So we put everything together
and we discover that OK,
00:45:56.070 --> 00:45:59.430
when we multiply
this by that, we get
00:45:59.430 --> 00:46:01.410
the element that goes up here.
00:46:01.410 --> 00:46:20.690
And so the element on top is
C1 T11 dagger plus C2 T12--
00:46:23.720 --> 00:46:27.920
I'm using T's and
daggers independently,
00:46:27.920 --> 00:46:29.610
and we do the same thing here.
00:46:29.610 --> 00:46:34.980
So we have a column
vector, and it's
00:46:34.980 --> 00:46:39.780
composed of the original state.
00:46:39.780 --> 00:46:45.030
And so anyway, when
we do everything,
00:46:45.030 --> 00:46:50.010
we discover that
using the solution
00:46:50.010 --> 00:46:54.990
to the two-level problem
from the Schrodinger picture,
00:46:54.990 --> 00:46:56.580
we know everything.
00:46:56.580 --> 00:47:08.620
So H twiddle, C twiddle plus
is H plus C twiddle plus.
00:47:08.620 --> 00:47:18.330
And that's just E plus 1, 0.
00:47:18.330 --> 00:47:35.180
And H twiddle C twiddle
minus is E minus times 0, 1.
00:47:38.355 --> 00:47:38.855
OK.
00:47:42.630 --> 00:47:47.200
This is all very confusing
because the notation
00:47:47.200 --> 00:47:49.890
is unfamiliar.
00:47:49.890 --> 00:47:52.770
But the important thing
is that everything
00:47:52.770 --> 00:47:55.800
you can do in the
Schrodinger picture you
00:47:55.800 --> 00:47:58.350
can do in this matrix picture.
00:47:58.350 --> 00:48:05.700
And you can find these
elements of this T matrix,
00:48:05.700 --> 00:48:09.600
and it turns out that what you--
00:48:12.810 --> 00:48:23.730
the matrix that diagonalizes the
Hamiltonian, the columns of T
00:48:23.730 --> 00:48:26.610
dagger are the eigenvectors.
00:48:29.700 --> 00:48:34.520
So you find this
matrix, it diagonalizes
00:48:34.520 --> 00:48:39.270
H. The computer tells
you the eigenenergies,
00:48:39.270 --> 00:48:42.520
and it also tells you
T dagger or T plus--
00:48:42.520 --> 00:48:43.020
TT.
00:48:46.260 --> 00:48:49.140
And so with that,
you know how to write
00:48:49.140 --> 00:48:53.970
in the original basis set
what the eigenbasis is
00:48:53.970 --> 00:48:55.280
for each eigenvalue.
00:48:58.340 --> 00:49:06.140
And so the next step, which
will happen in the next lecture,
00:49:06.140 --> 00:49:17.450
is the general unitary
transformation for two-level.
00:49:17.450 --> 00:49:22.310
Now, you expect
that there is going
00:49:22.310 --> 00:49:26.540
to be a general solution
for the two-level problem,
00:49:26.540 --> 00:49:28.910
because the two-level problem
in the Schrodinger picture
00:49:28.910 --> 00:49:32.360
led to a quadratic equation.
00:49:32.360 --> 00:49:35.300
And that had an
analytical solution.
00:49:35.300 --> 00:49:40.910
And so there is going to be
a general and exact solution
00:49:40.910 --> 00:49:42.800
the two-level problem.
00:49:42.800 --> 00:49:45.920
And this unitary
transformation is
00:49:45.920 --> 00:49:55.730
going to be written in terms
of cosine theta sine theta
00:49:55.730 --> 00:50:01.310
minus sine theta cosine theta.
00:50:01.310 --> 00:50:07.910
This is a matrix
which is unitary,
00:50:07.910 --> 00:50:12.440
and it, when applied
to your basis set,
00:50:12.440 --> 00:50:20.090
conserves normalization
and orthogonality.
00:50:20.090 --> 00:50:22.540
And so the trick is
to be able to find
00:50:22.540 --> 00:50:27.070
the theta that causes the
Hamiltonian in question
00:50:27.070 --> 00:50:28.930
to be diagonalized.
00:50:28.930 --> 00:50:32.790
And the algebra for that will
happen in the next lecture.
00:50:32.790 --> 00:50:33.800
OK.
00:50:33.800 --> 00:50:39.270
Remember, you're not
going to do this ever.
00:50:39.270 --> 00:50:44.730
You're going to use the idea
of this unitary transformation,
00:50:44.730 --> 00:50:48.180
and you're going to
use that to get--
00:50:48.180 --> 00:50:51.130
to diagonalize the matrix.
00:50:51.130 --> 00:50:53.710
And this will lead
to some formulas
00:50:53.710 --> 00:50:56.080
which you're going
to like, because you
00:50:56.080 --> 00:50:58.630
can forget sines and cosines.
00:50:58.630 --> 00:51:00.130
Everything is going
to be expressed
00:51:00.130 --> 00:51:03.160
in terms of things like this--
00:51:03.160 --> 00:51:07.590
V over delta.
00:51:07.590 --> 00:51:09.840
So we have matrix element--
00:51:09.840 --> 00:51:16.000
off diagonal matrix element
over the energy difference.
00:51:16.000 --> 00:51:20.290
And that's the basic form of
nondegenerate perturbation
00:51:20.290 --> 00:51:23.927
theory, which is applied
not just in 2-by-2 problems,
00:51:23.927 --> 00:51:24.760
but to all problems.
00:51:29.400 --> 00:51:34.380
And so you want to
remember this lecture as,
00:51:34.380 --> 00:51:37.570
this is how we kill
the two-level problem.
00:51:37.570 --> 00:51:40.860
Then we can discover
what we did and apply it
00:51:40.860 --> 00:51:43.320
to a general problem
where it's not
00:51:43.320 --> 00:51:44.760
a two-level problem
anymore, it's
00:51:44.760 --> 00:51:47.290
an infinite number of levels.
00:51:47.290 --> 00:51:51.560
And when certain
approximations are met,
00:51:51.560 --> 00:51:57.860
it applies, and it gives you
the most accurate energy levels
00:51:57.860 --> 00:52:00.530
and wave functions
you could want.
00:52:00.530 --> 00:52:03.710
And you know how accurate
they are going to be.
00:52:03.710 --> 00:52:08.870
And this is liberating,
because now, you
00:52:08.870 --> 00:52:12.510
can take a not
exactly-solved problem
00:52:12.510 --> 00:52:15.680
and you can solve
it approximately.
00:52:15.680 --> 00:52:19.410
And you can use the
solution to determine, OK,
00:52:19.410 --> 00:52:22.980
there's going to be some
function of the quantum numbers
00:52:22.980 --> 00:52:24.850
which describes
the energy levels.
00:52:24.850 --> 00:52:26.870
Well what is that function?
00:52:26.870 --> 00:52:31.050
And what are the coefficients
in that function?
00:52:31.050 --> 00:52:32.820
And how do they
relate to the things
00:52:32.820 --> 00:52:36.280
you know from the Hamiltonian?
00:52:36.280 --> 00:52:40.240
So it's incredibly powerful.
00:52:40.240 --> 00:52:44.590
And once you're
given the formulas
00:52:44.590 --> 00:52:46.990
for nondegenerate
perturbation theory,
00:52:46.990 --> 00:52:50.780
you can solve practically any
problem in quantum mechanics.
00:52:50.780 --> 00:52:55.360
Not just numerically,
but with insight.
00:52:55.360 --> 00:53:04.440
It tells you, if we make
observations of some system,
00:53:04.440 --> 00:53:06.720
we determine a set of energy
levels, which is called
00:53:06.720 --> 00:53:09.760
the spectrum of that operator.
00:53:09.760 --> 00:53:15.490
And the spectrum of the
operator is an explicit function
00:53:15.490 --> 00:53:19.630
of the physical constants--
the unique interactions
00:53:19.630 --> 00:53:21.830
between states.
00:53:21.830 --> 00:53:30.200
And so you will then know
how the experimental data
00:53:30.200 --> 00:53:34.002
determines the mechanism
of all the interactions,
00:53:34.002 --> 00:53:39.790
and you can calculate
the wave function.
00:53:39.790 --> 00:53:41.650
You can't observe
the wave function,
00:53:41.650 --> 00:53:46.900
but you can discover its
traces in the energy levels.
00:53:46.900 --> 00:53:49.790
And then you can reproduce
the energy levels,
00:53:49.790 --> 00:53:53.050
and if you have the energy
levels-- the eigenstates,
00:53:53.050 --> 00:53:58.390
you can also describe
any dynamics using
00:53:58.390 --> 00:54:00.530
the same formulas in here.
00:54:00.530 --> 00:54:07.640
So this is an
incredible enablement.
00:54:07.640 --> 00:54:11.310
And you don't have to
look at my derivations.
00:54:11.310 --> 00:54:12.750
I'm not proud of
the derivations,
00:54:12.750 --> 00:54:14.650
I'm proud of the results.
00:54:14.650 --> 00:54:20.070
And if you can handle
these results that
00:54:20.070 --> 00:54:22.416
come from perturbation
theory as well as you've
00:54:22.416 --> 00:54:23.790
done so far in
the course, you're
00:54:23.790 --> 00:54:30.840
going to be at the
research level very soon.
00:54:30.840 --> 00:54:33.680
OK, see you on Friday.