WEBVTT
00:00:00.000 --> 00:00:02.350
PROFESSOR: Find
our final solution,
00:00:02.350 --> 00:00:05.530
we just have to
match the equations.
00:00:05.530 --> 00:00:12.175
So Psi continues at x equals a.
00:00:15.000 --> 00:00:16.270
And what do we have?
00:00:16.270 --> 00:00:22.540
Well from two, you
have cosine of ka.
00:00:22.540 --> 00:00:31.290
And from four you would have
a equals e to the minus ka.
00:00:31.290 --> 00:00:34.080
This is the value of this--
00:00:34.080 --> 00:00:36.660
so the interior
solution at x equals
00:00:36.660 --> 00:00:40.760
a must match the value of
the exterior solution of k
00:00:40.760 --> 00:00:43.260
equals a.
00:00:43.260 --> 00:00:52.040
Psi prime must be continuous
at x equals a as well.
00:00:52.040 --> 00:00:55.750
Well what is the derivative
of this function?
00:00:55.750 --> 00:01:01.610
It's minus the sine of this.
00:01:01.610 --> 00:01:08.600
So it's minus k sine
of kx, that becomes ka,
00:01:08.600 --> 00:01:11.580
is equal to the derivative
of that one, which
00:01:11.580 --> 00:01:20.430
is minus Kappa A e to
that minus Kappa little a.
00:01:23.050 --> 00:01:25.730
Two equations, and
how many unknowns?
00:01:25.730 --> 00:01:31.140
Well there's A and some
information about Kappa and k.
00:01:31.140 --> 00:01:35.150
And the easiest way to eliminate
that is to divide them.
00:01:35.150 --> 00:01:42.210
So you divide the bottom
equation by this equation.
00:01:42.210 --> 00:01:44.150
So what do we get?
00:01:44.150 --> 00:01:47.510
Divide the bottom by the top.
00:01:47.510 --> 00:01:51.040
Minus k and and
the minuses cancel,
00:01:51.040 --> 00:01:53.875
we can cancel those
minus signs and you
00:01:53.875 --> 00:02:03.330
get k tan ka is equal to Kappa.
00:02:06.400 --> 00:02:12.530
But you already are
convinced, I hope, on the idea
00:02:12.530 --> 00:02:17.110
that we should not use
equations that have units.
00:02:17.110 --> 00:02:20.860
So I will multiply by
little a and a little a
00:02:20.860 --> 00:02:26.470
to get a [INAUDIBLE] size, and
therefore, the right hand side
00:02:26.470 --> 00:02:35.690
becomes Xi equals, and the
left side become Eta tan Eta.
00:02:48.740 --> 00:02:52.400
OK, I want to make a little
comment about these quantities
00:02:52.400 --> 00:02:52.900
already.
00:02:52.900 --> 00:02:58.720
So all the problem has turned
out into the following.
00:02:58.720 --> 00:03:03.250
You were given a potential and
that determines a number z0.
00:03:03.250 --> 00:03:06.970
If you know the width and
everything, you know z0.
00:03:06.970 --> 00:03:10.470
Now you have to
calculate Eta and Xi.
00:03:10.470 --> 00:03:16.120
If you know either Eta or
Xi, you know Kappa or k.
00:03:16.120 --> 00:03:21.250
And if you know either k or
Kappa, since you know v0,
00:03:21.250 --> 00:03:23.230
you will know the energy.
00:03:23.230 --> 00:03:28.690
So it's kind of neat to
express this more clearly,
00:03:28.690 --> 00:03:36.260
and I think it's maybe
easier if one uses Xi.
00:03:36.260 --> 00:03:47.310
And look at Xi squared is
Kappa squared times a squared.
00:03:47.310 --> 00:03:50.350
And what is Kappa
squared, it's over there.
00:03:50.350 --> 00:03:57.700
2m absolute value of e, a
squared over h bar squared.
00:04:01.370 --> 00:04:09.320
Now you want to find e, you're
going to get in some units.
00:04:09.320 --> 00:04:13.310
Even e is nice to
have it without units.
00:04:13.310 --> 00:04:19.050
So I will multiply
and divide by v0.
00:04:19.050 --> 00:04:28.040
2m v0 a squared over h squared,
absolute value of e over v0.
00:04:28.040 --> 00:04:31.510
After all, you probably
prefer to know e
00:04:31.510 --> 00:04:36.260
over v0, which tells you how
proportional the energy is
00:04:36.260 --> 00:04:38.240
to the depth of the potential.
00:04:38.240 --> 00:04:41.260
And this is your
famous constant z0.
00:04:41.260 --> 00:04:52.500
So e over v0 is actually
equal to Xi over z0 squared.
00:04:52.500 --> 00:04:54.650
And this is something
just to keep in mind.
00:04:54.650 --> 00:04:58.970
If you know Xi, you
certainly must know z0,
00:04:58.970 --> 00:05:01.140
because that's not
in your potential,
00:05:01.140 --> 00:05:04.820
and then you know how
much is the energy.
00:05:04.820 --> 00:05:08.030
All very convenient things.
00:05:08.030 --> 00:05:11.750
So punchline for solutions.
00:05:27.910 --> 00:05:29.080
So what do we have?
00:05:29.080 --> 00:05:31.170
We have two equations.
00:05:31.170 --> 00:05:33.525
This equation maybe
should be given a number.
00:05:39.110 --> 00:05:45.351
Xi equals 8 at tan Eta and Eta
and Xi squared giving you z
00:05:45.351 --> 00:05:45.850
squared.
00:05:45.850 --> 00:05:47.690
So how do we solve it?
00:05:47.690 --> 00:05:49.950
We solve it graphically.
00:05:49.950 --> 00:05:55.120
We have Psi, Eta,
and then we say,
00:05:55.120 --> 00:05:59.890
oh, let's try to plot
the two equations.
00:05:59.890 --> 00:06:01.820
Well this is a circle.
00:06:01.820 --> 00:06:03.440
Eta squared plus Psi squared.
00:06:03.440 --> 00:06:06.630
Now Xi and Eta must
be positive, so we
00:06:06.630 --> 00:06:09.960
look at solutions
just in this quadrant.
00:06:09.960 --> 00:06:13.860
Let's put here pi over 2.
00:06:13.860 --> 00:06:25.850
Pi 3 pi over 2, 2 pi, and
here is Eta and there is Xi.
00:06:25.850 --> 00:06:29.800
Well this is a circle, as we
said, but let's look at this.
00:06:29.800 --> 00:06:34.335
Xi equals Eta tan Eta.
00:06:34.335 --> 00:06:37.885
That vanished as Eta goes
to 0 and will diverge
00:06:37.885 --> 00:06:40.990
at Eta equals pi over 2.
00:06:40.990 --> 00:06:44.370
So this part, at
least, looks like this.
00:06:49.430 --> 00:06:52.640
And then it will go
negative, which don't care,
00:06:52.640 --> 00:06:59.670
from this region, and
then reach here at pi.
00:07:04.380 --> 00:07:08.370
And after pi, it will
go positive again
00:07:08.370 --> 00:07:14.050
and it will reach
another infinity here.
00:07:14.050 --> 00:07:21.960
And then at 3 pi, at 2 pi,
it will go again and reach
00:07:21.960 --> 00:07:25.090
not another infinity like that.
00:07:25.090 --> 00:07:27.770
So these are these curves.
00:07:27.770 --> 00:07:33.970
And the other curve, the
circle, is just a circle here.
00:07:33.970 --> 00:07:40.190
So, for example, I could
have a circle like this.
00:07:40.190 --> 00:07:45.990
So the radius of this
circle is radius z0.
00:07:48.750 --> 00:07:51.600
And there you go, you've
solved the problem.
00:07:51.600 --> 00:07:55.260
At least intuitively
you know the answer.
00:07:55.260 --> 00:07:59.490
And there's a lot of things that
come out of this calculation.
00:07:59.490 --> 00:08:05.680
If the radius z0 is 3
pi over 2, for example,
00:08:05.680 --> 00:08:09.500
and the radius z0 represents
some potential of some depth
00:08:09.500 --> 00:08:13.990
and width, there will
be just two solutions.
00:08:13.990 --> 00:08:16.620
These are these solutions.
00:08:16.620 --> 00:08:21.420
These points represent values
of Xi and values of Eta,
00:08:21.420 --> 00:08:23.470
from which you could
read the energy.
00:08:23.470 --> 00:08:27.570
In fact, you can look
at that state and say,
00:08:27.570 --> 00:08:32.549
that's the state of
largest Xi, and therefore
00:08:32.549 --> 00:08:36.990
it's the state of the largest
absolute value of the energy.
00:08:36.990 --> 00:08:39.849
It's the most
deeply bound state.
00:08:42.520 --> 00:08:44.820
Then this is next
deeply bound state.
00:08:44.820 --> 00:08:48.440
So there's two bounce
states in this case.
00:08:48.440 --> 00:08:54.110
Interestingly, however shallow
this potential might be,
00:08:54.110 --> 00:08:59.720
however small, z0, the
circle, will always
00:08:59.720 --> 00:09:02.270
have one intersection,
so there will always
00:09:02.270 --> 00:09:05.870
be at least one solution.
00:09:05.870 --> 00:09:07.790
That's the end of that story.
00:09:07.790 --> 00:09:13.100
Let me say that for the
odd case, odd solutions,
00:09:13.100 --> 00:09:14.330
I will not solve it.
00:09:14.330 --> 00:09:17.450
It's a good thing to do
in recitation or as part
00:09:17.450 --> 00:09:20.690
of the home work as well.
00:09:20.690 --> 00:09:28.360
The answer for the odd case is
that Psi is equal to minus Eta
00:09:28.360 --> 00:09:32.150
[? Cot ?] Eta.
00:09:32.150 --> 00:09:36.800
And in that case, I'll give
you a little preview of how
00:09:36.800 --> 00:09:40.040
the this [? Cot ?] looks.
00:09:40.040 --> 00:09:41.560
It looks like this.
00:09:41.560 --> 00:09:46.310
And then there are more
branches of this thing.
00:09:46.310 --> 00:09:51.110
So for the odd solution,
you have these curves.
00:09:51.110 --> 00:09:53.420
And if you have a
circle, sometimes you
00:09:53.420 --> 00:09:55.070
don't have a solution.
00:09:55.070 --> 00:09:57.320
It doesn't intersect this.
00:09:57.320 --> 00:10:02.060
So these odd solutions, you
will see and try to understand,
00:10:02.060 --> 00:10:04.220
they don't always exist.
00:10:04.220 --> 00:10:08.040
You meet a potential
that is sufficiently deep
00:10:08.040 --> 00:10:10.590
to get an odd solution.
00:10:10.590 --> 00:10:12.830
And then the odds
and even solutions
00:10:12.830 --> 00:10:16.070
will interweave
each other and there
00:10:16.070 --> 00:10:18.290
will be a nice
story that you will
00:10:18.290 --> 00:10:20.220
explore in a lot of detail.
00:10:20.220 --> 00:10:25.700
But the is, you've reviewed
the problem to unit free
00:10:25.700 --> 00:10:29.080
calculation, in which you
can get the intuition of when
00:10:29.080 --> 00:10:32.320
solutions exist and
and when they don't.
00:10:32.320 --> 00:10:34.890
But solving for the
particular numbers
00:10:34.890 --> 00:10:37.470
are transcendental
equations, and you
00:10:37.470 --> 00:10:40.130
need a computer to solve.