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00:00:00,499 --> 00:00:03,330
PROFESSOR: So, what is the
wave function that we have?

2
00:00:03,330 --> 00:00:05,280
We must have a wave
function now that

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00:00:05,280 --> 00:00:11,190
is symmetric, and built
with e to the k x, kappa x,

4
00:00:11,190 --> 00:00:12,940
and into the minus kappa x.

5
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This is the only possibility.

6
00:00:15,270 --> 00:00:22,850
E to the minus of kappa
absolute value of x.

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00:00:22,850 --> 00:00:26,830
This is psi of x for
x different from 0.

8
00:00:26,830 --> 00:00:30,310
This is-- as you
can quickly see--

9
00:00:30,310 --> 00:00:36,060
this is e to get minus
kappa x, when x is positive,

10
00:00:36,060 --> 00:00:41,480
A e to the kappa x,
when x is negative.

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00:00:41,480 --> 00:00:44,060
And, both of them decay.

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The first exponential negative
is the standard decaying

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00:00:47,690 --> 00:00:49,700
exponential to the right.

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00:00:49,700 --> 00:00:52,360
The one with
positive-- well, here x

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00:00:52,360 --> 00:00:55,130
is negative as you go
all the way to the left.

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00:00:55,130 --> 00:00:57,590
This one decays case as well.

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00:00:57,590 --> 00:01:03,260
And, this thing plotted
is a decaying exponential

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00:01:03,260 --> 00:01:06,380
with amplitude A, like that.

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And, a decaying exponential with
amplitude A, and a singularity

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there, which is what
you would have expected.

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So, this seems to be
on the right track--

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it's a continuous wave function.

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00:01:21,170 --> 00:01:24,470
The wave function cannot
fail to be continuous,

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that's a complete disaster to
show that an equation could not

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00:01:28,220 --> 00:01:29,850
be satisfied.

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00:01:29,850 --> 00:01:34,550
So, this is our
discontinuous wave function.

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00:01:34,550 --> 00:01:39,350
So, at this moment you
really haven't yet used

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the delta function-- the delta
function with intensity alpha

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down.

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I should have made a comment
that it's very nice that alpha

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appeared here in the numerator.

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00:01:53,110 --> 00:01:55,630
If it would have appeared
in the denominator,

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I would be telling you that
I think this problem is not

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00:01:59,020 --> 00:02:01,740
going to have a solution.

35
00:02:01,740 --> 00:02:02,850
Why?

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Because if it appears
in the numerator,

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it means that as the delta
function potential is becoming

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stronger and stronger,
the bound state

39
00:02:12,780 --> 00:02:14,610
is getting deeper and deeper--

40
00:02:14,610 --> 00:02:17,240
which is what you would expect.

41
00:02:17,240 --> 00:02:19,110
But, if it would be
in the numerator--

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00:02:19,110 --> 00:02:21,370
in the denominator--

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00:02:21,370 --> 00:02:24,090
as the potential gets
deeper and deeper,

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the boundary is going up.

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That makes no sense whatsoever.

46
00:02:29,270 --> 00:02:32,700
So, it's good that
it appeared there,

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00:02:32,700 --> 00:02:37,050
it's a sign that things are
in reasonable conditions.

48
00:02:37,050 --> 00:02:40,730
So, now we really have to
face the delta function.

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00:02:40,730 --> 00:02:43,470
And, this is a procedure you
are going to do many times

50
00:02:43,470 --> 00:02:44,200
in this course.

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00:02:44,200 --> 00:02:47,940
So, look at it, and
do it again and again

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00:02:47,940 --> 00:02:50,430
until you're very
comfortable with it.

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00:02:50,430 --> 00:02:56,730
It's the issue of discovering
what kind of discontinuity

54
00:02:56,730 --> 00:02:59,280
you can have with
the delta function.

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00:02:59,280 --> 00:03:01,740
And, it's a discontinuity
in the derivative,

56
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so let's quantify it.

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00:03:04,050 --> 00:03:07,211
So, here it is-- we begin
with the Schrodinger equation,

58
00:03:07,211 --> 00:03:07,710
again.

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00:03:11,590 --> 00:03:17,330
But, I will write now the
potential term as well.

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00:03:17,330 --> 00:03:27,810
The potential is plus v of x
psi of x equals E psi of x.

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00:03:27,810 --> 00:03:33,000
And the idea is to
integrate this equation

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from minus epsilon to epsilon.

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00:03:39,450 --> 00:03:43,830
And, epsilon is supposed to
be a small positive number.

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So, you integrate
from minus epsilon

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00:03:48,000 --> 00:03:51,440
to epsilon the
differential equation,

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and see what it does to you in
the limit as epsilon goes to 0.

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That's what we're
going to try to do.

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00:04:00,300 --> 00:04:01,830
So, what do we get?

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00:04:01,830 --> 00:04:07,820
If you integrate this, you
get minus h squared over 2 m.

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00:04:07,820 --> 00:04:11,790
And now, you have to integrate
the second derivative

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00:04:11,790 --> 00:04:14,940
with respect to x, which
is the first derivative,

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00:04:14,940 --> 00:04:20,550
and therefore this is
the first derivative

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00:04:20,550 --> 00:04:29,740
at x equal epsilon minus the
first derivative at x equals

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00:04:29,740 --> 00:04:30,700
minus epsilon.

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00:04:33,730 --> 00:04:38,050
This is from the first
term, because you integrate

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00:04:38,050 --> 00:04:44,200
d x d second d x squared
psi is the same thing

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00:04:44,200 --> 00:04:52,630
as d x d d x of d psi
d x between A and B.

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00:04:52,630 --> 00:04:56,150
And, the integral of a
total derivative is d

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00:04:56,150 --> 00:05:00,430
psi d x at B A--

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00:05:00,430 --> 00:05:02,440
I think people
write it like this--

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00:05:02,440 --> 00:05:07,280
A to B. Evaluate it at the
top, minus the evaluation

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00:05:07,280 --> 00:05:10,030
at the bottom.

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00:05:10,030 --> 00:05:18,050
Now, the next term is the
integral of psi times v of x.

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00:05:18,050 --> 00:05:20,790
So, I'll write it
plus the integral

85
00:05:20,790 --> 00:05:29,515
from minus epsilon to epsilon
d x minus alpha delta of x psi

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of x--

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00:05:31,390 --> 00:05:32,830
that's the potential.

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00:05:32,830 --> 00:05:35,796
Now, we use the delta function.

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00:05:35,796 --> 00:05:41,490
And, on the right hand
side this will be E times

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00:05:41,490 --> 00:05:50,970
the integral from minus epsilon
to epsilon of psi of x d x.

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00:05:50,970 --> 00:05:56,400
So, that's the differential
equation integrated.

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And now, we're going
to do two things.

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We're going to do some
of these integrals,

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and take the limit
as epsilon goes to 0.

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00:06:08,230 --> 00:06:17,010
So, I'll write this minus
h squared over 2 m limit

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00:06:17,010 --> 00:06:26,870
as epsilon goes to 0 of d psi
d x at epsilon minus d psi d x

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00:06:26,870 --> 00:06:31,660
at minus epsilon plus.

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00:06:31,660 --> 00:06:33,690
Let's think of this integral.

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00:06:33,690 --> 00:06:37,690
We can do this integral,
it's a delta function.

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00:06:37,690 --> 00:06:41,970
So, it picks the value of
the wave function at 0,

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because 0 is inside the
interval of integration.

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That's why we integrate it
from minus epsilon to epsilon,

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00:06:48,480 --> 00:06:51,060
to have the delta
function inside.

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So, you get an alpha
out, a psi of 0,

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and that's what
this integral is.

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It's independent of
the value of epsilon

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00:07:00,960 --> 00:07:03,510
as long as epsilon
is different from 0.

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So, this gives you
minus alpha psi of 0.

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And now, the last
term is an integral

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from minus epsilon to
epsilon of the wave function.

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Now, the wave function
is continuous--

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it should be continuous--

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that means it's finite.

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And, this integral,
as of any function

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that is not divergent from
minus epsilon to epsilon

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as epsilon goes to 0, is 0.

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Any integral of a function
that doesn't diverge

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as the limits of
integration go to 0,

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the area under
the function is 0.

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So, this is 0--

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the limit.

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And this thing goes to
0, so we put a 0 here.

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So, at this moment we got
really what we wanted.

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I'll write it this way.

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I'll go here, and I'll say
minus h squared over 2 m,

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and what is this?

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This expression says,
calculate the derivative

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00:08:31,530 --> 00:08:37,059
of the function a little
bit to the right of 0,

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00:08:37,059 --> 00:08:39,970
and subtract the derivative
of the function a little bit

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00:08:39,970 --> 00:08:42,030
to the left of 0.

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00:08:42,030 --> 00:08:48,150
This is nothing but the
discontinuity in psi prime.

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00:08:48,150 --> 00:08:52,080
You're evaluating for any
epsilon greater than 0--

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the psi prime a little to
right, a little too the left,

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and taking the difference.

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So, this is what we should call
the discontinuity delta at 0--

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00:09:04,100 --> 00:09:06,190
at x equals 0.

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00:09:06,190 --> 00:09:15,900
And, this and this is for
discontinuity of psi prime at x

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00:09:15,900 --> 00:09:23,130
equals 0 minus alpha
psi of 0 equals 0.

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00:09:23,130 --> 00:09:30,550
And from here, we discover
that delta zero psi prime

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is equal to minus 2 m alpha
over h squared psi of 0.

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00:09:42,900 --> 00:09:45,720
This is the
discontinuity condition

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00:09:45,720 --> 00:09:47,475
produced by the delta function.

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00:09:52,030 --> 00:09:59,120
This whole quantity is what
we call delta 0 of psi prime.

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00:10:03,670 --> 00:10:10,280
And, what it says is that
yes, the wave function

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00:10:10,280 --> 00:10:14,350
can have a discontinuous
first derivative if the wave

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00:10:14,350 --> 00:10:17,140
function doesn't vanish there.

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00:10:17,140 --> 00:10:20,485
Once the wave function
doesn't vanish at that point,

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00:10:20,485 --> 00:10:22,600
the discontinuity
is in fact even

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00:10:22,600 --> 00:10:26,575
proportional to the value of
the wave function at that point.

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00:10:26,575 --> 00:10:31,320
And, here are the constants
of proportionality.

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00:10:31,320 --> 00:10:34,580
Now, I don't think it's worth
to memorize this equation

152
00:10:34,580 --> 00:10:37,460
or anything like that,
because it basically

153
00:10:37,460 --> 00:10:40,640
can be derived in a few lines.

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00:10:40,640 --> 00:10:45,110
This may have looked like
an interesting or somewhat

155
00:10:45,110 --> 00:10:47,450
intricate derivation,
but after you've done

156
00:10:47,450 --> 00:10:49,340
is a couple of times--
this is something

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00:10:49,340 --> 00:10:52,250
you'll do in a minute or so.

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00:10:52,250 --> 00:10:56,060
And, you just integrate
and find the discontinuity

159
00:10:56,060 --> 00:10:59,150
in the derivative--
that's a formula there.

160
00:10:59,150 --> 00:11:05,920
And, that's a formula
for a potential,

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00:11:05,920 --> 00:11:08,410
minus alpha delta of x.

162
00:11:08,410 --> 00:11:10,720
So, if somebody gives you
a different potential,

163
00:11:10,720 --> 00:11:14,650
well, you have to change
the alpha accordingly.

164
00:11:14,650 --> 00:11:17,800
So, let's wrap this up.

165
00:11:17,800 --> 00:11:19,500
So, we go to our case.

166
00:11:19,500 --> 00:11:22,870
Here is our situation.

167
00:11:22,870 --> 00:11:24,880
So, let's apply this.

168
00:11:24,880 --> 00:11:25,980
So, what is the value?

169
00:11:25,980 --> 00:11:31,350
Apply this equation
to our wave function.

170
00:11:31,350 --> 00:11:35,100
So, what is the
derivative at epsilon?

171
00:11:35,100 --> 00:11:41,730
It's minus kappa A E to
the minus kappa epsilon.

172
00:11:41,730 --> 00:11:46,020
That's the derivative of
psi on the positive side.

173
00:11:46,020 --> 00:11:49,380
I differentiated the top
line of this equation

174
00:11:49,380 --> 00:11:55,230
minus the derivative
on the left side--

175
00:11:55,230 --> 00:11:57,070
this one, the derivative.

176
00:11:57,070 --> 00:12:04,960
So, this is kappa A E
to the kappa epsilon--

177
00:12:04,960 --> 00:12:08,350
no, kappa minus epsilon again.

178
00:12:11,980 --> 00:12:13,750
So, that's the left hand side.

179
00:12:13,750 --> 00:12:21,250
The right hand side would be
minus 2 m alpha h squared psi

180
00:12:21,250 --> 00:12:22,270
at 0.

181
00:12:22,270 --> 00:12:27,550
Psi at 0 is A, so
that's what it gives us.

182
00:12:27,550 --> 00:12:30,460
And we should take the
limit as epsilon goes to 0.

183
00:12:30,460 --> 00:12:37,020
So, this is going
to 1, both of them.

184
00:12:37,020 --> 00:12:43,110
So, the left hand side
is minus 2 kappa A,

185
00:12:43,110 --> 00:12:55,100
and the right hand side is
2 m alpha over h squared A.

186
00:12:55,100 --> 00:12:56,480
So, the 2--

187
00:12:56,480 --> 00:13:00,270
it's also minus, I'm sorry--

188
00:13:00,270 --> 00:13:04,170
so the 2s cancel,
the A cancels--

189
00:13:04,170 --> 00:13:07,830
you never should have expected
to determine A unless you tried

190
00:13:07,830 --> 00:13:10,020
to normalize the wave function.

191
00:13:10,020 --> 00:13:12,260
Solving for energy
eigenstates states will never

192
00:13:12,260 --> 00:13:15,920
determine A. The Schrodinger
equation is linear,

193
00:13:15,920 --> 00:13:21,150
so A drops out, the
minus 2 drops out,

194
00:13:21,150 --> 00:13:27,530
and kappa is equal to
m alpha over h squared.

195
00:13:31,390 --> 00:13:35,910
So, that said that's great
because kappa is just

196
00:13:35,910 --> 00:13:40,090
another name for the energy.

197
00:13:40,090 --> 00:13:42,540
So, I have kappa m
alpha over h squared,

198
00:13:42,540 --> 00:13:44,520
so that's another
name for the energy.

199
00:13:44,520 --> 00:13:47,530
So, let's go to the energy.

200
00:13:47,530 --> 00:13:54,870
The energy is h bar
squared kappa minus h bar

201
00:13:54,870 --> 00:13:59,380
squared kappa squared over 2 m.

202
00:13:59,380 --> 00:14:02,980
So, it's minus h bar squared.

203
00:14:02,980 --> 00:14:10,480
Kappa squared would be m squared
alpha squared h to the fourth,

204
00:14:10,480 --> 00:14:14,480
and there's a two m.

205
00:14:14,480 --> 00:14:15,920
All these constants.

206
00:14:15,920 --> 00:14:23,700
So, final answer.

207
00:14:23,700 --> 00:14:35,220
E, the bound state energy is
minus m alpha squared minus m

208
00:14:35,220 --> 00:14:38,144
alpha squared.

209
00:14:38,144 --> 00:14:43,750
The m cancels it over h
squared minus one half.

210
00:14:46,750 --> 00:14:53,020
So, back here the units worked
out, everything is good,

211
00:14:53,020 --> 00:14:56,756
and the number was
determined as minus one half.

212
00:14:56,756 --> 00:15:01,850
That's your bound state
energy for this problem.

213
00:15:01,850 --> 00:15:05,450
So, this problem is instructive
because you basically

214
00:15:05,450 --> 00:15:09,170
learn that in delta functions,
with one delta function

215
00:15:09,170 --> 00:15:10,580
you get a bound state.

216
00:15:10,580 --> 00:15:14,900
If you have two delta functions,
you may get more bound states--

217
00:15:14,900 --> 00:15:18,020
three, four-- people
study those problems,

218
00:15:18,020 --> 00:15:21,880
and you will investigate the
two delta function cases.