1
00:00:11,160 --> 00:00:18,119
welcome back to the lecture on the
2
00:00:13,830 --> 00:00:24,480
hydrogen atom in the last lecture we
3
00:00:18,119 --> 00:00:27,150
left at the point of the Schrodinger
4
00:00:24,480 --> 00:00:29,460
equation being written down using
5
00:00:27,150 --> 00:00:33,059
spherical polar coordinates for the
6
00:00:29,460 --> 00:00:35,280
hydrogen atom in this brief segment I
7
00:00:33,059 --> 00:00:38,489
shall tell you how the equation is
8
00:00:35,280 --> 00:00:41,040
separated into three component equations
9
00:00:38,489 --> 00:00:44,870
for the three variables that we proposed
10
00:00:41,040 --> 00:00:48,719
the radial coordinate the theta
11
00:00:44,870 --> 00:00:51,270
coordinate of the angular part and the
12
00:00:48,719 --> 00:00:55,200
Phi coordinate of the angular part as
13
00:00:51,270 --> 00:00:57,090
well the Phi coordinate solution will be
14
00:00:55,200 --> 00:00:59,899
identified immediately with the
15
00:00:57,090 --> 00:01:03,680
solutions of the particle in a ring and
16
00:00:59,899 --> 00:01:07,860
the theta coordinate will become the
17
00:01:03,680 --> 00:01:11,490
solutions earlier known in mathematics
18
00:01:07,860 --> 00:01:15,080
literature as due to Associated Legendre
19
00:01:11,490 --> 00:01:18,360
polynomials the radial part will be
20
00:01:15,080 --> 00:01:21,209
identified with laguerre polynomials and
21
00:01:18,360 --> 00:01:24,300
the hydrogen atom is a very good example
22
00:01:21,209 --> 00:01:27,750
of taking the mathematics to a slightly
23
00:01:24,300 --> 00:01:30,300
more rigorous level and showing that the
24
00:01:27,750 --> 00:01:34,289
analytic solutions for this particular
25
00:01:30,300 --> 00:01:37,259
real problem exists surprisingly that's
26
00:01:34,289 --> 00:01:40,619
it beyond this point all solutions
27
00:01:37,259 --> 00:01:44,670
become approximate so let's recap the
28
00:01:40,619 --> 00:01:47,610
equation the overall equation is
29
00:01:44,670 --> 00:01:49,200
displayed here from the last lecture
30
00:01:47,610 --> 00:01:53,759
this was the last part of the last
31
00:01:49,200 --> 00:01:56,099
lecture now you see that there is the
32
00:01:53,759 --> 00:01:57,780
radial derivative then there is the
33
00:01:56,099 --> 00:02:01,259
angular derivative and the Phi
34
00:01:57,780 --> 00:02:07,200
derivative first let me clarify a couple
35
00:02:01,259 --> 00:02:13,220
of notations here when you write dou by
36
00:02:07,200 --> 00:02:18,959
dou R or square dou by dou or say or
37
00:02:13,220 --> 00:02:22,740
theta fee what it means is a sum of two
38
00:02:18,959 --> 00:02:24,180
terms namely derivative with respect to
39
00:02:22,740 --> 00:02:24,720
order and derivative with respect to the
40
00:02:24,180 --> 00:02:27,680
first
41
00:02:24,720 --> 00:02:31,680
derivative therefore you have our square
42
00:02:27,680 --> 00:02:34,980
dou square by dou R square sigh the
43
00:02:31,680 --> 00:02:40,100
partial derivative of Y with respect to
44
00:02:34,980 --> 00:02:46,650
R and then the other term namely to or
45
00:02:40,100 --> 00:02:53,520
though sigh or theta phi by dou R that's
46
00:02:46,650 --> 00:02:57,090
what is meant by writing in a compact
47
00:02:53,520 --> 00:03:01,160
notation like this and second when you
48
00:02:57,090 --> 00:03:06,720
write one by sine theta in a similar way
49
00:03:01,160 --> 00:03:14,430
though by dou theta of sine theta no
50
00:03:06,720 --> 00:03:17,520
sigh no theta of or theta and fee this
51
00:03:14,430 --> 00:03:20,550
again means two terms namely the sine
52
00:03:17,520 --> 00:03:22,610
theta derivative being a cot theta here
53
00:03:20,550 --> 00:03:28,170
because it's cos theta by sine theta
54
00:03:22,610 --> 00:03:30,630
then you have those I by dou theta plus
55
00:03:28,170 --> 00:03:33,840
sine theta cancels when you don't take
56
00:03:30,630 --> 00:03:39,959
the derivative it is dou Square t by dou
57
00:03:33,840 --> 00:03:42,360
theta square so one must keep in mind
58
00:03:39,959 --> 00:03:46,459
that this is what is meant by writing
59
00:03:42,360 --> 00:03:49,970
derivatives in bracket ball okay now
60
00:03:46,459 --> 00:03:54,090
given this particular form of sigh or
61
00:03:49,970 --> 00:03:56,959
theta V and given the form of the
62
00:03:54,090 --> 00:04:00,030
differential equation our purpose was to
63
00:03:56,959 --> 00:04:03,299
solve this equation by separating the
64
00:04:00,030 --> 00:04:07,980
size into independent coordinate
65
00:04:03,299 --> 00:04:12,470
dependent functions namely writing sigh
66
00:04:07,980 --> 00:04:12,470
theta v function
67
00:04:14,730 --> 00:04:21,989
as the product of three functions a
68
00:04:17,880 --> 00:04:27,600
function of radial part only a function
69
00:04:21,989 --> 00:04:33,260
of theta coordinate only and a function
70
00:04:27,600 --> 00:04:35,940
of y coordinate only this separation is
71
00:04:33,260 --> 00:04:38,250
possible because of the particular form
72
00:04:35,940 --> 00:04:40,560
of the hydrogen atom equation namely
73
00:04:38,250 --> 00:04:42,660
that the potential energy is only
74
00:04:40,560 --> 00:04:45,150
dependent on the radial coordinates and
75
00:04:42,660 --> 00:04:48,990
therefore if you look at this particular
76
00:04:45,150 --> 00:04:57,000
equation here the radial terms that you
77
00:04:48,990 --> 00:05:04,770
have here the radial terms this and this
78
00:04:57,000 --> 00:05:06,389
and this will be separated out when you
79
00:05:04,770 --> 00:05:09,360
multiply the whole equation by r square
80
00:05:06,389 --> 00:05:11,820
you would see that these are the only
81
00:05:09,360 --> 00:05:13,800
terms which will depend on r and the
82
00:05:11,820 --> 00:05:15,360
other term will have the R square
83
00:05:13,800 --> 00:05:18,000
removed or square removed so they will
84
00:05:15,360 --> 00:05:19,590
depend on theta and fee therefore you
85
00:05:18,000 --> 00:05:23,070
will have a differential equation in
86
00:05:19,590 --> 00:05:24,960
which one part of the equation depends
87
00:05:23,070 --> 00:05:26,099
only on one coordinate the other part
88
00:05:24,960 --> 00:05:29,160
depends only and the other two
89
00:05:26,099 --> 00:05:32,280
coordinates and then you can immediately
90
00:05:29,160 --> 00:05:34,260
realize that these two independent
91
00:05:32,280 --> 00:05:36,680
quantities must be separately equal to a
92
00:05:34,260 --> 00:05:38,729
constant which will cancel each other
93
00:05:36,680 --> 00:05:41,190
therefore it's possible to separate this
94
00:05:38,729 --> 00:05:45,539
equation into independent coordinates so
95
00:05:41,190 --> 00:05:50,570
let's multiply the differential equation
96
00:05:45,539 --> 00:05:50,570
by or R squared
97
00:05:53,380 --> 00:06:17,170
d by r square and also divided the D by
98
00:06:06,910 --> 00:06:17,170
or of or theta of theta and Phi of Phi
99
00:06:17,800 --> 00:06:25,940
when you do that the resulting equation
100
00:06:23,260 --> 00:06:31,030
for the radial part in the angular part
101
00:06:25,940 --> 00:06:40,540
take this form minus H bar square by 2 m
102
00:06:31,030 --> 00:06:44,350
d by d r of r square b or by D or and
103
00:06:40,540 --> 00:06:48,500
since we have divided everything by the
104
00:06:44,350 --> 00:06:51,410
wave function itself you will have one x
105
00:06:48,500 --> 00:06:56,240
or because the theta fee will be
106
00:06:51,410 --> 00:07:06,830
cancelled likewise you have one x sine
107
00:06:56,240 --> 00:07:10,580
theta so let me do the following it is
108
00:07:06,830 --> 00:07:13,760
that and minus H bar square by 2 m is
109
00:07:10,580 --> 00:07:22,810
common to both so you have one x sine
110
00:07:13,760 --> 00:07:28,550
theta D by D theta of sine theta D theta
111
00:07:22,810 --> 00:07:33,200
by D theta and again this term will be x
112
00:07:28,550 --> 00:07:37,310
1 x of the stata function then you have
113
00:07:33,200 --> 00:07:45,380
the last term for the kinetic energy one
114
00:07:37,310 --> 00:07:51,470
x sine square theta d square phi by d
115
00:07:45,380 --> 00:07:56,030
phi square x 1 x 5 okay so this will be
116
00:07:51,470 --> 00:07:58,669
the radial part angular part of the
117
00:07:56,030 --> 00:08:02,960
kinetic energy term and then the
118
00:07:58,669 --> 00:08:04,760
remaining namely minus ze square or
119
00:08:02,960 --> 00:08:07,000
because we are multiplied everything by
120
00:08:04,760 --> 00:08:10,150
r square by 4 pi
121
00:08:07,000 --> 00:08:11,740
or not and will not have any function
122
00:08:10,150 --> 00:08:16,740
here because that has been divided out
123
00:08:11,740 --> 00:08:24,570
and also what is left over is the e or
124
00:08:16,740 --> 00:08:29,560
square is equal to zero okay so this is
125
00:08:24,570 --> 00:08:34,380
the form after doing the separation the
126
00:08:29,560 --> 00:08:38,550
division and then removing the the parts
127
00:08:34,380 --> 00:08:48,540
independently so what you have here or
128
00:08:38,550 --> 00:08:53,860
the radial part given by this term and
129
00:08:48,540 --> 00:08:57,220
these terms and everything else does not
130
00:08:53,860 --> 00:08:58,710
depend on radius the coordinate art but
131
00:08:57,220 --> 00:09:02,050
it ends only on the theta and fee
132
00:08:58,710 --> 00:09:04,600
therefore it's straightforward for you
133
00:09:02,050 --> 00:09:10,470
to write this as equal to some constant
134
00:09:04,600 --> 00:09:17,589
C in which case the remaining term this
135
00:09:10,470 --> 00:09:24,010
and this will be equal to minus C so
136
00:09:17,589 --> 00:09:26,680
that the sum of this is zero okay so we
137
00:09:24,010 --> 00:09:32,440
have two equations one for the radial
138
00:09:26,680 --> 00:09:39,280
equation minus H bar square by 2 m d by
139
00:09:32,440 --> 00:09:43,720
dr or square d capital r by b small r /
140
00:09:39,280 --> 00:09:49,120
r plus slow that's the kinetic energy
141
00:09:43,720 --> 00:09:51,670
term and then you have the potential
142
00:09:49,120 --> 00:09:58,300
energy contribution minus the d square
143
00:09:51,670 --> 00:10:03,400
or by 4 pi epsilon naught minus E or
144
00:09:58,300 --> 00:10:07,120
square minus C is equal to 0 because the
145
00:10:03,400 --> 00:10:09,010
salting is seeing and if you write it in
146
00:10:07,120 --> 00:10:11,830
the standard form you can now get rid of
147
00:10:09,010 --> 00:10:16,240
this or and multiply everything by the
148
00:10:11,830 --> 00:10:20,010
or so you have the or here and see this
149
00:10:16,240 --> 00:10:20,010
is the radial equation
150
00:10:25,510 --> 00:10:33,070
and the angular equation will be
151
00:10:28,870 --> 00:10:41,430
whatever is left over namely minus H bar
152
00:10:33,070 --> 00:10:46,470
square by 2 m sine theta 1 by sine theta
153
00:10:41,430 --> 00:10:46,470
sorry ordinary yes 1 by sine theta
154
00:10:50,300 --> 00:11:13,280
B by D theta what you have here is sine
155
00:11:01,970 --> 00:11:16,160
theta D theta by D theta plus 1 by sine
156
00:11:13,280 --> 00:11:19,610
square theta okay that's also one by
157
00:11:16,160 --> 00:11:26,510
capital of theta here 1 by sine square
158
00:11:19,610 --> 00:11:30,590
theta d squared phi by d phi square x 1
159
00:11:26,510 --> 00:11:36,200
by phi e all of this is equal to minus C
160
00:11:30,590 --> 00:11:40,430
therefore plus C is equal to 0 okay now
161
00:11:36,200 --> 00:11:44,840
this equation again can be separated
162
00:11:40,430 --> 00:11:48,100
into theta dependent part only and Phi
163
00:11:44,840 --> 00:11:48,100
dependent part only
164
00:11:54,089 --> 00:11:58,110
if you multiplied by sine squared theta
165
00:11:56,069 --> 00:12:00,779
the whole equation you will get c sine
166
00:11:58,110 --> 00:12:02,759
square theta and then the first term
167
00:12:00,779 --> 00:12:05,730
will contain all the theta dependence
168
00:12:02,759 --> 00:12:07,350
turn the second one will not have this
169
00:12:05,730 --> 00:12:10,680
theta dependent form it will be only
170
00:12:07,350 --> 00:12:15,930
five dependent part therefore when you
171
00:12:10,680 --> 00:12:26,519
multiply this by sine square theta
172
00:12:15,930 --> 00:12:33,749
throughout and equate the term d square
173
00:12:26,519 --> 00:12:37,740
Phi by D Phi square x 1 by Phi to some
174
00:12:33,749 --> 00:12:39,480
constant which by recognition of the
175
00:12:37,740 --> 00:12:44,459
particle in the ring problem we would
176
00:12:39,480 --> 00:12:46,800
equate that to a constant then the other
177
00:12:44,459 --> 00:12:49,110
term will depend on plus M square will
178
00:12:46,800 --> 00:12:52,910
be equal to plus M square so this is the
179
00:12:49,110 --> 00:12:52,910
Phi dependent equation
180
00:12:57,460 --> 00:13:03,500
and what you will have is for the theta
181
00:13:01,130 --> 00:13:10,190
dependent form or there is also a minus
182
00:13:03,500 --> 00:13:15,830
H bar square by 2 m here okay and then
183
00:13:10,190 --> 00:13:20,360
you have the theta dependent form which
184
00:13:15,830 --> 00:13:28,280
is minus H bar square by 2 m sine theta
185
00:13:20,360 --> 00:13:35,060
D by D theta of sine theta D theta by D
186
00:13:28,280 --> 00:13:39,730
theta and if we do the algebra carefully
187
00:13:35,060 --> 00:13:43,390
it will be plus C sine square theta
188
00:13:39,730 --> 00:13:43,390
times theta
189
00:13:53,390 --> 00:14:02,520
okay so that would be the C sine square
190
00:14:00,240 --> 00:14:06,720
theta minus M square theta is equal to
191
00:14:02,520 --> 00:14:09,440
zero so this would be the theta
192
00:14:06,720 --> 00:14:18,690
dependent equation and this would be the
193
00:14:09,440 --> 00:14:21,300
findin taqua shin okay so we are in a
194
00:14:18,690 --> 00:14:26,190
position to solve each one of them
195
00:14:21,300 --> 00:14:29,670
separately and obtain the formal answer
196
00:14:26,190 --> 00:14:32,910
the analytic solutions for these three
197
00:14:29,670 --> 00:14:34,680
quantities so what you do is when you
198
00:14:32,910 --> 00:14:36,570
solve these equations which I will not
199
00:14:34,680 --> 00:14:38,610
describe here when you solve this
200
00:14:36,570 --> 00:14:42,930
equation you will get a radial function
201
00:14:38,610 --> 00:14:46,500
you will get an angular function and you
202
00:14:42,930 --> 00:14:49,050
will get a Phi which is also and part of
203
00:14:46,500 --> 00:14:53,340
the angular function the product of the
204
00:14:49,050 --> 00:14:55,290
two together is known in not only
205
00:14:53,340 --> 00:14:58,650
hydrogen atom but in general for such
206
00:14:55,290 --> 00:14:59,790
equations it's known as spherical the
207
00:14:58,650 --> 00:15:05,970
solutions are known as spherical
208
00:14:59,790 --> 00:15:08,340
harmonics the radial function will
209
00:15:05,970 --> 00:15:11,180
contain what I've known as the log air
210
00:15:08,340 --> 00:15:11,180
polynomials
211
00:15:15,300 --> 00:15:21,040
spherical harmonics are constructed
212
00:15:17,830 --> 00:15:26,700
using the five functions and polynomials
213
00:15:21,040 --> 00:15:26,700
known as Associated Legendre polynomials
214
00:15:31,320 --> 00:15:38,610
all these things law gave Associated
215
00:15:34,600 --> 00:15:41,650
Legendre polynomial Bessel functions
216
00:15:38,610 --> 00:15:44,050
hermite functions which we will see in
217
00:15:41,650 --> 00:15:48,390
the solution of the harmonic oscillator
218
00:15:44,050 --> 00:15:51,490
her main function or hermite polynomials
219
00:15:48,390 --> 00:15:53,380
then she'll be shaved polynomial there
220
00:15:51,490 --> 00:16:01,810
are many ways by which this chebyshev is
221
00:15:53,380 --> 00:16:04,540
written chebyshev polynomials and so on
222
00:16:01,810 --> 00:16:06,990
they all form a group of polynomials
223
00:16:04,540 --> 00:16:10,780
well known in mathematics as orthogonal
224
00:16:06,990 --> 00:16:14,350
polynomials and these are important in
225
00:16:10,780 --> 00:16:19,450
the differential equation representation
226
00:16:14,350 --> 00:16:21,400
or a coordinate representation of the
227
00:16:19,450 --> 00:16:24,610
wave function in a suitable coordinate
228
00:16:21,400 --> 00:16:26,820
system and these polynomials are well
229
00:16:24,610 --> 00:16:30,460
known for more than 200 years
230
00:16:26,820 --> 00:16:32,290
Schrodinger saw that his equation mapped
231
00:16:30,460 --> 00:16:34,360
into the differential equations that
232
00:16:32,290 --> 00:16:36,700
were already known and therefore he
233
00:16:34,360 --> 00:16:38,680
immediately put forward the solutions
234
00:16:36,700 --> 00:16:42,270
from those differential equations and
235
00:16:38,680 --> 00:16:45,550
obtain the conditions let me summarize
236
00:16:42,270 --> 00:16:48,910
or let me conclude with the following
237
00:16:45,550 --> 00:16:51,250
statement that the radial function will
238
00:16:48,910 --> 00:16:58,630
depend on to coordinate to quantum
239
00:16:51,250 --> 00:17:01,090
numbers n and yell the the product of
240
00:16:58,630 --> 00:17:03,370
the two together will be called the
241
00:17:01,090 --> 00:17:06,070
spherical harmonics will depend on to
242
00:17:03,370 --> 00:17:09,130
quantum numbers l and young the L will
243
00:17:06,070 --> 00:17:11,860
be the same for both the radial and
244
00:17:09,130 --> 00:17:16,030
angular function for a given energy and
245
00:17:11,860 --> 00:17:18,700
therefore the overall solution will be
246
00:17:16,030 --> 00:17:21,760
the product of the two and that's equal
247
00:17:18,700 --> 00:17:27,670
to say with the three quantum numbers n
248
00:17:21,760 --> 00:17:29,770
L M or theta and fee I am NOT
249
00:17:27,670 --> 00:17:32,230
to describe how to obtain this radial
250
00:17:29,770 --> 00:17:34,210
and the angular parts but in the next
251
00:17:32,230 --> 00:17:36,880
part of this lecture I shall describe
252
00:17:34,210 --> 00:17:38,830
the forms of the radial parts and the
253
00:17:36,880 --> 00:17:40,990
forms of the angular part and we will
254
00:17:38,830 --> 00:17:42,700
see some pictures for the angular parts
255
00:17:40,990 --> 00:17:44,830
which are popularly known as the
256
00:17:42,700 --> 00:17:48,280
representations of the atomic orbitals
257
00:17:44,830 --> 00:17:50,050
in unison them in textbooks both in the
258
00:17:48,280 --> 00:17:52,630
high school and in college textbooks
259
00:17:50,050 --> 00:17:55,390
with the p orbital having two lobes in
260
00:17:52,630 --> 00:17:57,360
the z direction in the x direction and
261
00:17:55,390 --> 00:18:00,280
the y direction and the D orbitals
262
00:17:57,360 --> 00:18:01,990
having some other representation all
263
00:18:00,280 --> 00:18:03,730
these things are functional
264
00:18:01,990 --> 00:18:05,500
representations of the real and
265
00:18:03,730 --> 00:18:08,290
imaginary parts of the spherical
266
00:18:05,500 --> 00:18:10,090
harmonics on a spherical system on a
267
00:18:08,290 --> 00:18:12,610
coordinate system given by C Raquel
268
00:18:10,090 --> 00:18:14,800
polar coordinates the spherical surface
269
00:18:12,610 --> 00:18:17,200
we will see some of that and that will
270
00:18:14,800 --> 00:18:19,930
give us a clearer picture on what the
271
00:18:17,200 --> 00:18:22,000
solutions mean not necessarily how to
272
00:18:19,930 --> 00:18:24,190
obtain them that's part of the next
273
00:18:22,000 --> 00:18:25,540
level of mathematics course or next
274
00:18:24,190 --> 00:18:28,000
level of physics and chemistry course
275
00:18:25,540 --> 00:18:30,070
that you might take it's not part of
276
00:18:28,000 --> 00:18:31,900
this series of lectures you might find
277
00:18:30,070 --> 00:18:35,940
them elsewhere we will continue with
278
00:18:31,900 --> 00:18:35,940
this in the next part until then thank
279
00:18:52,000 --> 00:18:54,060
you