1
00:00:01,880 --> 00:00:13,300
Where shall we? Shall we begin? So last term we discussed the dynamics of a harmonic oscillator,
2
00:00:13,300 --> 00:00:22,270
which is a one dimensional system where the particle is bound by a potential V is a half CG squared.
3
00:00:23,140 --> 00:00:30,580
And we found a number of important results. Most obviously there was quantisation of the energy, so discrete ness of the energy levels.
4
00:00:30,910 --> 00:00:37,330
And very importantly there was the phenomenon of zero point energy that even when something was in its ground state,
5
00:00:37,600 --> 00:00:44,290
it had the uncertainty principle obliged to have non-negligible kinetic energy and potential energy.
6
00:00:44,290 --> 00:00:50,199
Indeed. So what we're going to do now is study a number of very simple one dimensional
7
00:00:50,200 --> 00:00:53,769
potentials motion in a number of very simple one dimensional potentials,
8
00:00:53,770 --> 00:00:56,940
which are themselves very artificial. These are step potentials.
9
00:00:56,940 --> 00:01:03,429
So they so they change this continuously at some values of X from one value to another value.
10
00:01:03,430 --> 00:01:10,149
So they're very artificial, but they the merit of them is that we can solve the governing well, the key equation,
11
00:01:10,150 --> 00:01:15,370
the time independent Schrodinger equation and obtain the states a well-defined energy from which we can,
12
00:01:16,000 --> 00:01:19,630
as we saw with the harmonic oscillator, recover the dynamics of the system.
13
00:01:20,440 --> 00:01:26,440
We can solve this equation, this important equation, simply for these rather artificial potentials.
14
00:01:26,440 --> 00:01:30,160
And we'll discuss at the end which of our results is artificial,
15
00:01:32,500 --> 00:01:39,940
reflects the artificiality of the potentials and which are generic and ones that we can we can believe in.
16
00:01:40,570 --> 00:01:43,740
So this is what we're going to start with is the square potential.
17
00:01:43,750 --> 00:01:51,760
Well, which is this structure, the potential energy being plotted vertically, position being plotted horizontally.
18
00:01:53,140 --> 00:01:58,630
Here is the origin. This is going to be a distance. A, this is a it's going to be symmetrical.
19
00:01:58,960 --> 00:02:04,420
This is going to be a potential level v zero and this is going to be potential level nothing.
20
00:02:04,630 --> 00:02:08,470
Right? So we set the zero as a potential energy to be at the bottom of the well.
21
00:02:08,710 --> 00:02:16,570
And then there's then there's when you're more than distance away from the origin, you have some potential energy.
22
00:02:16,570 --> 00:02:19,930
V0 is a constant. So this is highly artificial.
23
00:02:20,170 --> 00:02:26,030
But let's see what quantum mechanics has to say about motion in in here and in particular.
24
00:02:26,030 --> 00:02:31,960
So what we're going to look for is stationary states, that is to say, states of well-defined energy.
25
00:02:41,910 --> 00:02:44,970
Which we write like. That's right. So these are states of well-defined energy.
26
00:02:45,000 --> 00:02:52,139
We're interested in these because they enable us to to solve the time dependent Schrodinger equation trivially.
27
00:02:52,140 --> 00:02:59,580
Once we know what all these things are, we can write down. And we know once we know what these things are and we know how to,
28
00:02:59,580 --> 00:03:03,899
then we can express any arbitrary initial condition as a linear combination of these things,
29
00:03:03,900 --> 00:03:06,540
and we can time evolve it in a simple way, as we saw last term.
30
00:03:07,140 --> 00:03:11,910
So we're going to find these things and they're going to have wave functions which will call U of X.
31
00:03:12,360 --> 00:03:17,700
So this is X. E is the wave function.
32
00:03:22,690 --> 00:03:28,809
Now this potential is is symmetrical, is a is an even function of X, right?
33
00:03:28,810 --> 00:03:34,870
So we have that that V of minus X is equal to V of x.
34
00:03:35,290 --> 00:03:44,139
So it's an even function. The consequence of that is that we introduced the parity operation last term.
35
00:03:44,140 --> 00:03:51,700
We had that P if you remember, X, PFC P is an operator, the parity operator.
36
00:03:56,430 --> 00:04:07,860
Which makes out of a state of, say, the state that you would get or which has amplitude to be at minus x.
37
00:04:08,130 --> 00:04:13,920
Well, this, that. What is this? This is the amplitude to be it X when you are in the state that P makes out of a.
38
00:04:14,790 --> 00:04:18,720
And this was defined to be minus X upside.
39
00:04:18,900 --> 00:04:26,100
In other words, it was defined to be the amplitude to be at minus X when you were in the state of PSI.
40
00:04:26,130 --> 00:04:29,940
So the power is the operator makes a state which is the same as the state you first thought of.
41
00:04:30,240 --> 00:04:34,500
Except if if every point is reflected through the origin.
42
00:04:35,430 --> 00:04:44,520
We discussed this operator and because B is an even function of X, we have that PV.
43
00:04:45,780 --> 00:04:54,570
X PV psi is going to be equal to minus x.
44
00:04:59,880 --> 00:05:10,140
Minus X the upside, which because V is a is a is V is a function of position.
45
00:05:10,140 --> 00:05:16,730
It's it's a function of the position. Operator This is going to be V of x x of psi.
46
00:05:16,740 --> 00:05:21,090
In other words, it's going to be simply vivek's. Sorry, sorry, sorry, sorry.
47
00:05:21,090 --> 00:05:31,230
Right. I need a minus X and a minus x so it can be view of minus x times of psi of minus x.
48
00:05:37,540 --> 00:05:44,830
And that what's because the is an even function of x so v minus x is the same as v of x.
49
00:05:45,370 --> 00:05:53,390
This is equal to this is this and this could be written as p.
50
00:05:58,090 --> 00:06:03,310
So what does this mean? This means that because vs an even function of x, it computes with the operator.
51
00:06:04,180 --> 00:06:10,180
With the operator p. In other words, p v equals nought.
52
00:06:11,320 --> 00:06:14,950
Because this is rather a rigmarole.
53
00:06:14,950 --> 00:06:27,040
I have to admit p commands with the because the is an even function which is a way of saying that the potential.
54
00:06:30,200 --> 00:06:37,200
In symmetric. About the origin.
55
00:06:39,510 --> 00:06:48,240
This is the bottom line. If you have a potential which is symmetric about the origins and even function of X, then it commutes with a parity operator.
56
00:06:48,930 --> 00:07:05,230
The consequence of that is. So the Hamiltonian is of course is is as ever P squared over two m plus v of x p squared is an even function.
57
00:07:05,250 --> 00:07:12,090
If you write this in the position representation, it's minus h bar squared d two by the x squared.
58
00:07:12,090 --> 00:07:17,129
So this is an even function of x. So this commutes with the parity operator.
59
00:07:17,130 --> 00:07:21,780
We've just figured that this competes with the parity operator. So we have the P comma H.
60
00:07:23,430 --> 00:07:30,989
So this implies that peak homer H is zero. So the parity operator commutes with a Hamiltonian whose stationary states we would like to find,
61
00:07:30,990 --> 00:07:35,130
whose igen states that just say the stationary states we would like to find.
62
00:07:36,600 --> 00:07:38,740
So what does that imply when to operate his commute?
63
00:07:38,760 --> 00:07:48,480
Remember, the fundamental rigmarole is that this implies there's a complete set of mutual aid and states.
64
00:07:56,460 --> 00:08:15,500
Of P and H. That is to say we can if we wish is look for I can states of age which are states of well-defined parity and the
65
00:08:15,500 --> 00:08:23,330
way functions of these states are well defined parity will either be even functions of X if the parity is even,
66
00:08:23,540 --> 00:08:29,470
or there will be odd functions of x if the parity is odd. So we can.
67
00:08:29,480 --> 00:08:34,400
So what this means is we can insist or we can look for.
68
00:08:46,230 --> 00:08:50,340
So we can look for stationary states with wave functions.
69
00:08:57,370 --> 00:09:09,430
You have x meaning, of course, x e that are either even functions.
70
00:09:11,410 --> 00:09:20,350
You x equals equals you of minus x or odd functions.
71
00:09:23,970 --> 00:09:28,060
U of x is minus u of minus x.
72
00:09:31,180 --> 00:09:38,320
And this observation knowing what knowing that you're looking for an even function, say,
73
00:09:39,970 --> 00:09:43,480
makes it much easier to find that function than if you don't know whether it's even resolved.
74
00:09:43,660 --> 00:09:46,899
That's this is a this is a general observation.
75
00:09:46,900 --> 00:09:56,530
And we'll just find a concrete example of it in a moment. So here is all.
76
00:09:56,650 --> 00:10:02,560
Here is our potential. Again, a minus.
77
00:10:02,590 --> 00:10:06,579
Say, what do we have?
78
00:10:06,580 --> 00:10:11,740
We have here? What is the that the time independent Schrodinger equation.
79
00:10:13,840 --> 00:10:23,169
That is to say, what is the. What is that? That is the equation which shows that h of psi is equal to FC in the position representation.
80
00:10:23,170 --> 00:10:30,270
What is this equation important equation look like here. Well, the at this point here where the potential is zero,
81
00:10:30,700 --> 00:10:37,989
h is p squared over two M so at this location here it becomes this becomes piece
82
00:10:37,990 --> 00:10:47,290
going over to M which is minus H bar squared over to M D to u by the x squared.
83
00:10:48,430 --> 00:10:56,320
So this is I suppose I should change this sorry in this context to you probably know.
84
00:10:56,380 --> 00:11:05,170
Let's leave it with E that's what we were calling it the state e so this left hand part and that reduces
85
00:11:05,170 --> 00:11:10,510
to just this in the position representation because we only have the kinetic energy at this location,
86
00:11:10,510 --> 00:11:16,210
there is no potential energy. And on the right hand side, of course, we simply have e times u.
87
00:11:17,140 --> 00:11:21,010
So we're trying to solve this, this equation, and we know all about the solutions of this equation.
88
00:11:21,310 --> 00:11:24,370
This is just the simple harmonic motion equation of classical physics.
89
00:11:24,370 --> 00:11:34,990
Essentially, it tells us so we know that you is cos k x or you is sine provide solutions of this equation.
90
00:11:35,410 --> 00:11:42,220
Cos kicks is an even functions. So that must be the solution belonging to an even parity state.
91
00:11:42,430 --> 00:11:46,540
And sine k x is an odd function of x, so it's an odd parity thing.
92
00:11:46,960 --> 00:12:01,360
So this has solutions. U of x is equal to either cos k x or sine k x depending on parity.
93
00:12:02,860 --> 00:12:06,819
And we have that k, right.
94
00:12:06,820 --> 00:12:13,120
So when you double differentiate this, you're going to get minus K squared cos X.
95
00:12:14,350 --> 00:12:17,380
So the minus sign deals with that.
96
00:12:17,620 --> 00:12:21,580
We're going to have that minus K squared.
97
00:12:22,750 --> 00:12:28,090
E is equal to this stuff here. In other words, we're going to have a K is equal to.
98
00:12:36,430 --> 00:12:42,870
Yeah. No.
99
00:12:42,900 --> 00:12:46,110
This is wrong. Sorry. Sorry, but it's the wrong way up. What am I doing?
100
00:12:49,280 --> 00:12:55,470
Sorry. The derivative is here. Excuse me. So when we double differentiate, we'll get minus H plus squared k squared over to em.
101
00:12:56,360 --> 00:13:00,590
We'll get a minus coming from the differentiation which will cancel this and we will have
102
00:13:00,800 --> 00:13:11,150
that k squared is equal to 2me over each bar squared square root while k is is that.
103
00:13:13,670 --> 00:13:22,940
So we have determined what the what the what the wave functions are of the stationary states in that interval from minus eight away.
104
00:13:24,770 --> 00:13:29,990
And in terms of the energy, what we now need to do is.
105
00:13:33,760 --> 00:13:37,390
Is think about the state of affairs here.
106
00:13:38,590 --> 00:13:43,710
So, so. So this stuff is all true for monarchs.
107
00:13:44,380 --> 00:13:48,950
Less than a what about the case when monarchs is greater than a.
108
00:13:49,090 --> 00:13:52,480
So now the picture in this zone here.
109
00:13:52,720 --> 00:14:01,120
What is the time independent Schrodinger equation look like? It still has kinetic energy minus h squared of a to m d to u by the x squared.
110
00:14:01,720 --> 00:14:05,049
But now we have potential energy. How much? V zero.
111
00:14:05,050 --> 00:14:16,140
So v zero times u. So this is this is the Hamiltonian operator operating on you and that's equal to e u.
112
00:14:16,900 --> 00:14:19,479
And let's now say we're looking for bound states,
113
00:14:19,480 --> 00:14:28,930
that's to say states where the energy is less than the potential energy out here so that the classically the particle will be confined inside here.
114
00:14:29,320 --> 00:14:35,230
So we're going to look for bound states. You don't there are other states, too, but let's focus on the bound states.
115
00:14:40,600 --> 00:14:46,270
That means that that E is less than V0.
116
00:14:46,570 --> 00:14:50,500
So the particle classically is not allowed to get out of the the well.
117
00:14:52,120 --> 00:14:56,070
What happens then? Well, then this equation becomes, uh.
118
00:14:57,280 --> 00:15:01,780
D to you by the x squared is equal to.
119
00:15:02,200 --> 00:15:08,700
If we put this onto this side, since v0 is by hypothesis bigger than you,
120
00:15:08,710 --> 00:15:15,280
we have a negative right hand side and we can cancel the minus signs on the two sides.
121
00:15:15,670 --> 00:15:24,850
So we if we write it as we have to v v0 minus the energy over h bar squared times u.
122
00:15:25,870 --> 00:15:31,510
So again we have a double derivative is equal to some constant times the function.
123
00:15:31,900 --> 00:15:35,530
But the difference is now that we do not have a minus sign here.
124
00:15:35,710 --> 00:15:40,100
So instead of having sinusoidal solutions, we have exponential style solutions.
125
00:15:40,100 --> 00:15:47,649
So the solution to this equation is that you is equal to a constant times e to
126
00:15:47,650 --> 00:15:54,160
the plus or minus k times x where big k is the square root of this stuff here.
127
00:16:04,130 --> 00:16:07,190
What are we going to. What about this sign ambiguity here? Right.
128
00:16:07,190 --> 00:16:12,350
This there is a sign ambiguity here because it's the double derivative which has to be equal to a constant times u.
129
00:16:12,710 --> 00:16:17,240
If we go for the minus sign in taking the double derivative two derivatives, we get down to minus signs.
130
00:16:17,240 --> 00:16:21,560
We get a plus, obviously. So that's why this there's this ambiguity.
131
00:16:21,980 --> 00:16:23,690
What do we do about that ambiguity?
132
00:16:23,960 --> 00:16:34,040
Well, when X is greater than nought, we want the wave function to decrease as we head off to infinity, as X becomes larger and larger.
133
00:16:34,220 --> 00:16:40,340
Well, in fact. Right, because we would like to be able to normalise the wave function.
134
00:16:40,340 --> 00:16:42,829
We'd like to have the wave function mod squared integrated.
135
00:16:42,830 --> 00:16:47,780
Overall space comes to one and that's not going to be possible if we have an exponential divergence.
136
00:16:48,230 --> 00:17:00,440
So the consequence of that is that in this this zone here we take that U is proportional to E to the minus x because x is positive over there.
137
00:17:00,440 --> 00:17:05,210
And that means the bigger x gets more we move over here, the smaller the wave function becomes.
138
00:17:05,630 --> 00:17:10,460
If we're on this left side here where x is negative,
139
00:17:10,700 --> 00:17:26,179
then we want to take you goes like either plus or minus E to the to the plus k x because x is negative in this zone here we and
140
00:17:26,180 --> 00:17:36,860
the negativity of x gives the exponent of the of the of the exponential negativity so that the bigger that X becomes the modulus,
141
00:17:36,860 --> 00:17:41,450
the more we move over over here, the, the smaller the wave function becomes.
142
00:17:42,230 --> 00:17:49,820
And whether we want to take a plus sign, if we're looking for a state of positive, of even parity, then we want to take this plus sign.
143
00:17:49,820 --> 00:17:58,010
So the wave function over here has the same same numerical value as the wave function, the corresponding point over there.
144
00:17:58,220 --> 00:18:01,640
If we're looking for a state of odd parity, we take this minus sign.
145
00:18:01,880 --> 00:18:06,860
So the wave function at negative x becomes minus the wave function at the corresponding point at positive x.
146
00:18:08,330 --> 00:18:16,670
So, so that's, uh, that's where we are so far what we now need to do.
147
00:18:16,670 --> 00:18:25,280
So we've now solved what have we done? We've solved the time independent Schrodinger equation everywhere except an x equals A in x equals -80.
148
00:18:26,390 --> 00:18:35,120
But we haven't solved it at those points because at those points, if you go a bit to the, for example, at the point X equals A, if you go to the bit,
149
00:18:35,120 --> 00:18:41,690
to the left of that point, then the wave function is supposed to be causal sine x and if you go a bit to the right,
150
00:18:41,960 --> 00:18:46,970
it's going to be this exponential function. But at that point we must still have the time.
151
00:18:46,970 --> 00:18:51,860
Independent Schrodinger Equation Satisfied. So what?
152
00:18:52,190 --> 00:18:59,149
What does that require? Well, for the time independent SchrÃ¶dinger equation to mean anything.
153
00:18:59,150 --> 00:19:04,250
Even the second derivative of the wave function has to be well-defined at that point,
154
00:19:04,250 --> 00:19:08,660
because the time independent writing equation equates the second derivative to some stuff.
155
00:19:09,740 --> 00:19:32,330
So what we can say is that at x equals plus or minus a, we need that d to you by the x squared is defined or we can't solve the ties either.
156
00:19:33,440 --> 00:19:36,500
Satisfy that. Satisfy the ties either.
157
00:19:42,210 --> 00:19:45,570
Well, the rate of change. This is the rate of change of the gradient.
158
00:19:45,840 --> 00:19:50,340
That is certainly not going to be defined if the gradient isn't continuous.
159
00:19:51,690 --> 00:19:57,720
So that implies that do you buy the X is continuous?
160
00:20:04,730 --> 00:20:11,330
And that's a non-trivial requirement because at the moment we've got the wave function in pieces and there's no obvious reason.
161
00:20:11,730 --> 00:20:20,630
Thus we do do some engineering on our pieces. Why? The gradient is defined by one piece on one side of the barrier of the transition
162
00:20:20,870 --> 00:20:23,870
should equal the gradient from a completely different function on the other side.
163
00:20:25,010 --> 00:20:28,340
Okay. But so the gradient has to be continuous.
164
00:20:29,090 --> 00:20:37,370
The gradient is certainly not going to be even it's not even going to be defined unless the wave function itself is continuous.
165
00:20:37,670 --> 00:20:41,330
So we also require the same similar reasons.
166
00:20:46,500 --> 00:20:54,590
But you is continuous. At these points.
167
00:20:55,670 --> 00:21:10,460
So we have to insist. That you have a minus some tiny bit is equal to U of a plus some tiny bit.
168
00:21:11,210 --> 00:21:20,120
And we have to insist that the you by the X evaluated today minus the tiny bit is equal to the U by the x.
169
00:21:24,310 --> 00:21:28,930
Evaluated as a plus a tiny bit called Epsilon. That's what we've got to insist on.
170
00:21:29,770 --> 00:21:32,920
What does that amount to that amounts to here?
171
00:21:35,620 --> 00:21:49,959
If we're just to the left of a we have for the even parity solution we have that u is equal to cos k x an x is a minus epsilon,
172
00:21:49,960 --> 00:21:51,300
but epsilon as small as we like.
173
00:21:51,310 --> 00:21:59,080
So let's just make it equal to cos k that's got to equal the wave function on the right hand side which is some constant.
174
00:22:00,310 --> 00:22:11,540
We'll call it but call that constant big a times E to the minus big k, times x plus a tiny bit.
175
00:22:11,560 --> 00:22:21,430
Let's forget about the tiny bit because this is a continuous function. So we require that that's the continuity of U of x.
176
00:22:24,770 --> 00:22:31,130
Similarly, the gradient just to the left of of x equals A is given by the derivative of cos.
177
00:22:32,120 --> 00:22:45,110
So we're looking at minus K sine k and that's got to be equal to the gradient of a E to the minus X just to the evaluated at x equals A,
178
00:22:45,380 --> 00:22:59,440
so that's minus big k a E to the minus K and that's the continuity of you by the X and the nice.
179
00:22:59,450 --> 00:23:03,650
We also have to satisfy these continuity conditions that x equals minus AA.
180
00:23:04,220 --> 00:23:13,160
But the nice thing about choosing is deciding that you're going to look for a wave function of well-defined parity over an even
181
00:23:13,160 --> 00:23:19,160
functional odd function is that it's easy to persuade yourselves you probably want to sit down quietly and do this afterwards,
182
00:23:19,400 --> 00:23:26,390
that if you satisfies these conditions on the right hand side at x equals plus a of the origin,
183
00:23:26,570 --> 00:23:30,500
then you've also satisfied these conditions on the to the left of the origin.
184
00:23:31,340 --> 00:23:39,139
These equations suffice to fix up the arrangements at both of the discontinuities and you don't have to deal with them separately.
185
00:23:39,140 --> 00:23:43,700
That's the great advantage of of choosing way functions of well-defined parity.
186
00:23:45,970 --> 00:23:51,130
So what do we what do we have here? We have a pair of equations and we have a number of unknowns.
187
00:23:51,490 --> 00:23:58,059
As it stands, we do not know what little K is or Big K is, and we do not know what A is Big A is.
188
00:23:58,060 --> 00:24:07,060
Right? Those are all unknowns. We've two equations and we need fundamentally to determine these unknowns.
189
00:24:09,660 --> 00:24:19,170
Most important is to determine little K and Big K because they are related to the energy by formulae which are there's little K right at the top.
190
00:24:19,170 --> 00:24:25,740
There is the square root of 2me average bar squared and big k halfway up is to m v0 minus c.
191
00:24:26,070 --> 00:24:31,080
So little can big k are both related to the energy and once we found the energy
192
00:24:31,080 --> 00:24:34,920
will know what both Big K and little K on is energy we're fundamentally after.
193
00:24:34,920 --> 00:24:39,780
So that's what we want to focus on. Big is is of less interest.
194
00:24:39,780 --> 00:24:54,150
So let's get rid of big A by dividing this equation through by this equation, then we will find so we divide equation two basically by equation one.
195
00:24:56,310 --> 00:25:10,530
And that leads to the conclusion that minus let's do it down here minus k ten k from sign of a cosine is equal to minus big k nothing.
196
00:25:10,530 --> 00:25:15,690
Everything else goes because we have an a e to the minus big K in both equations.
197
00:25:17,280 --> 00:25:25,769
And let's ask what what this is. Let's try and relate this so big K and little K are both related to the energy
198
00:25:25,770 --> 00:25:30,540
from which it follows that I could express Big K as a function of little K.
199
00:25:30,750 --> 00:25:41,010
So let's do that. This is minus the square root of two and V0 minus E over bar squared,
200
00:25:43,140 --> 00:25:52,020
which is equal to minus square root now to m e average bar squared is actually from we've maybe just lost it, unfortunately.
201
00:25:52,230 --> 00:25:58,440
Let's bring it back into focus right at the top there. It's to me average bar squared is in fact K squared.
202
00:25:58,650 --> 00:26:01,860
So to me average plus squared is K squared.
203
00:26:02,370 --> 00:26:10,470
So what we want to do is write this as two and V nought over squared minus k squared.
204
00:26:13,230 --> 00:26:21,270
So here we have an equation now that this equals this which has only one unknown, namely K.
205
00:26:22,260 --> 00:26:25,950
So the left side is a function of K. The right side is a function of K.
206
00:26:26,610 --> 00:26:35,249
Any values of K for which you these two sides are equal are are provide solutions to our time independent
207
00:26:35,250 --> 00:26:39,570
Schrodinger equation and provide wave functions for stationary states for states of well-defined energy.
208
00:26:41,280 --> 00:26:45,540
To solve this to solve this equation, the way to go is to divide through by this.
209
00:26:45,720 --> 00:26:51,780
Well, obviously cancel the minus signs divide through by k both sides and this then becomes
210
00:26:51,780 --> 00:27:04,560
ten k is equal to the square root of 2mv nought over bar squared k squared minus one.
211
00:27:06,160 --> 00:27:14,770
And it's good now to multiply the top and the bottom of this by a squared and write this as
212
00:27:15,040 --> 00:27:29,830
the square root of w over k squared minus one where w is two and v0a squared over squared.
213
00:27:29,860 --> 00:27:37,840
Why do I want to do that? K is obviously dimensionless because this is a wave number.
214
00:27:39,820 --> 00:27:43,690
The wave function was sine qua. The argument of sine must always be dimensionless.
215
00:27:44,080 --> 00:27:50,580
So this. This is dimensionless. That's obviously dimensionless.
216
00:27:50,590 --> 00:27:52,170
Therefore this must be dimensionless.
217
00:27:52,180 --> 00:27:56,710
You can explicitly check that this is dimensionless for the reason I've defined w is that this thing is dimensionless.
218
00:28:00,920 --> 00:28:06,139
It's a dimensionless measure of the depth and width of the potential.
219
00:28:06,140 --> 00:28:10,190
Well, right. It depends on the depth of the potential. Well, it depends on the width of the potential well.
220
00:28:10,400 --> 00:28:18,290
And it's the dimensionless. The mathematics is telling us that this is how you quantify how you know what kind of
221
00:28:18,290 --> 00:28:21,320
potential well you've got where you've got a very deep one or a very shallow one.
222
00:28:23,970 --> 00:28:26,970
So how are we going to solve this equation here?
223
00:28:27,420 --> 00:28:34,020
Well, the way to go is is to plot both sides of the equation graphically.
224
00:28:34,050 --> 00:28:38,670
All right. So the tangent is is to plot both sides of the equation graphically.
225
00:28:38,850 --> 00:28:43,560
And see what you get, see at what points they meet.
226
00:28:44,220 --> 00:28:48,060
So if this is a plot, this is a being plotted this way.
227
00:28:48,720 --> 00:28:56,310
Then if I plot ten K, it starts off at zero and rises like this.
228
00:28:56,640 --> 00:29:01,860
And when K becomes equal to pi over two, it zooms off to infinity.
229
00:29:02,690 --> 00:29:07,500
Right. That's what tangents do. And down here, it goes in symmetrically.
230
00:29:09,060 --> 00:29:12,630
What does this thing do when K is nothing?
231
00:29:12,840 --> 00:29:17,670
That thing is obviously infinity because it becomes W over nothing square rooted.
232
00:29:17,940 --> 00:29:21,300
So this. This this is the left hand side. I better write that down.
233
00:29:21,310 --> 00:29:25,770
So this here is the left hand side. It's ten K.
234
00:29:26,430 --> 00:29:36,810
The right hand side is coming down from infinity and it's going to go to zero when K is equal to or k squared is equal to W.
235
00:29:37,140 --> 00:29:45,780
So this right hand side is cruising down from infinity and it's going to go to zero when K is root w.
236
00:29:49,260 --> 00:29:56,680
And from this we see it's obvious now that no matter what the value of W is.
237
00:29:56,700 --> 00:30:01,590
So as you change the width and depth of the potential, you move this point to the right of the left.
238
00:30:02,040 --> 00:30:07,230
But no matter where you put it from nowhere to infinity, it will cross.
239
00:30:07,560 --> 00:30:13,440
There will be this intersection here. These two curves always cross.
240
00:30:15,790 --> 00:30:23,830
And where they cross gives you a value of K and therefore a value of K,
241
00:30:24,130 --> 00:30:28,120
which is a solution to the to the time in the independent SchrÃ¶dinger equation.
242
00:30:28,570 --> 00:30:36,100
So what we've just learned is that this well, always.
243
00:30:38,410 --> 00:30:49,260
As a bound state. It doesn't matter how shallow the water is or how narrow the well is,
244
00:30:49,500 --> 00:30:57,050
it always has a bound state because these two curves always cross this tangent on the left.
245
00:30:57,060 --> 00:31:00,310
Right has has all the branches.
246
00:31:00,330 --> 00:31:06,840
There's also a branch. So tangent goes off, comes along here, goes off to pulse plus infinity.
247
00:31:07,110 --> 00:31:16,140
And then somewhere down here it's when, when, when K is equal to pie upon to plus a bit it becomes minus.
248
00:31:16,650 --> 00:31:20,280
It comes in from minus infinity and repeats itself.
249
00:31:22,420 --> 00:31:34,040
So this is the LHC second branch. And this second branch may or may not cut this curve of the right hand side again.
250
00:31:34,060 --> 00:31:38,560
Right. So if we made a smaller than here, this is where.
251
00:31:38,830 --> 00:31:44,500
This is what? This is where a. This is this is the place pie.
252
00:31:45,190 --> 00:31:49,749
If if we would make this less than that, we wouldn't get a second solution.
253
00:31:49,750 --> 00:31:57,250
But if we have a bigger than this, we do get a second solution. So it may have.
254
00:32:01,290 --> 00:32:04,440
Other even parity.
255
00:32:06,740 --> 00:32:26,190
The stationary states. So depending on the depth of the potential, we have one, two, three, four,
256
00:32:26,190 --> 00:32:32,310
etc. in stationary states with of even parity, we've only dealt with even parity case.
257
00:32:32,850 --> 00:32:37,679
If we want to look at the old parity states, let's just begin to do this.
258
00:32:37,680 --> 00:32:50,310
But this is basically an exercise for the problems. Then our conditions are a wave function for the odd parrot stage states.
259
00:32:50,320 --> 00:33:01,090
We've lost it somewhere but it's go away parity away function to the old parity states is is the sign
260
00:33:01,090 --> 00:33:08,860
kegs right up there right so in the middle it's sign kegs at the edge it's still a E to the minus X.
261
00:33:09,400 --> 00:33:24,310
So our continuity conditions become that sign K is equal to a E to the minus K, and that's the continuity.
262
00:33:27,640 --> 00:33:42,060
Of the wave function itself at x equals a the derivative of this is going to give me k cos k is equal to minus big k a e to the minus k.
263
00:33:45,390 --> 00:33:49,690
We divide this equation by this equation analogously to what we did before.
264
00:33:50,070 --> 00:34:00,570
We are going to get a k cotangent of k, a lot of a is equal to minus.
265
00:34:04,540 --> 00:34:12,219
Is equal to minus b k. So now we have only one minus sign.
266
00:34:12,220 --> 00:34:18,970
Previously we had a pair of minus signs which cancel. Now we have a one minus sign because we differentiate a sign.
267
00:34:18,970 --> 00:34:26,200
We didn't differentiate a cosine the and correspondingly we have a cotangent instead of a tangent.
268
00:34:26,710 --> 00:34:32,620
So this equation can be graphically solved. Uh, that's the exercise.
269
00:34:39,760 --> 00:34:58,450
And we find that we may get nought one two solutions for K and for every solution we have an odd parity.
270
00:34:59,560 --> 00:35:26,170
We have an old parity stationary state. Let's have a look at the uncertainty principle in this example.
271
00:35:32,600 --> 00:35:39,290
Just to remind you, last term, quite early on, we showed we considered the case.
272
00:35:40,100 --> 00:36:00,230
So this is last as a summary, last term. We consider the case where obsessive X is proportional to E to the minus X squared over four sigma squared.
273
00:36:01,370 --> 00:36:11,120
So we considered a wave function whose spatial form was a Gaussian such that when you mod squared this in order to get the probability distribution,
274
00:36:11,360 --> 00:36:15,229
you found that it was a Gaussian with dispersion sigma. So.
275
00:36:15,230 --> 00:36:20,480
So this thing here is the expectation of X squared.
276
00:36:21,300 --> 00:36:25,460
Right. That's what the sigma squared is from that formula. And what did we find?
277
00:36:26,090 --> 00:36:33,920
We found fundamentally by doing a for a transform, that if that's what the if that's what the wave function looks like,
278
00:36:34,220 --> 00:36:45,620
then the probability well, the amplitude to find this is.
279
00:36:46,220 --> 00:36:49,760
So this is x ABC. Right.
280
00:36:49,880 --> 00:36:55,310
The wave function looks like that. Then the amplitude to get a certain momentum,
281
00:36:55,310 --> 00:37:06,050
if you would make a momentum measurement was looking like E to the minus p squared over four sigma p squared.
282
00:37:07,100 --> 00:37:12,410
Right where this thing becomes the expectation value of p squared.
283
00:37:12,560 --> 00:37:17,660
So the variance, the expectation values momentum in this state is zero.
284
00:37:18,260 --> 00:37:23,210
The expectation of the square of the momentum is going to be this sigma p squared.
285
00:37:24,740 --> 00:37:32,389
And what we found was that Sigma Times Sigma P using wedge bar on to this was a concrete
286
00:37:32,390 --> 00:37:38,090
example of the uncertainty principle where the smaller the uncertainty in X is,
287
00:37:38,360 --> 00:37:40,400
the bigger the uncertainty in the momentum.
288
00:37:40,610 --> 00:37:46,700
And correspondingly the smaller momentum in P is, the bigger the uncertainty X has to be because the product is always the same.
289
00:37:46,730 --> 00:37:55,170
In this particular Gaussian example. So we have some idea that the, the, the uncertainty in X Times, the uncertainty in P, we,
290
00:37:55,340 --> 00:38:02,450
we say to ourselves is inherently never smaller than this number upon to it may be bigger than h bar upon two easily.
291
00:38:02,460 --> 00:38:08,840
It often is bigger than H, but usually is bigger than H bar to. But this is kind of as small as it gets.
292
00:38:10,190 --> 00:38:15,139
So that's that was what we did last term. So let's have a look at it in this particular case.
293
00:38:15,140 --> 00:38:21,240
Right. So let's look at the ground state.
294
00:38:22,200 --> 00:38:28,200
Remember this? Oh, yeah. Another point to remind you is that the ground state wave function of the harmonic oscillator.
295
00:38:28,380 --> 00:38:35,250
So the the ground state of a harmonic oscillator actually has a wave function, which is this Gaussian.
296
00:38:35,430 --> 00:38:39,390
The time that we did this calculation. This Gaussian was just picked out of the air.
297
00:38:39,630 --> 00:38:45,500
But for the ground state of the harmonic oscillator actually does have this this thing here.
298
00:38:45,510 --> 00:38:51,660
This fits this example. So let's have a look at the wave function of the ground state of this of this that we have here.
299
00:38:55,950 --> 00:39:09,030
What you might argue to yourself, you might say, okay, so what we can say is that P is less than or on the order of.
300
00:39:13,010 --> 00:39:18,440
Well, peace. Quiet over to him. The energy is less than zero, right?
301
00:39:18,950 --> 00:39:23,420
Because it's a bound. It's bound. So it's less than zero.
302
00:39:23,810 --> 00:39:33,350
So what we can say is that P. P squared is less than two and V zero.
303
00:39:35,110 --> 00:39:41,970
Yeah. Sorry. That's meant to be a less than seems reasonable, doesn't it?
304
00:39:42,720 --> 00:39:50,070
The particle can't have any more kinetic energy than the energy it requires to escape because we know it's bound.
305
00:39:51,180 --> 00:39:56,820
The next thing you might say is, well, so what's the what's what's what's x squared?
306
00:39:57,600 --> 00:40:01,950
What's the uncertainty in X? Well, you might say to yourself, Well, look, this particle is trapped in this potential.
307
00:40:01,950 --> 00:40:05,100
Well, that goes from plus A to minus A.
308
00:40:05,640 --> 00:40:10,709
So what's the uncertainty in in X, in x squared?
309
00:40:10,710 --> 00:40:13,920
Well, this must be on the order of a squared.
310
00:40:16,650 --> 00:40:18,840
Seems reasonable, doesn't it. Less than on the order of.
311
00:40:20,140 --> 00:40:28,440
Well I'll put it in a factor two or four or just to be sure that where the lesson is is, is holding because the thing is trapped by potential.
312
00:40:28,440 --> 00:40:33,030
Well it extends only from plus A to minus A, so it has a range of two ways they can run in.
313
00:40:33,450 --> 00:40:39,660
So we say X squared is definitely less than two squared. It's almost certainly it's clearly less than that, significantly less than that.
314
00:40:40,920 --> 00:40:46,560
So what does that give me? And and this this square is the momentum is so.
315
00:40:47,100 --> 00:40:51,630
So we have that that X squared.
316
00:40:53,130 --> 00:41:01,080
P squared. I mean, I can put expectation values around that, too, I think is less than on the order of.
317
00:41:03,550 --> 00:41:08,500
Whatever it is. 8 a.m. v not a square.
318
00:41:14,930 --> 00:41:18,740
But that is w is somewhere. Yeah. Look, w.
319
00:41:22,130 --> 00:41:25,370
This is basically w because that was to Venus.
320
00:41:25,460 --> 00:41:35,210
So this Venus squared is uh, something like 4mv not a squared is is h bar squared.
321
00:41:53,780 --> 00:41:59,270
If I done this right, because I'm getting an answer, which I'm. Thomas W Yes, exactly.
322
00:41:59,300 --> 00:42:05,150
That's right. The dimensionless number. W Right. The point is right that W could be made as small as you like.
323
00:42:05,150 --> 00:42:06,680
We've agreed. And you slope have bound state.
324
00:42:08,370 --> 00:42:14,880
So this naive argument suggests that we're going to violate the uncertainty principle because we can make this as small as we like.
325
00:42:20,290 --> 00:42:23,680
So what's the problem? This argument's bogus. What's bogus about it?
326
00:42:25,000 --> 00:42:31,430
A hole in the wall. It's not all in the well as you make w smaller and smaller.
327
00:42:31,800 --> 00:42:33,860
Let's draw what the wave function looks like.
328
00:42:37,790 --> 00:42:49,820
So if we have let's, let's have a nice big value of w a big value w means a nice spicy well with right high walls.
329
00:42:50,330 --> 00:42:56,330
And then we will have a ground state wave function, which is a cosine and some little bits like this.
330
00:42:57,560 --> 00:43:05,270
And indeed the probability to find the particle outside the well outside, plus or minus, they will be not very much.
331
00:43:05,780 --> 00:43:16,880
But as you make this smaller and smaller and smaller, you bring this in making a smaller and and or you lower these making this curvature smaller.
332
00:43:17,300 --> 00:43:24,350
So that curvature of that wave function is a reflection of the kinetic energy, which is a reflection of of nought.
333
00:43:24,770 --> 00:43:30,050
We we come to the we have a nice we have a narrow, a techy witchy thing like this.
334
00:43:30,530 --> 00:43:34,579
Then you have barely you have an almost a straight line across here.
335
00:43:34,580 --> 00:43:39,320
I can't draw it. Well, almost a straight line across here. And then enormous long exponential decays.
336
00:43:42,050 --> 00:43:48,050
And the particles almost certain not to be in the potential. Well, that's a very remarkable conclusion to come to, right.
337
00:43:48,980 --> 00:43:55,940
That you can trap a particle with a potential well, which is almost it has very little probability of actually being in.
338
00:43:58,340 --> 00:43:59,480
But that's what the theory says.
339
00:44:10,390 --> 00:44:19,360
Something else we can we can show is we can we can say we can argue as follows supposing we given some other potential.
340
00:44:19,360 --> 00:44:23,469
Well, here is some other potential. Well, I'm sorry.
341
00:44:23,470 --> 00:44:26,840
It's meant to be it's meant to be an even function of X, right?
342
00:44:26,860 --> 00:44:31,930
So it's meant to be the same on both sides. Symmetrical and somebody also.
343
00:44:32,050 --> 00:44:36,280
So does this potential well have a bound state then?
344
00:44:36,280 --> 00:44:39,700
You reason as follows let us inscribe a nice square potential.
345
00:44:39,700 --> 00:44:43,270
Well, there are many square potential worlds you can inscribe into it. Right. But there's one.
346
00:44:44,500 --> 00:44:50,290
Then you can argue that this potential well is narrower and shallower than the one you were given.
347
00:44:50,740 --> 00:44:57,700
So less likely to trap a particle. But this potential world has a state has a bound state.
348
00:44:58,210 --> 00:45:04,040
So this. Well. Shallower.
349
00:45:07,070 --> 00:45:15,080
Narrower. But it has a bound state.
350
00:45:15,470 --> 00:45:19,190
At least one bound state because it's a square. Well, and we understand about square wells.
351
00:45:19,970 --> 00:45:24,680
So you reason that this wider and deeper well also has a bound state.
352
00:45:26,720 --> 00:45:31,850
So we we're making an inference from this very special, rather artificial squared.
353
00:45:31,850 --> 00:45:43,940
Well, potential about all one dimensional square wells. What happens now?
354
00:45:43,970 --> 00:45:47,150
Let's let's conclude with a special case.
355
00:45:47,150 --> 00:45:56,170
An important special case. Which is the infinitely deep.
356
00:45:56,170 --> 00:46:05,659
Well, woops. So now we let V0 become arbitrarily large and we have a, we have a,
357
00:46:05,660 --> 00:46:09,560
well it looks like this, it just goes up and up and up and up and up and up and up and up.
358
00:46:11,510 --> 00:46:16,150
What happens then? Well, let's go to our graphical solution.
359
00:46:16,160 --> 00:46:19,459
What have we done? We've made we've left a constant.
360
00:46:19,460 --> 00:46:27,770
I mean, we've left at some values and we've allowed V zero to become arbitrary large, which means we've made W arbitrarily large.
361
00:46:27,890 --> 00:46:31,610
What's the implication of that for our solution of this equation?
362
00:46:32,420 --> 00:46:38,330
So we're where we're solving this equation for w arbitrarily large.
363
00:46:40,670 --> 00:46:49,220
So so we're saying that the roots are going to become where ten K the even parity roots are going to be where ten K equals infinity.
364
00:46:49,850 --> 00:46:55,160
And we know what that means. We know that K must be pi upon two, three, pi upon two, etc.
365
00:46:55,880 --> 00:47:09,020
So we've done, we've made W goes to infinity, which implies that the governing equation, the equation that determines K becomes ten K equals infinity,
366
00:47:09,500 --> 00:47:22,399
which implies that K is equal to either pi upon two or three pi upon to, etc., etc., etc. and all of these values are going to be okay.
367
00:47:22,400 --> 00:47:25,790
Graphically, what's happening is that this curve, right?
368
00:47:25,790 --> 00:47:28,520
This was the curve of the right hand side for finite.
369
00:47:28,850 --> 00:47:40,129
For finite W is going to just become a line along here at infinity and it's going to cross these various branches of the tangent at infinity,
370
00:47:40,130 --> 00:47:46,970
which means that these pi upon two, three pi upon two points, and it's going to cross every single one of them all the way, all the way out.
371
00:47:47,000 --> 00:47:54,320
Right? So we can have an infinite number of state of even parity stationary states and they're going to have those values of K.
372
00:47:56,000 --> 00:48:03,200
And if when you when you study the corresponding problem for the AUD, for the AUD parity states,
373
00:48:05,180 --> 00:48:09,170
which we just vaguely discussed here, it's going to be the same deal.
374
00:48:10,490 --> 00:48:26,240
We're going to be solving an equation which says that Cot K is going to be infinite and that means that K is going to be is going to be pi to pi.
375
00:48:26,300 --> 00:48:32,220
Three pi. So on. So these are going to be the even parity states.
376
00:48:35,640 --> 00:48:46,470
And then we will find okay is equal to pi, two pi, three pi, etc. for the quad parity states.
377
00:48:49,110 --> 00:48:52,530
So the infinitely deep well will have an infinite number of solutions.
378
00:48:52,950 --> 00:49:05,020
And what do the wave functions look like. Well the wave functions are going to be cos K X where, where K is equal to some number of odd pi by twos.
379
00:49:05,280 --> 00:49:08,970
So the wave function of the even parity states is going to vanish.
380
00:49:10,530 --> 00:49:20,310
This is going to be cos k x and the wave function is going to vanish here and here by virtue of these conditions there.
381
00:49:21,600 --> 00:49:25,679
And the same thing will happen for the odd parity states.
382
00:49:25,680 --> 00:49:34,530
For example, the first odd parity state is going to have a wave function which which looks like this.
383
00:49:34,890 --> 00:49:38,310
It's going to it's going to be sine.
384
00:49:40,040 --> 00:49:45,320
K x. The best piece of chalk.
385
00:49:47,060 --> 00:49:54,889
It's going to be sign kegs where K is equal to a number of pies.
386
00:49:54,890 --> 00:49:58,850
And therefore the the sign vanishes. So it's going to vanish.
387
00:50:07,470 --> 00:50:09,450
So this is a concrete example,
388
00:50:09,810 --> 00:50:28,830
but it leads to the general it inspires the general principle that a wave function vanishes adjacent to an infinite region of infinite potential.
389
00:50:39,140 --> 00:50:50,790
But. Physically.
390
00:50:50,790 --> 00:51:00,029
What's happened is that is that this well, this big cake has grown bigger and bigger and bigger and bigger,
391
00:51:00,030 --> 00:51:06,330
and therefore this exponential has grown steeper and the curvature of it has grown larger and larger and larger and larger.
392
00:51:06,720 --> 00:51:09,570
So we've gone through, as we raise the walls,
393
00:51:10,050 --> 00:51:19,260
the transition at the edge of the well goes from being sort of nice and nice and easy with a small value of K.
394
00:51:21,980 --> 00:51:30,459
Through. Big a value about a bigger value of K to ultimately an infinite value of K,
395
00:51:30,460 --> 00:51:36,370
which allows it to continuously go from a finite slope right round to two no slope.
396
00:51:38,450 --> 00:51:44,089
And that's why it's a general property of the solutions of the time independent Schrodinger
397
00:51:44,090 --> 00:51:48,950
equation that as you approach the edge of an infinite region of infinite potential,
398
00:51:49,160 --> 00:51:53,330
the wave function vanishes and in anticipation of that infinity.
399
00:51:53,840 --> 00:52:00,740
So another strange aspect, you know, another physically strange thing that the theory is telling us is that the wave function,
400
00:52:01,160 --> 00:52:07,940
that the particle has negligible probability of being found in the neighbourhood of this region where it's strictly forbidden.
401
00:52:08,950 --> 00:52:12,280
It anticipates the fact that it's going to be forbidden.
402
00:52:14,510 --> 00:52:20,090
And you won't find it even near this dangerous place. Okay, it's time to stop.