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Okay. So these this is we were at work on the harmonic oscillator, which, as I said,

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is the single most important dynamical system in physics, a model for a huge number of physical applications.

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We had this Hamiltonian, which is which is basically P squared plus X squared.

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The crucial thing is that it's quadratic. It's obviously quadratic in P, but especially it's also quadratic in X.

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That's his peculiar, peculiar characteristic of its of its potential energy.

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We went the plan for what we're trying to do is find the stationary status.

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In other words, we're trying to find the states here which are eigen functions of the Hamiltonian,

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because they will enable us to to follow the dynamics of the system.

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Oh, they're crucial tools.

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We went after them by defining this thing a the annihilation operator, which is a dimensionless operator made up of X and P.

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And we found that it the we evaluated what we found to it.

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This thing had two properties.

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First, it a dagger a is very nearly the Hamiltonian a dagger ray plus a half h bar omega is the Hamiltonian first important property.

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Second important property that it's commutative with its emission joint is.

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Is minus one when done this way round. So using those properties we said, suppose we have a state, a stationary stage of energy.

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If we multiply this equation through this defining equation of that state through by

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a dagger and do some swapping of the order of operators using the results above,

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we were able to show this is where we finished that h on the state that you

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get by using a dagger on E is equal to E plus a bar omega plus h bar omega,

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times that self-same state. In other words, this state here, this state here is essentially a state.

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Well, it is a state, a stationary state of energy, increased by age by omega.

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And because we could repeat this,

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it follows that the energy is if there is an energy E then there must be energies e plus a h bar omega e plus two bar omega.

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And as we can see, e plus any amount, any number of bar omegas, any number of omegas are.

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What remains at this point is to find out what the number e is that we first thought of.

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And we do that by applying not a dagger to that equation, but a to that equation.

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So then we have if we take if we we start with now e is equal to h e and we multiply both sides of the equation by a, not a dagger this time.

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Then we swap the order here, just as we did before into h, h, a, and then we have to add in the commentator a comma h.

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All operating on e still, we yesterday this is absolutely a repeat of what we did yesterday.

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This is H.A. Now we replace that H as advertised up there by a dagger, etc. We observe.

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So this should become the this is going to be replaced by a dagger, a plus a half h bar omega.

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We can forget about the half because we're inside a commentator and a half commits with everything.

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We can take the bar omega outside the commentator and then what we're looking at is plus a comma, a dagger, a h bar, omega, close brackets,

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e we use our usual rules for taking the commentator of products,

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which which is to say it should be the commentator of this with this without that standing idly by.

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And then in principle the commentator of this with this with this standing idly by with the second commentator vanishes.

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And this commentator, by the result up there is equal to plus one.

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Okay, because the turned around there a dagger is minus one.

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So this commentator a comma, a dagger must be plus one.

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So this is saying this becomes H plus a H bar omega.

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E So we have an a e with no operator in front on this side of the equation, but we have the same thing on that side of the equation.

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So we take these together and onto this side of the equation and we then have the E minus.

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This H bar omega goes to the other side and becomes a minus H bar omega e is equal to h e,

44
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which establishes on the face of it what we might have hoped,

45
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which is that the stage you get that you have after you use a on this stationary

46
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state e is another stationary state with an energy lower than E by each power omega.

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So a dagger raises your energy. It increases the excitation of the harmonic oscillator and evidently lowers the energy of the harmonic oscillator.

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Now, you have to at this point start to worry, because by the previous rhetoric it would follow that if there is E,

49
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then there is also E minus, h, bar, omega and so on down through E, minus n, h bar omega.

50
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And it looks as if you can find states of lower and lower energy without limit.

51
00:05:42,350 --> 00:05:47,240
And you should be worried about that because thinking about the Hamiltonian up there,

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because classically it looks manifestly positive, because p squared should be positive, an x squared should be positive.

53
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You can't trust such things in quantum mechanics, but what we can do is we can work out and we can we can say,

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look, e is the expectation value of the Hamiltonian in the state.

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E Obviously because H on E is equal to E comes outside and and this on this is one.

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So that's a self-evident equation. Replace what's inside here.

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This becomes over to m of e p p e plus m squared omega squared e.

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X x. E. And because these are observables and therefore emission, I can put a dagger there if I want to.

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It doesn't make any difference because P dagger is p and I can observe that this is now manifestly p e mod squared.

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This thing is the mod square of that of the catch you get by using P on e and this is m squared omega squared of x e mod squared it's delivered to m,

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which is boring and this is manifestly greater than or equal to zero.

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Even in quantum mechanics, there can't be any argy bargy. This has to be greater than equals zero.

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So all the energies, all the allowed energies, the entire spectrum of the Hamiltonian has to be positive.

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But we've apparently established that by using a we can get states which by successively using a and a and a and a,

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we can get states of lower and lower energy because we take H Bar Omega off.

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Every time we use E three, we use a. So there's a problem.

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And the problem is the assumption. So what let's just look carefully at what we've established here.

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What we've established is, is this equation here?

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This equation is copper is absolutely copper bottom.

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It's beyond it's beyond criticism. It's true. It was obtained by totally legitimate operations.

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And. It. It establishes that this thing is an eigen function with this lower energy.

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Provided this thing is non-zero. But if at any stage in this chain of applying A's, this thing here would vanish.

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So if when you get to a certain energy, the lowest energy you will and you apply a simply kills it.

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It produces a cap of no length at all. Then we haven't got we won't have a state of even lower energy.

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And since it's clear that this chain of operations it is not it's it's logically

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impossible to go on creating states in lower and lower and lower energy.

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This chain of results of applying extra factors of a has to stop somewhere.

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And the only way it can stop since this equation is true, is by this thing becoming zero.

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Then because h on zero is equal to any number you like on zero then.

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So. So if this is zero, this equation does not establish that this is an eigenvalue of h, right?

81
00:09:13,240 --> 00:09:19,300
But that's the only circumstance in which this equation would not establish that this was an eigenvalue of h.

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So for the lowest energy, the ground state energy, it has to be that A kills it.

83
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So, so. That's the ground state.

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We must have. But a on a zero equals nought.

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00:09:56,880 --> 00:10:01,110
Right? So I'm using this symbol now to indicate the lowest. Energy.

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00:10:07,300 --> 00:10:10,540
So let's. So what does this equation mean to give, to give, to give?

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00:10:10,540 --> 00:10:16,600
More precise meaning to it? What we so what we say we say is that this case, as you get here, has no length squared.

88
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So let's just evaluate the model length squared.

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So so we're saying that A is zero month squared, which is equal to E, a dagger.

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A E is nothing but this thing.

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00:10:36,040 --> 00:10:39,340
We have it somewhere up there. It's g ust.

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Yeah, it's just in range. H is equal to a dagger plus a half h bar.

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So this thing is this is the expectation value of H over H by omega minus a half.

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So and this one, these ones have zeros on it. Sorry. These one have zeros because we're talking about the ground state energy, not any old energy.

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Now we're just talking about that one special lowest energy, the ground state energy this equation is valid for.

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And so what we're discovering is that E0 over H Bar Omega is equal to a half.

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In other words, the ground state energy is zero is a half h bar omega and now we know what the general energy is

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because we know that we can make states of higher energy by applying a dagger to the ground state.

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Moving up by H Bar Omega. Each time the energy must be N plus a half.

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H Bar Omega. So we have found the allowed energies.

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And the next item on the agenda is to find the corresponding find a way functions of the corresponding Afghan state stationery states.

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But let's move over here. So let's look at let's ask about wave functions.

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Of the stationary states. Oops. Now we've found the allowed images.

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00:12:29,840 --> 00:12:38,750
Actually. But for I do that, it's probably good to generalise this calculation here.

105
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So what we've established is. So let's just talk about normalisation as a sort of preliminary.

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I think it's better to do it just now. Normalisation.

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What we've established is that is that a dagger on and I'm going to use a new notation.

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Right. We're going to say what previously I'd call E is going to go to N.

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So this is the state. With E equal n plus a half.

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H Bar Omega. Right.

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00:13:19,870 --> 00:13:23,830
We could. We could write an end in here. But everybody writes just end.

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00:13:23,860 --> 00:13:27,099
It's just saves energy. It works well.

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00:13:27,100 --> 00:13:31,870
Right now. We've discovered what the energies are and that they're labelled by an integer n nought.

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One, two, three, four. So this is and equals nought, one, two, and so on.

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00:13:38,140 --> 00:13:43,870
It makes sense to have a nice compact notation into and to call our states, the stationary states, the state nought.

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The ground state, the state one. The first excited state. The state to the second excited state.

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00:13:50,420 --> 00:13:57,950
And what we've established is that a on N is some constant I'll call it K times N plus one.

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Right, because we've discovered that when we used A on the state e we got the state.

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E We got, we got we got a state which was an Oregon catch of H with eigenvalue e plus h bar omega,

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which would be in the new notation the state n plus one. If if this e was n plus a half h for omega, this will be n plus three halves h bar omega.

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00:14:22,960 --> 00:14:26,800
And the issue arises. So what value does this thing have?

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00:14:28,970 --> 00:14:38,480
So just just to be clear, we so when we have we've had relationships like this, we had a corresponding one up there.

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Yep. There we go. We arrived yesterday.

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We derive this equation and that equation establishes that this thing is an organ of age.

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It doesn't establish that as a properly normalised dog and kids of age, I want it to be properly normalised.

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And so I'm asking about what's the normalisation constant I have to use after I have applied a dagger?

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00:14:59,230 --> 00:15:05,320
It's easily found. Found because we just take the mod square of this side of the equation, of the monster of that side of the equation.

128
00:15:05,620 --> 00:15:13,280
So taking the mod square. So taking. Mud square on both sides.

129
00:15:15,650 --> 00:15:29,180
We get an a a dagger and is equal to k mod squared times one because that's by definition going to be correctly normalised.

130
00:15:30,710 --> 00:15:35,720
We we can take K to be a real number. We can take K to be a complex number if we determine to.

131
00:15:35,720 --> 00:15:39,770
But why don't we just agree to take it to be a real number? And this just this just becomes k squared.

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We just trying to get the thing normalised. We don't care about the face. And let's ask ourselves what this is.

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If I swap this over, I want to swap this over because then I can relate it to the Hamiltonian.

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Then I have to add on a comma, a dagger.

135
00:15:58,640 --> 00:16:01,880
Whoops. Commentator. So that's.

136
00:16:03,380 --> 00:16:10,430
That's that rewritten. This is the Hamiltonian minus is the Hamiltonian overage bar omega.

137
00:16:10,760 --> 00:16:16,379
Where is it? Where is it? Where is it? Yeah, it's up there.

138
00:16:16,380 --> 00:16:21,180
It's the Hamiltonian minus. It's the Hamiltonian overage bar, omega minus a half.

139
00:16:21,180 --> 00:16:26,729
So this is H overreach bar omega minus a half.

140
00:16:26,730 --> 00:16:33,690
That's that, that's this thing. And what's this? This is plus one, I think because other the commutation, the other way round was minus one.

141
00:16:33,690 --> 00:16:46,800
So this is plus one. So that's that that h on n is equal to n plus a half h bar omega by definition.

142
00:16:47,250 --> 00:16:53,729
Right? So this on this produces n plus a half by omega divide by the omega.

143
00:16:53,730 --> 00:16:58,020
And we have n plus a half take away a half. We have an add one, we have n plus one.

144
00:16:58,380 --> 00:17:04,950
So this is in fact equal to one plus one. So what?

145
00:17:07,430 --> 00:17:10,560
Yeah. So. So. So what follows this?

146
00:17:10,580 --> 00:17:14,840
So comparing. Comparing this with this.

147
00:17:14,840 --> 00:17:27,410
We find that k squared is equal to n plus one. Going back to up there, we have that n plus one is equal to one over the square root.

148
00:17:28,220 --> 00:17:34,400
Sorry, sorry. Yeah, that's right. Square root of N plus one. Of of a degree.

149
00:17:38,030 --> 00:17:47,780
This is a very important equation. If we would repeat this, if we would go through this rigmarole.

150
00:17:49,620 --> 00:17:55,979
The same logic using a on end being some other constant times n minus one we would we would

151
00:17:55,980 --> 00:18:04,049
find that n minus one is equal to one over the square root of n of a operating one n.

152
00:18:04,050 --> 00:18:10,650
We know that operating on n is going to produce n minus one. It depletes the energy by omega and therefore reduces n by one.

153
00:18:11,070 --> 00:18:17,070
The normalisation constant turns out to be one over the square root of n. I would recommend you check that after that, after the lecture.

154
00:18:17,280 --> 00:18:19,440
It's a precise repeat of this logic here,

155
00:18:20,610 --> 00:18:28,499
and it's worth remembering these rules because this is such an important relationship and they're very easy rules to remember.

156
00:18:28,500 --> 00:18:33,570
The normalisation constant is one over the square root of the biggest number that appears in the equation, right?

157
00:18:34,080 --> 00:18:39,030
So in this equation, the biggest number appearing is N plus one. That's what you use in this equation.

158
00:18:39,030 --> 00:18:42,210
The biggest number that appears is N, that's what you use.

159
00:18:45,610 --> 00:18:49,120
So these are these are two very important equations which physicists remember.

160
00:18:53,130 --> 00:18:58,770
Okay. So so what we have to do now we're trying to find a way functions fundamentally.

161
00:18:58,770 --> 00:19:07,260
This is just the tedious details of the normalisation. Although those equations are of bigger use than just with wave functions.

162
00:19:07,860 --> 00:19:18,360
Let's find the ground state wave function. What is it?

163
00:19:18,720 --> 00:19:21,830
It's we will call it use zero of X.

164
00:19:22,560 --> 00:19:25,710
And it is, of course, x nought.

165
00:19:25,890 --> 00:19:30,820
It's the it's the function was defined by this, the amplitude to be it x.

166
00:19:30,840 --> 00:19:35,820
If you were in the ground state also called this right, these are just two notations for the same thing.

167
00:19:37,860 --> 00:19:45,660
And this satisfies a nasty equation because what we know is that a operating on nought is equal to nought.

168
00:19:45,700 --> 00:19:49,980
Right. The ground state is defined to be it comes into the world as the state,

169
00:19:50,480 --> 00:19:55,200
the one and only state that the operator, the destruction operator, a kills stone dead.

170
00:19:56,670 --> 00:20:03,060
So if we multiply this equation on the left by X, we still have a valid equation.

171
00:20:05,070 --> 00:20:13,500
And this. And let's write in what a is.

172
00:20:16,670 --> 00:20:20,330
So a is an omega x.

173
00:20:23,660 --> 00:20:30,110
Plus I.P. In principle, it's over the square root of two M by Omega.

174
00:20:30,110 --> 00:20:38,450
But because I'm about to put this stuff equal to nought, I can I can I can neglect the factor on the bottom.

175
00:20:38,450 --> 00:20:42,170
I can multiply through both sides of the equation the factor on the bottom and get rid of the garbage.

176
00:20:42,190 --> 00:20:46,970
Right. Clean things up. Why don't we write this?

177
00:20:47,240 --> 00:20:50,809
I mean, we're more or less committed to writing this in the in the position represent.

178
00:20:50,810 --> 00:21:03,260
Well, no, I mean, we have it there. Right. This tells me that m omega x and omega x x nought because this is the position

179
00:21:03,710 --> 00:21:08,120
operator we this is an observable I can imagine that it operates backwards on this.

180
00:21:08,870 --> 00:21:12,890
This is the eigen function of that operator. So those are both operators.

181
00:21:13,310 --> 00:21:16,969
That's not an operator. This this operates on this.

182
00:21:16,970 --> 00:21:20,240
If I want backwards, this is its eigen function.

183
00:21:20,450 --> 00:21:27,230
So I get an X the number times this this then meets that and produces this, which is our wave function.

184
00:21:27,680 --> 00:21:46,700
And then we also have plus i x p nought and the momentum operator was defined by saying that this thing is minus i d by the x of x.

185
00:21:47,530 --> 00:21:56,000
All right. This was the this was the definition for any wave function of how of how P operated with an H.

186
00:21:56,240 --> 00:22:00,139
Thank you. Thank you. Absolutely. So I need to write this stuff down.

187
00:22:00,140 --> 00:22:03,410
I need to write. And most of this is equal to zero. This is equal to zero.

188
00:22:03,410 --> 00:22:07,610
Provided I put in the m omega x x.

189
00:22:09,410 --> 00:22:16,430
If I rewrite that in in wave function notation becomes m omega x,

190
00:22:17,030 --> 00:22:31,370
you note of x is equal to and it's clean this stuff up I get a sorry it's not equal to plus h bar d by the x of u nought of x equals nought.

191
00:22:31,380 --> 00:22:32,270
So what is this?

192
00:22:32,270 --> 00:22:40,370
This is a first order linear differential equation, and it's in fact an ordinary differential equation because there are no other variables.

193
00:22:40,370 --> 00:22:48,290
The next present I've written, this is a partial derivative consistency with what we do in more than three dimensional cases,

194
00:22:48,290 --> 00:22:51,649
I suppose, but it is a first order linear differential equation.

195
00:22:51,650 --> 00:22:56,660
There is no sort of differential equation that's more friendly, user friendly that we encounter.

196
00:22:58,580 --> 00:23:03,139
We solve such equations by using an integrating factor just to get this into standard form.

197
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I would write this as I would actually write this, as do you note by the x plus and omega over h bar x u nought.

198
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This is the sort of standard form for a first order linear equation, which you should remember from Professor Ross's course or whatever.

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In the first year it has an integrating factor which is e to the integral of the coefficient of the linear of the constant well,

200
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the not derived term this term each of the integral am omega over h bar x d x, which is clearly e to the omega x squared over to h bar.

201
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That's its integrating factor. And the equation is then if you multiply it by the integrating factor,

202
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the equation says that the x of the integrating factor times you nought is equal to the right side, which happens to be zero.

203
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Ergo, this quantity here is a constant.

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Ergo you note is equal to e to the minus and omega x squared over two, which I want to write is e to the minus x squared over four l squared.

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Why do I want to do that? I want to do that because the probability associated with X,

206
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the probability of finding your particle at x which is equal to you nought mod squared will then be e to the minus x squared over to l squared.

207
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So this is this is a normal distribution. With dispersion.

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L So the reason I want to I want to get this into this form is in order because because I can identify that is the Gaussian width of the distribution,

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the width of the Gaussian distribution.

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Also, once I say that this is the probability distribution for my knowledge of statistics and stuff, I know what the normalising constant has to be.

211
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I know that it's two pi l square root to pile squared.

212
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Right, because that's the factor you need to normalise a Gaussian distribution.

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Oops, I'm missing a minus sign. Crucial, right. Crucial.

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So what does l have to be in order that this form is the same as that form?

215
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It. It's, I think, simple algebra to show that l has to be, uh, h bar over to omega.

216
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What l squared has to be that. Let's check that this has the right dimensions.

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00:25:48,170 --> 00:25:53,209
One should always be checking one's dimensions in physics. This has the dimensions and momentum times distance.

218
00:25:53,210 --> 00:25:56,780
So this has dimensions of. M. X. P.

219
00:25:57,910 --> 00:26:02,020
A. So.

220
00:26:02,030 --> 00:26:07,519
So this is sorry. This is m x v says what I should have said.

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Momentum times and the x right.

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00:26:12,770 --> 00:26:16,159
Talking about yes and the x right.

223
00:26:16,160 --> 00:26:20,450
And the x is the dimensional structure here of.

224
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Dimensions here. So indeed this thing has dimensions of length squared.

225
00:26:33,070 --> 00:26:37,840
So L is indeed the length. So it's it's. It's okay. So what did we learn?

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We've learned that the ground state of a harmonic oscillator has a wave function, which is a Gaussian with a characteristic width given by this.

227
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And crucially, what we've already studied, right?

228
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We've already studied distributions of particles which are Gaussian, have way functions which square up to Gaussians.

229
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And we now know what the amplitude distribution would be or the probability distribution is for momentum from what we did before.

230
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We will have the probability of the momentum will also be a Gaussian.

231
00:27:19,240 --> 00:27:30,639
It'll be each of the minus P squared over two, should we call it sigma p squared over some square root to pi sigma P squared.

232
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And remember, we have the uncertainty principle which said that the dispersion in x times,

233
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the dispersion in momentum is equal to h for over two for the Gaussians.

234
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This was a result we established when talking about the free particle.

235
00:27:48,220 --> 00:27:57,730
So that tells me that sigma P is equal to h bar over two L so, so, so there's some,

236
00:27:58,690 --> 00:28:06,220
there's some characteristic width, uh, there's some characteristic momentum that the particle has.

237
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Now this. This set in the ground state.

238
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A particle is not stationary, it is moving with it with a characteristic amount of energy.

239
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It it's also not at the bottom of the potential. Well, it has a characteristic amount of average potential energy.

240
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So we've come to the conclusion we have an example here of one of the most important,

241
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perhaps the most important prediction of quantum mechanics, which is the existence of zero point energy.

242
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The basic issue is that if we're trying to minimise the energy of the particle,

243
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which clearly the ground state by definition does minimise the energy of the particle is the state with the lowest energy by definition.

244
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If we're trying to minimise this energy, we want to get the potential energy, energy to be as small as possible.

245
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That clearly means moving towards the origin and getting as close to the origin as you can.

246
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But there is isn't because of the uncertainty relationship,

247
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because the more narrowly confined you are in real space, the more uncertain your momentum has to be.

248
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If you if you restrict yourself in too much in position to be too much at the bottom of the potential,

249
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well, you will have a large uncertainty in your momentum and you'll have kinetic energy.

250
00:29:28,690 --> 00:29:32,110
So. So in practice in the ground state,

251
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there's a compromise between being being reasonably close to the origin and having a reasonably small kinetic energy.

252
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So because of the uncertainty relation, quantum mechanical systems in their lowest states have a finite extent spatially.

253
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Even though they even though really it's a point particle,

254
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but because of the there's a there's a finite extent in which you'll find the particle and a finite kinetic energy.

255
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And this is a totally. I hope you see that this is a specific example of a totally general phenomenon.

256
00:30:04,050 --> 00:30:09,330
We have to expect to occur always when we are considering particles trapped in some kind of potential wells.

257
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And this is this is enormously important because it's exactly this physics

258
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we will see that is exactly this physics which determines the size of atoms.

259
00:30:18,720 --> 00:30:24,600
Electrons. So atoms here are typically pretty close to their near enough in their ground states.

260
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And the size of these atoms is determined by the electron if if the atom gets any smaller and indeed you, you know,

261
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you take you take a piece of steel or something and you stuff it into a press and squeeze the thing down.

262
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It will get smaller, but it will resist violently. You'll have to do work.

263
00:30:40,170 --> 00:30:42,180
You have to increase its energy to make it smaller.

264
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And what happens is that in order to make it smaller in real space, you have to give it more kinetic energy by the uncertainty principle.

265
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And that's the work that you do, squashing it down. And you can see that idea worked out quantitatively in one of the later chapters of the book.

266
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It so the size of atoms is determined by this by this zero point energy business and the uncertainty principle.

267
00:31:06,150 --> 00:31:12,330
And interestingly, the mass of protons and neutrons is not entirely but is overwhelmingly accounted for by the

268
00:31:12,540 --> 00:31:17,790
kinetic energy of the quarks and gluons inside there which are moving around relativistic li.

269
00:31:17,970 --> 00:31:26,850
They're in a very deep potential. Well and because and very narrowly confined right into ten to the -15 metres in order to be confined,

270
00:31:26,850 --> 00:31:29,970
even though the fairly massive particles into this very small space,

271
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they have to have a lot of kinetic energy and that and the the mass associated with that kinetic energy accounts for most of the mass,

272
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you know, of us. That's what it mostly is. So this zero point energy phenomenon is extremely general and enormously important.

273
00:31:45,060 --> 00:31:47,880
And here we have the simplest, the classical example.

274
00:31:48,690 --> 00:32:03,720
In fact, you see the if we say H is equal to one over two M of P squared plus m m squared, omega squared, x squared.

275
00:32:04,710 --> 00:32:14,040
And we put in the uncertainty relation, we say that that x squared if we if we say that x squared is sorry,

276
00:32:14,040 --> 00:32:25,350
if we say that p squared is equal to h while squared over two x gets right sorry over four because I've squared it over for x squared.

277
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Right. So this is so for the ground, say x squared is essentially the uncertainty in x.

278
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So it's associated with the uncertainty momentum. In the same way if you stuff that into this, you put this relationship in,

279
00:32:37,410 --> 00:32:49,800
you find that H is one over two M is going to be whatever it is h by squared of a for x squared plus m squared, omega squared, x squared.

280
00:32:50,100 --> 00:32:55,739
This now is a function. Maybe I shouldn't call it x squared. Maybe we should call it maybe we should call it l.

281
00:32:55,740 --> 00:33:05,570
Actually, perhaps that would be more helpful. So I'm saying that x squared is on the order of L squared and p squared is on the order of this.

282
00:33:05,900 --> 00:33:09,020
So here we have a function of L and the minimum.

283
00:33:09,020 --> 00:33:14,130
If you if you ask yourself what, what's the, what value of L, does this function have a minimum?

284
00:33:14,150 --> 00:33:17,720
The answer is it's that value that we gave up there from quantum mechanics.

285
00:33:18,050 --> 00:33:25,610
So it's really true that the that L is being chosen to minimise the energy given the constraints imposed by the uncertainty principle.

286
00:33:34,640 --> 00:33:40,520
Okay but we so that we found the ground state way function. It would now be useful I guess to show how we should calculate.

287
00:33:41,550 --> 00:33:47,340
The. Sort of a natural order here.

288
00:33:49,090 --> 00:33:53,770
Yeah. So what we want to do, let's get the, let's get the first excited state as an example.

289
00:33:54,220 --> 00:34:01,370
So first excited state. Excited state wave function.

290
00:34:02,420 --> 00:34:07,830
So you won of X, which is by definition x one.

291
00:34:08,890 --> 00:34:14,450
What do we know? We know that one is equal to one over the square root of one.

292
00:34:14,510 --> 00:34:22,890
Time's a dagger working on the ground state. So if I bra through by X, that's that.

293
00:34:23,000 --> 00:34:31,740
That tells me that you won of x, which is equal to x one is equal to one over the sorry the one of the square.

294
00:34:31,830 --> 00:34:35,930
One doesn't need to be written any more, but this is one over the square root of N plus one.

295
00:34:35,930 --> 00:34:39,740
And here is note of a dagger.

296
00:34:41,180 --> 00:34:51,060
Which is. And Omega X minus II minus IP.

297
00:34:51,090 --> 00:34:57,780
Yes. Over the over the square root of two and H bar omega.

298
00:35:10,000 --> 00:35:20,200
Now what we want to do is it's helpful actually to find out what is to rewrite this in terms of l this characteristic length there.

299
00:35:24,170 --> 00:35:31,240
So let's just say what a dagger is in terms of l to m h bar omega.

300
00:35:31,250 --> 00:35:37,160
Well was. So if you if you if you take that equation up there that defines l.

301
00:35:39,040 --> 00:35:48,340
And you multiply both sides by the square root of two m omega and then you multiply through by a square root of h.

302
00:35:50,660 --> 00:35:53,690
You find. So I need to write this down.

303
00:35:53,900 --> 00:36:00,950
So l is equal to the square root of H bar over to omega multiplying through.

304
00:36:00,950 --> 00:36:12,860
By this I find that the square root of two and omega is equal to the square root of h of l if I multiply through by h bar.

305
00:36:13,430 --> 00:36:22,580
That's bar by the way. Sorry I multiply through by h bar. I find the square root of 2 a.m. h bar omega is in fact equal to H over l.

306
00:36:23,060 --> 00:36:29,660
So this factor here is equal to h over L. So I can say that a dagger is equal to.

307
00:36:31,680 --> 00:36:44,360
This is equal to an omega x on the bottom is h over l so an L here and an h bar there minus i.

308
00:36:45,930 --> 00:36:52,860
I need to. I have an H bar on the bottom and an L on the top.

309
00:36:53,570 --> 00:36:57,100
Times p. Let's keep working.

310
00:36:57,700 --> 00:37:02,409
This stuff is looking remarkably like El all over again.

311
00:37:02,410 --> 00:37:10,600
Right? If I would multiply this by two on the top and two on the bottom, then this would become one of l squared.

312
00:37:10,870 --> 00:37:17,030
So this is is is is equal to the L squared would would cancel this and I would have that.

313
00:37:17,050 --> 00:37:20,140
This was X over to L minus.

314
00:37:21,160 --> 00:37:25,450
Now let's put this into the position representation. In the position representation.

315
00:37:25,600 --> 00:37:29,170
This is minus h bar D by the x.

316
00:37:30,960 --> 00:37:36,870
So the. Yes.

317
00:37:37,110 --> 00:37:42,870
So the eyes get together and make a minus sign and the minuses cancel each other.

318
00:37:42,870 --> 00:37:48,810
So we have an overall minus sign. The H bars cancel and this becomes l d by the x.

319
00:37:49,980 --> 00:37:53,219
A dagger was billed as being dimensionless.

320
00:37:53,220 --> 00:37:56,820
Is it dimensionless? Yes. Because they have an X over L and an L over x.

321
00:37:57,900 --> 00:38:01,590
So this is a handy formula for future reference.

322
00:38:05,390 --> 00:38:09,379
So let's find out what? Sorry, I shouldn't have written that complicated expression up there.

323
00:38:09,380 --> 00:38:18,590
It wasn't helpful that you one of x is equal to this baby, this animal working on.

324
00:38:22,590 --> 00:38:27,149
What does it work on? It works on you. Zero of x. But what is you?

325
00:38:27,150 --> 00:38:39,870
Zero of x? Its its business end is easier than minus x squared over four l squared.

326
00:38:40,200 --> 00:38:44,880
And under here I have to have a two pie l squared to the quarters power.

327
00:38:45,960 --> 00:38:51,450
This factor comes in. I said what the normalisation constant was.

328
00:38:52,350 --> 00:38:59,130
So just to see where that comes from, I needed P of x to have this.

329
00:38:59,970 --> 00:39:06,690
So we obtained u as I should have set a constant k times this.

330
00:39:06,690 --> 00:39:13,979
Right? There was an arbitrary constant in this in this arbitrary constant of integration, which is in fact going to be the normalising constant.

331
00:39:13,980 --> 00:39:17,700
So I know now that the wave function is behaves like this,

332
00:39:18,090 --> 00:39:24,600
which gives me a probability that looks like this, the correct normalisation of the probability is this.

333
00:39:25,080 --> 00:39:34,920
So what I need to do now is to say that in order to get things to work out well, I should replace that K with the quarter power of what's inside here.

334
00:39:34,920 --> 00:39:42,080
So when you square it up. So when you square it up, you find the right normalising constants of the probability.

335
00:39:42,860 --> 00:39:47,330
So. So we've got the ground state now at last properly normalised.

336
00:39:47,960 --> 00:39:56,840
It has that quarter power and this now this is going to come out because we've we've paid proper attention to normalisation.

337
00:39:56,850 --> 00:40:01,400
This will come out correctly normalise. And it's what happens is very simple.

338
00:40:01,700 --> 00:40:12,799
We when we do this differentiation, we're going to bring down a minus X over over to L squared.

339
00:40:12,800 --> 00:40:21,590
And this L will cancel that and we'll be in and cancel this and we'll have an X over to L coming from here we've got an X over to L coming from there.

340
00:40:21,980 --> 00:40:28,100
So the whole thing at the end of the day is one over this two pi l squared.

341
00:40:28,820 --> 00:40:36,290
One quarter of power x of l e to the minus x squared over four l squared.

342
00:40:36,320 --> 00:40:48,830
That's the first excited state wave function. To find the second excited state wave function would use this selfsame operator.

343
00:40:49,420 --> 00:41:00,319
Not going to do this, but let's just see what it would look like. You two would be x over two l minus l d by the x one over two factorial.

344
00:41:00,320 --> 00:41:10,129
Sorry, one over the square root of two. That's the one over square root of n plus one operating on you one which is x of l e of

345
00:41:10,130 --> 00:41:18,740
the minus x squared over four l squared over two pi l squared two one fourth is power.

346
00:41:22,200 --> 00:41:26,040
And you can see that what's going to happen is we're going to get an X squared term times.

347
00:41:26,040 --> 00:41:29,280
These are the garbage we're going to get from this differentiation.

348
00:41:30,000 --> 00:41:35,000
We're going to get an X term from this from the diff.

349
00:41:35,580 --> 00:41:43,320
We'll get various things. We're basically going to get an x squared term and when differentiating away this will get a term in X to the nothing.

350
00:41:44,460 --> 00:41:48,090
And when we bring this down, we'll get another x squared term as well.

351
00:41:48,090 --> 00:41:54,780
And we multiply this and it's going to be squared. So we're going to get terms in x squared, an x to the nothing times e to the minus thingy.

352
00:41:55,650 --> 00:42:10,030
So this is going to be. A poly polynomial of degree to.

353
00:42:13,650 --> 00:42:16,590
It goes by the name of a hermit polynomial. It doesn't matter.

354
00:42:18,180 --> 00:42:22,830
And every time we use this operator, we're going to think we're going to get a more and more elaborate polynomial.

355
00:42:23,430 --> 00:42:25,800
Can you see that? That's that's what's going to be the consequence?

356
00:42:26,160 --> 00:42:31,080
Well, wave functions are all going to be this Gaussian that came came with the ground state.

357
00:42:31,560 --> 00:42:34,830
And then they're going to be times polynomials, which are going to be of order.

358
00:42:34,830 --> 00:42:46,860
N So the general state UN of X is going to be my polynomial h and of x e to the minus x squared over four l squared.

359
00:42:46,860 --> 00:42:51,419
I'm not paying proper attention to the normalisation at the moment, and it's straightforward to find out what these are.

360
00:42:51,420 --> 00:42:56,850
You only have to differentiate. Don't do any clever, any things, just differentiate and they will all drop into your lap.

361
00:42:58,050 --> 00:43:06,750
Something important to notice is that the ground state wave function is an even function of x right e to the minus x squared.

362
00:43:06,960 --> 00:43:14,280
The first excited state wave function is an odd function of x because this operator is odd, right?

363
00:43:14,280 --> 00:43:17,669
It has one power of x in both places is numbers. It changes sine.

364
00:43:17,670 --> 00:43:24,299
If you turn x to minus x, this one is going to be an even function of x because you because we're going to apply

365
00:43:24,300 --> 00:43:28,410
another odd operator to a thing that's odd and we'll end up with an even result.

366
00:43:29,160 --> 00:43:34,620
So. So the ground state. Well, ground state.

367
00:43:36,930 --> 00:43:42,600
And isn't even a function of X.

368
00:43:45,360 --> 00:43:51,510
First excited. It's an odd function.

369
00:43:54,780 --> 00:44:03,090
And the second excited. So in fact you kn of x is even.

370
00:44:04,680 --> 00:44:09,620
If N is even. An odd. Otherwise.

371
00:44:13,330 --> 00:44:19,980
We'll meet this phenomenon in other in other cases that it's it's very often the case.

372
00:44:20,020 --> 00:44:26,170
The ground state is even in the first exciting resort and the next one is even in the next one is odd and so on like that for similar reasons.

373
00:44:27,640 --> 00:44:34,540
And quantum mechanics has its own jargon for this. It says that this is an odd parity state sorry, an even parity state.

374
00:44:34,540 --> 00:44:43,270
This isn't even parity state parity. Parity just means is it even or is it all the wave function and this isn't even parity state.

375
00:44:45,700 --> 00:44:52,360
We'll have more to say about parity in a general context later on when we're covering the material in Chapter four.

376
00:44:52,750 --> 00:44:56,970
So I think both. Sorry.

377
00:44:57,420 --> 00:45:01,680
First, though, this one is odd. Excuse me. I was not sure which line I was on.

378
00:45:01,890 --> 00:45:12,600
This one is even the ground state is even. The first excited state is odd and this is is has a parity.

379
00:45:13,170 --> 00:45:22,530
We say this has a parity minus one to the end so that you know that jargon is used.

380
00:45:22,530 --> 00:45:29,979
Sometimes I wouldn't worry about it. It turns out to be useful to know we'll find it's useful to know whether your wave

381
00:45:29,980 --> 00:45:36,910
function is even or old and then enables you to short circuit various computations.

382
00:45:40,700 --> 00:45:45,780
Okay. We can.

383
00:45:46,200 --> 00:45:50,870
Yeah. All right. Round a border and.

384
00:45:57,160 --> 00:46:02,050
Let's. Yeah.

385
00:46:02,060 --> 00:46:09,160
Let's have a go at this. Let's work out the expectation value in the excited state of x squared.

386
00:46:10,180 --> 00:46:15,570
So we want and it will be nice now to build some connection to classical physics.

387
00:46:15,580 --> 00:46:19,059
Can we connect these results to two classical physics?

388
00:46:19,060 --> 00:46:22,540
Classical physics? As I've said several times, it's all about expectation values.

389
00:46:22,550 --> 00:46:27,760
The connection from quantum mechanics to classical physics occurs through expectation values.

390
00:46:28,240 --> 00:46:32,770
So let's work out this expectation value, which is going to enable us, for example,

391
00:46:32,770 --> 00:46:39,460
to work out what the mean potential energy is because the potential energy is proportional to x squared in the excited state.

392
00:46:41,260 --> 00:46:46,090
Now. How do we work this out? What we do is we observe that a.

393
00:46:49,470 --> 00:46:53,460
Now. We had a dagger. We went to some trouble.

394
00:46:53,880 --> 00:46:57,180
Yeah, we got. We got. We got. Here is.

395
00:46:57,180 --> 00:47:16,059
Here is a dagger. A is going to be L3 is going to be x over to l plus i l upon h bar p and a dagger.

396
00:47:16,060 --> 00:47:23,740
We've got there. We've got there is x over to L minus i, l p over h bar.

397
00:47:24,430 --> 00:47:32,950
So if you add these two equations, you discover that a times l sorry, a plus a l a plus.

398
00:47:32,950 --> 00:47:37,270
A dagger is equal to x.

399
00:47:38,380 --> 00:47:47,440
So a handy this is a very handy relationship. Expresses the x operator as a sum of annihilation and creation operators.

400
00:47:47,650 --> 00:47:52,630
Times L. Right, I've just added these two equations the momentums, the momenta of cancelled on each other.

401
00:47:53,020 --> 00:47:55,710
These have added up to give us an x over l here we've had a plus,

402
00:47:55,730 --> 00:48:02,800
a dagger I've multiplied through by L so when I want to work out this expectation value.

403
00:48:07,390 --> 00:48:10,510
I can replace each of those Xs with this thing here.

404
00:48:10,540 --> 00:48:20,050
L squared comes out because it's only a number and I have an into a plus eight dagger squared and.

405
00:48:22,380 --> 00:48:38,160
Let's multiply this out. It's l squared into an x squared plus a dagger squared plus AA dagger plus dagger a.

406
00:48:48,110 --> 00:48:53,270
Now what's this a squared applied to that is proportional to when plus two.

407
00:48:53,840 --> 00:48:59,690
Right. Because sorry n minus two because each of these A's takes away a unit of excitation.

408
00:49:00,050 --> 00:49:05,750
So a squared times that is proportional to n minus two with n minus two is orthogonal to that.

409
00:49:06,050 --> 00:49:08,840
So that makes no contribution to this expectation value.

410
00:49:08,840 --> 00:49:16,159
Similarly, a dagger squared on this produces some multiple of N plus two, which is orthogonal to that.

411
00:49:16,160 --> 00:49:20,930
So that doesn't contribute. So these two terms don't contribute to the expectation value.

412
00:49:21,800 --> 00:49:24,830
These terms jolly well do contribute to the expectation value.

413
00:49:25,070 --> 00:49:32,750
We know that a on this produces root and minus root n times, root n times.

414
00:49:33,740 --> 00:49:38,030
What we've at times end minus one and this produces routine.

415
00:49:38,840 --> 00:49:43,700
Let me write that out. So this is going to be l squared of rn.

416
00:49:47,510 --> 00:49:54,860
Hey now wants a dagger on this is going to produce root and plus one of end plus one.

417
00:49:59,200 --> 00:50:04,510
Right this this a dagger working on that will produce route end plus one times and plus one.

418
00:50:05,320 --> 00:50:08,590
And then uh. And now I've left the other way to be done.

419
00:50:08,740 --> 00:50:18,490
And here we going to you we're going to say that this is equal to n a dagger times the result of a working on that which is root n times n minus one.

420
00:50:18,730 --> 00:50:22,540
Remember that's the square root of the largest integer occurring in the equation.

421
00:50:24,130 --> 00:50:31,390
And then this a is going to produce a root and plus one times N which will couple with this and produce produce one.

422
00:50:31,390 --> 00:50:35,350
So this is going to be L squared, right?

423
00:50:35,350 --> 00:50:39,340
This is going to lower N plus one down to n, which will produce a one when it meets this.

424
00:50:39,550 --> 00:50:42,520
And we'll have another of these square roots. So we're going to have N plus one.

425
00:50:43,360 --> 00:50:51,430
And when we use this on this, this N minus one is going to be raised back to N, which will produce one when it meets this.

426
00:50:51,700 --> 00:50:55,780
But we'll get another square root of N in the process. So we'll have plus n.

427
00:50:56,380 --> 00:51:01,690
So in other words, it's equal to two N plus one L squared.

428
00:51:03,770 --> 00:51:06,330
So my time is up. What I want to do.

429
00:51:06,350 --> 00:51:15,260
So we've discovered we've discovered something interesting, which is the expectation value of x squared is two and plus one l squared.

430
00:51:16,160 --> 00:51:19,570
We already knew. I think about what? Did we already know about it. No, we didn't.

431
00:51:19,580 --> 00:51:23,390
In a certain sense, yes, we did. We use for the ground state and equals nought.

432
00:51:23,690 --> 00:51:30,349
We knew that it was a we knew the probability distribution was a Gaussian and we knew that the dispersion of that Gaussian,

433
00:51:30,350 --> 00:51:34,130
i.e. the expectation value of X squared was L squared.

434
00:51:35,360 --> 00:51:46,400
We've now discovered how the how the dispersion increases when we when we add excitations and the probability distribution gets broader.

435
00:51:46,700 --> 00:51:49,970
But tomorrow, I want to connect this to classical physics.