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So, now we proceed with the technical 
solution of the quantum harmonic 

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oscillator problem. 
And I should mention that there exists 

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many equivalent ways to solve this 
problem. 

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But today I am going to present a purely 
algebraic solution. 

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Which is based on so called creation and 
annihilation operators. 

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I'll introduce them in this video. 
And as you will see the harmonic 

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oscillator spectrum and the properties of 
the[UNKNOWN] functions will follow just 

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from an analysis of this creation 
annihilation operators and their 

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commutation or leashes. 
But we're not going to even write down 

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the Schrodinger's equation as the 
differential equation, we're not going to 

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worry too much about any boundary 
conditions. 

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So these operators, and their commutation 
relations are going to be sufficient for 

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us to determine pretty much everything we 
want to know. 

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And the problem that we're actually 
solving is the Eigen value problem for 

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this operator H with this quadratic 
potential corresponding to the harmonic 

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oscillator. 
So again the Eigen value problem is H psy 

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is equal to E psy. 
Now there are only three dimensional 

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quantities that appear in that problem. 
On which our final results, in particular 

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the energy spectrum, can possibly depend, 
and these quantities are the particle 

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mass, the frequency of the oscillator, 
and the blank constant, which is always, 

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which is always there. 
And it turns out that you can construct 

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just one parameter out of this guy, which 
has the physical dimension of energy, and 

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this parameter has each times Amanda. 
So this has the physical image of energy. 

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So whatever our spectrum is going to look 
like. 

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So it must be it must scale as H omega. 
It must be proportional to H omega. 

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So motivated by this let me write down my 
Hamiltonian H divided by this H omega and 

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its going to be simply I first, I first 
will write the potential energy term, m 

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omega x squared over 2 h bar plus the 
kinetic energy p squared over 2 m omega 

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over h bar. 
So I just switched the order, and each of 

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these two terms appear in this 
Hamiltonian. 

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And the, the next step I'm going to use 
is also going to be pretty strange from 

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any point of view. 
So, I'm going to just write the first 

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term as the square root of m omega over 2 
h bar x squared plus the second term 

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likewise is going to be p over 2m omega h 
bar squared. 

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And now so the reason I rooted this way 
is because I want to use an analog of the 

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following expression that we know from 
elementary calculus namely that if we 

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have two variables let's say A squared 
minus B squared. 

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We can write it as A minus B times A plus 
B or an equivalent expression to this 

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involving complex numbers, if we have A 
squared plus B squared so we can write it 

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as A minus iB, A plus iB. 
And when we square the imaginary constant 

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its going to give rise to the plus sign 
here. 

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So in this expression I'm going to 
associate the first term with A and the 

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second term with B and so what I'm going 
to write is going to be the following. 

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So I will represent my Hamiltonian 
divided by h Amanda. 

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This energy scale and the problem as well 
this guy square root of m omega over 2 h 

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bar x minus i momentum divided by this 
whole thing times the same thing but with 

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a plus sign here. 
So m omega over 2 h bar x plus i, p, over 

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2m omega h, 1. 
At this stage, I have to admit that I 

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have cheated very seriously in this 
derivation. 

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And this equation contains, an error and 
as a matter of fact there is some missing 

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terms that I have omitted. 
And in the next video quiz you are 

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supposed to catch me and tell me where 
exactly I have cheated. 

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So hopefully most of you have figured out 
where the problem is. 

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And of course the problem is in that the 
objects that appear in our quantum 

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mechanical problem are not just some 
variables A and B. 

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These are operators and for the operators 
in particular for operators x and p, it 

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actually matters in which order they 
appear in an equation. 

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So x times p is not equal to p times x. 
So in other words, these operators do not 

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commute. 
And so if we take a if we try to 

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re-derive this identity for these 
operators, we're going to see of course 

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that there are going to be cross terms 
appearing. 

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And so therefor there is an additional, 
there is an addition to this, to this 

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guy. 
So let me let me write it explicitly. 

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So this terms that, these terms that I 
have omitted, so it's equal to minus i 

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over 2 h bar. 
the commute of x and p. 

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So this commute of x and p is equal to xp 
minus px. 

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You can verify that this term indeed 
appears just by expanding this product 

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term by term and making sure that we 
reproduce the Hamiltonian in the left 

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hand side. 
One other thing that we can verify by 

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looking at this expression is that the 
first bracket is at Hermitian conjugate 

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of the second bracket. 
So indeed x and p are physical operators, 

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and as such they are Hermitian operators, 
so x dagger is equal to x, and p dagger 

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is equal to p by definition. 
Therefore, if I commission conjugate 

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let's say, the second bracket of x will 
remain x, b will remain b. 

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All the Gaussian here are real and the 
only thing that's going to happen is i 

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will become minus sum. 
So we'll reproduce the first bracket in 

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this expression. 
Based on this fact let me introduce a new 

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operator. 
So we'll just call the second bracket an 

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operator a. 
And the first bracket is going to be 

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Hermitian conjugate to a therefore a 
dagger. 

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I am going to be ahead of myself, let me 
mention that this operators a dagger and 

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a are called creation and annihilation 
operators. 

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And the following discussion and 
derivation will contain, a proof that, 

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these operators, these guys indeed, 
deserve the names of creation, 

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annihilation operators. 
But to explain the reason, behind this 

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terminology right away, let me advertise 
the main result before we actually derive 

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it. 
And the main result here, is the energy 

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spectrum, or the[UNKNOWN]. 
Which we'll see, contains a series of 

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equidistant energy levels. 
That is energy levels such that any 

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neighboring pair of levels are separated 
from one another by the same energy. 

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And this energy happens to be each omega, 
exactly the energy scale which is cast 

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previously. 
And the energy of the ground state, the 

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lowest energy state is each omega over 2, 
and by the way this h omega over 2 comes 

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exactly from this, additional term that 
we have in this identity. 

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And so, the importance of creation 
annihilation operators are the following. 

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So let's say if we prepare our, our 
quart, quantum oscillator, in the ground 

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state. 
So let me sim symbolically represent it 

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by dot here. 
So let's say we have an oscillator in the 

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ground state. 
And if we apply a creation operator a 

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dagger, it will create essentially, 
quantum of energy h omega, by promoting 

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this oscillator from the ground state to 
the first excited state. 

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If we apply it again, so then, we will go 
from the first excited state to the 

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second excited state etcetera, etcetera. 
So if we apply, let's say, a dagger 10 

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times, we will go from the ground state 
to the 10ths excited state. 

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And you can probably guess that the 
action of the operator a is opposite to 

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it. 
So if we have say you have a quantum 

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state with n equals 1. 
So this is the first excited state. 

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By applying a, we're going to go back to 
the ground state, okay. 

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So essentially this operator is a and 
they [UNKNOWN], they move us between 

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these states in the in the[UNKNOWN]. 
And in the next video, we're going to 

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prove all these statements and this 
proof. 

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We'll rely, in a very essential way, on 
various commutation relations between the 

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operators involved here, x, p, and then a 
dagger.