So, now we proceed with the technical 
solution of the quantum harmonic 
oscillator problem. 
And I should mention that there exists 
many equivalent ways to solve this 
problem. 
But today I am going to present a purely 
algebraic solution. 
Which is based on so called creation and 
annihilation operators. 
I'll introduce them in this video. 
And as you will see the harmonic 
oscillator spectrum and the properties of 
the[UNKNOWN] functions will follow just 
from an analysis of this creation 
annihilation operators and their 
commutation or leashes. 
But we're not going to even write down 
the Schrodinger's equation as the 
differential equation, we're not going to 
worry too much about any boundary 
conditions. 
So these operators, and their commutation 
relations are going to be sufficient for 
us to determine pretty much everything we 
want to know. 
And the problem that we're actually 
solving is the Eigen value problem for 
this operator H with this quadratic 
potential corresponding to the harmonic 
oscillator. 
So again the Eigen value problem is H psy 
is equal to E psy. 
Now there are only three dimensional 
quantities that appear in that problem. 
On which our final results, in particular 
the energy spectrum, can possibly depend, 
and these quantities are the particle 
mass, the frequency of the oscillator, 
and the blank constant, which is always, 
which is always there. 
And it turns out that you can construct 
just one parameter out of this guy, which 
has the physical dimension of energy, and 
this parameter has each times Amanda. 
So this has the physical image of energy. 
So whatever our spectrum is going to look 
like. 
So it must be it must scale as H omega. 
It must be proportional to H omega. 
So motivated by this let me write down my 
Hamiltonian H divided by this H omega and 
its going to be simply I first, I first 
will write the potential energy term, m 
omega x squared over 2 h bar plus the 
kinetic energy p squared over 2 m omega 
over h bar. 
So I just switched the order, and each of 
these two terms appear in this 
Hamiltonian. 
And the, the next step I'm going to use 
is also going to be pretty strange from 
any point of view. 
So, I'm going to just write the first 
term as the square root of m omega over 2 
h bar x squared plus the second term 
likewise is going to be p over 2m omega h 
bar squared. 
And now so the reason I rooted this way 
is because I want to use an analog of the 
following expression that we know from 
elementary calculus namely that if we 
have two variables let's say A squared 
minus B squared. 
We can write it as A minus B times A plus 
B or an equivalent expression to this 
involving complex numbers, if we have A 
squared plus B squared so we can write it 
as A minus iB, A plus iB. 
And when we square the imaginary constant 
its going to give rise to the plus sign 
here. 
So in this expression I'm going to 
associate the first term with A and the 
second term with B and so what I'm going 
to write is going to be the following. 
So I will represent my Hamiltonian 
divided by h Amanda. 
This energy scale and the problem as well 
this guy square root of m omega over 2 h 
bar x minus i momentum divided by this 
whole thing times the same thing but with 
a plus sign here. 
So m omega over 2 h bar x plus i, p, over 
2m omega h, 1. 
At this stage, I have to admit that I 
have cheated very seriously in this 
derivation. 
And this equation contains, an error and 
as a matter of fact there is some missing 
terms that I have omitted. 
And in the next video quiz you are 
supposed to catch me and tell me where 
exactly I have cheated. 
So hopefully most of you have figured out 
where the problem is. 
And of course the problem is in that the 
objects that appear in our quantum 
mechanical problem are not just some 
variables A and B. 
These are operators and for the operators 
in particular for operators x and p, it 
actually matters in which order they 
appear in an equation. 
So x times p is not equal to p times x. 
So in other words, these operators do not 
commute. 
And so if we take a if we try to 
re-derive this identity for these 
operators, we're going to see of course 
that there are going to be cross terms 
appearing. 
And so therefor there is an additional, 
there is an addition to this, to this 
guy. 
So let me let me write it explicitly. 
So this terms that, these terms that I 
have omitted, so it's equal to minus i 
over 2 h bar. 
the commute of x and p. 
So this commute of x and p is equal to xp 
minus px. 
You can verify that this term indeed 
appears just by expanding this product 
term by term and making sure that we 
reproduce the Hamiltonian in the left 
hand side. 
One other thing that we can verify by 
looking at this expression is that the 
first bracket is at Hermitian conjugate 
of the second bracket. 
So indeed x and p are physical operators, 
and as such they are Hermitian operators, 
so x dagger is equal to x, and p dagger 
is equal to p by definition. 
Therefore, if I commission conjugate 
let's say, the second bracket of x will 
remain x, b will remain b. 
All the Gaussian here are real and the 
only thing that's going to happen is i 
will become minus sum. 
So we'll reproduce the first bracket in 
this expression. 
Based on this fact let me introduce a new 
operator. 
So we'll just call the second bracket an 
operator a. 
And the first bracket is going to be 
Hermitian conjugate to a therefore a 
dagger. 
I am going to be ahead of myself, let me 
mention that this operators a dagger and 
a are called creation and annihilation 
operators. 
And the following discussion and 
derivation will contain, a proof that, 
these operators, these guys indeed, 
deserve the names of creation, 
annihilation operators. 
But to explain the reason, behind this 
terminology right away, let me advertise 
the main result before we actually derive 
it. 
And the main result here, is the energy 
spectrum, or the[UNKNOWN]. 
Which we'll see, contains a series of 
equidistant energy levels. 
That is energy levels such that any 
neighboring pair of levels are separated 
from one another by the same energy. 
And this energy happens to be each omega, 
exactly the energy scale which is cast 
previously. 
And the energy of the ground state, the 
lowest energy state is each omega over 2, 
and by the way this h omega over 2 comes 
exactly from this, additional term that 
we have in this identity. 
And so, the importance of creation 
annihilation operators are the following. 
So let's say if we prepare our, our 
quart, quantum oscillator, in the ground 
state. 
So let me sim symbolically represent it 
by dot here. 
So let's say we have an oscillator in the 
ground state. 
And if we apply a creation operator a 
dagger, it will create essentially, 
quantum of energy h omega, by promoting 
this oscillator from the ground state to 
the first excited state. 
If we apply it again, so then, we will go 
from the first excited state to the 
second excited state etcetera, etcetera. 
So if we apply, let's say, a dagger 10 
times, we will go from the ground state 
to the 10ths excited state. 
And you can probably guess that the 
action of the operator a is opposite to 
it. 
So if we have say you have a quantum 
state with n equals 1. 
So this is the first excited state. 
By applying a, we're going to go back to 
the ground state, okay. 
So essentially this operator is a and 
they [UNKNOWN], they move us between 
these states in the in the[UNKNOWN]. 
And in the next video, we're going to 
prove all these statements and this 
proof. 
We'll rely, in a very essential way, on 
various commutation relations between the 
operators involved here, x, p, and then a 
dagger.