In this video we are going to use the 
creation and angulation operator 
[UNKNOWN] introduced in the previous 
segment. 
To generate the energy spectrum of the 
harmonic oscillator. 
And this calculation as we discussed is 
going to involve in very important way 
calculating various commentators of the 
operators. 
So what we just proved in the previous 
video was that the harmonic oscillator 
Hamiltonian can be represented as so. 
As the sum of two terms. 
The first term is, a, dagger, a, where, 
a, dagger is this creation operator and, 
a, is this annihilation operator. 
And also there was a second term which is 
proportional to the commutator x of x and 
p position and momentum. 
And by the way this commutator of x and p 
is one of the so callled canonical 
commutation relations in quantum 
mechanics. 
And we actually should have discussed it 
in more details earlier in the course, 
but better later than never. 
So let me just first give you this result 
for this commutator. 
And this commutator in fact is 
proportional just to so-called c number, 
or in the, the quantum, early quantum 
mechanical jerk one. 
Or in other words to the identity 
operators. 
So in, in other words, this commutator 
effects and p, is not an operator at all. 
Its actually just a constant and to see 
this we can go to the position 
representation. 
And let's see in postition representation 
x times p acting on a wave function psi 
is equal to x. 
Which is just multiplication operator, 
followed by minus i, h bar d over dx, 
acting on psi. 
And so what we're going to have is minus 
i h bar psi x psi prime, so this is the 
result of the action of these two 
operators in this order. 
Now if we consider a different order of 
the operators. 
If we first put you know x is going to 
act first and then we're going to have 
the momentum so in this case we can write 
it as false. 
It can be be minus i h bar g over dx x 
times psi and in this case we have to use 
the chain rule to calculate the 
derivative of both terms in this product. 
So its going to involve therefore minus i 
h bar derivative for facts is just 1. 
Psi minus i h bar x, psi prime and now, 
if we if we put everything together and 
apply this commutator to an arbitrary 
function psi of x. 
So let me actually write it in red, so if 
we now have x commutator of x and p, sort 
of acting on the wave function psi. 
So what it means is that we want to act 
by the following operator xp minus px on 
psi and so we essentially have to 
subtract result from this result. 
And so this two guys are going to cancel 
each other out and the only thing which 
should going to remain is plus ih bar 
psi. 
So the action of the commutator position 
momentum on an arbitrary function psi x 
is equivalent to multiplying this 
function by i h bar. 
Which is exactly the result that I 
advertised well in the beginning and this 
is again perhaps most of you or many of 
you already knew it. 
So this is again one of the canonical 
commutation relations in quantum 
mechanics. 
Now to continue this algebraic exercise 
in calculation commutators let me now 
calculate a very useful commutator for 
the following. 
Namely the commutator of, a, and a dagger 
and this is another so called canonical 
commutation relation. 
Let me write the result first and then 
I'm going to prove it. 
So and we will see that the commutator 
of, a, and, a, dagger is going to be 
equal to 1. 
So to prove this we use the definition 
of, a, and a dagger from the previous 
video and this commutation relation of 
position and momentum. 
So if we calculate a, and, a, dagger, the 
commutator of, a, and a dagger so we will 
have basically two types of terms. 
So here we're going to have m omega 2 h 
bar x plus i p 2 m omega h bar and the 
second term in this commutator is going 
to be the same thing. 
But with a minus sign in front of the i. 
So here it is. 
And we'll see that we have commutator of 
x and x. 
We have commutation of p and p. 
But of course, a commutator of an 
operator with itself is equal to 0 
because of course, for example, x x minus 
x x is equal to 0, right? 
So any operator commutes with itself. 
So the only commutators that are going to 
be non-trivial are going to be these 
cross-commutators of x and p and p and x. 
And you can convince yourself that the 
commutator of x and b is of course equal 
to the minus of commutator of b and x. 
Because we simply switched the order in 
which the operators appear. 
So therefore we can write it so this 
result, so lets just go back to the blue 
fonts. 
So it's going to be a twice, let's say 
the commutator of this guy, and this guy. 
So m omega over 2 h bar, from this term, 
minus i, 2 m omega h bar from this guy. 
And here we're going to have x and p. 
But this is something we know, we just 
calculated this. 
So this is this commutation relation of 
position momentum and its equal to, i a h 
bar. 
And so indeed if we put everything 
together we will see that well m and 
omega cancel here. 
So h bar and h bar under this score would 
cancel here so i times minus i gives us 
1, and 2 over 2 is equal to 1. 
So oh, no, it's equal to 1[UNKNOWN] . 
So and this eh, eh, you know completes a 
very important iteration. 
Namely the iteration of the Folic 
identity. 
And it's an operator identity. 
It doesn't really matter what 
representation we use, what kind of 
methods we want to solve. 
We have proven that the harmonic 
oscillator Hamiltonian, so let me switch 
to red. 
The harmonic oscillator Hamiltonian 
that's written here can be represented as 
H omega, a, dagger a plus one half, so 
this is from this commutator. 
Where the, a and a dagger commute to 1. 
Or in other words you may see it so this 
annihilation and creation operators 
satisfy the economical commutation 
relations. 
So the second term in this expression for 
the Hamiltonian as we discussed is just 
proportional to a number. 
So it's not, it can not possible affect 
the solution of the ion value problem. 
Namely if we want to solve the equation h 
psi is equal e psi and if psi happens to 
be an ion function of, a dagger a. 
So it would indeed be ion function of the 
whole Hamiltonian. 
h with a slightly shifted energy because 
the identity operator just reproduces the 
function. 
So and therefore in order for us to find 
the spectrum of the operator h in order 
for us to solve this eigenvalue problem. 
And find the eigenvalue functions and 
eigenvalues, we have to focus on the 
eigenvalue problem for this operator, a 
dagger a. 
Now comes a very important part where 
we're going to generate the entire energy 
spectrum of the Hamiltonian. 
From just a single eigenstate of it, or 
more precisely, from an eigenstate of 
this guy a dagger a. 
And to generate the spectrum, we would 
need just two pieces of information. 
The fact that we can represent the 
Hamiltonian as so, and the commutation 
relations between the ca-, ca. 
Annihilation and creation operators that 
we just derived. 
So at this stage, let me assume that I 
know an eigenstate of this guy, of a 
dagger a. 
And let me call this eigenstate n, and 
the corresponding eigenvalue will also be 
n. 
So this would also of course imply that 
the state n is also a solution to the 
time. 
And depends I show in your equation for 
the harmonic oscillator Hamiltonian. 
Or in other words, it's an old Synagon 
state of the Hamiltonian with the energy 
h omega n plus 1 half. 
So this is the energy of this state. 
Now let us examine what happens with this 
state we assumed exists. 
Under the action of this creation and 
anihilation operators. 
And of course what we're interested in is 
whether or not construct new eigenstates 
out of an existing eigenstate. 
So first let me add on this n by the 
operator a dagger and lets see how this 
operator behaves with regards to this 
guy, a dagger a. 
So what we can do here, we can take 
advantage of the canonical commutation 
relations that we derived in the previous 
slide. 
And using this commutational relations we 
can arrive a dagger a, a dagger as a 1 
plus dagger a. 
So here, we are going to have so 
basically exchanged the order of these 
two operators. 
So we are going to have instead of a 
dagger a, a dagger we are going to have a 
dagger, 1 plus a dagger a, acting on n. 
But, so this is just a number, and this 
guy, we already know how it acts on n. 
So by assumption, this is an eigenstate 
of a dagger a. 
So therefore, we can write it as a dagger 
n plus one half. 
I'm sorry plus one and since this is 
operator and this is just a number we can 
just exchange the order and so we have n 
plus one a dagger n. 
Which is exactly the same function as in 
the left hand side. 
So what we have proven Is that if n is a 
nagging function of the operator a dagger 
a. 
And consequently if it's a nagging 
function of the operator h, the 
Hamiltonian, then a dagger is also a 
nagging function. 
But with a nagging value n plus 1. 
So it raises the energy of the 
oscillator, by one energy quantum, if you 
want, of h omega. 
In the very similar way, what we can 
prove is the following. 
So we can prove that if n is indeed an 
eigenfunction of the Hamiltonian then the 
action of the annihilation operator on n. 
So this function is also an eigenfunction 
of the[UNKNOWN] . 
but it is an eigenfunction with a smaller 
energy, which sort, sort of justifies the 
terminology of this being an annihilation 
operator. 
So, while if we were to repeat this 
exercise that we just did for a dagger. 
And if we were to apply a dagger a on 
this function, we would, we would see, 
and I'll leave it for you as an exercise. 
That this guy is also an eigenstate of a 
dagger a. 
But now with a smaller eigenvalue, n 
minus 1. 
And here I will have the same function. 
So by knowing an arbitrary eigen function 
of the Hamiltonian, I can construct an 
infinite number. 
What seems like an infinite number of 
other eigenfunctions, with different 
eigenvalues. 
And by applying this creation and an, 
annihilation operator. 
And so we can also construct the spectrum 
of the Hamiltonian. 
But we haven't yet answered the question 
of what n could be. 
But with ultimate thinking we can 
convince ourselves that n here must be an 
integer going from 0 up to infinity. 
0,1,2,3 etcetera. 
And, a way to see this is essentially by 
uh,science check, if you want, of going 
to the physical problem we're trying to 
solve. 
So remember that, what we're doing do 
here, we're trying to find the energy 
levels in these harmonic oscillator 
potential. 
Which is simply the probably potential 
proportional to x squared. 
So this is via fax proportional to x 
squared. 
Now if we assume that we know a and 
certain a I can function with an energy. 
E sub n, as so and this is, let's say 
this corresponds to the state n. 
Then part, by applying the creation 
operator a dagger. 
We can construct a series of higher 
energy states. 
So n plus 1, n plus 2 n plus 3 etcetera. 
And there is no problem with that. 
But also by applying a, we can lower the 
energy. 
And we can do so seemingly indefinitely. 
But, clearly we have, this procedure has 
to stop somewhere, because we cannot go 
below 0. 
So in other words, we cannot go below the 
minimum of the potential, which in this 
case is 0. 
And this also implies that therefore, 
this procedure of generating new eigen 
functions, using this annihilation. 
op, operator, must terminate, and so 
there must exist, therefore, basically, 
the ground state. 
For which the action of the annihilation 
operator simply is equal to 0. 
Okay? 
Or in other words so this guy the vacuum 
state is the eigenstate of this guy a 
dagger a with the eigenvalue being 0. 
Or using this the result. 
So the energy of ground state therefore 
is equal to h omega over two which is the 
so called vaccum energy of the harmonic 
facilator. 
But this is really important and this 
tells us that this is really the property 
of the ground state. 
Okay? 
And this is by the way the simplest way 
to find an analytical expression for the 
ground state. 
We simply have to find a function that is 
annihilated by this operation. 
And finding this function is really the 
only remaining step in completing the 
iteration. 
Because here we essentially have done 
everything which needs to be done. 
We found a series of energy levels so we 
just write it again. 
So E sub n, the energy levels of the 
harmonic oscillators is equal to h omega 
n plus 1 half where n is equal to 0,1, 2, 
et cetera. 
And this was constructed under the 
assumption that eigen state exists. 
But we haven't yet shown that it is 
indeed so. 
We haven't yet, constructed explicitly, 
the eigenstate. 
And this is really the last piece of the 
puzzle, that we will put together in, in 
the next video. 
Where we also will summarize all the 
findings about the harmonic oscillator. 
And go a little closer to physical 
reality from this abstract algebraic 
derivation. 
Which, however, I hope shows the power of 
this algebraic method. 
So here, again, we are solving initially, 
what looks like a differential equation. 
But we get the results, and the 
quantization of the energies, without 
even calculating any derivatives. 
So all based on this commutation 
relations. 
And well I find it very elegant, and, I 
hope some of you also enjoy this 
derivation, but there is more to do.