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.