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.