So in this last video of this lecture, 
I'm going to derive the harmonic 
oscillator wave functions. 
But I will do so without even writing 
down the Schrodinger equation. 
we will rely entirely on the solution, 
that algebraic solution I presented that 
took advantage of the creation and 
annihilation operators. 
First, it turns out that they are very 
helpful in determining the explicit form 
of the oscillator wave functions of psi 
of x. 
So but before we write down these wave 
functions, we derive these wave 
functions. 
Let me summarize what we know so far, 
which is actually a lot. 
So first of all, we found a way to 
represent the harmonic oscillator 
Hamiltonian in this compact form, in 
terms of this creation annihilation 
operators here. 
And using the algebra or the commutation 
relations of this harmonic oscillator 
operators, we derived the energy spectrum 
of the oscillator, which has this form, h 
omega n plus 1 half, where n is equal to 
zero, one, two , etcetera. 
So and again, I reiterate that this 
spectrum is special in that the level 
spacing between the neighboring levels is 
always the same, so it doesn't change as 
we go higher up in energy. 
So another thing we derived, and here, 
I'm being a little more specific is that 
given in arbitrary Eigenstate of the 
Hamiltonian with the quantum number and, 
as here. 
So we can generate the entire spectrum by 
applying the creation operators and 
annihilation operators. 
The creation operators raise the energy 
of the system by each omega So we go from 
n to n plus 1, and in the case of 
normalized eigenfunctions there appears 
this coefficient. 
I haven't derived it in the previous 
video, but this is rather straightforward 
to do so using the same methods and it's 
done in many textbooks, so I'm not going 
to spend any time on that. 
And likewise, given the same eigenstate 
n, we can find the eigenstate first 
pointing to a lower energy with one less 
quantum of h omega by applying this 
annihilation operation . 
But what we don't know at this stage, we 
don't know an example of any 
wavefunction, explicit wavefunction. 
So, and they preferably would like to see 
a wave function within real space which 
is more intuitive than other sort of, 
let's say abstract space that have been 
operating with so far in this problem. 
So in order to find the in, the explicit 
form of an eigenfunction, let me actually 
focus on the ground state, that is the 
lowest energy state in our problem. 
And this eigenstate as we discussed in 
the previous video is characterized by 
the fact that the action of this A dagger 
a on it use simply 0. 
So it gives us, its an eigenstate with an 
eigenvalue 0 or the eigenenergy of the 
state is equal to[UNKNOWN] over 2. 
So in other words, so the annihilation 
operator acting on this state must 
produce 0, and we can you know, write 
this annihilation operator explicitly in 
real space. 
So we, we use the definition, if you 
want, of this a and so, and here, what 
I'm going to do, I'm going to just write 
down explicitly the form of the momentum 
operator and position operator. 
So in real speeds the position operator 
is really simple, it's just the 
multiplication operator while the 
momentum operator as we know is just 
minus ih bar d over dx. 
So therefore you know, here, I will have 
simply, I will just simply remove the 
measuring constant. 
So all in all the equations, so I should 
have written perhaps here, si not, 
because I'm specifically talking about 
the ground state of the harmonic 
oscillator. 
So explicitly, the equation for this psi 
naught, therefore, becomes the full 
length. 
So we'll have this square root of m omega 
over 2 h bar x plus h over 2m omega h bar 
in the denominator d over dx. 
So, which is coming from this momentum 
operator and this acts on psi naught, and 
this whole thing must be equal to 0. 
So here, what I'm going to do, I'm 
going to well simplify this thing a 
little bit, so I'm going to put the 
square root here on top, I'm going to get 
rid off the factor two. 
And so, what I see here is that I can 
introduce a new parameter. 
Let me call it x naught, which is the 
square root of h divided by m omega and 
it appears both here and here. 
And so, this, so you can check that the 
physical dimension of this square root of 
h over m omega is the is a length scale. 
So it, so it has a physical dimension of 
length. 
So this is, in some sense, a natural 
length scale for the harmonic oscillator 
problem. 
And so, it is convenient, therefore, for 
me to rewrite my equation as above in the 
full length form. 
I'll write it as x over x naught plus x 
naught times d over dx. 
And this whole this thing x on the wave 
function [UNKNOWN] and we want to produce 
0. 
So now let me just rewrite this equation 
in a slightly different form. 
So I'm going to write d over dx is psi 
prime, so this is going to be psi prime 
and in with this guy in the left-hand 
sign. 
So I'm going to have minus x over x 
naught squared times psi and so I just 
want to solve this differential equation, 
and you, you can check you can solve it 
formally. 
Or you can just convince yourself that 
the solution to this equation is an 
exponential, it's a Gaussian basically, 
minus x squared over 2x naught squared. 
So and this is let me just put zero, 
zeros in, in here which correspond again 
to the ground state and this is pretty 
much the result. 
So c here is a coefficient which is 
determined by the normalization of this 
wavefunction. 
So again, this wavefunction describes a 
particle in the ground state of the 
harmonic oscillator, and so therefore, on 
the, you know, total probability for me 
of finding a particle somewhere in space 
must be equal to 1. 
So therefore, this integral for minus 
infinity to plus infinity of sine r 
squared. 
So don't bother here with the complex 
size because everything is real so far. 
So this must be equal to 1 and this gives 
me the value of the way of the 
coefficient that multiplies this ground 
state wave function. 
So I can just, let me just write it down, 
so the c is equal to m omega over 2 I'm 
sorry, pi, that is pi h bar to the power 
1 4th. 
So this is the normalization coefficient. 
But notice that we have found the ground 
state wavefunction without actually 
solving the Schrodinger equation. 
We actually solved a much simpler 
equation, this first order differential 
equation just by by requiring that our 
wave function is in eigenstate with an 
eigenvalue of 0 of this annihilation 
operator. 
Okay, so this is great. 
So we can now work with this wavefunction 
if we want, but what about the higher 
energy state? 
So what we have just done was we 
determined this guy, the lowest energy 
state. 
So here, what I'm showing, let me just 
write down. 
So this is first of all the potential of 
this [UNKNOWN] x, so this is x, and these 
these levels here correspond to the 
energy levels of the harmonic oscillator. 
And the lowest energy has the wave 
function as so, which is usual Gaussian. 
So this is a typical Gaussian and high 
energy states have a more complicated 
sort of texture, so how do we get this 
eigenstate? 
Well, so here, we can take advantage 
again of the creation operators now, and 
so for, as we discussed in the first 
slide in this video, so the relation 
between the lowest, low energy states and 
high energy states. 
So let's say we can go from n to n plus 
1, we can simply apply the creation 
operator divided by n plus 1, square root 
of n plus 1. 
So if we apply it to our eigen state n, 
we're going to get oops, there is no 
square root here, it's just n plus 1, a 
normalized state which corresponds to 
high energy. 
And so what we can do, we can apply these 
operators, let's say to aroused state, so 
we can apply this guide as many times as 
we want to the ground state, for 
instance, we can generate this 
wavefunction. 
So we are just applying this derivative, 
essentially to the Gaussian and this will 
produce a certain coefficients in, in 
front of the Gaussian. 
And this well, not coefficients better I 
say polynomials which are functions of x, 
and these guys, these polynomials they 
are called Hermite polynomials which are 
special functions. 
And you can Google them or you can look 
them up in Wikipedia and in many books. 
So this sort of, this solution using the 
explicit form of the wavefunctions and 
the explicit form of the Hermite 
polynomials appears in pretty much any 
book on quantum physics. 
And, I'm not going to spend time on 
discussing these details. 
I will simply notice that, well, you can 
actually solve the Schrodinger equation, 
starting from the spectrum up to the 
ground state and up to the, the excited 
states using only the creation and 
annihilation operator. 
So there is very little in, insight that 
is actually needed from Schrodinger 
equation which is in fact quite 
remarkable. 
So in your homework, you will need to 
play around a little bit with this 
equation, annihilation operators and with 
this ground-state wavefunction. 
And I hope that you found this solution 
illuminating and of course, I encourage 
you to look into the literature to see 
more about this very important problem of 
quantum physics.