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.