So in this last video of this lecture, I'm going to derive the harmonic oscillator wave functions, but I will do so without writing down the Schrödinger Equation. We will rely entirely on the solution that algebraic solution I presented that took advantage of the creation and annihilation operator, it turns out that they are very helpful in determining the explicit form of the oscillator wave functions, si of x. So 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 and operators here. And using the algebra or the computation relations of these harmonic oscillator operators, we derived the energy spectrum of the oscillator, which has this form. H omega n + one half. Where n = zero, one, two, etc. 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 so here I'm being a little more specific. Is that given an arbitrary icon state of the Hamiltonian with a quantum number n as here, so we can generate the entire spectrum by applying the creation operations and alienation operators praise the energy of the system by h omega. So we go from n to n + 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 not in many textbooks so I'm not going to spend time on that. And likewise, given the same eigenstate n. We can find the state corresponding to a lower energy with the one less quantum of h omega by applying this alienation operators. But what we don't know at this stage, we don't know an example of any wave function explicit wave-function. So. And preferably we would like to see a wave function that's in real space, which is more intuitive that other sort of, let's say abstract spaces that we have been operating with so far in this problem. So in order to find 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 eigen state as we just discussed in the previous video is characterized by the fact that the action of this a dagger e on it gives simply zero. Gives us it's an eigen state with an eigen value zero. Or the eigen energy of this state is equal to him over two. So in other words, so the angulation operator acting on this state Must produce zero and we can write this operator explicitly in real space. So we use the definition if we want of this a as so and here what I'm going to do is I'm going to just write down explicitly the form of the momentum operator and position operator. In real space the position operator is really simple. It's just a multiplication operator, while the momentum operator as we know, is just minus i h bar d over dx. So therefore here I will just simply remove the measuring constant. So, all in all the. Equation, so I should have written perhaps here psi0 because I'm specifically talking about the ground state of the harmonic oscillator, so explicitly the equation for this psi0, therefore, becomes the following. So we'll have this square root of m omega over two h bar x plus h over two m omega h bar in the denominator d over d x so which is coming from this momentum operator. And this acts on si not and this whole thing must be equal to 0. So here what I'm going to do, I'm going to simplify this thing a little bit so I'm going to put the square root here on top. I'm going to get rid of the factor two so what I see here is that I can introduce a new parameter. But we 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 you can check that the physical dimension of the square root of H. Over m Amanda is a length scale. So it has a physical dimension of length. So this 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 will write it as X over x not plus x not times d over dx. And this whole thing acts on the wave function. So it effects and we want to produce 0. So now, let me just rewrite this equation in a slighty different form. So I'm going to write dx over dx sine prime, so this is going to be sine prime and put this guy in the left hand side. So I'm going to have minus x over x 0 squared times sine and so I just want to show this differential equation. You can check, you can solve it formally or you can just convince yourself that the solution to this equation is Is an exponential, it's Gaussian, basically. Minus x squared over 2 x not squared. So, and this is, let me just put 0's 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 wave function. So again, this wave function describes a particle in the ground state of the harmonic oscillator and so therefore, the total probability for me of finding a particle somewhere in space must be equal to one. So therefore, this integral from minus infinity to plus infinity of sine r squared. So I don't bother here with the complex sines because everything is real so far. So this must be equal to one. And this gives me the value of the wave of this coefficient that multiplies this ground state. A wave function so let me just write it down. So the C is equal to m omega over two, 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 wave function without actually solving the Schroedinger equation. We actually solved a much simpler equation. This first order differential equation just by requiring that our wave function is in state with an value zero of this operator. Okay, so this is great. So we can now work with this wave function if we want, but what about the higher energy states? 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 v of x if this is x, and this 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 these eigan states? Well, so here, we can take advantage again of the creation operators now. So as we discussed In the first slide in this video so the relationship between the low energy states and high energy states, so let's say go from m to m + 1, we can simply apply the creation operator divided by n + 1. Square root of n+1. So if we apply it to our in state n, we're going to get, oops there's no square root here, there's just n+1, a normalized state that corresponds to high energy. And so what we can do, we can apply this operators. let's say to our ground state. So we can apply this guy. As many times as we want to the ground state. For instance, we can generate this wave function So by just applying this derivative essentially to the Gaussian. And this will produce certain coefficients in front of the Gaussian and this, well not coefficients, better to 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 wave functions. 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 you can actually solve the Schrödinger equation starting from this spectrum, up to the ground state and up to the excited state, using only the creation and annihilation operators. So there's very little inside that is actually needed from the equation. Which is in fact quite remarkable. So in your homework, you will need to play around a little bit with this equation. And with this grounded state function 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.