 So if we have the harmonic oscillator version of the Schrodinger equation, and we've convinced ourselves that functions of this form ought to be solutions to the Schrodinger equation, let's convince ourselves even further that that is the case, and plug this wave function in both to the kinetic energy term and the potential energy term and on the right side of this equation. So to do that we need derivatives. So we'll take the derivatives of this function, the first derivative of the wave function, derivative of an exponential is an exponential times, so the same constant out front, times the derivative of the exponent. So the derivative of x squared is 2x, and I've got a negative alpha and a divided by 2. So the 2 from the x squared and the 2 from divided by 2 cancel each other, and I've got a total of, leaving room for my negative sign, minus 2 alpha n times x e to the minus alpha x squared. So I've pulled down an extra factor of minus alpha x by taking the derivative of this exponent, minus alpha, not 2 alpha. So minus n times alpha times x e to the minus alpha x squared. Second derivative, the derivative of this derivative, again the derivative of the exponential is again going to pull down another factor of minus alpha times x. So I've got now negative times negative is positive, alpha squared, x squared times n, still times the same exponential. But I've also got this x out in front, taking the derivative of that one, leaving the rest of it alone, gives me a minus alpha n e to the minus alpha x squared over 2. So that should be the derivative, the second derivative of our hypothesized wave function. If I look a little more closely at that, I see that n e to the minus alpha x squared over 2 shows up in this derivative. So let's simplify that a little bit by writing it as alpha squared x squared times the original wave function, minus alpha times the original wave function. That'll help when we go to plug the second derivative into Schrodinger's equation. And when we do that, Schrodinger's equation becomes negative, some constants h squared over 8 pi squared mu. Second derivative is alpha squared x squared minus alpha times psi plus 1 half kx squared times psi. And that should be equal to energy times the wave function. So we've got constants. In fact, let me go ahead and write, let's collect some terms. We've got constants times an x squared times psi, constants times an x squared times psi, and other terms that just look like constants times wave function, constants times wave function. So if we collect the terms with similar powers of x, the first term in parentheses here, negative constants times alpha squared, that becomes minus h squared over 8 pi squared mu alpha squared. That's going to multiply x squared and the wave function. But I have another term that multiplies an x squared and a wave function. That's this term here. So I also have plus 1 half k, also multiplying x squared and the wave function. So I've collected the two terms that look like x squared times the wave function. In addition to that, I have a term that looks like constants times wave function. So negative times negative leaves a positive sign, h squared over 8 pi squared mu alpha times wave function, and all that should equal energy times wave function. So is it true that the stuff on the left equals the stuff on the right? Constants times wave function could easily equal constants times wave function if we have the appropriate value for e. But this equality is only true if the stuff in parentheses, these two terms cancel each other out, leaving the term in parentheses equal to 0, so that the x squared times psi term is 0. So if this whole first term is 0, then in fact, the Schrodinger's equation is satisfied. So we need for it to be true that this term in parentheses needs to equal to 0. So how can we guarantee the term in parentheses is equal to 0? The only term in here that we have any control over, we've been given this function for some molecule that behaves as a harmonic oscillator. It has some mass. It has some spring constant. We don't get a choice over what the value of k and mu are or h and pi. The only constant here that we have any control over is the constant that's in the wave function that we are trying to write down, that we're trying to write down the solution for the Schrodinger equation. So we get to choose the value of alpha. If we rearrange to make the term in parentheses equal to 0 solving for alpha, we'll find that, let's see, first we need that h squared over 8 pi squared mu alpha squared has to be equal to 1 half k. So if we rearrange and solve for the alpha, alpha squared, putting everything else over on the right, I'm going to get an 8 pi squared mu on top, h squared on the bottom, and I've still got a 1 half k that was on the right all along. That'll simplify a little bit to 4 pi squared over h squared times k times mu. I don't want alpha squared, I want alpha. If I take the square root of both sides, the square root of 4 and pi squared and h squared are easy to do, and then k and mu just give me the square root of k times mu. So what I've determined here is specifically what the value of alpha must be, so not just any old e to the minus alpha x squared over 2 will be a solution to the Schrodinger equation. Only if alpha is equal to this collection of constants will the x squared times wave function term from the kinetic energy cancel the x squared times wave function term from the potential energy. And if that cancellation happens, then this function will solve Schrodinger's equation. With that particular value of alpha, we've also in the process of inserting this wave function into the Schrodinger equation, we've also determined the value of e. This is a solution to Schrodinger's equation. Only if e is equal to this collection of constants, so I need this collection of constants to be equal to e. So the energy of the wave function that we've now determined is h squared over 8 pi squared mu times alpha. But now that we know what alpha is, that's h squared over 8 pi squared mu times 2 pi over h square root of k mu. So there's a lot of cancellation that's going to happen here. Let's see, we've got a 2 divided by 8, so that leaves me a 4 in the denominator. I've got a pi divided by a pi squared, so that leaves me a pi in the denominator. We've also got h squared divided by h, so that's one factor of h in the numerator. I've got a square root of k. There's no k cancellation that takes place, but mu, I have a square root of mu here and I'm dividing by mu here, so I have 1 over square root of mu after some cancellation with the mutes. So if I rearrange that a little bit, I'm going to write that as 1 over 4 pi. I'm going to write as 1 half times 1 over 2 pi. For reasons that'll become clear in just a second, I've got a factor of h and I've got square root of k over mu. That should look a little familiar. 1 over 2 pi square root of k over mu, that quantity is what we called for the classical harmonic oscillator, the equilibrium vibrational frequency. So that quantity, 1 over 2 pi root k over mu, that's the frequency with which this harmonic oscillator vibrates. So I can write this now as 1 half times h times nu. The quantity in pink I've just rewritten as the vibrational frequency nu. So the energy of this wave function that we've written down when alpha takes on this particular value is 1 half times Planck's constant times the vibrational frequency of that same oscillator. So that combination of this wave function and this constant and this energy now compose a solution for the harmonic oscillator. This wave function solves the harmonic oscillator wave function with this particular energy. You may already be wondering is this the only solution? Is this like other quantum mechanical problems where there's more than one solution and you probably won't be surprised when I tell you there are more than one solution. This is one solution but there's a whole family of solutions that are somewhat similar to this function. So the next step will be to explore what those other solutions to the Schrodinger equation look like.