nanoHUB U Organic Electronic Devices/Lecture 2.2: The SchrÃ¶dinger Equation
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[Slide 1] Hello and welcome back to nanoHUB-U's course on organic electronic devices. The last time we chatted we saw that atomic and molecular orbitals give rise to bonding and anti-bonding structures in all types of organic materials, and included in that class is organic semiconductors. And during that time, we saw different pictures of what atomic orbitals look like, and during that lecture I told you that we would come back and discuss how we can predict and define what those atomic orbitals look like theoretically, and that's what we'll do today, and actually we can measure those experimentally as well. OK? The rationale behind all of that is really in the Schrodinger equation. So right now we're going to walk through a derivation of the Schrodinger equation, explain certain highlights throughout it and then use it in a couple of examples that will let us talk about solving for different energies of electrons and how do those energies of electrons can be used to describe the electronic structure of materials. Once we have the electronic structure of materials, we'll be able to talk about the electronic transport of materials and how to utilize these organic semiconductors in electronic devices. So let's get started.
[Slide 2] So what we'll be talking about today is to define a term that I'll call the wavefunction. And I'll show you that it has a variation both in time, a temporal component, and in space, a spatial component. What we'll talk about today is, though, is we'll take away the time-dependent property, and we'll derive the time-independent Schrodinger equation. Once we've derived that time-independent Schrodinger equation, we'll use it to solve a specific problem that's of great interest in terms of moving on into talking about band structure future in this course. At the end of this presentation, I've hoped we've hit the following learning objectives. We should be able to derive the Schrodinger equation and label the terms that are associated with the kinetic, the potential and the total energy of the system. You should be able to show how discrete energy levels arise from something called a particle in a box, or an electron in a box situation.
[Slide 3] So what do we have is, is we're going to have a wavefunction and it's going to be one-dimensional, and this wavefunction is going to be symbolized by the Greek letter psi. And in 1-D we'll call the one dimension x, and we'll call the time component t. So we're going to have psi of x and t. And this is going to be a wavefunction that describes our electron. So we know from a classical wave perspective, that if I take the partial derivative of psi x of t, that's going to be equal to 1 over u squared, where u is the speed of the wave, times the partial derivative with respect to time of psi x of t. So we have d squared psi dx squared. Those are the partial derivatives, is 1 over u squared times d squared psi divided by dt squared. So that's the partial with respect to time. We can break this wave equation, this wavefunction, into two parts. There's psi x of t. We can actually have psi, only in terms of x, multiplied by psi, only in terms of t. So now we've broken it down in both the spatial and temporal parts of the wave. And in fact, what we'll do is we'll say that psi of t is equal to cosine omega t. Well, what is omega? Omega is just going to be defined to be 2 times pi, so a constant, times the frequency of the wave. And you know that the frequency of the wave is important, and we've seen this before in your physics class, where E is equal to h times the frequency, where h is Planck's constant. So you know that frequency, and thus in turn omega, is going to be related to the energy of the electron described by psi. We've now broken this down into psi of x times a temporal target.
[Slide 4] If I substitute that second equation that I just had on the screen into the first equation, the left-hand side stays the same. We just have the second-order partial derivative with respect to psi. The constant for the speed of the wave stays the same, but now we have psi of x cosine omega t. Now I can take that second derivative, and if you take the second derivative of cosine omega t, we'll end up pulling out a negative omega squared, times psi of x over u squared. OK? So now we have something in terms of psi of x omega squared. If we have an equation that's solely a function of position, so now we have the partial with respect to x equals psi of x over u squared omega, I can rearrange that such that it's the second derivative with respect to x of psi plus psi of x over u squared omega squared equals 0. So now I've reduced this equation into something that's only in terms of position, or in terms of space, and that's how we get away with the time-independent Schrodinger equation.
[Slide 5] If I substitute in, that omega is just 2 pi times nu, the frequency, then I'll pull out the exact same thing over here, but now I'll have 2 times pi squared is 4 pi squared, and then I have the nu squared term. So that's still all equal to 0. But a frequency times a wavelength is just a velocity. Remember, frequency has units of something like inverse seconds? Wavelength has something like units of meters. So if I take meters times inverse seconds, I get a meter per second. So that looks a lot like a velocity, or a speed unit. So then I can rewrite that as lambda is just equal to u over the Greek letter nu. Or alternatively, nu squared over u squared is just 1 over lambda squared. So if I substitute this part of the equation right into here, I'll get 4 pi squared over lambda squared right there. Alright? And that's exactly what we get, and that's all equal to 0. So nothing's changed yet. We're just rearranging variables into things that are convenient for our final term.
[Slide 6] Now a little bit of quantum physics for you. OK? We have two properties that we need to discuss. The first is that the energy of the wave can be expressed in terms of the potential and the kinetic energy, and we'll talk about that right now. The second is that the wavelength is related to momentum through Planck's constant, and that was really the insight of de Broglie. So the first one is just that the total energy, E, is equal to the kinetic energy, plus the potential energy, and here the kinetic energy is expressed as the momentum squared, so p squared, divided by 2m. Remember that momentum, p, is just equal to mass times velocity. So that would be m squared V squared over 2m, and that looks a whole lot like the classical physics term 1/2 mV squared. This potential energy term is something that will vary as a function of position, and you'll have to describe it on your own, and we'll do specific cases, where we do describe it here in just a second. OK? The second says that the wavelength of the particle is just going to be equal to Planck's constant, which I've listed right here, divided by the momentum. So now I have these two definitions readily handy to me.
[Slide 7] So if I plug those two together I'll see that if I solve for momentum, I just get the momentum is equal to 2m times E, the total energy, minus the potential energy to the 1/2. OK? Once I have that, I can plug it into the de Broglie wavelength equation by direct substitution and see that lambda is equal to this. So that's just simply plugged in from the last slide. If we substitute that whole expression in for lambda, in the bottom of our initial equation, we'll see that now this term has replaced lambda. So it's still lambda squared, but this is now lambda. So we're getting there. Trust me. I know we're going through quite a bit of math, and we're doing a lot of substitutions. What this will end up doing is allow us to explain exactly how the wavefunction corresponds to the energy of the material.
[Slide 8] If we keep going, I can reduce and arrange the algebra such that we have the partial derivative with respect to x of psi plus 4 pi squared psi over this now lambda squared term written in terms of energy, and if I rearrange this, I'm just going to flip this guy to the top, I'm going to end up getting out this d squared psi over dx squared plus 2m now, times E minus v of x divided by h-bar squared now, and this h-bar term is just Planck's constant h divided by 2 pi. This way I don't have to carry around a 2 pi squared everywhere with me. I'll just call it h-bar squared. So this is just another symbol change right now. But now we have this equation, and it's starting to look simpler. Maybe not simple, but simpler than what we started with. If we rewrite it, if we multiply by h-bar squared over 2m, we're going to end up pulling out the negative h-bar squared over 2m times the second derivative of psi plus the potential energy times psi is equal to the total energy times psi. And this is the commonly used way we write the time-independent Schrodinger equation. OK? So you can see that if we break this down, that first term there is the kinetic energy term. And that's what we associate with the kinetic energy of the wave. The second term is associated with the potential energy of the wave, and if you sum up the kinetic and the potential energy, the third term must be the total energy. So now we have a way, using partial differential equations, if we know what psi of x is, we can get at the kinetic, the potential and the total energy of the system. Let's see how we can do that.
[Slide 9] So we're going to look at a very classical example in quantum mechanics, and sometimes it's called a particle in a box, or a free electron in a box, and the reason it's called that is because we're going to have some one-dimensional box that ranges from x equals 0 to x equals L. OK? And within this box, the potential energy everywhere within the box, anywhere between 0 and L, the potential energy is equal to 0. So that Vx term, we're going to be able to negate it this whole time. Outside of the box, the box walls here at x equals L and x equals 0, the potential energy there is extremely high. We'll say it goes to infinity. What does that do for us? Well, that means that trying to find the particle outside of the box is impossible. So we've now put a bound on the probability of where our particle can be. It must be somewhere between 0 and L, because this energy outside of L or outside of 0 is just too high. OK? If we do that, we can reduce the Schrodinger equation to basically eliminate that Vx term. So that used to be a Vx term there, and this used to be the E term, and I just added the E term over. So now we have d squared dx squared of psi plus 2mE over h-bar squared psi of x equals 0, and now we have to talk about the boundary conditions. So we have a second-order partial differential equation. Right? If we have that, we need two boundary conditions with which to solve it, and the good news is that if we know that if we need the particle to be within 0 to L, that psi at L and psi at 0 must go to 0. So our boundary conditions are psi at x equals 0 equals psi at x equals L, and that's equal to 0.
[Slide 10] The general solution to a problem that looks like this is the following. It says that psi of x is just A cosine kx plus B sine kx, and in this case, for our specific example, k is equal to these values right here: 2mE to the 1/2 divided by h-bar. OK? So let's go ahead and solve for psi of x and solve for A and B. We can do this by using our boundary conditions. So if we plug psi of x equals 0 in, we get that A cosine of 0 plus B sine of 0 is equal to 0. Well, this term reduces to 0 right away. Right? Because sine of 0 is 0, and then we have A times 1, because the cosine of 0 is 1, A times 1 plus 0 is equal to 0. Therefore, A must be 0. That's the only way that can be possible, and we know that psi then is just B sine k times x. So now we just need to solve for B.
[Slide 11] So now we'll use our other boundary condition, and we'll plug in L and say that B sine of k times L must equal 0. Now, the easiest way to get this done is to say that B equals 0, right, because that would make the whole term go to 0 as well. However, that's what we call the Trivial Solution. We don't want to do that because that basically says that psi is equal to 0, which we know would satisfy our conditions but wouldn't be of that much interest. However, you can get sine kL to equal 0 if k times L is equal to some integer n times pi. So you know that every pi cycles that that sine term is going to go through 0. So as long as n times pi is equal to k times L, that will always be true. So we can rearrange this and say that psi -- and now I've actually included this n just to remind you that it's a function of this integer n. We can say that psi n of x is equal to sine n pi over L times x. So now we have a description for psi of x. If we plug that into our original equation, we'll be able to solve for E. So here's where we're taking the second differentially, the second differential expression here, and we'll pull out the negative n squared pi squared over L squared times sine n pi L over x. But remember this term right here is just the same as psi of x. So we put in psi of x there. You might want to do that on your own and prove to yourself that that's actually the case.
[Slide 12] But if we have that, then we know that we have the expression that we just solved for, and we put it in right here, plus 2mE over h-bar squared psi of x. You can put that in there. And that's still equal to 0. So I can divide the whole equation by psi of x. So we can eliminate the psi of x, and now we can solve for E sub n, and when we solve for E sub n, what we find is that E sub n is just equal to n squared times h squared over 8mL squared. Notice that I've dropped the h-bar now, because some of the pi's canceled out. So I can go back to h instead of h-bar, and that's the classical expression for energy as a function of n, and we'll call that the Quantum Number. So now we know that the energy will increase as the square of n, and go ahead and on your own prove that algebra from this line to this line. I think you'll be able to do it.
[Slide 13] So what I hope we've shown you today is that the Schrodinger equation is a very powerful tool, and one that lets us define quantum mechanics in a very elegant manner. OK? We started from a very simple equation that was more or less derived for the position of a vibrating string. We applied that to the atomic orbital theory and looked at electrons in a box, but we were just solving a classical PDE system, but when we do that, quantum mechanics allows us to pull out very discrete energies in a very elegant and simple manner. And what we'll find in future lectures is that we can actually use this to describe real molecules. OK? When we look at our particle in a box, we end up pulling out this energy term that goes as the square of the quantum number and inversely with the length of the box squared. So that tells me the longer that box is, the longer that box is, the lower the energy will be and in a few lectures I hope what you see is that it's directly related to what we talked about when we talked about making low bandgap polymers, and spreading out the length of conjugation, increasing L, lowered the bandgap energy. It's not a one-to-one comparison, but it's something to think about. So with that I'll thank you for your attention. I'll let you know that we're going to use these materials again and apply them in a particular situation and then expand the Schrodinger equation into three dimensions, and once we expand the Schrodinger equation into three dimensions, we'll really be able to take off and talk about how we develop electronic structures into large-scale materials, so beyond just the simple atomic and molecular approach. Thank you for your attention. I look forward to seeing you again on the next lecture