Hello everyone, welcome back to Exploring Quantum Physics. I'm Charles Clark. And in this part of the lecture, we're going to take a venture into the realm of perturbation theory, very powerful tool. And particularly, we're going to look at it in the context of understanding electrical properties of neutral matter. Okay, just to remind you, we're looking at the properties of a sort of a test case potassium rubidium molecule, which is of interest in multiple common [INAUDIBLE] of physics. And at the end of the previous part, we got to the realization that, no, the ground rotational state of the molecule, its dipole moment, is oriented randomly. And so there's no net dipole moment, we want to orient it. And how do we do that? Well, we're going to apply an electric field to the system and see what changes that causes according to the picture of quantum mechanics. The tool we're going to use is called perturbation theory. It's a very powerful tool. Actually, in the form that we use it now, it was original developed by Schrodinger. At least, it was used in almost the very same form that we use it today in his early papers on the hydrogen atom, in which he deduces not just the spectrum of a hydrogen atom, but the response of that spectrum to an electric field. It's a very versatile and powerful tool, of quantum theory, and also if you do a search on the phrase perturbation theory, you'll find that is a very powerful to applied mathematical analysis in a more general context. So I think the lessons that you learn from working examples in perturbation theory are widely useful no matter what your analytical interests are, because here's the basic idea. Let's suppose that we're solving an equation of promotion of a very complex form perhaps is given by a Hamiltonian which has a lot of complexity and we don’t know how to solve the equation. So the idea of perturbation theory is, that you try to find an approximation to that real Hamiltonian in terms of a Hamiltonian whose properties you know how to solve. And where the equations of motion that you're really interested in can be reached To some degree by an expansion in a small parameter. And so the idea is that you take a known Hamiltonian and its spectrum. And then you seek to develop, in a systematic way, the description of the new equations of motion. In terms of the power series in a set of functions that can be computed iteratively. I should say that very often it is, it's not necessary to go much higher than 1 or 2 in this series in order to. Actually learn very useful things. So we're going to work a couple examples. Here's the basic formulas, and as I showed you before, we say, well, the Hamiltonian consists of the two part, the known one and then something else. We're now going to focus on solving the time-independent Schrodinger equation by this technique. So we want to find the energy associated with this composite Hamiltonian. Now we do that by expanding both the wave function psi here and the energy E in terms of a formal power series in lambda. And then we just plug these two expressions back into the equation and we match the power supply in a setting order. So again I hope you can see that, so we have H naught plus lambda v naught times psi naught plus lambda psi 1 plus etc etc, equals E naught, plus lambda E 1 plus etc psi 0 plus lambda psi 1 etc. So let's match the powers of lambda to the 0. So that means we take h0 times psi 0 from here. There's no other match and then on the other side we have E on sign on there. So let's just run a reality check and see if that makes sense to you. So I hope your persuaded by looking at it that at least this first equation makes a lot of sense because it just gives one back the spectrum which one presumably knows how to control. Well now, let's see, how do we go up? Let's go to the next order. So, we're just taking powers of lambda through the 1 on each side. Lambda the 1, so from the left hand side, that's h0 times psi 1 + V times psi 0 = E0 times psi 1 and then E1 times psi 0. Okay, well, so far there's not much, not much going on just it's symbolism so now let's, what we do now is we take the matrix element, the bra of size zero with respect of this equation. So zero on this, on both sides. And so we now get this expression here and I hope it's clear to you that when we look at this term here. That's exactly equal to this term by the formation conjugation process. So, we can drop these two terms. And then we get this simple equation for the first order energy correction. It's just the expectation value of the potential over the 0 wave function. My good friend and colleague, Mark Edwards, call this the most used equation in quantum theory which I think very well may be right. It's not the most written equation quantum theory, which is the Schrodinger equation, but this is one which is probably solved more often than anything else because it really gives one a very good guidance at an early stage on what you have to do. And it really just requires taking an integral of something rather than solving a spectral problem. So we're going to do some of that solving ourselves. Okay, so I guess you can see that there's some sense that you get in making some, looking at simple considerations of standard problems when, the trivial sort of attempts that we apply we see the perturbation theory. Describes that case well. Okay, so now, we're going to go to the real problem, which is to try and understand the properties of a rotating polar molecule in the presence of an electric field. So here's our Hamiltonian you see. It's the original rotational Hamiltonian with another term which is the interaction of an electric dipole moment, d, with the electric field. This is the interaction from classical electrostatics and so we can write this in simply in polar coordinates in this way. And now I'm using these temperature equivalents just for convenience. So there's a sort of, keep in mind this is just a temperature equivalent saying well, you have an energy and you divide it by both cost and that gives a temperature but it does describe the scale in a useful way. As you may recall, this equivalent temperature of the rotational motion in the ground state is something like 55 millielectron, millikelvin. And now with using a sort of a dimensionalist parameter which is a field strength of 1 volt per meter, then there's an equivalent electrical energy, which it turns out is very small compared to the rotational energy. So this is a sign that a perturbation theoretic approach will be useful in this problem because we are using it to treat something that's small compared to the underlying rotational motion, the molecule. And you might well imagine from a physical standpoint. If you have a molecule that's freely rotating and you're just, you just have a small break to apply to try to slow it down. That when that break is weak, it's not going to be that effective. So here we go on our way to solving this problem. We have the E0 vanishes, here's the 0th order. Wave function sign on, it's just the ground state of rotating molecule. So now we're going to evaluate the most used equation in quantum physics, the first order energy It's the expectation value potential over the ground state wave function. And now you see this is just, the potential is just the cosine of the angle with respect to the axes. So our expectation value takes this form. Now, I hope that through the exercises that you've done throughout this course, you've learned that sometimes, you don't actually need to go through the mechanics of evaluating integral. It vanishes because of symmetry reason, and this equals 0 for a very simple reason and that is, it's an integral over all space of the cosine of an angle with respect to some axis. The cosine of this angle here and well obviously the cosine of that angle whatever point you have here there's another point where the cosine has it an ultimate sign so this vanishes by symmetry Which is quite a common occurrence when evaluating first order energy corrections of this type, as we'll see.
