 All right then. Today we're going to deal with bound state perturbation theory. I think I hardly need to tell you how important perturbation theory is in physics. That's because everywhere in physics there are problems that are close to solvable problems. We need to make small perturbations around them. The Earth is merely a sphere but not exactly. The Sun has most of the mass in the solar system. The period of molecular vibrations is small compared to the... or rather small. It's a large period compared to the period of electronic motions and the molecule. And then the height of the anatomy affects the spin and all the... this relativity are small compared to the effects of the Coulomb potential. So these are just all examples of perturbation theory. So as I said, today we're going to be dealing with a bound state perturbation theory and we're going to be dealing with the effects of perturbation on the bound states of the discrete... that's the discrete spectrum of the bound states of the quantum systems. So let's assume we have a Hamiltonian for a system of 1 by h0 plus a correction term of what is lambda h1. Let's assume that h0 is a Hamiltonian that we can solve and know its energy levels and eigenvalues. In fact, let's have a note that is in the following way. Let's call them k alpha like this. Let's call the energy levels epsilon and k. The notation here is that k is the label of the energy level which we allow to be degenerate. Altruism index is also degeneracies so alpha runs 1, 2 and so on like this. So these are the unprocured energy levels of the amount of epsilon and k. Here h1 is a perturbation. This is an interpolation parameter that we use to switch on the perturbation so that the lambda is 0. We have just the integer of the Hamiltonian to the lambda is 1 and we have the full interterm system. It's often times convenient to introduce this. The effect of lambda becomes a bookkeeping device for the order of the perturbations. Now, to give you a qualitative idea of what can happen, allowing you to draw an energy axis here that corresponds to the unprocured system h0. Let me just schematically draw in what some of the energy levels might look like. This is actually not very realistic for real problems. Let's call these energy levels epsilon and k like this running up and down. Let's focus our attention on just one of these energy levels which is epsilon is again at the index n. For the duration of this discussion, let the index n be fixed. We're going to be interested in perturbations just at one level. Now, the other term system may also have a continuous spectrum. Frequently does in practice. If so, it's probably going to be a higher energy and a higher energy. That's important to keep in mind especially later when we start writing our resolutions of the identity and so on. Those have to include the continuum. Nonetheless, the attention of our perturbations there is going to be placed in the bound states there. In particular, this one, that's the one at the end. Now, when we turn on the perturbation to get h0 plus h1, the energy levels, the underturbed energy levels will change around. And if they're degenerate, they will in general split. So for example, our epsilon n level that we're interested in might do this. It might split into three levels like this. Of course, if you did, it means that it was at least three-fold degenerate. And it might be more than that because the resulting levels that are produced when the perturbation is turned on might themselves be degenerate. And this is a general case, a general sort of situation that happens in perturbation. Let's denote one of the exact energy levels by capital E when it grew out of the epsilon n here. And let's denote the exact energy I can state by psi. So for the derivative equation we wish to solve is that h psi, which is the same thing as h0 plus lambda h1, acting on psi, should be e psi like this. So psi here is the exact energy I can state, and e1 is the exact energy. As we vary the lambda parameter between 0 and 1, let's say we start at 1. Speaking of the energy level E, it goes from the exact level E to the exact value of the underturbed level epsilon n in a continuous manner. Likewise, the exact energy I can state psi out here at lambda equals 1. If we bring lambda back down to 0, it tracks back down. It becomes some particular state which lies in the underturbed I can state or find an I can value epsilon n. This is a degenerate I can space and some vector that lies in that space. And which one it is depends on which of these branches we follow back. So this is the basic setup of the problem. Now, in the following, it will help to have some of the geometry and the Hilbert space in mind. Begin with, let me define epsilon, script E n to be the underturbed energy I can space. So E n is the span of the underturbed energy I can state in alpha where alpha runs in a space and alpha runs over a range of runs over like that. It allows us to be multidimensional. By the way, when this E n is multidimensional, we refer to a degenerate perturbation theory. And when it's only one dimensional, so the underturbed value is non-degenerate, then we refer to non-degenerate perturbation theory. We'll see that in a little more detail later on. But for right now I'm being general and assuming that E n could be multidimensional. So we have the degeneracies of the underturbed system. Alright. Now, Hilbert space is infinite dimensional, but let me name a very schematic sketch in which I draw one axis as E n like this. And the other axis as E n corresponds to the orthogonal subspace where the Hilbert space splits up in two directions like this. In practice, E n is often times as a small dimension whereas E n perp is often times infinite dimensional. So this horizontal axis is kind of the easy axis to deal with. Now the exact dimension of the eigenstate psi is presumably close to some underturbed, it's presumably close to some state that lies in the underturbed energy I can state E n. So on this diagram here we can imagine it schematically is a factor which is almost lying in E n. That's just what we mean by perturbation theory. It doesn't move too far away from the underturbed eigenspace. This picture, this is the state psi and it's a schematic diagram. This picture encourages us to think about the projection operators that project onto the underturbed eigenspace in the opposite direction. Let's call P the one that projects down and let's call Q the one that projects to the left like this. So that P here, explicitly P is the sum on alpha of the outer product of N alpha within alpha. This is just a projector on the hand of the underturbed energy I can state E n, epsilon n. And Q is a complementary projector, so it's 1 minus P. And that can be represented as the sum on all k which you're not equal to N and the sum on alpha of the outer product k alpha k alpha like this. And so these are the two projectors. These projectors have the usual properties of projectors. They first of all, their item properties of the square of P is equal to P and the square of Q is equal to Q. There are also orthogonal projectors, so QP equals PQ equals 0. You can find the two of the different ones in succession. They just kill each other. Their sum is equal to 1 because of complementary projectors. These are the main properties of these projection operators. Now if we take the projection operator P and apply it to our exact eigenstate psi, it'll project it down like this and give us a vector that looks like this that lies in the underturbed eigenspace. That's why this is P psi. And what's left over is the complementary vector that points like this. This is Q psi. This obviously has to be small because this is a small deviation away from an underturbed eigenstate. In the following, we're going to think of P psi as the easy part of this problem. We'll think of Q psi as the hard part. The reason for that is that P psi actually is an underturbed eigenstate. And we know what the underturbed eigenstates are because we're assuming that we can solve the underturbed system. Usually there's not very many of them. So this P psi, whatever it is, is a linear combination of usually some small numbered states that we know. Whereas the Q psi is in the environmental direction. This is an infinite dimensional space of many times of practice it is. And so this is in some sense a hard part of the problem. So this is just a way of thinking about these two parts, P psi and Q psi. All right. Now to proceed, allow me to take the eigenvalue eigenvector problem which looks like this and rearrange it by bringing the lambda H1 term over to the other side. And so I'll write it like this as E minus H0 applied to psi. That is equal to lambda times H1 applied to psi. Now in analyzing this equation, it turns out to be useful to consider the inverse of the operator D minus H0. In a sense we want to divide through by this to solve for psi. That's very roughly the idea. Actually I'm not doing a very good job of motivating interest in the inverse. I could do a better job of it but it would take a lot of time. And I think that it's better just to go ahead with it. But let's just say that dividing by E minus H0 is a useful thing to do. So let's consider this operator E minus 1 over E minus H0. This is an example of a function of an operator. E here is just a number. This is the exact energy eigenvalue to remind you that H0 is a number. We don't know what E is yet, but at least it's just a number. And H0 is the temperature of Hamiltonian, which we do know all about. So this is, as I say, it's an example of a function of an operator. And you know the way you define a function of the operator. You write down the projectors and then you just replace the eigenvalues by the functions of the eigenvalues. So this is the sum of all k and alpha of the outer product k alpha, k alpha. Those are being entered into eigenstates, unperturbed into eigenstates. Divided by a denominator which is E minus X1 is k. The E here in the two sides is just a number that I copied. And the S1 of k is a value of H0 on the state k alpha which is the numerator. Anyway, that's the inverse of this operator. Of course, normal operators have inverses. And there's a question whether this one does. Actually, moreover, there's another worry, too, whether or not it really does have an inverse. There's another worry which has to do with the terms k equals n. And I'll just say let's worry about those terms and put a question mark here. The reason we worry about the terms k equals n is because the denominator there is E minus X1n. But from this diagram over here, you can see that E is the exact energy level that grew out of X1n. So presumably it's close to it because we're doing perturbation theory. So the difference between E and X1n is small. And those these denominators here, well these denominators for those terms are going to be small. This is going to get large terms here. It's even worse than that if you left lambda equal to zero because then E approaches X1n and then that denominator goes to zero and this thing is not defined. So these are problematic terms in this sum. And what we will do to handle that is just to exclude them. Let's consider a different sum which is really the same thing except we exclude the k equals n terms. The formula for this is the same. It's out of product k alpha k alpha divided by E minus X1k. However, this is no longer the inverse of E minus X0. It's a different operator now because we exclude some terms. But the point is this is now well-behaved and none of these denominators are small. Let's give this operator a name and I'm going to call it r here. And r here is saying it's really a mnemonic for a resolvent. The resolvent operator is something that appears in the theory of Green's functions and this operator here is actually very close to what's called a resolvent. Now, before I go on, there is a further worry here which is that we excluded the terms k equals n because those are the terms that are obviously small when you draw a diagram like this. What we need to ask is if possible that any other terms for k not equals to n might also be small. In particular, suppose, what I do is not true, but suppose there was another other term in general that was very close to e n. Let's call it epsilon n. Let's call it epsilon m is very close to it. This sort of thing happens very frequently in practice because giving systems have a multiple and some closely aligned energy levels. And so when we turn on this perturbation, it makes epsilon n grow when you're spread out. It's going to bump into the epsilon n if it's close enough. If the perturbation is strong enough, if that can make the perturbation strong enough, it will start devouring levels on either side and make it big enough. There's a question about when there's some other values in the case should be excluded as well. What I'm going to do for now is to assume that there are no energy levels close enough for such a thing to happen. In other words, all other overtured energy levels are far enough away and or the perturbation is weak enough so that this spread is not too large so that we don't have to worry about colliding with neighboring levels. I'll come back at the end of this lecture, this part of the lecture, and what happens if that's not true. For now, let's just assume that's true and that means that this operator is actually well behaved and in particular, there's not any large terms in it. All right. Now, what about the meaning of this operator R? What does it actually mean? We can get this, an idea of this. If you take R, let's multiply it on the left by E minus H0. Apply the R. E minus H0 is going to happen with K alpha and it's going to bring out E minus epsilon K. And that will cancel out the denominator. So all you've got left is the sum of the projectors with the K equals N terms excluded. Well, that's the same thing as the Q projector you see right here. So this is actually equal to Q. It's the Q projector. And by the way, you may notice it worked the same in the other direction that if I took R times E minus H0, E minus H0 acting from the right but after those bras would bring out the same factors that killed the denominates just as well. So R is an operator which when multiplied by E minus H0 on either side gives us the operator Q. That's why it's not one of the inverts of E minus H0 because then you get the identity there instead of Q. However, there's an interpretation of that because the operator Q, which is the projector on the orthogonal subspace can be regarded as the identity operator on that subspace. It doesn't do anything to factors that are in that space. So that's an identity as far as factors in this space are concerned. That's what R is. R can be understood this way. It's the inverse of E minus H0 on the space on the orthogonal subspace E in current. As far as what R does to the energy eigenspace E in is an annihilation because these k-alpha's are orthogonal to all the vectors of y in the integer eigenspace. So R is 0 in this space but an inverse of E minus H0 in that space. All right, anyway, that's an interpretation of R. All right, now let me show you how this gets used in perturbation. You take the formula of the upper right corner of the board which is basically just the energy eigenvector problem and allow me, since it's way up there, you can use my pointer here. Allow me to multiply by R, let's say the left-hand side so we have R times E minus H0. There is this expression here, it converts that into Q and then we get a factor of R on the right-hand side. So to copy this down, what we have is Q and psi is then equal to lambda times R times H1 psi. And this is already a nice result because it's giving us the hard part of the perturbation as this part sticks up like that. It shows you that it's small, it's a border lambda, in terms of our resolved operator H1 on psi. Now we don't actually know what psi is, of course, yet we haven't solved the problem. But we do know that psi is, and on the contribution psi is P psi. Q psi is a small correction and P psi is something that's being easy. So if we plug that in here, we get the hard part in terms of the easy part and at least that's progress. Now, actually, I'm not going to do that because I don't need to make approximations yet. We'll see it in the following way. Let's take P psi and add it to both sides of this equation. On the left-hand side, T plus Q projectors is one so we just get psi. And on the right-hand side, we get P psi plus this term, M, R, H1 psi. And so here's the P psi part that we think is easy and here's the hard correction of the Q psi written in that form. This equation can be solved explicitly for psi in the total psi in terms of P psi in the following way. Bring this, let me have a term over to the left-hand side and write it this way. 1 minus lambda R H1 applied to psi is equal to P psi. Okay? And then in a moment, I'm going to divide by this operator and we'll generate a series and that's supposed to be the basic series for perturbation theory. Before I do that, however, I'm going to say some things about P psi. P psi is, again, the projection of the unperturbed eigenstate onto the unperturbed eigenstate. Since P psi lies in the unperturbed eigenstate in the end, it must be a linear combination of the unperturbed eigenvectors that span that space. These are the vectors in alpha, you see it here. So it must be possible to write P psi in its manner as a sum on an index beta times the unperturbed eigenvectors in beta with some expansion coefficients called C beta. It's a long way to plug that in like that. These expansion coefficients have also an interpretation. If I take the scalar product of both sides of this equation with n alpha, one of the unperturbed eigenstates inside the unperturbed eigenspace for our lovely end, then on the left-hand side, we get n alpha P psi, obviously. But the projection of P acting on n alpha, so it goes to the left, doesn't do anything to it because n alpha's already inside that unperturbed eigenspace. So this is just the same thing as n alpha scalar product psi. But applying n alpha to the right-hand side picks up just one term of this sum and so we get C alpha. So you can see the coefficients C alpha are actually the expansion coefficients of the exact eigenstate in terms of the unperturbed eigenvectors that lie inside the end. Well, in any case, let me go back here to this P psi and I'll write this as a sum on beta, of n beta times the coefficient C beta, like this. We're going to need to know these coefficients C beta if we want to know what the P psi, the projection of the exact eigenstate on the unperturbed eigenbase of the eigenspace is. All right, now, the next step is to divide it through by this operator 1 minus lambda r h1. There's an identity in the ordinary calculus which I'm sure you all know is this. 1 of 1 minus x is equal to 1 plus x plus x squared plus x cubed and so on. It's a Taylor series expansion of 1 over 1 minus x. And in dividing it through by this creator of the inverse operator, you can use the same expansion series. Now it becomes a series for operators so it works in the same way. So dividing through by this will be obtained as an expression that says this. P psi is equal to the inverse of this operator times the sum. But you write it this way, right? It's a sum on beta. We'll put the inverse of this operator which is 1 plus lambda r h1 plus lambda squared r h1 r h1 plus sum on, you write up the parameter easily, times n beta times multiplying c beta. Okay? So this is now an expansion in powers of lambda giving us the exact eigenstate in terms of a bunch of operators here acting on the unimproved eigenstates in the unimproved eigenstates in the unimproved basis. Now, excuse me, in the unimproved energy with the energy eigenstates, e n, e n. Okay, now, this is one of the basic expansion on which we can structure the derivation. When I do this division or whether I rather when I do this expansion here, the expansion really necessarily converges. I'm not claiming that this is an exact series anymore. It's now a formal series and it's been carried out. In fact, this series may not converge. Frequently, it is not as a matter of fact. The perturbation theory frequently gets rise to asymptotic series. Even though they don't converge, however, they're often times useful for getting corrections and approximations. So we don't really need to converge in a series to proceed with this theory. Well, the next thing to do is to give expressions that will allow us to find the energy levels. And we will do this this way. We're going to combine this equation with something on the upper board down here just to copy down e minus h0 acting on side is equal to lambda h1 acting on side. That's the exact energy eigenstate and asymptotic problem. Let's take both sides of this equation here and let's multiply by I didn't leave myself room for this. Let's multiply first of all by e minus h0 on both sides. This is the case of some e minus h0 side. You see here. I hope you can see this here. Probably this is not so much as well in the movie. Let's then take a scalar product of n alpha e minus h0. So let's apply this multiplicative factor if you can see that on both sides of this equation. So on the left-hand side, what we get is n alpha times e minus h0 acting on side. Now the e minus h0 acts to the left on n alpha because n alpha is an exact eigenstate of the unperturbed problem that h0 gets replaced by epsilon n and it's just a c number which comes out. This becomes e minus epsilon n. What's left over is the scalar product n on the other side. But as we saw above, that's the same thing as the expansion coefficient c alpha. So we get e minus epsilon n times c alpha. That's doing this thing on the left-hand side here. Now, as far as the right-hand side is concerned, let's take the right-hand side that's going to be, it's going to be raw n alpha times, let's see, yes, yes, so, yes, so, here's what we've computed so far as is n alpha e minus h0 psi, that's the left-hand side that we're done over there. And the right-hand side has to be lambda n alpha is the scalar product of h1 psi. And so, so let's take lambda n of h1 psi and let's take the psi and let's take the series before we get right there. You see, the series is summed in data there's a bunch of operators. Here's a vector. So the h1 is going to multiply the series. We get this h1 there, and it's h1r h1r h1r h1r h1r h1r h1r And so doing that, then what we get is that this whole thing now becomes equal to, I'll write this down again as e minus epsilon again times c alpha is equal to this, is equal to somewhat of theta. I think we've got n alpha, and then there's an operator, which is this, it's lambda squared h1r h1 plus in solid lambda cubed h1r h1r h1, and then we have n beta, and then we've got coefficients c beta like this. I'm going to box this equation because this is a basic equation used for finding the energy of this. Now, to proceed with this, I'm going to take, first of all, a special case of non-generic perturbation theory. Of course, a non-generic perturbation theory is this epsilon n level is non-generic, so you don't get a splitting like this all the year. You could get that one level moving, not returning it to some other energy. That's the sort of thing that would happen. This is stuck over here. Up here, we need to make changes. This looks like a change of notation. For the non-generic case, by the way, when we talk about the non-generic case, the general, when I talk about the non-generic case, all I mean is that the energy level of these perturbations that we are considering are non-generic. The other levels may still be degenerate. It's just this one. So the non-generic case, in the first place, the arguments say n alpha. We don't need an alpha index anymore, so they just turn into a single underturn that I can state n. Also, we don't need the alpha and beta indices. So the coefficients c, alpha, and c, beta, and so on just turn into c. There's just one index now. The sum goes away. And if we do this, we just write this out. So for the non-generic case, the energy equation, which is that box equation up there, becomes e minus x1n times c, because there's only one c. The sum of the data goes away. The n alpha just becomes n. Then we have lambda h1 plus lambda squared h1r h1 plus dot dot dot is the operator. And then we've got n on the right. And then we've got c beta over there just turns into c again. Like this. And so the expansion coefficients c, I can direct with the same on both sides, and we'll erase them and we obtain this equation. And now we're bringing the epsilon in over on the other side. We can write this like this and we see that e is equal to epsilon n. The exact energy level is equal to the underturn one. Plus lambda, the first term is the expectation value. And the underturn is Hamiltonian with respect to the underturn of eigenstate. And the second one is lambda squared. I need the operator h1r h1 plus h1r h1 with the sum on alpha. Excuse me. Okay, not in someone's alpha. How does the product behave? Because h1, h1, how does the product behave? How does the product behave? h1 divided by the energy of an air, e minus epsilon n. e plus epsilon n. Using the definition of r, which I think might not disappear, but this is what it gets if you hear it gets. Now we sandwich that between the n and the n and the two sides. So the answer can now be written this way. There's the sum on k not equal to n, sum on alpha. Putting the n at the left side here and the n at the right side there, you get two matrix elements in the numerator, which are much more complicated than each other. And so it turns into this. It's tn, h1, k alpha, absolute value squared by e minus epsilon n sub k. And this goes on in higher order terms. So writing out the non-degenerate energy equation explicitly through second order, this is what we obtain. Now, this is not exactly a solution for the exact energy of the system, which is this capital E, because the capital E appears not only on the left-hand side of the equation, but it also appears as a denominator of the second order term. However, in the second order term, this is already second order. So if I replace E by 0 for a correction, which is the unperturbed eigenvalue, that doesn't change that this is still a second order term. So if we second order act and replace this with E by epsilon n, and we want to obtain these energy denominators which are the energy differences between unperturbed energy levels or the numerator which is strictly positive. So this is the problem with the most important result in the case of non-degenerate derivation here is the result is used a lot. The simplest thing to say is that the first order energy corrections, this is worthwhile remembering, you probably know it already, is that the first order energy shift for a non-degenerate level is just the expectation value perturbing to the Hamiltonian with respect to the unperturbed energy eigenstate. The second order is a more complicated formula, but it involves the sum over all states which are not equal to the unperturbed eigenstate. So these are the percentages of states. This is an infinite sum in general, and by the way, if there's a continuum, the sum really here is a schematic that represents a regular sum over the discrete levels but also a integral over the continuum. But in any case, there is a numerator here which is a certain positive that's a square of a matrix element and a denominator, I mean, like this, just this way, absolute n. There's a denominator which is an integer denominator like this. Notice, by the way, if you're dealing with a special case of the perturbation of the ground state of the system, suppose it's non-degenerate, so this is the formula you use. Now let's suppose the first order perturbation is managed, so you need to go in the second order. What you can see is that the second order perturbation, in this case, epsilon n is the ground state, that's the minimum energy. So the epsilon k's, which occur here since k is not equal to n, are always greater than epsilon n. And the result is that the ground state energy can only get lowered by the perturbation. This is a special case for your recursive equation, this practice. All right. Yes. It seems like you would be, I guess I'm not sure why, but it seems like you could use the previous order of perturbation. You should use like E plus... Well, this was the exact energy e in the formula we derived right a few steps ago before I started erasing it. And so the result is that this is not an explicit solution for these recurs now that you're also. But this is only a second order term. And so if I just substitute in the zero-order approximation for capital E, which is just epsilon n, this will still be second order. It will just further change to the third order, and the third order terms will become more complicated. I have to expand this out. But the second order is actually quite easy to do this and just replaces epsilon n. And then we have an explicit solution through second order. All right. Yes. You mentioned that the series representing the wave function could be diverging. Yes. But is the series representing the energy almost converging? No. No, in general it doesn't converge either. So then how do we gain that we have a correction if it's not converging? Well, the asymptotic series oftentimes give you, as you add more and more terms, they frequently give you better and better approximations to the answer you're interested up to a certain point and then it starts to go bad and it starts to diverge again. And it's actually fairly common in perturbation expansions if they behave that way that they really are asymptotic series. Actually, and then I will well tell you about the stark effect which is kind of interesting because that's an example when you turn into perturbation, the bound state energy level is actually strictly speaking if you continue on. You don't see any evidence of that in perturbation. So if you put a hollow million in it and you carry out the series behind that order, that would start to make some of this presence known in the incident of perturbation where it's not converging. So the terms decrease but they don't decrease asymptotically. They decrease up to a point when they start growing again. Asymptotic series typically have to behave that it's something like epsilon to the n times n factorial. So if epsilon is, you know, 10 to the minus 6 or something, you need to go out to about a million terms before the factorial starts. Yes? Could you explain again why you were saying like it lowers the chromatic energy like in what circumstances? It's a special case. It's when the first order term vanishes. Suppose the first order term vanishes and the ground state is zero, which happens sometimes. Is that what happens in anti-parametricism? I don't know. I don't know. But it happens in this arc-effective microchip which we'll talk about in a little while. And this is a negative number. n is the ground state and this is a negative number because dn is the e0 and c is the n equal to zero. This is as long as the n is the case. Because the case is summed over all the other states, in this case, with all the other side states. All right. Okay, so this is the basic energy equation for non-generic perturbation theory. Now, what about the wave functions? I think the wave functions will go back to our wave function formula which is right here. Now, it will make the same changes for the non-generic case. So the sum goes away and the c beta here just turns into a single c and this beta index disappears. And so what we have this is we have side and I'll put the c up front like this. It's been an operator one plus lambda r h1 plus, not my thought, applied to the unperturbed state n. This is what we get. Now, to write this up more explicitly this is c times the unperturbed state n plus lambda in certain expressions of r it's the sum on k not equal to n and the sum on alpha. And then what we've got is k alpha and then we have k alpha made for the sum at h1 with n divided by energy denominator which is e minus epsilon sub k goes back to my thought which is writing this up explicitly. Now, once again the energy eigenvalue appears here but the lowest order that's the same thing as epsilon sub n so I can put that in and I get the same approximate energy denominator so we get the second order of correction to the energy levels. These are the non-linear styles appearing in the first order of corrections to the wave functions. So notice that the exact wave function here is given explicitly as the unperturbed eigenfunction in n and then the corrections are all in the orthogonal direction because the sum excludes the case k equals n. So you can make it a vector of the small it's really the same as the side this is the p side and that's the q side turn look at this. Now as far as the constant c is concerned you can do this by squaring both sides. If we assume that side is normalized then when you square it you're going to get one on the left-hand side you're going to get c squared on the right-hand side also c is real because it's a facing convention for the state side. Now squaring things in the parentheses the n-scatter product of itself is one and if you look at the cross-terms in the first two terms those are zero because n times k often is zero they're orthogonal states so there is no ordered lambda direction and the next correction that's ordered lambda squared like this and so through ordered lambda which is all we've got here c is we can write this by taking the square root we can say c is equal to one plus order lambda squared and so if you're only interested in going through one to erase the parentheses here and then what we obtain is the expression for the perturbation in the eigenfunction in non-generic case and I'll put in an epsilon in here instead of a v and this is the this also is a quite useful result in non-generic perturbation we're giving us the direction of the eigenstates let's do the degenerate case this equation is quite general and covers the degenerate case as well as the non-generic form but maybe there's some space here and not a race of things that I really can't read in the degenerate case well to recapitulate slightly in the non-generate case we've got the corrections to the energy levels through the second order but only the wave functions through the first order, yes I don't know if it's ordered lambda squared or c to the square root it's still that's how you name it hope we shouldn't proceed because if I have 1 plus lambda 1 plus 3 lambda squared if I take the square root of that that's equal to 1 plus 3 half lambda squared it's still not the square root okay so so yes as I was saying in non-generic theory we carry the corrections to the energy levels through the second order and the corrections to the wave function only in the first order in the degenerate case I'm just going to carry the corrections to the energy levels through the first order and as far as the wave functions go we only get them in the zero order one can go to higher orders but in practice this is what you mostly need most of the time so as far as the energy levels are concerned let's take this equation and just carry it through the order of lambda h1 and then let all the rest of the terms zero we do this and what we get is d minus f1 n times c alpha is equal to the sum on beta we make it so the n alpha and n beta sandwiched around lambda h1 some of the lambda up front sum on beta of n alpha h1 n beta times the coefficient of the n which is c beta and this is a basic rule for for the lowest order corrections to the energy in the degenerate preservation theory what you can see is that the e minus epsilon n's we go back to the diagram this one because in the degenerate case it will in general split in different levels so that diagram is supposed to indicate that an original level epsilon n can split into several different energies capital D but what are those energies well the e minus epsilon n is the amount of splitting it's different to train the exact and the immature level you can think of the c's as an eigenvector and it's an eigenvector of an equation of a matrix which of the matrix elements of the perturbing Hamiltonian inside the degenerate eigenspace of the unperturbed system that's just the end here it's the fixed energy level so here's the rule if you form the matrix of the perturbation inside the eigenspace of the unperturbed system and you find it's eigenvalues and eigenvectors the eigenvalues are the energy shifts when the eigenvectors the c's are the coefficients which are used erase this already the coefficients that are used to expand c sides to some of the beta and beta times c beta the c's are the coefficients of the unperturbed energy eigenspace inside that degenerate eigenspace whose linear combinations give you the projection of the exact state onto that space the q-side direction is smaller, it's the first order so the 0th order, this is all you need this shows you which vector in the unperturbed eigenspace the given energy level code if you have an exact eigenstate it goes to if you have to take the lambda back down to 0 so there's a linear algebra problem in diagonalizing matrix in the degenerate perturbation theory this takes care of the energy levels it takes care of the wave functions to the 0th order because the 0th order this is equal to the exact state psi plus the order of lambda which is the first order of correction order of lambda being q-sci which we don't care about if you weren't aware of the story so that's the summary of the rules for degenerate perturbation theory now there's only one more thing that I want to address and that's the question of how nearly degenerate perturbation theory I don't know if anybody else uses that terminology but here's the issue to go back to this diagram let's suppose that there's another level called an epsilon again which is close enough to the level of epsilon again that when you turn it into perturbation the spreading out of the perturbation has been a cause of collision with an epsilon again here now what that means is is that we can't use this r-sum that we used before that I don't have before because there's now going to be small denominators involving epsilon here and the question is what can we do well the answer is actually fairly simple there's actually more than one answer but it'll give you one way of doing this we'll take the problem that will rewrite the problem in an ordinary degenerate perturbation theory let's take the h equals h0 plus h1 of our original problem and let's write h0 in terms of its trajectory so it's the sum of all k and alpha of epsilon k times the other product k alpha k alpha and so I think we've got these two levels epsilon again and epsilon again that we're worried about let's choose some let's choose some level called an epsilon bar which is somewhere in here it might be the average of the two or it might be one of the two epsilon in or epsilon in and let's define h0 prime in the following way it's first of all the sum of all k which are not equal to n and m and alpha the same sum that's about epsilon k k alpha k alpha and then for the n and m terms sum of k is equal to n again but we replace the epsilon k by the epsilon bar which is a pretty good approximation and this is the other product k alpha k alpha so that's h0 prime but as far as h1 prime is concerned it's equal to h1 plus terms that make the answer turn out to be the same as the prime sum okay so it's h1 plus a sum of k is equal to n come again and alpha of now epsilon k minus epsilon bar times k alpha k alpha the effect of doing this is to merge these two nearly degenerate underterm levels epsilon in and epsilon in again into the same level epsilon bar it doesn't change the energy of these states they're still the same but the eigenvalue is now epsilon bar that means that h0 prime is now really degenerate the degenerate level and then from the correction terms in the h1 it's okay because in order as the original perturbation of h1 at least they will be of h1 because of the degree of the size such as to mix these two levels together so once you've done this this reduces to a probability of degenerate perturbation as I say things like this are quite common in practice because of the existence of multiple close line that you very interested in in generations are well that's all for this beginning that's the basis of perturbation it includes the most important cases of practice and I'm not only going to make a beginning on the first application of this which is to the stark effect in pipers and in multiple atoms stark effect concerns the effect on electric fields on atomic energy levels the stark is the physicist who first carried on the experiments he won his Nobel Prize for doing it he was a he was obviously a brilliant man he anticipated by about 10 years the work of Compton on the Compton scattering he was one of the first people to take seriously Einstein's ideas about the existence of particles of light people didn't call them photons yet but that's of course what they were everybody thought Einstein was crazy which stark didn't in fact he proposed experiments about 10 years earlier than Compton that were very similar to what Compton actually did carry out later on and when Compton finally did carry out his experiments in 1923 that was what really convinced people that photons were real nevertheless he didn't get along very well with Einstein they had serious conflicts and in fact later on stark became a Nazi so he was a complicated character we'll say in any case being a stark effect we'll talk about the stark effect in the case of hydrogen and alkali alkali atoms to make this simple I'm going to use an electric standard model for these atoms so the under-term Hamiltonian will just simply be a kinetic plus potential energy a second force potential for the atom in the case of hydrogen of course the under-term potential B0r is equal to minus E squared over r in the case of an alkali the under-term potential B0r is equal to well there's no formula for it you can't it's really a simple formula for the potential in the case of an alkali but physically you picture this as being due to a screened Coulomb potential and as far as a picture goes both of these potentials are qualitatively the same I've got a potential engine for hydrogen of course that passes from C0 as r goes from infinity and goes from minus infinity as r goes from zero the alkali potentials qualitatively do the same thing they have a stronger there's a stronger electric force here than infinity so they go to zero faster out of the larger area they go to zero about the same speed as hydrogen instead of plotting this as a function of the radius r you plot as a function of the coordinate z along the z axis so that we can talk about the potential also in the negative z axis what we do is a symmetric curve that looks like this and then there are bound states in this and this well here which are the bound states that we have alright now a basic question that we can ask before we proceed to do perturbation theory is how accurate are the results going to be this is an electrostatic model which I'm choosing for simplicity but that means it's neglecting all kinds of small effects such as the fine structure effects of spin or refuting electrons and spinless but of course in the atoms the spin is really there and it has an effect and so the question is going to be how accurate and how realistic are our answers going to be in the perturbation theory the answer to this is if the electric field that they planted the atom is strong enough our answers are going to be good but if the electric field is weak it won't be good but the reason is that in these nominal levels that come from the electrostatic model that are due to fine structure in the spin small splittings if the electric field is weak then the effect of the electric perturbation is smaller than the splittings therefore qualitatively the answers are not going to be right if we ignore fine structure on the other hand if the electric field becomes strong enough then it can in effect overwhelm the small splittings that are already there present in the amplitude system and so for strong electric fields the answers we get by this level are actually realistic and so I think since it's a hard time we'll have to retake it that next time