 the review of what I covered in the last few minutes in the last lecture. We are thinking about the stark effect in hydrogen and alkali atoms. Stark effect concerns the behavior of atoms in the presence of electric fields. And we're taking the case of the hydrogen and the alkali atoms because these are single electron atoms. So it's a matter of simplicity to do that. Of course, in the case of the alkali, that's only an approximation. We're approximating the core electrons by means of a charge cloud, which we assume is a rotational invariant. But in any case, this gives us a central field in Hamiltonian for both cases. Where the potential, we call it potential, being on a bar, meaning it's the unperturbed potential. In the case of hydrogen, of course, we have a formula for the potential that comes from the coulomb potential in the nucleus. In the case of the alkali, there is no formula for it, no simple formula anyway. But the idea is that there's a screen charge, which is a function of position. And it goes to the pole as unscreen charge of the nucleus as r goes to 0. And it goes to a charge of just plus 1 as the electron moves to infinity, leaving just the ion behind. OK, so I think that's the basic setup, which I discussed last time. Now, in both cases, whether it's the hydrogen potential or whether it's the alkali potential, the potential energy curve has a function of radius, let's qualitatively the same. In both cases, it goes to 0 as the radius goes to infinity, and it goes to minus infinity as the radius goes to 0. Actually, here I plot the potential not as a function of radius, but rather as a function of the coordinate z along with z-axis. So there's a negative side to it as well, and the potential is insymmetric of altitude. And I've indicated schematically some of the bound states that send this in as potential. All right, so that's kind of the set of the unperturbed system. Now, before we start doing perturbation theory, under any circumstances, it's always a good idea to understand the unperturbed system untherally. That means knowing the energy levels, its eigenfunctions, and the degeneracies. So let me just summarize all of that. If we talk about two cases, hydrogen and the alkali, they're slightly different. For hydrogen, but the Hamiltonian, well, the energy eigenfunctions are central force eigenfunctions. So I'll write them in Catalan, which is NLM, the three quantum numbers. By the way, we're neglecting the spin here, so there's no spin quantum number. In the case of hydrogen, the energy level depends only on the principal quantum number, so that comes out. And that energy level, again, is the usual hydrogen energy level. It's 1 over 2 N squared. And then the energy itself, the factor that gives the dimensions of energy is E squared over A naught. That's about 26 electron volts. So that's what it is in hydrogen. There's also a degeneracy, these are the degenerate, and the order of the degeneracy is N squared in the case of hydrogen. In the case of the alkali, if we let the Hamiltonian act on the eigenstate, again, an energy that depends on both the N and L quantum numbers, although naturally not the magnetic quantum number, there is no formula for energy E and L, you can't write a formula for it. But you can certainly see what the degeneracy is, it's 2L plus 1, coming from the fact that the energy doesn't depend on the magnetic quantum numbers. I think all this should be kind of reviewed. If we talk about hydrogen, we have a typical, the standard interdevelopal diagram for that, because the 1S level is the ground state, 2S is the first excited state, is degenerate with 2P, above that is a 3S, degenerate with 3P, which is degenerate with 3D. So this is, it goes on, so this is what it looks like in the case of hydrogen. In the case of an alkali, I'll give you just an example of sodium, the ground state is a 3S, someone above that is a 3P, and then above that is a 3P. Somewhere over here is a 4S, and then there's a 4P, and here's somewhere in the cell line, it makes a problem with a book like this. This is some of the low line levels of sodium, just like that. And you see the sodium levels, the energies are due dependant on the angle of the opponent over L. All right, so that's kind of the center of the unperturbed system. Now, next we want to turn to the perturbation. The perturbation, we do it over here. The perturbation is that there's going to be an electric field in the z direction, and when you use the letter f instead of e for electric field, so it's not confusing with energy. Let's write f is equal to the magnitude of f in the z direction, so this is the electric field. This means the electrostatic potential is minus f times z, and it means the potential that appears in the Hamiltonian, I'll call it v1, you can also call this h1, it's the same as the perturbing Hamiltonian, which is charged in an electron, which is minus e times phi, and so this becomes plus e fz, which is the perturbation. Okay, so that's just writing on the formula for it. Now, if I sketch the perturbation on z axis, it's just a straight line in z, and it kind of comes through like this, and the origin goes down like that. So this is the v1. And if we have the unperturbed potential to the perturbation, and the dotted line, since v0 goes to the zero, as z goes to infinity, it asymptotes to the v1 line, and then it comes down like this. So the idea is that a small radii, the coolant potential dominates, but if you go out far enough, the actually v1 is kind of not a perturbation anymore, it's larger than v0, if you go far enough out, because v0 goes to zero. On the other side, it's going to do this, it'll add in these two curves together, you get a curve that, again, asymptotes to the v0 line, and the maximum like this is going to go down like this, which is the coolant potential. So I hope you can see my dotted lines there. You can see that the perturbation is introduced in asymmetry between plus z and minus z, and that in the downhill side here, there's a hill, there's a potential, a hill like this, where the maximum is right about there. The maximum potential is a place where the force is zero, but physically what that point means is that it's a place where the electric force pulling the electron towards the nucleus is exactly balanced by the force from the applied field, which is pushing it the other way. So the force is an unstable fixed point for the position of the electron. Now, there's a question about the orders of magnitude of the various electric fields that appear here. Of course this depends on how strong an applied electric field may apply, but let me just say some things about that. If you ask for a typical electric field produced inside an atom, in effect an atomic unit of an electric field, a reasonable choice for that is the electric field seen by the electron in the hydrogen atom in its ground state, that's to say at a more radius. If you work that out, let's call this the electric field produced by the nucleus or the atomic electric field, let's call it the atomic, the epithomic curve. What is that equal to? If you work out the numbers, it comes out to about 10 to the 9 volts per centimeter. This is actually easy to see because the ionization potential is 13 electron volts, and that's the energy the electron has to gain in moving approximately an angstrom, so putting the numbers together comes to about 10 to the 9 volts per centimeter. This is a very strong electric field. By ordinary standards in laboratory, you hardly ever get 1 by 10 before the 10 to the 5th volts per centimeter. We can take the applied electric field for the same thing in laboratories, being something like 10 to the 5th, maybe 10 to the 5th max, multiple centimeter. The main point is that it's much less than the atomic field. What that means is that in such an applied laboratory field, down here for these low-lying energy eigenstates, they're actually not affected very much by the applied field, just a small perturbation. However, you can see since the energy eigenstates pile up as an infinite number and then pile up here at 8 and 0, you can see that some of the, in fact, an infinite number of them are going to be above this top of this barrier here. In fact, they're not going to even exist as bound states anymore when you turn to the perturbation because they're free just to go on out because there is no potential binding anymore. Actually, evenly bound states, even the ground state, which is somewhere way down in this well, even that no longer is strictly speaking a bound state when you turn to the perturbation. The reason is that this potential energy curve, if I can continue to not rely on it down here, goes arbitrarily far down, maybe a weak field compared to the atomic field, but if you go far enough out, you can build an overall voltage drop as large as you want. And so, for example, this bound state, the underturbed system, now has the ability to tunnel through this barrier, come out into the real world here and get traveled away going on out. And what that means is that when you turn on this perturbation here, turn on this perturbation H1, is it's strictly speaking all the bound states of the hydrogen atom actually become low resonances, and there's no continuous spectrum. This is really like a scattering problem of the titans that was analyzed in an earlier homework problem in WKB theory where you had a wave coming in, a barrier, and bound states inside. However, the time that it takes to tunnel through this barrier here may be very large. In WKB theory, you find out that the tunneling probability is exponentially small in the action, and it should go deeper and deeper down, or if the field gets weaker and weaker, so this curve becomes more and more shallow and it's got a longer distance to tunnel through. And for the ground state, it's something that might be beyond the age of the universe before it would ever tunnel out and escape. In any case, you can see a lot of things from just a qualitative standpoint, but just growing in a potential energy curve without doing any calculations. One of the interesting features about this tunneling is that there's no evidence of it in the perturbation theory that we're going to be doing. The perturbation theory doesn't show that tunneling. You have to do a different kind of analysis to get that. Anyway, this board here is kind of the set of the problem for the stark effect in these two... these two times the system's hydrogen and all the watts. All right. Now, we're going to do a bound state perturbation theory studying, amongst other things, the shifts in the energy levels due to this perturbation. The perturbation theory is roughly divided into the degenerate and the non-degenerate types of which the non-degenerate is simpler. So let's talk about the non-degenerate perturbation theory first. I better bring this board back again. That means that we need to study unreturnable energy levels which are non-degenerate. This would include the ground state of hydrogen, which is non-degenerate because it's n equals 1, so the n squared is 1. It would include any of the s levels of the alkalis if 5s is up here, too, for which 2l plus 1 is equal to 1. It does not include the higher decided s levels of hydrogen because they're degenerate with a higher angle of n space. This is extra degeneracy in hydrogen here. So just to take a case, let's take the ground state of hydrogen and just analyze that first sort. Or maybe later we'll look at these s states and alpha ones. So we can do non-degenerate perturbation theory and see what happens. So the basic result in non-degenerate perturbation theory, let's say we're talking about the ground state of hydrogen. Let's call it delta p, delta 1, 0, 0. Those are the nlm quantum numbers of the state that we're perturbing. So it's the ground state here. And I'll put a 1 here to indicate first order perturbation theory. The basic result is that the energy shift is the expectation value of the perturbing Hamiltonian with respect to the temperature of guidance state. So here it's going to be 1, 0, 0. Same sort of perturbation, which you see up there is E, F, Z, like this 1, 0, 0. And E and F are constants which we can take out and basically what's left is the matrix element of the coordinate Z with respect to the ground state. Well, this matrix element manages. And it does so basically because of parity. The ground state is an eigenstate of parity because self-enforced eigenfunctions are always eigenstates of parity with the parity is minus 1 in the L. It's a useful rule to remember for atomic problems. But in any case, it's the same parity in both sides. Whereas Z is the component of the position operator which is odd under parity. And so the position operator can only connect together states that have the opposite parity and the result is 0. And so there is no energy shift at first order in the ground state of hydrogen. If we go to the S states of the alkalis, it's going to be the same story. Now the states are going to be in 0, 0 instead of 1, 0, 0. But the basic matrix element is the same, sandwiching Z amongst between those two. And the basic factor makes that the two states on the two sides are eigenstates of parity. And so the answer has to be 0. So there is no, as we say, there is no linear stark effect. Linear stark effect means a energy shift at first order perturbation theory. There is no linear stark effect in any of these non-degenerate states here and here and there, like that. Okay. Now, to give you a little bit of a physical interpretation of this, let me say some things about electric dipole moments. In classical electrostatics, a dipole moment vector, let me call it d, is determined by a charge distribution. And it's basically just a position weighted, just a charge weighted position. So if I have a continuous charge distribution, a row of r, I'm just multiplied by the vector r in the inner right over all space. And this is the definition of classical dipole moment vector. Now, to interpret this in a quantum problem, we might want to take the charge density of the classical problem and replace it by the charge of the electron, which is minus e, multiplied by the probability density of the electron, like this. If you just make that substitution in this interval, then you get a new interval, which can be written like this. It becomes v cubed r. And then we've got psi of r, complex conjugated. And then the middle we have minus e times r. And then we've got psi of r on the other side. That's just making the substitution of the integral. And you see in doing this, it becomes an expectation value of an operator, which is minus e times r. For the quantum mechanics, we call this operator minus e times r. We call this dipole moment operator as distinct from the dipole moment vector of numbers that appears in classical mechanics. It's a very similar idea, obviously, because here's a charge and there's a position. It's a charge weighted position. I might add that if you have more than... This is just for a single electron. If you have a system with more than one charge particle, then this gets replaced by the sum of the particles and the charges times the position vectors like this. It's a charge weighted position. In any case, what we have here then is the expectation value now with respect to the state psi of the dipole moment operator. Now, I notice that if we take the z component of this, we get the expectation value of z, which is the same expectation value that's been appeared here in this calculation of the first-order energy shift, which turned out to be zero because of parity. And so, again, in my parity, if this state psi of r is an energy eigenstate, which is also an eigenstate of parity, then the answer will be zero. And as we say, there would be no linear stark effect. Another way to say the same thing is is to say that there's no permanent electric dipole moment. If there's a state psi of r, normally one thinks of an energy eigenstate here, if there's a state psi of r such that this integral is nonzero, then you can say there is a permanent electric dipole moment and say nonzero expectation value of this. So these two things go hand in hand to linear stark effect in existence of a permanent electric dipole moment. Now, we can be more general in just these special cases of non-degenerate states, the hydrogen 1s and the alkali ns states. If we can generalize this to... I mean, one might ask whether this result of the vanishing of this expectation value is due to the simplicity of our lawmaker where we can elect and scan and multiply the distinct facts and all kinds of things. Or maybe if we took a more complicated system with multi-particle systems, maybe they had nuclear forces, you know, all kinds of things, maybe if we find this expectation value to be nonzero. Well, if we restrict ourselves to non-degenerate states, the answer is that won't happen. The reason is that if you have a non-degenerate energy eigenstate and we're dealing with a Hamiltonian, a Hamiltonian H0, which is isolated, then that Hamiltonian begins with parity, begins with a parity operator, in fact, to a very good approximation. The only approximation is that they've elected the weak interactions, which in ordinary circumstances are really extremely small. So for almost all practical purposes, H0 can be used with parity. And if we have a non-degenerate eigenstate of H0, then we have a theorem that went back to a very first lecture or the very first week of lectures and said that it's going to also, because of parity in use of H0, it will also begin eigenstate of parity. And so whatever the state psi is, it's an eigenstate of parity. If the parity is the same on both sides, this is out of their parity when the thing is zero. But this applies actually to much more general situations than just these two hydrogen and alkali that we're talking about. At least for non-degenerate eigenstates. All right. And the basic rule here is that there's no permanent electric dipole moment either for these states. Now, it seems evident from this analysis so far that if we want to find a first-order energy shift, we're going to have to look at degenerate states. We're going to start with the degenerate perturbation theory. So there's two obvious choices for degenerate states. We're going to take some of these alkali states, the 3P, for example, and 3D, which have a degeneracy of 2L plus 1. Or we can take some of the hydrogen states which have a higher degeneracy, namely n squared. So for simplicity, let's take the alkali states first because the degeneracy is not as big. So when I think of a 3P state of the alkali, for example. Now here in the degenerate perturbation theory, the basic result is that the shifts in the energy levels are the eigenvalues of the perjurian Hamiltonian inside the degenerate eigenspace of the unperturbed Hamilton. What is the degenerate eigenspace of, let's say, the excited space of the alkali, 3P, for example? These have the form n, l, m, and the degeneracy comes because of the magnetic quantum number. They don't depend on m. They do depend on n and l. And so the matrix elements that you want to consider of a perjurian Hamiltonian, h1, sandwiched between states that look like this, n, l, m on one side, n, l, m on the other side, and put a prime on one of the n's. Notice where the primes occur. The n and the l are not prime because they identify the unperturbed eigenspace of the unperturbed Hamiltonian. The n and the m prime are prime because they represent the basis inside that eigenspace. So, letting n prime run over all possible values, this is a matrix, which as I mentioned in 2,0 plus 1. And then we have to diagonalize that to find the shifts in the energy levels in the first order. That's the rule of the degenerate perturbation theory. Well, before we start to worry about diagonalizing a big matrix, we don't need to actually do very much work. And the reason is that for central force Hamiltonians, these central force eigenstates are again eigenstates with parity. It depends only on the 12th quantum number. It's minus 1 to the l. But in particular, it's going to be the same on both sides. h1 in the hand is still the same h1. It's ez, and z is odd in the parity. And so, in fact, the entire matrix is 0 for the same reason as before. It's actually my parity once again. So, parity kills this. And the result is that there's no linear stark effect leading to the excited states of the alkalis. So where are we going to find a first order stark effect? Well, it's clear that we're going to first of all have to have a degenerate energy level. And secondly, there have to be one that mixes together combines together states of opposite parity. And where do we find such a thing? The answer is we only find it one place in the kind of real universe of problems, and that's the hydrogen atom. The degeneracy between the different l values. For example, you see the parity is a function of l. The s states are even and the p states are odd. The p states are even and so on. And the parity. So just looking at the n equals 2 levels, we've got even and odd states that are degenerate with each other. So therefore, we can't expect a first order stark effect than the n equals 2 levels of hydrogen. And that's what I might analyze now by degenerate perturbation theory. So yes, I have to erase some things now. They ought to have all physics lecture halls with boards of the whole way around so that somebody could just start drawing what they are. All right, anyway. Anyway, let's take a look at this level here. This is the n equals 2 level of hydrogen that has 2 squared equals 4 states. There's one here and there's 3 there. And what are these states? So not this way. This is going to tend to be in the notation nl down. So the 4 states are first of all 2, 0, 0. That's the 2s state. And then there's the 2, 1, 0. And 2, 1, 1. And then 2, 1 minus 1. These are the 3 states in the 2p level like this. So we're going to have a 4 by 4 matrix. So let me just set it up like this. I'll write kets across the top row and I'll write bras down a column here. So this is 2, 0, 0. 2, 1, 0. 2, 1, 1. And a 2, 1 minus 1 like this. So we're going to have a 4 by 4 matrix like this. And we want to sandwich around this the perturbation, which is the exe. This is the h1. All right. So here's what our matrix column is going to look like. And then it'd be nlm. And then we've got z and I'll take the ef out. This is the constant. And then nlm on the other side. And I'm going to prime the l and m on the first side because those are allowed to be different from the l and m on the other side. I want the n is the same because that labels the eigenspace, the degenerate eigenspace of the temperature of the system. All right. So there's 16 making supplements to fill out here. And they all have this form. So what you should do in a case like this is to start looking at selection rules to see when the matrix element is going to be 0. And in fact, what you've got here is something which the bigger echart there applies because the z is in fact the, it's a t10. It's a q equals 0 component of a tensile operator, namely the vector operator for the position value. And so this has got a q equals 0. So as usual with the bigger echart there there's two types of selection rules. One involves the angular momentum. And the angular momentum, according to the bigger echart there, we've got a, we'll have delta, delta difference between l and l prime. Can either be equal to 0 or plus or minus 1. Those are the rules for addition of angular momentum. There's also a magnetic quantum rule that says that m is equal to m prime, or m prime equals m because the q is 0. So those are the selection rules and making supplements guaranteed advantage unless those conditions are whole. This is coming from the bigger necromancer in which concerns the symmetry under, under proper rotations. If you throw parity into the next time, there's an additional selection rule which says that delta l has to be odd because you have to change the parity of the states. And so the 0 goes away, and what we end up with is just delta l equals plus or minus 1 and m equals m prime. Those are the net selection rules of everything coming from proper as well as improper rotations. This is the standard selection rules for dipole matrix cylinders which actually I've been owning before already. The matrix cylinders occur in a stark effect of the same ones that occur in the radiative transitions, which I talked about earlier in getting an example. This is the bigger record, and it's the same example appearing again. All right, so in any case, it means the l and the l prime have to be different. And so we're certainly going to have 0 as far as the diagonal of this matrix because the l and l prime are equal to the diagonals. Also, these matrix elements, the matrix is for mission, because the matrix is for mission operator. And in fact, it's actually real so in fact it's symmetric matrix, so I only need to do one half of it and fill it in. So running across here, 2, 0, 0, 2, 1, 0, the l has got a delta l equals 1. That's okay, and the m's are equal. However, 2, 0, 0 with 2, 1, 1 and 2, 1 minus 1 have to be 0 because the magnetic quantum numbers don't agree. So these vanish. On the second row, 2, 1, 0, 2, 1, 1, 2, 1, 0 can they have to be 0 because of magnetic quantum numbers? As far as 2, 1, 1 and 2, 1 minus 1, that's also 0 for the same reason, magnetic quantum numbers. So now we'll just fill in the other side using the symmetry of the matrix as it looks like this. And you'll see of the 16 elements, there's only 2 that can be non-zero and in fact, there are people with this other, it's only only one matrix solid that's non-zero. They're equal because it's a symmetric matrix and it's symmetric because it's a real emission matrix. It's real because the wave functions are real and you can see it's real. So if I block-diagonize this matrix like this, there's a 2 by 2 block up here which is just these 2 off-diagon elements and everything else is 0. Now, I don't know if I want to erase this diagram. I might need it later on. Well, maybe not, I guess. So, what can we say now? So the energy shifts that are the eigenvalues of this 4 by 4 matrix. So this is the delta P. This is the energy shifts in the beginning that's 2 levels of hydrogen due to this perturbation. You can see 2 of the eigenvalues are 0 and the other 2 eigenvalues are going to be the eigenvalues of this upper 2 by 2 matrix. Well, let me just do some work and actually calculate matrix elements there. That just means an integer could be non-zero. Actually, the integer itself is e times f times the scalar product of 2, 0, 0 from the last coordinates e in the middle and then 2, 1, 0 on the right. And this turns out to be a negative number. So I'll write it as minus w so that w is simply a positive number. And I know this because I plugged in the wave functions, the yln's and the radial integral so I did the integral. And if you do that, what you find is that w is equal to 3 times the charge e times f times the whole radius a, 0. w is an energy and you can see it's an energy which is at the order of magnitude necessary to move the electron from one side of the atom to the other in the external field. Here's the charge. There's the distance of the electric fields and that's the force. It's only times a factor of 3 which comes mainly because these are equals to wave functions which are larger and more radius. But anyway, if you do the intervals, this is the answer you get. Let me just call it w. It's easier to write. And as far as the x here, then I replace these two things by minus w over minus w here. And w is defined that way. And now if we compute the eigenvalues of this 2 by 2 matrix, we find that they're both plus w and minus w. And so those are the energy shifts now computed by the generic perturbation theory. Make a diagram. We started with 2s and 2p levels in terms of generate under the Hamiltonian h0. And then we go over to the full Hamiltonian h0 plus h1. We get a splitting diagram and we're going to do it for this. We get a splitting diagram that looks like this. There are two levels that remain the same. That's the 0, 0. And there's one level that goes up by plus w and another level that goes down by minus w like this. As far as the two levels that remain the same, you can see which ones they are. The 2, 1, 1 and the 2, 1 minus 1 levels. Because that's this little block here. So this is the 2, 1, 1 and the 2, 1 minus 1 levels. These are not affected by the perturbation. This is the 2p states for n equals plus and minus 1. All right. So that's the energy level diagram under the perturbation. Now I'll come to the wave functions in a moment and before I do that I want to address some questions of symmetry that appear here. This is the result of first order perturbation theory. What we see is that the 4-fold degeneracy is only partially lifted by this perturbation. There's two same ones, but there's still a double left over so there's still some degeneracy left. This is the kind of thing that happens very frequently in studying perturbations of energy levels. But something like this happens. You are led to wonder whether this degeneracy at first order perturbation theory is something that might be lifted if you went to higher order. First order perturbation theory is only an approximation. It's a first approximation. But so you wonder would this be possibly lifted in higher order or would this degeneracy persist to all orders? Well, if the degeneracy persists to all orders it almost certainly means that there's some symmetry of the full Hamiltonian, the full H0 plus H1. This would be, this would be the full Hamiltonian which would be p squared over 2m plus rv0 of r plus rvfc. This is the full Hamiltonian now. The, as I said some degeneracy persists to all orders. It's almost certainly a symmetry, represents a symmetry of the full Hamiltonian and the full perturb Hamiltonian. It's a basic fact that in quantum mechanics the degeneracies are associated with symmetries. That is your key to recognizing symmetries and very important in discovering symmetries. It's fact. So let's take a look at the symmetries both in H0 and in the full Hamiltonian H which includes this extra term for the S term of electric field. Let's see if we can, we can guide you from that. So if I talk about H0 and H0 plus H1 let's look at the symmetries. So first of all, what are the symmetries of H0? Well, it's a central force Hamiltonian so it's got four rotational symmetries that's SO3, the dimensional rotational symmetry. What else does it have? It's symmetric in comparity because all central force Hamiltonians are commutative parity. And furthermore, it commutes with time reversal because any kinetic plus potential Hamiltonian commutes with time reversal. It doesn't have to be a central force potential with Hamiltonian to do that. Remember, the way to break time reversal is magnetic fields. What about H0 plus H1? Well, it doesn't commute with all rotations anymore because the electric field is not giving us a preferred direction, which is in the C direction. There's no breaking of that symmetry. So in the Z direction, maybe about this, we'll have. However, we still have a rotational symmetry about the Z direction. So that means that SO2, two-dimensional rotations about the Z axis. What about parity? Parity is certainly not a good quantum number anymore because the electric field indicates a preferred direction but plus E and minus E is not the same physics. So there's no parity anymore when this one goes away. And finally, what about time reversal? Well, time reversal, any kinetic plus a potential Hamiltonian doesn't have to be a central force that commutes with time reversal. So we still have time reversal. So the symmetry of the perturb system is less than the symmetry of the emergent system, but it still has some symmetry. SO2 here, by the way, is generated by LZ. The Z component of any momentum which generates rotations about the Z axis. So in terms of the initial operators of LZ and theta for the fully perturb system. So let's try to do an analysis of the fully perturb system understanding symmetries without making many approximations. So what we've got is a full Hamiltonian H and we've got an operator LZ that commutes with. There aren't any others actually. There's not any obvious ones anyway. In fact, there aren't any others. These are the obvious ones and all there is. You know, it's usually true if you find the obvious symmetries that they're the only ones that there are. Occasionally there are some sneaky hidden ones that you wouldn't have suspected. And the most outstanding case of that is hydrogen where there's actually an SO4 symmetry instead of an SO3. It has more symmetry than an accurate average of a central force problem. There's also extra symmetry to isotropic harmonic oscillators. Except for that it's hard for me to think of any examples of whether it's extra or hidden or hidden symmetries. So the normal rule is what is obvious is actually all there is. And that's the case here. These really are the only... LZ is really LZ and time reversal are the only two symmetries in this whole Hamiltonian task. Well in any case, since these two operators commute they of course possess simultaneous eigenstates. Let's diagonalize LZ first. And if we do it, let's call this one number M and we'd think of M equals 0, M equals 1, M equals minus 1, and so on, ranging from plus and minus infinity. I'll just... I'll just draw these first three, yeah. Can I ask you a little bit? Is there a question? Yeah, a question. What's the fourth? What's the fourth? The SO4? Well this is... The SO4 applies to hydrogen. It actually only applies to the electrostatic model of hydrogen. When you start putting other details it goes away. The... Its rotation is in a four-dimensional space but it's... obviously it's not no physical space so it's an abstract space that's constructed by the operators in the hydrogen atom. The hydrogen atom has extra operators in the Hamiltonian it's called the Rondel Inspector. There's nothing that corresponds to the Rondel Inspector for your average central force problem. It's only for hydrogen. Yes? How... Is there any way to, like, know that those are the only symmetries that apply to the problem? Yeah, that's an interesting question. It's how do you know if there's real symmetries? Actually, a classical mechanism is ways of finding out you can... you can... you can prove that there's no extra conserved quantities beyond a certain... a certain obvious set, or a certain set. In quantum mechanics, it's a slightly... it's a different... it's a somewhat different question there. Maybe it's best to re-discuss it in office hours but it probably has to do the matter of definition of what you actually need by symmetry. Anyway, it's a practical rule what I'm saying is the way it works. All right. So, thinking of this exact system suppose we diagonalize LZ first. What that gives us is a set of... breaks the Hilbert space up and then the eigenspaces, which would be eigenspaces of LZ corresponding to different quantum numbers. In each one of these eigenspaces then we can diagonalize the Hamiltonian and let me make a vertical axis to indicate energy levels so we get where they do that. And so if you get energy levels here then I'll just draw the kind of... I'll draw the numbers if they're random. They aren't random actually but just to make it go out we can just put some in. You get energy levels like this. So let's sequence them starting with... let's call it U equals 1 U equals 2 starting with more energy going on up. So what we get is energy levels and let's call it U and M like this and we have the fact that the Hamiltonian acts on this and brings out an energy which in general must depend on both U and M like this. So those are our energy eigenstates dedicated to schematic meetings of these diagrams here. These are also eigenstates of LZ that LZ acts on U and brings out M. Now the energy... in fact, the energy would depend on M. The energy doesn't depend on M in rotational invariant problems because M indicates the orientation of the system. But this is no longer rotational invariant and we wrote that by putting it in the external field. So now we do expect the energy to depend on M. In fact, if you look for the resultants who they do is A equals 0 and this is A equals less than 1. All right. In any case in any case this is a... this analysis of the exact system. Now the... unfortunately I didn't give myself enough room here to really do this. Let me do this. But I'll do it just below over here. So now let's ask what is time reversal? Let me remind you that if you have an energy eigenstate of a Hamiltonian that can enter time reversal the time reversal map of the energy eigenstate into another... into another state which is an energy eigenstate with the same energy. So in the present case this means if I take theta and let it act on one of these new M states this is an energy eigenstate with the same energy. So in particular the Hamiltonian act and that brings an equal amount. That's what energy does. Now, what about LZ? Let's take LZ and apply the theta Lm and you'll have this. LZ and theta don't communicate any communication because any momentum are odd under time reversal. So this turns into minus theta times LZ acting on new M. That brings out LZ brings out the M the result is this is minus M times theta acting on new M. So this theta acting on new M is an eigenstate of energy with the same eigenvalue as new M itself. However the magnetic quantity number is changed and the LZ is changed with the offset itself. What that means is in this diagram up here is that if we take one of these energy levels let's say the new equals one here with new equals one and we apply theta to it it's got to map it over into an eigenstate of the same energy on the new equals minus one. In other words it must be an energy level here that reads from that one over there By the way this level could have been degenerate say it's two-fold degenerate and theta applied to both of those and they get two levels over here with exactly the same energy. They're quite subversive. So the result is although I didn't draw it this way the energy levels on this new equals one must be exactly the same as the energy levels on this new equals minus one. And we draw this very well but that's what they're about to be. So the energy action only depends on the absolute value again and the degeneracies are always even in particular at least they're at least equal to two there's at least two-fold degeneracy. Now this argument doesn't work for an equal zero because if I've got an equal zero level then theta maps it into another equal zero level it could be the same level. And so the argument does not require that there be degeneracies of an equal zero only for the values of them which are greater than zero. And if you look at our results in the first order of perturbation theory this is exactly what we've got. We have a degeneracy here between a equals plus one and a equals minus one. And this argument shows that this is exact so although we've got an interest order of perturbation theory it actually applies to all orders. On the other hand the W states are eigenstates of they're eigenstates anything from zero and these are non-generative. There's plus W minus W means there's two different energies so on a diagram like this we look like this here's the minus value and here's the plus value and the point is these are non-generative and they're conserved. All right. So those are just considerations of symmetry of the results of the exact non-generative. All right. The energy levels on the n equals one and n equals minus and I'm always going to have four things each other like not just to the new equals one like you drew but also to the new equals two. Oh yes always. You take any one of you if I fade into it you're going to get another energy eigenstate with the same energy but the energy changes so it's got to be a level over this side with the same energy. Right. So then for each new do you have a plus and minus W? So are there twice as many states on the n equals zero over the drama? Do you mean? No. No. I see. You're saying are you saying because there's two here and there's two there which is that big? Yeah. As far as I can see right now I don't think that follows from this analysis although I agree it's what side of this example. So what would the new equals like so would you have? I guess what would the state above plus W be under the n equals zero? Well this was the analysis of the n equals two level of hydrogen. Right. If we analyze n equals three level we get another diagram that would be more complicated because it's in the right state now. Okay. But in there in those nine states there would be n equals one levels and they would have to be there would have to be two-fold genera. So we already have to use n equals two levels and they're going to have to be two-fold genera too. The n equals zero levels don't have to be in fact they don't like them to be their proper sequence. Okay. This is supposed to be related to what's called lambda dumbling in molecular physics but since this is not chemistry I won't tell you about that but it's all lambda dumbling. All right. Okay. So okay so now all right so look back at this perturbation analysis then what I've done so far is to find the shifts in the hydrogen levels which are here and indicated by this diagram this diagram and there's another diagram. We may also be interested in the wave functions of the eigenstates and just again to remind you the results of the genera perturbation theory you get the eigenstates we need to find the eigenvectors of this matrix. The eigenvectors give us the linear combinations of the unperturbed eigenstates. These are just the basis these are just the basis of states of the unperturbed eigenspace and certainly the combinations of them will be the eigenstates of the perturbation system. Well it's easy to find the eigenvectors of this two by two matrix. The eigenvectors of this two by two matrix will be the eigenvectors of this there's two of them to normalize and it's one of the square root of 2 1 1 and the other one's one of the square root of 2 1 minus one easy enough to get. This first one corresponds to the energy minus value of the second one corresponds to the energy plus value. And so what we can get then what we can save is the states plus and minus value are linear combinations of these two states two zero zero two one zero and look like this is one of the root two times two zero zero and it's a minus or plus two one zero and so these are the exact eigenstates well not exact but these are the first sort of perturbations where these are the eigenstates in the presence of the perturbation the linear combinations of these two states will be zero and they have different energies. For lack of time I won't bother doing sketch making sketches of the wave functions but let me just say if you write out the wave functions for these two states and add and subtract the wave functions and then square them to get probabilities or charge densities you'll find in the state where the plus w has a charge moved against the electric field and it's shifted against the electric field basically the atom becomes polarized by the electric field either with or against it depending on which of the two signs you have and this would come out of a contour plot of the wave function the net wave function of the preserved system has to have a diagram so you can see that alright now so that's just some remarks about the energy and hydrogen states let me say something we've got now about some interesting physics connected with these n equals 2 levels to do on this board here in hydrogen we have the 1s we've got the 2s and the 2p and then the 3s and the 3p and the 3d like this and so on in the higher states the dipole radiated transition rules require delta l equals plus and minus 1 so you get a transition like this and one like this and you get one like this and you get one like this and you get one like this and you get one like this but there's nothing to taking it from the 2s to the 1s and so if you take a population of hydrogen atoms and you do something to knock them into a into a collection of excited states you could do this with the electron beam let's say up through n equals 3 or 4 then what will happen is they'll start undergoing radiated transition rattling down following these arrows like a paint like a pinball machine and they act for a while they'll either end up in the 1s state which is the ground state or also end up in the 2s state which is metastable the time transition time for the 2p to 1s is a dipole transition chosen about 10 to the minus 9 seconds and the 2s state is stuck there and it has to go by a higher order transition we'll talk about later on next semester and it takes a much longer time it's about 10 to the minus 1 second about 100,000,000 times longer than the 2p to 1s transition now the 2s state if I write it in terms of the NL and quantum numbers this is the 2-0-0 state and this is an NL but if we turn on the electric field it is not an eigenstate of the detergent system because those eigenstates are plus and minus w which since it's now covered up I have to rewrite it it's the 2-0-0 minus 2-1-0 like this so the presence of the electric field with 2-0-0 is no longer an energy eigenstate it's a linear combination of plus w and minus w these two components don't evolve at the same frequency because they have different energies there's a delta e energy between the two which is equal to twice w and so if you look at the relative time evolution of the two components plus and minus w of this state it goes by b to the minus i is delta b of 2-0-0 h1 and the result of this is that you get an oscillation and probability between the 2s and the 2p it swings back and forth as a function of time but more exactly that's what it would do if there were no radiated transitions but since there are radiated transitions as soon as as soon as some of the 2p excuse me some of the 2s state it shifted over to the 2p but rapidly in a period of 10 minus 9 second nanosecond it rocks down into the ground state remaining photons so the result is if you create this population of med stable 2s hydrogen and then you turn the switch to click on an electric field what happens is all of a sudden you get a lot of photons and then all of a sudden the ground state three person photons comes out this is some of some of my physics connected to this alright now the now then let me go back to this I want to go back to the ground state of hydrogen which is here as a matter of fact we found that the first load energy shift was zero there's no but the question is if there's no energy shift what about the wave function is there a shift in the wave function so let's look at that question now what is the I mean if you just think of it physically it's clear there should be because the normal state of hydrogen is a symmetric distribution of charge like this and if you put an electric field in one direction and the electrons push that way and the proton that way looks like this and the plus sign again and minus sign there so it turns in particular dipole moment vector that looks like this you see pointing in the same direction as the electric field so this dipole moment vector this is called an induced dipole moment as I explained earlier there is no permanent electric dipole moment in the ground state but in the presence of an external field it requires one and that's what's called an induced electric field and the personality constant is called alpha and it's going to convey polarizability it's got mission polarizability it's just a proportionality factor okay so let's try to analyze this from the standpoint of quantum mechanics according to first order perturbation theory the shift in the the perturbed wave function this is well the the original wave function is the ground state 1 0 0 and applying the formula for the first correction what you get is a sum over all states N L M which are not equal to the ground state 1 0 0 all the states except for the 1 that should be perturbed and you get a linear combination of states N L M these are states orthogonal to the original state and this is the original state h h h the ground state and then you have an energy denominator which is d 1 minus d n like this because the energy is only dependent in the principle of quantum physics so this is just a formula straight from perturbation theory now what about the now the question arises what about the dipole moment of the state in the presence of d 1 so let's compute psi sandwiched around the dipole moment operator which is minus d I'll just take the z component of it minus d z psi like this what about this expectation now so what I need to do is take this sum here as a ket on this side and this sum is broad on that side and you'll see the before terms this sum here however both sides gives us zero that was this calculation of the box so that doesn't contribute there's going to be two cross terms they're actually equal to each other because these are all real matrix solvents and then the last one would be a second order where I've got the sum on both sides and I'll ignore that because I've just worked the first order so the result is there's a factor of two because of the two cross terms there's a factor of EF which I'll take off that's another constant as far as E1 minus En is concerned let me write this as minus En minus E1 because the energies are increasing functions again in that way I want to positive denominator so you take like the minus one that comes from there and then what's left over let me call it S which is a sum so S here is a sum and this is equal to the sum and again what we need to do is 1 0 0 as a bra I need to put in the minus EZ or just Z now because everything else is taken out so it turns into this it becomes 1 0 0 Z NLM times NLM Z 1 0 0 same matrix only written backwards divided by E N minus E1 let's just take that as a sum so this turns into true E squared F times the sum S and I'm sorry I heard the bell let me just do one more thing and I'll let you go and that is this definition of polarizability we take the Z component of it as EZ and it's equal to FZ FZ is the same thing as where I'm going to just wrap here you see the polarizability now can be read off this is the the Z component of the dipole moment to the polarizability that's everything else except for that here so what we get is alpha is equal to twice E squared times this sum S where S is going to be near by this sum and this is an example of how atomic polarizability is going to be calculated for quantum mechanics this easily leads you to the dielectric constant of the material it's a little more realistic in this case because nobody has a gas of hydrogen atoms you always have hydrogen molecules but at least it will take the principle of how electric susceptibility is going to be calculated for its principles okay so I'll see you tomorrow morning and we'll let continue under this cycle I hope to finish our video