 We have talked about ground stage of many electron atoms, helium and lithium to be more precise. Now we talk about the excited states. Why is it that we must talk about excited states? First of all, let us not forget that one of the beginnings of quantum mechanics is in spectroscopy and spectroscopy involves transitions between ground and excited states. Secondly, well, maybe not secondly, another way of saying it is that all the color that we see all around us, we live in a colorful world, all the color that we see is because of transitions involving ground and excited states. Finally, it is not only color, but a lot of reactions that take place. The very fact that we see is something that involves excited state. Photosynthesis, something that we are all familiar with involves excited state processes. So, it is not sufficient to restrict our discussion to ground states, we must understand excited states as well. Of course, building descriptions for excited states is much more complicated than ground states. But what we achieve to what we try to do in this course, our goals are not very, very high. We will be able to do it and get an idea of what exactly is done. And on a lighter note, we often do things better when we are excited. Same holds for atoms and molecules. Many times, they do things better whatever they are supposed to do when they are excited by light by absorbing light. So, that leads to all the photochemistry around us. Let us not neglect excited state, let us learn how to study that. But before going there, what we have learned so far is about spin orbitals. Spin orbitals, okay. Let me for once define a spin orbital because I do not think I have said it so far. A spin orbital is a one electron wave function incorporating the spin part. Well, it is as simple as that. It is not rocket science, but it is important to not forget that an orbital is for the n plus 1 th time one electron wave function. A spin orbital in addition to having the special coordinates for this one electron wave function also has the spin part, right. So, for 2 electron systems, we have learned that we can have 4 spin functions alpha 1, alpha 2, beta 1, beta 2, but we cannot have alpha 1, beta 2 and beta 1, alpha 2 by themselves because we cannot distinguish between electron number 1 and electron number 2. So, we cannot say for sure that it is electron number 1 and not electron number 2 that has alpha spin. We cannot say for sure that it is electron number 2 and not electron number 1 that has beta spin. So, the best we can do is we can take linear combinations and while taking linear combinations, there is no reason why we should stick only to plus. A linear combination connecting the 2 terms via minus sign is equally acceptable. So, we have to take that as well and we have said that if you use exchange operator just interchange the labels, then the wave function with plus is symmetric with respect to exchange, the wave function with minus is anti-symmetric with respect to exchange. The reason why we worry about whether the functions are symmetric or anti-symmetric with respect to exchange is poly principle or 6 postulate of quantum mechanics, which says that for fermions like electrons, the complete wave function of the system must be anti-symmetric with respect to exchange. So, psi of 1 and 2 must be equal to minus psi 2 1. If you interchange the labels 1 and 2, the wave function must change sign. So, what we are saying then is that if the 2 electrons in 1s orbital had same spin, then we would end up getting the function psi s 1 psi s 2 psi 1 s 1 psi 1 s 2 multiplied by alpha 1 alpha 2, which gives us psi 1 1 2 would be equal to psi 2 1. It is not allowed. So, that is what leads to poly exclusion principle. And we have sort of taken this idea forward a little more by looking at stator determinants of bigger atoms as well. This is where we stopped in the last module. Now, let us talk about the excited states of helium. So, let us start with an electron configuration. What is a state? A state at least as a first approximation is determined by the electron configuration. So, we are going to have some excited state corresponding to say 1s 1 2s 1 configuration. As we will see, a particular configuration can give rise to different states. That we will see in a moment. But let us start. How do I write a wave function for a state with electron configuration 1s 1 2s 1? I can write 1s then 1 in brackets multiplied by 2s 2 in brackets. The problem is what this implies once again is that it is electron number 1 in 1s orbital, electron number 2 in 2s orbital and not vice versa. Who has told you that this is going to be the case? As we said earlier, electron numbers 1 and 2 are indistinguishable, 1 and 2 are labels that we are using to formulate the mathematical problem. So, if we just write 1s 1 2s 2, then our description would be incomplete. We must write 1s 2 2s 1 as well. And then as we have done in the earlier scenario, we have to take plus and minus combinations of these two wave functions. So, since we cannot say which is which again this is becoming a cliche, we have to consider this wave function with a plus sign and also this wave function with a minus sign. So, right away you see that corresponding to a single electron configuration for helium excited state 1s 1 2s 1, I have generated two wave functions only while considering the spatial part. I have not even talked about spin yet. Only by spatial part, I have constructed two wave functions and it is not very difficult to see that the first one is symmetric and the second one is anti-symmetric. Now, knowing poly principle our job should be simple. We know that the total wave function has to be anti-symmetric. Now, see a symmetric function multiplied by another symmetric function gives you a symmetric function. And anti-symmetric function multiplied by an anti-symmetric function gives you a symmetric function. A symmetric function multiplied by an anti-symmetric function or vice versa gives you an anti-symmetric function. So, now see in order to ensure that the total wave function is anti-symmetric, the symmetric spatial part must be multiplied by an anti-symmetric spin part. Now, you might remember what the spin wave functions are for a two electron system. Remember, we had said you can write alpha 1, alpha 2 or you can write beta 1, beta 2, you can write 1 by root 2, alpha 1, beta 2 plus beta 1, alpha 2 and you can write 1 by root 2, alpha 1, beta 2 minus beta 1, alpha 2. Great. Now, these three wave functions are symmetric with respect to exchange. This lone wave function is anti-symmetric with respect to exchange. And here the spatial part is symmetric. So, in order to have an anti-symmetric total wave function, it is imperative that the spin part has to be anti-symmetric. So, the only choice we have for this spatial part is this alpha 1, beta 2 minus beta 1, alpha 2 function. This is called the s equal to 0, ms equal to 0 state, we will come to that shortly. But before that, let us see what happens for this spatial part where there is a minus sign. This itself is anti-symmetric. So, we have to multiply it by one of the symmetric functions. We can multiply it by alpha 1, beta 1. No, what am I saying? We can multiply it by alpha 1, alpha 2 or beta 1, beta 2 or alpha 1, beta 2 plus beta 1, alpha 2 multiplied by 1 by root 2. All these three functions can be multiplied. So, what do you do? What do you get? From the symmetric spatial part, we generate a single wave function. From the anti-symmetric part, however, since we have three symmetric spin parts, I will generate three wave functions by multiplying this single spatial part by one of the three different symmetric spin parts of the wave function. And then this is going to be called the collection is called s equal to 1. The first one is called s equal to 1, m is equal to 1 state. Second one is s equal to 1, well, capital S equal to 1, capital M is equal to 0 state. Third one is called capital S equal to 1, capital M is equal to minus 1 state. What does all this mean? Well, before saying all this means, let us have a look and let us see whether it rings a bell. I have two quantum numbers a equal to 0, b equal to 0. Then the next one is a equal to 1 and for a equal to 1, b is equal to 0 and plus minus 1. Have you encountered this situation somewhere? Of course, we have. Remember J and M rigid rotor. Remember L and M hydrogen atom, same thing. When L equal to 0, ML has to be equal to 0. When L equal to 1, ML has to, ML can be 0 plus 1 or minus 1. So, same thing is there. So, what this capital S stands for really is the total angular momentum and capital MS stands for the z component of this total angular momentum. Once again, this is a quantum mechanical phenomenon completely. However, as I told you, it is strange, but it is true that when you talk about spins, you can actually get most of the things done by using a very simple classical vector model, where you just draw the angular momentum as an arrow of appropriate length. So, how do I depict this S equal to 0 state? Well, first of all, I draw this arrow, up spin. Now, remember the angle, remember the angle. The angle is same for all electrons. Now, when electron number 1 has alpha spin, electron number 2 has to have beta spin and vice versa. We have to look at individual terms. So, how do I depict this individual first term? When electron number 1 has up spin, electron number 2 has down spin. And in order to get S equal to 0, S you can think is a vector sum of the angular momentum of the 2 spins. So, these 2 have to be disposed along a straight line. So, what I am saying is that if this is the angular momentum vector for electron number 1, this is the angular momentum vector for electron number 2. Do not forget that their 5 values are not defined. In the classical picture, they are assumed to be precessing like this. I do not like to confuse students by invoking this precision business too much. But if you understand better, what I am saying is that it is both the up spin and the down spin of electron numbers 1 and 2 are actually delocalized over all possible values of phi. But then they are delocalized in a correlated manner. It is important to understand. Suppose you could freeze and measure, you would always find that the difference in phi is between the angular momentum vector of 1 and angular momentum of vector of 2 is 180 degrees. Suppose I for this situation, I take projections on the xy plane. Do not forget where the xy plane is, draw something like this. This circle is in the xy plane. Suppose I draw a projection of the spin angular momentum of electron number 1 and I draw a projection of spin angular momentum of electron number 2. These two projections are going to be opposite in direction in the xy plane. That is what it means. They cancel each other. So, of course, if you take z components, then also 1 is up, 1 is down, they will cancel anyway. That is why s equal to 0. And if s equal to 0, what is ms? ms is also equal to 0 because if the length of the arrow that you get by vector sum is 0, what will the z component be? It has to be 0 and nothing else. Next, let us look at this s equal to 1 business. Here the vector model is something like this. s equal to 1 means what? It means that the total length has to be 1. They have to add up. So, the first one alpha 1, alpha 2 is like this. Beta 1, beta 2 is like this. What would be the z component of this case? You have added 2. Remember, this length is root 3 by 2 and the z component is plus half. This length is root 3 by 2, z component is plus half. Half plus half, what do you get? You get plus 1. Of course, you might ask if this is root 3 by 2, this is root 3 by 2, you just add them, you get root 3 and not 1. So, that problem is actually there. But do not forget, this is a very qualitative classical picture. But z component is easy to understand. Z component is definitely going to be 1. In this case, it is not difficult to see that z component is going to be minus 1. Now, if you draw like this, then you will get some component along the x direction or whatever plane these 2 vectors are in. So, s will still be equal to 1, but m s has to be 0. Because the vector sum is going to be along say x axis or y axis or somewhere in the x y plane, z component will be 0. So, this is roughly, we have not really done the vector sum in very great detail. But roughly, this is what the picture is. This is why you call them capital S equal to 0, capital S equal to 1 and so on and so forth. We are going to come back to this later on in a little more detail manner. But for now, we have without saying it, learned a very, very important concept. See, when the spatial part is symmetric, as we said earlier, the single anti-symmetric part has to be multiplied by it to get the total wave function. Since you get only one wave function for the spatial part, this is called the singlet wave function. The state described by it is called the singlet state. What about this? Here, the spatial part has minus sign, fine, but you still have 1s and 2s orbitals. Does it matter whether electron number 2 or electron number 1 is in orbital 1 or orbital 1s or orbital 2s? It does not. So, this energy of this state where the sign here is minus is going to be exactly the same as the energy in this state. Unless you want to consider the energies of, unless you consider that there is interaction between spins and all that, that is not there. So, we do not have to worry about that. Just think of the spatial part, think of Schrodinger equation. If you could solve Schrodinger equation, here you have one electron in 1s, one electron in 2s orbital, same thing here. Look at the configuration, they are the same. So, energies are same. So, energies of these things are same. And then the energies of these three wave functions is definitely going to be the same because the sign here is different from what it is here. Are we saying that the energy of singlet and triplet are same? Actually no. From whatever we have discussed so far, maybe you think the answer is yes. And from what I said. But one thing we have not mentioned so far is what is the effect of spin to be considered when talking about energy. For that, let me remind you of once again something that we have studied in high school. Remember, Hund's maximum multiplicity rule states that have a greater multiplicity, have a lower energy, everything else being the same. Here you see which one has greater multiplicity? This one has greater multiplicity. You might have studied multiplicity as 2s plus 1. And here that 2s plus 1, this is the s that you talk about. So, 2s plus 1 in this case is 1. 2s plus 1 in this case is 3. Not very difficult to see, this is what you have learned. But what I am saying is that this singlet and triplet actually has got to do with the total number of wave functions associated with these states. For a singlet state where capital S equal to 0, you have only one single wave function. For a triplet state where capital S equal to 1, Ms varies, you have 3 wave functions. That is why it is called triplet. And remember triplet has a lower energy because of Hund's rule. It will suffice if you say that right now. It is actually a little more complicated than that. But right now we do not have to get into it. So, we have learned this very important concept of singlet and triplet states. And perhaps this takes our understanding a little beyond this 2s plus 1 business. Now, let us try to write the wave functions. And let us see whether we can use our old friend stator determinant to write the wave functions of the excited states. To start with, let us focus on one of the triplet wave functions. I start with this because the answer is simple. And since we are not doing it for the first time, we know already that the answer is going to be simple. The answer here is going to be more difficult because you see there are 2 terms. So, here I have 2 terms and I am multiplying it by another factor with one term. So, I am going to get 2 terms. Here I will get 4. So, let us start 1 by 1. 1s1 2s2 minus 1s2 2s1 multiplied by alpha 1 alpha 2. We will have 4 terms. Sorry, we will have 2 terms. Each term will consist of 4 factors. Let us work out the first 1s1 2s2 alpha 1 alpha 2. So, we can write it as a determinant. We will write it as a determinant and we will write it in such a way that the determinant we write is going to be slated determinant. So, what happens in slated determinant? What changes when you go from left to right in the in a row? What changes when you go from left from top to bottom in a column? I hope you remember in one case the label changes in one case the spin orbital changes. So, let me write what will the 1 1 element be the term I have is 1s1 2s2 alpha 1 alpha 2. So, we start with the lowest energy. So, we will write 1s1 alpha 1 here and in the diagonally opposite one we will write 2s2 alpha 2. Since we are going diagonal your label as well as spin orbital both change. So, this is what we will write 1s1 alpha 1 2s2 alpha 2. What will I write here and what will I write here? What changes when you go from left to right in slated determinant? What changes when you go from top to bottom? When you go from top to bottom we write the same spin orbital change the label. When you go from left to right you change the spin orbital and keep the label. So, this is the slated determinant form of the wave function for the 3 1 state in which capital S equal to 3 and ms equal to 1. Now, I would like you to pause the video work out the determinant for this one. Actually, I do not want you to pause the video without pausing you should be able to tell me what the answer is. What would the answer be exactly the same treatment? Instead of alpha we have to write beta. So, the answer is similar determinant wherever there is alpha we write beta that is your psi 3 minus 1 state wave function for the state in which capital S is 3 so triplet and the z component of the angular momentum total angular momentum is minus 1. Second one now we are going to have not 2 but 4 terms. So, we are going to have to write a sum of 2 determinants. But then in every term we have we still have 4 factors. So, I can still write 2 by 2 determinants maybe let us see what we get if you try to write this as a sum or difference of 2 determinants. Now, I want you to pause the video really and see whether you can write it in such a way that you get a linear combination of 2 slated determinants. Can you do it please? I hope you have done it. I hope you had stopped this video and you have done it and let us see if you have got the right answer. This is the answer 1s1 alpha 1 2s2 beta 2 minus 1s2 alpha 2 2s1 beta 1 plus half into 1s1 beta 1 2s1 alpha 1 1s2 beta 2 2s2 alpha 2 state of determinant. When you go from left to right you keep the label when you go from top to bottom you change the label. So, this is what you get for the triplet state for the alpha alpha beta beta functions you get a stated determinants for the alpha beta plus beta alpha function you get a sum of 2 stated determinants. To conclude this discussion let us see what we are going to write for the singlet wave function. As you see the singlet wave function is almost identical almost identical is an little oxymoronic the singlet wave function is very very similar to the s equal to 1 ms equal to 0 wave function. What is the difference? The difference is instead of plus here there we had a minus instead of minus here we had a plus that is all. So, this will also be a linear combination of 2 determinants this is what it is going to be please work out yourself these things you can understand only when you practice by yourself. Now let us compare the wave function for the 1 0 state to that of the 3 0 state what do we get right why am I written 1 0 it is actually 0 0 sorry 0 0 psi 0 0 state and psi 3 0 state same stated determinants in the psi 0 0 case you have minus sign between them in psi 3 0 case you have plus sign between them that is what makes a difference. So, what we have managed to learn is that we can write spin orbital wave functions of multi electron atoms has slated determinants or there it is better to say linear combinations and this is if an extremely useful way of writing wave functions and it is also the starting point of more sophisticated treatments of multi electron systems. So, that is what we are going to discuss now and that is where we are going to enter the next phase of quantum mechanics because all these interactions are going to make things so complicated that we have to invoke approximations. So far we have used approximations but they have been very very simple ones next we are going to talk about little more rigorous applications where again rigorous application sounds strange we are going to talk about more systematic detailed thorough approximation methods that are commonly used in quantum mechanics. We are going to talk about the variation method and we are going to discuss perturbation theory somewhere down the line if time permits it fits in I really would like to talk a little more about angular momentum until then.