 Welcome to module 42 of chemical kinetics and transition state theory. In the last module, we discussed transition state theory in micro canonical ensemble. And what we find in that theory is we need to calculate density and sum of states. Now, in this module, we will discuss a little bit more details on how this density of states and total number of states is calculated. So, a quick recap, the transition state theory estimate comes out to be this formula where W is the total number of states. So, W dagger is a total number of states calculated at transition state excluding the reaction coordinate, E a is the activation energy and G is the density of states at energy E computed in the reactant region only. So, with this formula we derived in the last module. So, today we are going to discuss how to calculate this G and W. G is the density of states at energy E and W is the total number of states with energy less than equal to E. Within classical mechanics, we also argued like in 2 modules ago that this W can be calculated as the volume occupied by the surface of H equal to E where H is the Hamiltonian divided by H to the power of 3m, this little H. So, this is really the way we typically will calculate W, at least if classical mechanics holds true. And to calculate G, we just differentiate W with respect to E. So, this is going to be the main strategy. So, let us see how this works. First thing before we calculate these G and W, we will play our regular tricks and separate rotations from vibrations. Because as it turns out, we can analyze both of them individually very well, but combined it becomes somewhat harder. So, the common trick that we again do is, let us say I am trying to calculate the total number of states. So, if I want to calculate G of E, total energy and let us say the total energy is the sum of vibrational energies plus rotational energies. Translation can always be separated out exactly. So, I am not considering that right now, translation is there, but that is exactly separable anywhere. So, the G of E is nothing but the total number of states at that energy E or the density of states at that energy E. So, what I can do is, I can sum over vibrational states nu, find the number of states at that energy E nu. So, I am thinking I have a total energy E and then I can partition this into E nu and E r. So, let us say the total energy E is this and this is let us say E nu and this is let us say E r. So, I am saying if my vibrational energy is E nu, well the total number of states then will be the density of states at E nu multiplied by the density of states at E r and then as sum over all possible combinations of E nu that I can have. Instead of this E nu, I might have a different combination where E nu is equal to this and E r is equal to this. So, I sum over all possible possibilities of E nu. Of course, E nu is less than E. So, this I just modify a little bit and this state this is equal to sum over nu g nu E nu and g r E minus E nu. So, this is of course, an approximation. What is the approximation really here? The approximation is that the rotational states and vibrational states are independent of each other. However, vibrations you have excited one quanta of vibrational excitation or two quanta of vibrational excitation, the rotations are unaffected by it. That is why I have been able to write it as a product form. Otherwise, more accurately I should include vibrational state as well into the definition of g here. So, this is ignored and that is the approximation really and that is a regular approximation we have made on many, many times that rotation and vibration can be decoupled. It is usually a good approximation except for a few cases when it does not. So, our goal becomes in calculating these rotational density and vibrational density separately which is always good. So, I will show you today. Let us start out by looking at these rotational density first. So, let us look at it in 1D. 3D is the same concept it is just the math becomes slightly harder. So, we will do 1D here. So, what we have to do? We have to find, we will do W e, calculate W e first and W e is nothing but 1 over h into the volume of q pi by the surface h equal to e. I am saying 1 over h only because it is 1D only. So, it is h to the power of 1. So, what is h? h for 1D rotor is nothing but p phi square over 2i where my particle is let us say constrained on a circle and this is phi. So, this is how Hamiltonian for this system is given and phi goes from 0 to 2 pi. So, we want to find this is equal to e. I want the contour corresponding to h equal to e and find the volume occupied by this contour. So, let us make a plot of phi versus p phi. Phi goes from 0 to 2 pi. How much this p phi goes from well minus infinity to plus infinity, but we have a constraint. A constraint is h equal to e. So, h equal to e implies p phi equal to root 2i e plus or minus. That is why if I equate these two, this is easy to see. So, I have root of 2i e here, I am assuming energy to be positive e here minus and phi goes from 0 to 2 pi. So, effectively you get a box like this. So, w is nothing but the area occupied by this rectangle divided by h. So, w equal to area of rectangle the shaded region divided by h. Well, area is easy to calculate. This length is 2 pi and this side is what 2 into root 2i e. So, area of rectangle is 2 pi into 2 root 2i e divided by h. So, this is nothing but 2 over h bar root 2i e. So, this is w of e. I can also calculate g of e now as dw over de. So, this is going to be a strategy for classical mechanics at least. And remember that for rotations classical mechanics is usually very good. So, this is a pretty good approximation on what we are doing. So, this will be equal to 2 over h bar root 2i into half of 1 over root e. So, this is nothing but 1 over h bar 2i over. So, I am not doing the quantum version here right now, but classical mechanics is usually holds very good anywhere. So, now let us look at vibrations. Vibrations classical mechanics is not good. One must do a quantum treatment. So, in this module I will try to also convince you that that is in fact correct. But let us start with the quantum version. And then we will do it do the classical version as well and see how different they are. So, let us say I have I have to first get my pen. Let there be n oscillators with frequencies omega 1, omega 2 till omega n. So, what is the energy of this? Let us say the corresponding quantum number. So, the energy for this set of quantum numbers will be equal to sum over i h bar omega i ni plus half i equal to 1 to capital N. So, energy of one oscillator is h bar omega into n plus half. So, I am simply summing over all possible different modes, different oscillators. So, the idea is very simple. W of let us say E naught is nothing but total combinations of n 1, n n such that E of this n 1 to n n is less than E naught. So, you really literally count it for all possible n 1s to n ns such that the energy, the harmonic oscillator energy is comes out to be less than the desired energy. So, we will see one example. So, that things become clear. So, the question statement is that we have that is the only three oscillators with frequencies 800, 1000 and 1400 wave numbers. Calculate the total number of states with energy less than or equal to 2500 wave numbers above the zero point energy. So, let us exclude the zero point energy. So, 2500 wave numbers above the zero point energy. So, my energy is going to be h bar omega 1 plus h bar omega 2, I am sorry I have forgotten my n 1s. So, I have three quantum numbers here n 1, n 2, n 3. This will be h bar omega 1, n 1 plus h bar omega 2, n 2 plus h bar omega 3, n 3. I have not written plus half because that is a zero point energy. So, this is the energy above the zero point energy. So, let me just put in the numbers here 800 n 1 plus 1000 n 2 plus 1400 n 3. So, we have to literally calculate all possible combinations of n 1, n 2 and n 3 such that the energy remains less than or equal to 2500 wave numbers. So, we are going to manually calculate it. There is no formula now n 1, n 2, n 3 energy and we will count only those combinations for which e is less than 2500. So, the lowest one is of course, zero. I can put one quanta here. If I put n 1 equal to 1, n 2 equal to zero and n 3 equal to zero, I will get 800 wave numbers. I can put a 0 1 0 that is 1000 wave numbers, still less than 2500, 001, 1400 wave numbers, still less than 2500 wave numbers. Then let us start putting two quanta. Let me put two quanta here. Then I will get 1600 wave numbers. What can I do next? I can put 1 and 1 and 0. That will give me 1800 wave numbers. What do I do next? Let me put 101 as sure that is 2200 wave numbers. I have actually forgotten one. So, let me make another table here. So, it is laborious. You have to just go ahead and do it. I can also put this way. I would have gotten 2000 wave numbers. Finally, I can do 3000. That is 2400 wave numbers or I can do 011. That is also 2400 wave numbers. These two are degenerate, but they represent different states nonetheless. So, we count them equally. So, the total number is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. W of E or W of 2500 is equal to 10. So, this is how you calculate W quantum mechanically. Often when you have large number of oscillators or large amount of energy, what you can do is to use a program to do this for you. You do not have to do it manually. So, you can easily construct a program or you can find a program written by somebody else, which will do this exact job for you. But this is what will always happen, this mathematics. Now, let me also emphasize upon you what would have happened. If I calculated this W classically, maybe I am being a fool, maybe I am doing all this laborious work for no reason. Classically, I can get an analytical expression and just use that. So, let us see what would have happened. So, classically my H now is what? It is well p 1 square over 2 m plus half. Let me write 1 omega 1 square x 1 square plus p 2 square over 2 m 2 plus half m 2 omega 2 square x 2 square plus dot, dot, dot p some I have n oscillators x n square. So, again I will play the same trick. I will say H is equal to E and this is some n dimensional space, a huge space. But it does not matter, I have this huge space in which I have a contour of H equal to E and I simply have to find the volume occupied by this contour. That is still my general recipe. However, large n is even if I cannot draw the figure, even if I cannot imagine the figure. For this, let me just rewrite this in a different form. The form is p 1 square over 2 m 1 E root 2 m 1 E square plus x 1 square divided by I am of course, looking at this formula and trying to get it in that particular format square root square plus dot, dot, dot plus p n square divided by root 2 m n E square plus x n square divided by 2 E by m n omega n square, square root. Oh my God, I cut it off. This was a brutal beheading of 2 E is equal to 1. I have forgotten to write this is equal to 1. So, for this ellipsoid, the volume is given to be this. It is a mathematical formula that we can assume. So, all we have to do is to use this formula. So, volume g is volume divided by x to the power of n. So, volume will be pi to the power of n into this a 1 into a 2 into a 3 where a 1 a 2 are here. So, this is my a 1, this is my a 2 dot, dot, dot, this is my a. This is a 2 n. I have 2 n variables. I have p 1 comma x 1 comma p 2 comma x 2 total p n comma x n. So, I have 2 n. So, that is why I have a 2 n here. So, pi to the power of n into root 2 m 1 e into root 2 e over m 1 omega 1 square dot, dot, dot root 2 m n e root 2 e divided by m n omega n square divided by n factorial. Masses happily cancel. So, I get pi to the power of n. I forgot my h. Planck will be angry with me. And here I get 2 e to the power of n. Yeah, I have 2 e to this, this thing is 2 e square root of 2 e into square root of 2 e n times. So, 2 e to the power of n divided by product of i of omega i. So, I have omega 1 square here, omega 1 square square root is omega 1 and it is a simply a product or divided by h to the power of n n factorial. So, I will just write it in a form that is more amenable to us. This I will write as e to the power of n divided by n factorial into product of see carefully what I am doing h bar into omega i. So, I notice I have here this is pi into 2 by h to the power of n. You see this constant h 2 pi I have taken them out written it in this form. This is nothing but 1 by h bar to the power of n and I put 1 h bar for each omega i omega i is still lonely. I have to provide them with h bars. So, I get this expression. So, this is what g of e finally is. While I am at it, I forgot I have put s here instead of n. I have been using n so far. So, let me just in my practice I was using s. While I am at it, let me also just provide you what g of e will be. g of e is nothing but dw over de. This we will need in the next module. So, I am just doing it right now. So, this is equal to just an extra fact for you that we will use in the next module. So, let us calculate the same problem. We have 3 modes with frequencies 800, 1000 and 1400 and I ask you now calculate the classical number of states with energy less 2500 wave numbers. So, you remember we got 10 for quantum. Let us use this formula now and see what we get now. So, w will be 2500 cube here s or n is 3 divided by 3 factorial into product of the frequencies 800 into 1000 into 1400. So, note that this is dimensionless. I have 2500 wave number cube and these are all wave numbers. So, this cancels out that is nice. And then I can simply plug it on our calculator. I did it earlier and I got 2.32. So, you see w quantum was 10 almost 5 times or 4 times more than the classical version. So, the classical w is not a good idea. We should do a quantum w. We will discuss a little bit more of this in the coming modules. This is a famous mistake actually made by RRK. So, by R and R by Rice and Ramsberger Kessel actually corrected it. So, in summary what we have looked today is how to calculate this density and sum of states. We start out by separating out the rotational and vibrational density of states. For 1D rotational we showed that the classical mechanics is typically good and you can calculate w as 2 over h by root 2 i e. Quantum vibrational quantum version is that you simply count the number of states. It is laborious, but that is the right way of doing it. The classical version nonetheless is e to the power of s divided by s factorial into product of frequencies. Thank you very much.