 We are going through a little tedious but so far not very difficult treatment of trying to find an analytic solution for Schrodinger equation of a harmonic oscillator by using the power series method. This is where we had stopped in the last module. Just to do a quick recap what we are doing is we are proposing the solution of h xi, please do not forget where h xi comes from or where xi comes from xi is simply root over m omega by h square multiplied by x. The wave function is h which is a function of omega sorry function of xi multiplied by e to the power minus xi square by 2. Now what we are trying to do is we are trying to find out what this h of xi is if you remember for very large values of xi it is a constant but for all values of xi what is the general expression to know that we have expressed is as a power series h of xi is sum over j equal to 0 to infinity aj xi to the power j. And then what we have done is we have written this recursion formula aj plus 2 in terms of aj and there we have put in j minus v hence we have proved that j max is equal to v you cannot have aj value what is j value j is the number of terms in the summation for h of xi. So in the most general form it is an infinite sum so what we are saying is actually it is a finite sum you do not need to go beyond the v th term so which means suppose v equal to 0 then 0 th term is enough 0 th term does not mean no term 0 th term means for j equal to 0 what will it be xi to the power j will be 0 xi xi to the power j will be 1 and you will be left with a 0. Suppose your v equal to 5 that means you have to sum from j equal to 0 1 2 3 4 5 that is enough you do not have to go to j equal to 5 32,097 that is what is the meaning of j max equal to v then we said that now the thing is this this recursion formulae connect alternate amplitudes. So depending on whether j is even or odd all amplitudes for odd or even terms respectively are going to be 0 because you will not be able to access them that is what we say in this line. So what we have been able to do is we have been able to write instead of 1 2 separate series and this time terminated series for the function h of xi for odd values of v h of xi is equal to a 1 multiplied by xi plus a 3 multiplied by xi cube plus a 5 multiplied by xi to the power 5 so on and so forth the last term is a v multiplied by xi to the power v for even values of v we have h of xi equal to a 0 plus a 2 multiplied by xi square plus a 4 multiplied by xi to the power 4 so on and so forth last term once again is a v multiplied by xi to the power 5 v. So we have 2 series 2 kinds of wave functions 2 kinds of functions for one for odd one for even and now as I said earlier even when the harmonic oscillator problem was being worked out using quantum mechanics for the first time the mathematics was already developed for this kind of differential equations right that is how science progresses you do things perhaps without even thinking that there can be an application you do it because it is there and then applications often come out eventually. So it was known that this kind of polynomials are actually were called Hermite polynomials. So I will show you some examples of Hermite polynomials these are the Hermite polynomials H v for v equal to 0 1 2 3 4 and 5. So you see what you have for v equal to 0 the polynomial is just 1 for these are normalized Hermite polynomials for v equal to 1 it is 2 xi for v equal to 2 note you have something in xi to the power 0 and something in xi to the power 2 for v equal to 3 there is no xi to the power 0 term you have minus 12 xi plus 8 xi cube that xi square and xi to the power 0 are missing similarly for v equal to 4 you have xi to the power 4 xi to the power 2 xi to the power 0 and it goes on. So these are the Hermite polynomials how do you find the wave function by multiplying each Hermite polynomial by this common exponential common Gaussian function e to the power minus xi square by 2 using Hermite polynomials has it advantages. In fact, even if you cannot prove it I mean sure what you can do is you can at least verify what I am going to write now it was known already that this Hermite polynomials I have this relations with each other we have already discussed one kind of recursion relation here we have something that is related this is what it is I have written it in a little different form than that of the group that that of the book I have written 2 xi multiplied by h v of xi is the sum of the polynomial after h v plus of in xi plus some constant well this should be 2 v I am sorry this should not be n should be v 2 v multiplied by v minus 1 th Hermite polynomial in xi what does this mean let us take anything let us take xi 3 multiply xi 3 by xi what do you get you get a term in xi to the power 4 you get another term in xi square now look at the polynomial after you have something in xi to the power 4 something in xi square and you have a constant look at the polynomial before it is 4 xi square minus 2 so what I am saying is that once you multiply any polynomial by xi what you get is a linear combination of the polynomial before and polynomial after it becomes dimensionally consistent as well so this is a very important recursion relation of Hermite polynomials which obviously also relate the vibrational wave functions and this is very handy in deriving the selection rule for spectroscopic transition for a harmonic oscillator well the selection rule as many of us might know for harmonic oscillator is delta v equal to plus minus 1 you can only jump one step at a time how that comes will not dwell into it further whoever is interested can see this recordant module and perhaps the module after this which is available on youtube this is from our NPTEL course on molecular spectroscopy from last semester right now we can conclude our discussion on quantum harmonic oscillator we have not really done everything there are certain things that are left but we are going to give some of it as a tutorial problem to you assignment problem to you but let us summarize what we have learned first thing we have learned is that when we use quantum mechanics very naturally vibrational energy of a harmonic oscillator turns out to be quantized the quantum numbers range from 0 to in principle infinity and the energy of the vth level turns out to be v plus half multiplied by h cross omega which leads to a corollary that a harmonic oscillator can never be addressed its vibration energy can never be exactly equal to 0 because as we said several times already but let me remind you once again because if it comes to rest then it means that position is the mean position x equal to 0 plus minus 0 and momentum is also 0 plus minus 0 uncertainty in position uncertainty in momentum both are equal to 0 so uncertainty principle is violated uncertainty principle cannot be violated right so it makes sense that a quantum harmonic oscillator can never be at rest even at 0 galvin the 0 point energy half h cross omega is there and in fact once again perhaps we will give it to you as a an assignment at room temperature for most most molecular harmonic oscillators it is the v equal to 0 level that is only occupied v equal to 1 level is usually not occupied at room temperature because the energy gap is too large if you use the simple distribution then Boltzmann distribution you will see that for an energy gap of 1000 centimeter inverse only 8 molecules would reside in v equal to 1 level for 1000 molecules in v equal to 0 level so practically at room temperature even v equal to 1 is not occupied forget about v equal to 2 and 3 and so forth okay so 0 point energy is an fundamentally important concept for quantum harmonic oscillators then we have worked out the general expression for wave functions and it is beautiful it is beautiful because it is a product of a constant a Hermite polynomial and a Gaussian function the Gaussian factor ensures that no matter how wise I you can go to the wave function falls off to 0 not too much beyond the potential energy surface why it can even go beyond the potential energy surface remember is your homework problem you have to figure it out right and the Hermite polynomials if you remember are for v equal to 0 it is just 1 for v equal to 1 it is a first order polynomial so first order polynomial means what something like xi so for xi equal to 0 where is xi equal to 0 for x equal to 0 midpoint then what will happen you will get a node the wave function will change sign if you go to v equal to 2 that is a second order polynomial in xi so equate that to 0 you will get 2 roots and these are the 2 roots where the wave function will become not only become 0 but change sign please remember a node is a point or later on a plane or surface or something where wave function changes sign it goes through 0 but it has to change sign for example if you go here at x equal to infinity or xi equal to infinity there also the wave function is equal to 0 but xi equal to infinity or x equal to infinity is not a nodal point because the wave function becomes 0 asymptotically it does not change sign so as you go higher up the energy ladder you get more and more and more nodes that leads to this rule of thumb that more the number of nodes higher is the energy associated with the wave function and the nodes here come from the Hermite polynomial being equated to 0 whatever is the order of the polynomial will be equal to the number of nodes that is why we get more and more nodes as we go higher up then we have learned that recursion relation is there among connecting these wave functions so just multiply a wave function by xi you will be able to get a linear combination of the wave function before and wave function after so that is all for a simple harmonic oscillator this as we said earlier has very profound application in vibrational spectroscopy next we move on to rigid rotor which also has profound application in vibrational sorry rotational spectroscopy but also one more reason why we study it is that the develop the methods that we develop there are going to come handy when we talk about the first molecule hydrogen sorry the first atom hydrogen atom remember this course is on quantum chemistry of atoms and molecules we have already learned the quantum chemistry of vibrating molecules now we will learn the quantum chemistry of rotating molecules and that will enable us to get into hydrogen atom in which Schrodinger equation is exactly solved okay that is what we will take up in the next few modules