 Hello, welcome to the lecture number 32 of the course, Quantum Mechanics and Molecular Spectroscopy. In the previous lecture, we were looking at the harmonic oscillator, because harmonic oscillator is a good approximation for a chemical bond at the equilibrium distance. And we will use harmonic oscillator to derive the selection rules. Rather we will use harmonic oscillator wave functions to derive the selection rules for the vibrational spectroscopy. In the last class, I told you the solutions of the harmonic oscillator are of the form psi n equals to nn, some normalization constant hn of x root alpha, this is the Hermite polynomial multiplied by a Gaussian function exponential minus alpha x square by 2 is the Gaussian function and this is your normalization. Now, where n will take the values of 0, 1, 2, 3, etc. Now, and your nn is given by alpha by pi to the power of 1 4th 1 over 2 to the power of n factorial. Now, if I use this and then I can write a wave function psi 0 will be equal to alpha by pi to the power of 1 by 4 exponential minus alpha x square by 2. Here psi 1 will be equal to 4 alpha cube divided by pi to the power of 1 4th x exponential minus alpha x square by 2 and psi 2 will be equal to alpha by 4 pi to the power of 1 by 4 to alpha x square minus 1 exponential minus alpha x square by 2. Now, what you can see this exponential alpha x square by 2 is going to be common for all wave functions. So, essentially the shape of the wave function will be dictated by the Hermite polynomial this one. So, in this case this is into 1, in this case this is x and this this is 2 alpha x square minus 1. So, one can plot the harmonic oscillator wave functions as this is n is equal to 0 as something like that equals to 1 something like that and n is equal to 2 will be. So, they look very similar to the wave functions similar to particle in a box wave functions in appearance only for the appearance sake they look. Now, one of the most and of course, there is a energy value E n is equal to n plus half h nu. Now, we must always remember that the wave functions will look very similar to the particle in a box, but they are not exactly mathematically very same because these are Gaussian functions multiplied by Hermite polynomials and those are sinusoidal functions. Now, when we are now looking at spectroscopy, so what we will need is a transient moment integral. So, for selection rule what we need is transient moment integral. Now, transient moment integral basically connects two states with two different quantum numbers. So, one can think of TMI will be nothing but integral in the case of harmonic oscillators these will be minus infinity to place infinity. Now, the wave function that we will have is psi m mu psi n theta. So, that will be my transient moment integral where m and n will be different quantum numbers when m is not equal to n. TMI is equal to I can write in terms of bracket notation this will be psi m mu psi n where m is not equal to n. Now, we know that mu itself can be written as operator mu itself can be written as mu naught plus d mu by dx evaluated xe into x plus half d square mu by dx square evaluated xe plus x square plus etc. So, this is nothing but your Taylor series expansion of dipole moment. Now, you will realize this is nothing but your permanent dipole moment and this is a constant. So, mu naught for a molecule is constant the permanent dipole moment is constant and this is nothing but the dipole moment derivative. Now, what we will do is we will ignore higher order terms and we will take just mu will be equal to mu naught plus d mu by dx evaluated xe into x. Now, one thing that I must say there is x is just a generic coordinate I could call it as x could be called as q or r depending on the r is nothing but radial vector connecting 2 atoms. So, when I have a and b bond this distance I could call it as x or q or r does not really matter they are just variables. So, in the previous classes if I have used r I can interchangeably used r and x. So, my TMI will now become integral psi m mu psi n which will be nothing but psi m mu naught plus d mu by dx into x evaluated xe. So, the operator now has two terms I can evaluate this as two integrals that is nothing but psi m mu naught psi m plus psi m d mu by dx evaluated xe multiplied by x. Now, I told you mu naught is a constant. So, I can always bring out the constant out of the integral. So, this will be nothing but mu naught psi m psi n plus okay when you look at this particular derivative that is nothing but the integral or sorry the differential first derivative of mu with respect to x evaluated at xe that means d mu by dx whatever that derivative is if I evaluated xe. So, I will call it as mu prime xe or rather I can call it as mu prime evaluated xe and this again is a constant mu prime evaluated xe psi m x psi n. So, you have two integrals. So, essentially now my transition moment integral. So, my transition moment integral now will be equal to mu naught psi m psi n plus mu prime evaluated xe psi m x psi n okay and I told you this is a constant which is nothing but d mu by dx evaluated okay. Now, I have two integrals and these two integrals must be evaluated. Now, one thing we know is that H harmonic oscillator psi n is equal to En psi n where this is the harmonic oscillator Schrodinger equation and I told you that whenever we have a Schrodinger equation all the solutions that means the solution psi n form a complete set set and integral psi m psi n will be equal to delta m n that means they are mutually orthogonal. So, the first if that is the case then your TMI will now become mu naught psi m psi n and this integral will be either 0 or 1 okay it will be 0 when m is not equal to n and will be equal to 1 when m is equal to n. So, when m is not equal to n it will go to 0 because we are looking for transition from state m 1 to 1 state to other state that means we are only considering this option when you consider this option this integral will go to 0 and will not contribute to the TMI. So, now your TMI is left with only one integral that will be nothing but mu prime at xe psi m x psi m this is x okay. Now this even integral needs to be evaluated okay, but fortunately for us this is not a very difficult to integrate because there is something called recursion relationship. Now what is this recursion relationship in the case of Hermite polynomials one can write 2 times x by alpha of h n will be equal to h n plus 1 plus 2 times n h n minus 1 okay. So, I will slightly rewrite this okay. Now what I am going to do is I am going to multiply this equation in a slightly different way sorry not rearrange this. So, this times x times h n will be equal to alpha by 2 h n plus 1 plus n alpha h n minus 1. So, this relationship I want to remember okay. Now let us go and write the transition moment integral. So, now time moment integral TMI this will be equal to mu prime evaluated at xe psi m x psi m okay. And what is your psi m psi m is equal to n m h m e to the power of exponential minus alpha x square by 2 and your psi n will be equal to n n h n exponential minus alpha x square by 2. And we also know that x times h n will be equal to alpha by 2 h n plus 1 plus alpha times n h n minus 1 okay. Now we will use these three formulae or these three equations and plug it in this okay. Let me write this in a slightly different way that is my TMI is now equal to mu prime xe that is the first dipole moment derivative of mu with respect to x and evaluated xe. Now psi m will be equal to n m h m exponential minus alpha x square by 2 multiplied by x n n h n exponential minus alpha x square by 2 right. So, that is my TMI. So, TMI now will be equal to okay let me rewrite that equation mu prime xe integral n m h m e to the power of minus alpha x square by 2 x n n h n e to the power of minus alpha x square by 2 okay. So, this will be nothing but mu prime xe. Now you see n is nothing but n m is n m and n n are nothing but you are nothing but you are this normalization constants and they are constants so one can bring them out. So, n n n m okay. Now rest is integral h m e to the power of minus alpha x square by 2 h x h n e to the power of minus alpha x square by 2. So, this is equal to mu prime xe n m n n okay. Now I will write in terms of integral okay integral h m x h n e to the power of minus there is a 2 of them. So, this will be minus alpha x square okay. So, that is my integral that I need to evaluate. So, this is nothing but now we know x h n this is a generating function or the recursion relationship and that is equal to mu prime xe of x n m n integral h m okay into x into h n will be give me alpha by 2 h n plus 1 plus alpha n h n minus 1. So, that is what I have written down in terms of generating recursion e to the power of minus alpha x square t tau okay or one could write it as from here one could write it as this is equal to mu prime xe n m n n integral h m e to the power of minus alpha x square by 2 to x h n e to the power of minus alpha x square by 2 t tau. Now I can plug in the recursion relationship here. So, this will be equal to mu prime x e n m n n integral h m e to the power of minus alpha x square by 2. Now this one will be alpha by 2 h n plus 1 plus alpha n h n minus 1 e to the power of minus alpha x square by 2 t tau. So, one can rewrite this integral t m i equals to mu prime x e n m psi m e to the power of minus alpha x square by 2 into n n alpha by 2 h n plus 1 e to the power of minus alpha x square by 2 this is one form integral d tau plus mu prime x e integral n m e to the power of minus alpha x square by 2 n n alpha n h n minus 1 e to the power of minus alpha x square by 2 d tau. So, it will come out as two integrals. So, this will be nothing but mu prime of x e if I take common. Then you see this is they are all constants. So, this is nothing but your psi m multiplied by this is nothing but sum because the normalization constant is slightly different than n plus 1. So, what I will call it as n n n plus 1 prime I do not know what that is what is sum number psi plus 1 d tau plus similarly psi m this is integral into n prime n minus 1 psi n minus 1 d tau. So, we will have two integrals. So, which means if this so this n plus 1 is just a constant similarly this n prime n minus 1 is also a constant. So, these two are constants therefore, mu prime of x e if this should work then psi m and psi n must. So, integral so some constant n prime n plus 1 times integral psi m and psi n plus 1 d tau similarly plus n prime n minus 1 psi m psi minus 1 d tau. If these have to be then m must be equal to n plus 1 or m must be equal to n minus 1. So, then if I bring it rather m minus n must be equal to 1 and m minus n is equal to minus 1. So, this is nothing but your delta n is equal to 1 and delta n is equal to minus 1. The T m i will be not equal to 0 when m minus n is equal to 1 or m minus n is equal to minus 1. Of course, both of them simultaneously cannot happen. So, that means your delta n that is a change in the quantum number should be equal to plus minus 1. This is nothing but changing. Apart from that your mu prime x e must not be equal to 0 because if this goes to 0 then you are multiplying everything with 0. That means d mu by d x evaluated x e should not be equal to 0. That means dipole moment derivative must be non-zero. So, this is the vibrational selection rule. So, summing up we will get delta n or delta mu depending on what the symbol that use this must be equal to plus minus 1 and d mu by d x evaluated x e must be non-zero. So that finishes the vibration selection rules. We will continue in the next class. We will stop it here. Thank you.