 Welcome to lecture number 28 of the course quantum mechanics and molecular spectroscopy ok. In the last class we are looking at the transient movement integral for the rotational transitions. So, what T m i was was equal to y j prime m prime theta phi mu naught and anyway it is y j double prime m double prime theta and phi ok. Now I told you that y j prime m double prime the double prime will belong to the initial state y j prime m prime will belong to final state. So, single prime represents a final state and double prime represents the initial state. Now of course, if you consider the molecule AB and this is some direction R then it could be R could be pointing in any direction along x, y, z. So, your mu naught will be equal to mu naught along x axis into i vector which is a unit vector along x axis plus mu naught to y along y axis where j is the unit vector plus mu naught z to k vector that is the unit vector. So, this i, j, k are unit vectors along x, y, z axis ok. Now so your T m i will now become y j m theta phi mu naught x i plus mu naught y j plus mu naught z k along the electric field vector y j prime m double prime. Now the problem here is these are in viles ok, these are in spherical polar coordinates, these are in Cartesian coordinates. So, one has to express your mu naught in terms of Cartesian, spherical polar coordinates. Therefore, one can write mu naught is equal to mu naught into x i plus mu naught y j plus mu naught z k will be equal to mu naught into x is nothing but sin theta cos phi in spherical coordinates x is equal to sin theta cos phi y is equal to sin theta sin phi and z is equal to cos theta. So, this is from the conversion of spherical polar coordinates into Cartesian coordinates cos phi to i plus sin theta sin phi j plus cos theta ok. Now this is the modified mu naught mu naught in terms of mu naught terms of spherical polar. Now therefore, your T m i can now be written as y j prime m prime theta phi. So, your mu naught is now sin theta cos phi sin theta sin phi plus cos theta k whole thing along the epsilon y j prime m prime theta and phi. So, that is your transition moment integral ok. Now but this integral has to be integrated over spherical polar coordinates. So, if I write the integral this will become integral actually double integral 1 over theta and 1 over phi y j prime m prime theta and phi of theta and phi multiplied by sin theta cos phi i plus sin theta sin phi j plus cos theta k y j double prime m double prime theta and phi sin theta d theta d phi. Because the volume element in polar coordinates for theta and phi will be volume element in polar for theta, phi will be equal to sin theta d theta. So, we have these transition moment integral will look like this. So, this transition moment integral of course, you know depending on whether your dipole moment is along x axis or y axis or z axis and the electric field could be along x axis y axis z axis we will get 3 integrals T mi will be now be equal to it will take mu naught as common. By the way there is a mu naught there is a mu naught that I forgot because you know everything that this multiplier. So, mu naught. So, this will be equal to mu naught into integral phi d phi integral theta ok phi will go from 0 to 2 pi theta will go from 0 to pi y j prime m prime theta and phi will have 3 integrals 1 will be sin theta cos phi ok there is an I but I is inconsequential sin theta sin phi and cos theta whole thing multiplied by y j m double prime ok. Now, what I am going to do is that I am going to rewrite this as 3 separate integrals. So, that will be equal to T mi mu naught equals to phi 0 to 2 pi theta is equal to 0 to pi y j prime m prime theta phi y j double prime m double prime theta phi this sin theta and that sin theta will be sin square theta cos phi d theta T phi. So, that is one integral ok plus mu naught will be phi theta 0 to pi y j prime m prime theta phi y j double prime m double prime theta phi this is sin theta that sin theta again sin square theta sin phi d theta T phi plus the third integral phi is equal to 0 to 2 pi theta is equal to 0 to pi y j prime m prime theta phi y j double prime m double prime theta phi this is cos theta that is sin theta. So, cos theta sin theta d theta T phi. So, you will have 3 integrals ok. Now, for the transient moment integral to become non-zero or have a selection rule any one of the integrals must be non-zero out of the 3 integrals at least one of the integrals must be non-zero ok. Now, when you evaluate these integrals ok in terms of spherical harmonics. So, your transient moment integral T mi in a compact form will be mu naught integral over phi integral over theta y j prime m prime theta comma phi y j double prime m double prime theta comma phi multiplied by sin theta cos phi sin theta sin phi cos theta into sin theta d theta T phi ok. Now, one of this integral must be non-zero. So, there are 3 integrals 1 2 and 3 and one of them must be non-zero and when you evaluate what appears is you get delta j is equal to less minus 1 and delta m is equal to 0 ok. So, that is the selection rule ok. So, essentially one has to put the wave functions of these y j m are nothing but your spherical harmonics and once you plug in the spherical harmonics one can evaluate this integral. So, the function of this integral is tedious but not it is possible to evaluate it will take about you know an hour or so to evaluate each of these integrals. So, which is a which you can do it as a homework. Let us just look at one integral 2 integrals. So, at the end when you have that when you have the selection. So, what you will get is delta j is equal to less minus 1 ok. That means transition from j is equal to 0 to j is equal to 1 will be allowed. Similarly, from j is equal to 1 to j is equal to 2 will be ok etcetera and your delta E rotation will be equal to what was our delta E rotation? E rotation was nothing but H B j into j plus 1 ok. You can go back and check or we call it as B equilibrium ok. So, this will be now equal to E j is equal to 1 minus E j is equal to 0. So, if I do that this will be H B when j is equal to 1 you will get 1 into 1 plus 1 2. So, this will be 2 minus when j is equal to 0 this will go to 0. So, that will be 0 that will be nothing but 2 H ok. So, similarly if you have delta E rotation is equal to E j is equal to 2 minus E j is equal to 1. So, this will be equal to when j is equal to 2 this will be 2 plus 1 3 3 into 2 6. So, this will be 6 H B minus 2 H B. So, this is equal to 4 H B. Similarly, when you have delta E rotation j is equal to 3 minus E j is equal to 2 I am looking for. So, this is from 0 to 1 this is from 1 to 2 and this is from 2 to 3. So, when you have 3 this will be 3 plus 1 4 4 into 3 12. So, this will be 12 H P E minus this is when is equal to 2 is a 6 H B. So, this is 6 H V E. So, this will be nothing but 6 H P E. Now you can see a geometric progression this is 2 H P E this is 4 H P E. So, this is 6 H B E. So, what appears is that. So, when you have this one. So, the first line will come at 2 H P E ok. The second line will come at 4 H B E and third line will come at 6 H. Now we will see the difference between these lines and it is also possible to show that it is the fourth line will come at 8 H B E ok. So, this is nothing but energy. So, this is will be from 0 to 1 this will be from 1 to 2 this will be from 2 to 3 and this is from 3 to 4 value of J's ok. Now if you have this now this will again be this is 2 H B E this also 2 H B E this also 2 H ok. So, the rotational transitions separated by 2 H P E ok. By the way in some textbooks this also written as this H is absorbed into B E and then you can also you also get as 2 B E that is a B E prime ok where B E prime is equal to H B E. So, they are equally spaced rotational lines ok and each of them will be a geometric progression. Now, when you have such rotational spectrum. So, you will think that you will get a rotational spectrum which is like this, but unfortunately the rotational spectrum does not look like this what it looks as a cyclically different. Now, this rotational spectrum will also the intensity of the transition will also depend on the population. Now, you know rotational states have a degeneracy of degeneracy of 2 J plus 1 ok. So, the 0th state that is J is equal to 0 state has degeneracy of 1 is non-degenerate while J is equal to 1 state has a degeneracy of 3 and so on that 2 J plus 1 ok. So, now what happens is that your N J by N 0 will be equal to population will be equal to 2 J plus 1 multiply e to the power of minus delta E by K T. So, this is nothing but 2 J plus 1 delta E ok. Now, if I want to go from J into J plus 1 J is or 0 to J to the power of minus H B E J into J plus 1 divided by K T ok where T is your absolute temperature and K is the Boseman constant. So, in that scenario now, if I plot N J by N 0 as a function of J then this will look something like this. So, this is 1 and this value I will call it as J max. So, this will be 0 1 2 in this case 3 could be J max 4 5 6 3 3 happens to be J max, but this J max will depend on 2 things ok. So, J max will depend on 1 temperature rotational constant or value of B E N J by N 0 is given by 2 J plus 1 into e to the power of minus H B E J into J plus 1 by K T. Now, you see this is e to the power of minus B E. So, as B E increases this decreases and as. So, J max J max decreases as B E increases also J max increases as T increases ok. So, depending on the molecule and the temperature the J max will keep changing because for different values of temperature and the rotational constant B will have different J max. Therefore, the rotational spectrum will look like something like that. So, it will have intensities which will go up and come down something very similar to this. So, this is 1 0 to 1 1 to 2 2 to 3 3 to 4 4 to 5 5 to 6 6 to 7 7 to 8 to 9. So, this profile will reflect population of rotational states. Now, there is one more thing we say that delta J is equal to plus minus 1. So, this is all will reflect to delta J is equal to plus 1. So, increasing, but there is a another set of lines which will have a mirror image will go like that will come to have delta J is equal to minus 1. So, this will be from 1 to 0 2 to 1 3 to 2 4 to 3 etcetera ok. So, there will be a mirror image. So, this is called this is called P branch this is called R branch and there is something right in the middle called Q branch where is delta J is equal to 0 is called Q branch ok, but this line is missing in the rotational spectrum. So, that is 0 to 0 1 to 1 2 to 2. So, those transients are not allowed. So, they will be missing. So, your rotation spectrum will look like this ok. We will stop it here and continue in the next lecture. Thank you.