 We are studying rigid rotor and at the moment we are trying to set up Schrodinger equation for a rigid rotor. So far we have reached a situation where we have written the Hamiltonian well we have arrived at the same Hamiltonian in two different ways. Like the last discussion we had in the previous module was that we started with the L square operator we realize that the Hamiltonian is bound to be this L square operator divided by 2i hence we got minus h cross square by 2 mu r 0 square multiplying 1 by sin theta del del theta operating on sin theta del del theta plus 1 by sin square theta multiplying del to del phi 2. And now we will define the wave function in a particular fashion and then we will try and separate the equation into individual equations in theta and phi. To do that first thing we do is we write the wave function as a product of a theta dependent part and a phi dependent part. Let me just paraphrase this expression to you because generally for us Greek letters are intimidating there is nothing to be intimidated it here on the left hand side instead of psi I have written y. Now yj what is j what is m let us wait a little bit I will tell you what they are they are basically quantum numbers their functions of theta and phi that we have written as a product of capital theta which is a function of theta. So capital Greek letter here is a wave function small Greek letter is the variable or coordinate. So capital theta is essentially the theta dependent part of the wave function and capital phi is the small phi coordinate dependent part. Why do we write it like this because we cannot write it as capital theta plus capital phi can be I mean that would be dimensionally inconsistent we can multiply. So for separation of variables in situations like this it makes perfect sense if we take the wave function to be the product of wave functions in independent coordinates that is a very standard technique of separation of variables. Now what I need to do is I know the Hamiltonian already I want to take the Hamiltonian and make it operate on the wave function and here since I am going to separate the equation into two different equations we have as discussed already written everything in theta in blue except the sin theta and everything in phi in green. So our purpose will be to get all the blues on one side all the greens on one side all theta dependent terms on one side all phi dependent terms on the other side that is what we will try to do. So first of all look at this well look at the second term that is easier 1 by sin square theta del 2 del phi 2 operating on capital theta capital phi capital theta is a function of theta and not phi. So as far as phi is concerned it is going to be constant and capital phi is a function of small phi not a function of theta. So what is forget about this 1 by sin square theta for the moment what is del del 2 sorry del 2 del phi 2 of capital theta capital phi capital theta is constant goes out and you are left with del 2 phi del phi 2. In fact it makes perfect sense to write d phi d 2 phi d phi 2 because there is no theta anymore capital phi is exclusively a function of small phi. So there is no point of writing delta anymore I have continued with delta in the subsequent discussion but I hope you understand that delta is not even required d is fine. So the second term will be capital theta by sin square theta multiplied by d 2 d phi 2 d 2 phi d 2 capital phi d small phi 2 what about this one here capital phi will be constant. So we will get capital phi by sin theta multiplied by del del theta operating on sin theta del capital theta del small theta. So this is what I get right what are you trying to do we are trying to separate the equation. So first of all let us multiply by 2 mu r 0 square divided by h cross square that is very simply to get rid of the coefficient here and we will also multiplied by 1 by capital theta capital phi why because once we do that here we have capital phi in the numerator if you multiplied by 1 by capital phi capital theta then phi and phi cancel you are left with capital theta. So you will get capital theta multiplied by 1 by sin theta well 1 by capital theta multiplied by sin theta then del del theta operating on sin theta d theta d theta d capital theta d small theta. So in that first term there will be nothing in phi anymore what about the second term in the second term we are going to have well capital this capital theta will go 1 by capital phi sin square theta d 2 capital phi d phi 2. So sin square theta will be there but in the next step we will get rid of it. So this is what we get 1 by capital theta sin theta del del theta sin theta del del theta del theta del theta del theta remember this del is not even required d is fine plus 1 by capital phi sin square theta d 2 capital phi d phi 2 equal to 2 mu r 0 square by h cross E. So I have got rid of the wave function from the right hand side but even though the first term in the on the left hand side is sorted the second is not. So to get rid of that it is not very difficult to see what I have to do I have to multiply by sin square theta right. But before I do that let me rewrite this minus 2 mu r 0 square multiplied by E divided by h cross square as minus beta why minus beta because that is how it is written in solution of Schrodinger equation for hydrogen atom whatever treatment we are doing now is exactly the same thing that is done for the angular part of Schrodinger equation for hydrogen atom right that is why it is so attractive other than it describing a rotating molecule in the first place we will write it as minus b. So what do we do we multiply by sin square theta and then we can rearrange right because once you multiply by sin square theta the first term becomes sin theta by capital theta d d theta of sin theta d theta d theta second term becomes 1 by capital phi d 2 capital phi d phi 2. So nothing in theta anymore on the right hand side you have minus b multiplied by sin square theta you are multiplying throughout by sin square theta remember. So if I bring that minus b sin square theta to the left hand side then left hand side and if I take this 1 by capital phi d 2 capital phi d phi 2 to the right hand side then the left hand side will be entirely in theta right hand side will be entirely in phi something like this sin theta by capital theta del del theta of sin theta del theta del theta plus beta sin square theta is equal to 1 by capital phi well minus 1 by capital phi d 2 capital phi d phi 2 please remember the capital letters denote wave functions small letters denote coordinates okay and I have not written capital phi and then small phi in bracket because the notation is such that it is not very difficult to understand alright. So almost done but now what we see is on the left hand side we have an expression in theta on the right hand side we have an expression in phi and they are equal to each other how is that possible these variables are different they can only be that can only be possible if both are equal to a constant yeah then it is possible then it does not depend on theta does not depend on phi either so we equate this to m square now y m square y not m y not k well once again the answer is you are not doing it for the first time somebody else has already done we know very well that the subsequent treatment becomes not only easy but also meaningful if you write m square here that is why we write m square and not anything else okay what do we have we have two equations now we can write sin theta divided by capital theta d d theta of sin theta d d theta plus beta sin square theta is equal to m square that is an equation that is entirely in theta and we can write 1 by capital phi d 2 capital phi d phi 2 is equal to minus m square that is an equation that is entirely in phi and not theta so we have achieved the separation of variable that we had set out to do okay now we are going to solve the phi part because very very simple we are not going to solve the theta part because it is very non trivial but then the solution was known already when this rigid rotor business came into discussion so we will discuss the solutions they are very important for now let us try to see whether it is possible to solve the phi part whether it is possible and it is very easy so 1 by capital phi d 2 d phi 2 is equal to minus m square this is what we have got so let us simplify a little bit we can write d 2 phi d phi 2 and now I have well written D instead of del d 2 capital phi d phi 2 is equal to minus m square capital phi okay what is the solution to this let us use the trial solution the trial solution we will use is phi equal to a e to the power plus minus i m phi now one might say that the most general solution is actually a 1 e to the power i m phi minus a 2 e to the power minus i m phi why are we not taking that solution why are we taking this we are taking this because well wave functions tell us a story wave functions has a meaning that meaning comes out very nicely if we take a part and not the whole and the part is as valid a solution of the differential equation than what the whole is so we will just work with this one at a time we will work with either plus or minus sign in the exponent so this is a trial solution please satisfy yourself that if you differentiate it twice you do get something like minus m square capital phi you get it differentiate once you get d phi d phi is equal to plus minus i m a e to the power plus minus i m phi differentiate once again you get d 2 capital phi d phi 2 is equal to plus minus i m was there multiplied by another plus minus i m a e to the power plus minus i m phi okay that is an eigenvalue equation the eigenvalue would be the value of whatever this d 2 capital phi d phi 2 might be the operator form okay so this is the equation that we are going to use to solve you know the trial solution already now see if you remember phi actually ranges from 0 to 2 pi okay that leads to a very important consequence if you remember born interpretation remember one of the things that we studied from born interpretation is that a wave function must always be continuous what does that mean okay let us say we start from some point okay in the x y plane we go a full circle and come back to the same point let us say this point is associated with a coordinate phi small phi if you go around then the coordinate that we reach is phi plus 2 pi but then it is the same point yeah so we should well single value not not continuity I made a mistake sorry at the same point the wave function must have the same value so from here what we can say is I do not know why we have been continuous same single value is the correct answer so capital phi at small phi plus 2 pi essentially is equal to capital phi at phi okay since this occurs with the period of 2 pi this is called periodic boundary condition see what we have got here is a boundary condition remember particle in a box we are encountered boundary conditions there in your harmonic oscillator also we talked about boundary condition here the boundary condition arises out of not really continuity but a single valuedness and it is called periodic boundary condition the value repeats after periods of 2 pi okay that is what is going to lead to quantization let us say now if capital phi at phi plus 2 pi is equal to capital phi at phi then we can write a multiplied by e to the power plus minus phi plus 2 pi is equal to a multiplied by e to the power plus minus i m phi right so what is the solution first solution one might think is e to the power plus minus i m phi multiplied by e to the power plus minus i m 2 pi is equal to e to the power plus minus i m phi e to the power i m phi e to the power i m phi cancel you are left with e to the power i m 2 pi is equal to 1 and it is not very difficult to figure out what will be from there but what we do is we use the trigonometric relationship we know that the exponential term is really equal to cos 2 pi m plus minus i multiplied by sin 2 pi m right so we write this to be equal to 1 so it turns out that cos 2 pi m is equal to 1 so this is going to be true not only for m equal to 0 but also m equal to plus minus 1 plus minus 2 plus minus 3 plus minus 4 and so on and so forth right so we have obtained this quantum number capital M which can go from 0 to infinity with increasing with integral increase integral steps of increase and now we are curious to know what is the story that this capital phi tells us what kind of information does capital phi contain and to say that we can use something that we know already something that is rather simple actually what is that see what is phi phi is essentially the angular displacement between the x coordinate and the projection of the position operator in the x y plane so suppose the projection starts from x axis and starts going round phi increases right so phi corresponds to circular motion in the x y plane from our knowledge of classical rotational dynamics we know that if you have circular motion like this then the angular momentum is like this angular momentum is normal to the plane of circular motion so here any change in phi since it is tantamount to rotational motion in the x y plane we can expect that the information that will be associated with this wave function capital phi should be the angular momentum in z direction what is angular momentum in z direction nobody has said that the rigid rotor is actually rotating in x y plane it can rotate you know whichever way possible right but what we are saying is that we talk about phi then we only consider the motion in x y plane part so the angular momentum associated with that would be normal to the x y plane or in other words it will be the angular momentum along z axis this is called L z or z component of the angular momentum so since phi is associated with circular motion in x y plane the information it might contain is the information about the z component of angular momentum right and we can convince ourselves that that is really the case by starting from the basic definition of angular momentum we will have to come back to this in the next module as well so L is given as r cross p so r is written as x i plus y j plus z k i j k essentially are the unit vectors along x y and z axis respectively p will we have written as p x i plus p x j plus p sorry p x i plus p y j j plus p z k okay now when we take a cross product then what happens something like this we can write it as a determinant i j k in the first row x y z as the second row and p x p y p z as the third row okay what is the consequence you get something like this i multiplied by y p z minus z p y j multiplied by z p x p z minus z p x well there is a minus sign before that k multiplied by x p y minus y p x remember p x p y p z these are operators in fact x y z are also operators so this quantity y p z minus z p y is essentially called the commutator between y p z y and sorry no mistake please delete that I got carried away there so x p y minus y p z okay that is not the commutator it is just two operators operating simultaneous well one after the other this is what we get now tell me what is the z component of angular momentum from this expression along z you have k vector so the z component of angular momentum will be x p y minus y p x that is your that is your l z so how do I construct l z operator it will be the same thing x operating x dot p y where without forgetting that these operators minus y hat dot p x hat this is l z similarly one can work out that p y is equal to h cross by i del del y p x is equal to h cross by i del del x very simple expressions you can even remember them but not required but you will see we never talk about p x and p y hardly ever talk about it you only talk about l square the square of angular momentum and we talk about the z component of angular momentum why please wait until the next next time we meet for the answer okay so this is l z now knowing that the operator for p y what it is we can work out this h cross by i x del del y minus y del del x this is the l z operator so after little bit of hard work we get l z equal to h cross by i del del phi very very simple straight forward relationship h cross by i del del phi that is l z not surprising because remember what is the operator for linear momentum is h cross by i del del x or del del y for del del z depending on whether the motion is along x or y or z axis similarly now motion is along phi axis you can think so l z hat is equal to h cross by i del del phi okay now the question that we want to ask is we are surmised that the phi dependent part of the wave function might contain information about l z so that is the case then it should be an eigen function of l z and it should yield some value of l z as its eigen value when the l z operator operates on it let us do that and l z operates on capital phi what do you get you get once again plus minus i m multiplied by a e to the power plus minus i m phi do not forget that there is this h cross by i as well so you multiply by that then you get h cross remains h cross and this i in the denominator cancels the i that was coming here from the differentiation m is left so you are left with l z operating on phi is analogous to well it gives you m h cross multiplied by phi so capital m h cross is the eigen value for your z component of angular momentum m is a quantum number what is the range of n m right now we have said from 0 to infinity later on we will see that some limit actually does come in so m h cross is indeed the z component of angular momentum what does it mean what does it mean that z component of angular momentum is quantized what does it mean by that what it means is that if you think of this rotation of the rigid rotor let us say rotates in this plane then the plane can have only certain given orientations why because the arrow can only point in certain directions let us take an example let us say m is equal to well let us say half m equal to half so what is the orientation that it will take half h cross into well half h cross well half h cross is going to be the eigen value that will be the z component from there you can work out what the angle will be so only certain angles are allowed and it takes us back to the familiar concept of space quantization that had arisen during Bohr-Sommerfeld theory of hydrogen atom so what we see here is that by using quantum mechanics we see that for a rigid rotor the angular momentum can only point in certain specific directions and that means that the plane of rotation can only take up certain specific directions space quantization so this is what we have well next part of the story I will just tell you that we said that the total wave function is theta dependent part minus multiplied by phi dependent part we have worked out the phi dependent part one thing that we have not worked out is a normalization constant that we will do during the assignments and the solution of the theta dependent part we are not going to do it believe me when I say that the solution is again some kind of a polynomial but a polynomial in cos theta it is not x square or y square or z square or even theta it is something in cos square theta and we show you specific values when we talk about hydrogen atom and again these polymers these polynomials are actually very special polynomials they are associated legendary polynomials which means that one can write a similar recursion relations among these polynomials as one could for the harmonic oscillator wave functions so the crux of the matter is that the wave function that is called spherical harmonics is given by a product of a phi dependent part and a theta dependent part there should be a normalization constant here the phi dependent part is a to the power plus minus i m phi and the theta dependent part it is enough if you remember that it is a polynomial in cos theta okay so finally well here we have put in the constant and this is what the wave function is we have the Hamiltonian we have the wave function the next step is to apply the Hamiltonian on the wave function extract energy