 Welcome back to the lectures on Quantum Chemistry or Quantum Mechanics. The series of lectures is being given to you through the support and the help of the National Program on Technology Enhanced Learning NPTEL and I am Mangala Sundar. I am from the Department of Chemistry IIT Madras. I am a professor of Chemistry and I teach subjects in physical chemistry, particularly Quantum Mechanics, Spectroscopy, Group Theory and so on. This is a continuation from the lectures that we have so far had. With a slight change, we will study today the transformation and the change of coordinates from Cartesian to polar spherical polar systems. We shall follow that for the purpose of solving the hydrogen atom later on, but an appreciation of the chain rule in differentiation and its application in arriving at the suitable forms for the Hamiltonian operator, particularly the kinetic energy operator. That is important and some clarity is needed. So, let me start with the coordinate transformation on two dimensional systems. Cartesian to polar, we can do other systems in a similar way, but let me limit my attention to that. In a Cartesian system, of course, when you write x and y axis and a point as x, y, then the polar representation of that point is given by its radial distance from the origin which is r and r is given by the vector sum that you are familiar with from the unit vectors multiplied by the magnitude of the projection added to the unit vector multiplied by the magnitude of the projection on this y axis. So, r is given by square root of x square plus y square where x and y are the magnitudes in the respective directions. And then there is a theta which is the angle from the chosen x axis, then you have x is given by r cos theta and y is given by r sin theta and therefore, theta is tan inverse y by x or tan y by x ok. So, these are the simple transformations and the corresponding if this is the transformation, then this is the inverse or in theta in terms of x and y. If any function is expressed in terms of x and y and we want to change that to x is equal to r cos theta, y is equal to r sin theta, the functional form will be slightly different and we will write that as g r theta. An example of that is say x square y, if you do that then x square is r square cos square theta r sin theta which is r cube cos square theta sin theta. If we want to do the derivative dou f by dou x or dou f by dou y or dou squared second then further derivatives dou squared f by dou y square dou squared f by dou x dou y etcetera if we want to do that. Then we want to convert those things into r and theta, the corresponding expression you should be familiar with is dou by dou x is given by the chain rule dou r by dou x dou by dou r, x is a function of r, x is a function of theta. So, dou theta by dou x dou y theta and likewise for the partial derivative operator dou by dou y it is dou r by dou y dou by dou r plus dou theta and likewise for the second and third derivatives. Now, we can calculate these because these are keeping the other variables constant and dou r by dou x you can calculate using the relation r is equal to square root of x square plus y square. So, if you do that r is equal to that x square plus y square therefore, dou r by dou x is going to be x by square root of x square plus y square which is r cos theta by r and therefore, that is cos theta yeah likewise we can derive the derivative dou theta by dou x, we know theta is tan in this y by x therefore, dou theta by dou x will give you 1 by 1 plus y square by x square times minus y by x square which simplifies to minus y by x square plus y square r when you put that in terms of r and theta it is r sin theta divided by r square gives you minus sin theta by r. Therefore, the derivative equivalent of dou by dou x which is in Cartesian coordinates in terms of the polar coordinates is now given by the cos theta times dou by dou r minus sin theta by r dou by dou theta ok. Now, exactly what is meant by this you can very easily verify that a given function for example, x squared y which in the polar form is this whereas, which in the Cartesian form is this in the polar form it is r cube cos theta sin theta. If you take the derivative of this function with respect to x dou by dou x you will get to x y which if you substitute in terms of x and y you will get the polar form. To get that same form from here you must use this derivative expression and therefore, the the equivalence is that x squared y and dou by dou x or equivalent respectively to r cube cos square theta sin theta and cos theta dou by dou r minus sin theta by r dou by dou theta that is what it is you have to transform everything carefully. Now, this is only with respect to the derivatives when we do integration if we write an integral d x d y and suppose we do the full integration minus infinity to plus infinity minus infinity to plus infinity some function x and y. If you want to replace that using the polar coordinates then it is important for this area element d x d y to be x plus in terms of d r and d theta and that is also given by the standard Jacobian determinant ok which tells you that the quantity d x d y is the determinant without the sign of dou x by dou r dou x by dou theta dou y by dou r dou y by dou theta d r d theta ok x y x y r theta r theta in the denominator easy ways to remember those things ok. Now, that is easy to calculate again using the transformation rule since x is equal to r cos theta dou x by dou r is cos theta ok dou x by dou theta is minus r sin theta and y is equal to r sin theta therefore, dou y by dou r is sin theta and dou y by dou theta is r cos theta. So, if you put that substitute that here in the determinant form what you get is dou x by dou r which is cos theta then dou x by dou theta which is minus r sin theta then you have dou y by dou r which is sin theta and you have dou y by dou theta which is r cos theta d r d theta and that when you do this is r d r d theta ok. The last thing is the transformation of the limits with that our idea of doing this for the two dimensions is taken is taken care of. So, the limit x tending to x from minus infinity to plus infinity and y from minus infinity to plus infinity essentially means integration over the entire plane of x and y and you can using the r and theta if you choose any value of r ok and if you do the theta integration from 0 to 2 pi then you integrate over the line and as you vary r you will and for each value when you integrate this over theta and phi you reproduce this entire area of integration through circles of different radii r and the theta being from 0 to 2 pi. So, that the limit becomes r goes from 0 to infinity theta is from 0 to 2 pi this with d x d y is equal to r d r d theta and therefore any function f of x comma y if you have to do that from x is equal to minus infinity to plus infinity y is equal to minus infinity to plus infinity when you express f of x in terms of g of r and theta by substituting x and y the integral is r is 0 to infinity and the integral theta is from 0 to 2 pi g r theta r d r d theta this is the equivalence with respect to integration. The derivative is dou by dou x I have written you can write a similar derivative for dou by dou y the moment you know dou by dou x and dou by dou y you can construct dou square by dou x square by having those two operators next to each other and being careful enough to recognize that the derivative will not only the dou by dou r will not only act on the dou by dou r and dou by dou theta but also on any function of r that appears in the derivative. So, if I have to write dou square by dou x square this whole thing acting on itself will have to be done carefully because when dou by dou r acts on this part in its following form it will have not only a second derivative of dou square dou r dou theta but also dou by dou r acting on minus sin theta by r because it is an r here. Making sure that those things are taken care of algebraically this is essentially all that we need to do in transforming the coordinate system from one representation the Cartesian coordinate representation to another coordinate representation the polar coordinate representation. Just to complete to this let us carry this exercise out for x square y and just to do the differentiation is trivial but still let us do that dou by dou x or x squared y is 2 x y which is 2 r square sin theta cos theta dou by dou x on x squared y is cos theta dou by dou r minus sin theta by r dou by dou theta acting on the polar form of x squared y which is r cube cos square theta theta. So, if you calculate this derivative dou by dou r gives you cos cube theta dou by dou r is 2 r square sorry 3 r square cos cube theta sin theta minus sin theta cos square theta sin square theta sorry dou by dou theta and sin theta will give you cos theta. So, you have r square cos cube theta ok which is 1 minus what do we want to do we want to show that it is sin theta cos theta. So, let us represent this by this is 3 r square cos cube theta sin theta and you have the derivative oh there are two things dou by dou theta on this. So, there is one more term which is stop you have to go back in that in the Armikinon sorry dou by dou x on x squared y is on x squared y is cos theta x on x squared y minus sin theta by r dou by dou theta acting on the polar form of this which is r cube cos square theta and sin theta ok. So, dou by dou r on r cube gives you 3 r square. So, it gives you 3 r square cos cube theta sin theta minus sin theta by r of course the r will get cancelled you will have r square dou by dou theta acting on the product. So, you have two terms namely 2 cos theta minus sin theta. So, it is minus sin square theta and then dou by dou theta acting on sin theta gives you cos cube theta. So, you have cos cube theta cos cube theta sin theta 3 r square 2. So, if you simplify that you will get the answer 2 r square sin theta cos theta because there will be a cos square theta here there will be a sin square theta here there will be a plus sum goes to 1 and 3 cancels with one of this to give you 2. So, this is the form therefore, you see the equivalence the derivative acting on the function is equivalent to the derivative in the coordinate changed coordinate representation acting on the function changed representation. These are things that you have to remember for 3 dimensions and also in general for anycoordinate transformation process. Now,when non orthogonal systems are involved the second derivative expressions for the kinetic energy from Cartesian to the polar form gets very messy, but that is a process that one has to follow by carefully following the derivative formula. These are all done for second 2 coordinates 2 dimension systems. Now, what about the 3 dimensional spherical polar coordinates to write. So, I will leave you with an exercise namely before we move on to the 3 dimensional coordinate system. So, I will write dou square by dou x square plus dou square by dou y square which is a Laplacian into dimensions usually it is written as del squared write this using r and theta dou by dou r dou by dou theta dou square by dou r square dou square by dou theta square and dou square by dou r dou theta if any of them are involved ok using that do the transformation it is standard I will not do that here, but it is very easy to follow if you have followed the algebraic details namely dou square by dou x square is this acting on itself dou square by dou y square is a corresponding expression for dou by dou y maybe let me just write down dou by dou y alone as a help. So, it is dou r by dou y dou by dou r plus dou theta by dou by dou theta and you can calculate this since r is square root of x square plus y square you can easily say that dou r by dou y is y by square root of x square plus y square and that is sin theta and the expression theta is tan inverse y by x therefore, dou theta by dou y would be 1 by 1 plus y square by x square times 1 by x which will give you x square in the denominator. So, I guess it will give you x by x square plus y square therefore, it gives you cos theta by r. So, the expression for dou by dou y should be sin theta dou by dou r plus cos theta by r dou by dou theta ok cos theta by r dou by dou theta phi. So, with that you can calculate dou square by dou x square dou square by dou y square and then dou the sum ok. So, that is a two dimensional transformation. Now, let us do the three dimensional transformation, but before we do that we will see this one animation which is an illustration of the three dimensional coordinate system. Now, what you have here is the spherical polar coordinate system well it looks like a sphere, but you can see the axis system right handed coordinates and any point or any point which is now represented by the sphere of that radius x, y, z just do that. So, any arbitrary point with the coordinate x, y, z has in the polar form you see theta is the polar coordinate with reference to the z axis and the projection of the vector r on theta gives you r cos theta. The projection of the vector r on the x, y plane gives you r sin theta, but it may not be in any alignment with the x axis or the y axis. So, if the arbitrary angle if the angle of this projection with the x axis is given as phi then you can see that it is r sin theta cos phi and the other is r sin theta sin phi. So, this is the transformation that one has to be familiar with namely x is equal to r sin theta cos phi, y is equal to r sin theta sin phi and z is equal to r cos theta. Now, the two dimensional representation that you had was assuming theta to be 90 degrees because it was only on the x y plane. So, if you were to look at that then it is easy the same thing you can see it here. So, when this whole thing is reduced to theta is equal to 90 degrees you see that it is on the x y plane and since it is sin theta r cos theta is 0, since it is sin theta you have r cos phi and r sin phi which is what you had used for the two dimensional representation ok. So, let us now get back to the algebraic details ok about how we do this and this transformation is important for the study of hydrogen atom by standard methods and for understanding the mathematical basis for the orbitals that we have used earlier in the elementary chemistry courses and so on ok. So, one has to write this as r sin theta cos phi, y as r sin theta sin phi, z as r cos theta ok. Now, the rest of the lecture I do not want to derive the relations it is not possible for us to do that, but I shall indicate how to do it and there is a handout which is there in the NPTEL website associated with my course where these things have been written out fairly carefully. If you find mistakes please do let me know and the colors have been given. So, that the three coordinates r theta phi are given three different colors to see how the transformation is done it is just to help those of you who are not certain of the algebra to hand hold you through that part, but I am only going to summarize the results here ok. Now, the transformation equations are r is obviously the squares of all the three added together x square plus y square will give you r square sin square theta plus z square will give you r square cos square theta therefore, r is square root of x square plus y square plus z square ok. These are the inverse transformations and for getting theta if you take the squares of these two add them up you get r square sin square theta. So, the square root of this sum gives you r sin theta therefore, theta is tan inverse square root of x square plus y square by z ok and phi is of course, the tan y by x gives you I mean y by x gives you tan phi therefore, phi is tan inverse y by x. So, these are the three inverse relations if you start from x y z therefore, the corresponding derivatives expression which we will need for the hydrogen atom and any function of x y z then you substitute x in terms of r cos theta r sin theta cos phi y is equal to r sin theta sin phi and z is equal to r cos theta if you do the substitution this will give you a functional form g r theta phi ok and the differentials if you have to write dou by dou x dou by dou y and dou by dou z have to be also written in terms of dou by dou r dou by dou theta dou by dou phi and that is just an extension of one more coordinate from what we did earlier we had r in terms of theta sorry we had x in terms of r and theta, but now we have in terms of r theta phi. So, dou by dou x is dou r by dou x dou by dou r plus dou theta by dou x dou by dou theta plus dou phi by dou x dou by dou phi because x is a function of all three and likewise for the derivatives dou by dou y which is dou r by dou y dou by dou r plus dou theta by dou y dou by dou theta plus dou phi by dou y dou by dou phi and the last dou by dou z is dou r by dou z dou by dou r plus dou theta by dou z dou by dou theta plus dou phi by So, these are the derivatives that you need to calculate these ok and they are fairly straightforward to do. So, I shall give you just the resultsand these need to be substituted in order to get the hydrogen atomthe kinetic energy operator which is in terms of the minus h bar square by 2 m that p square p square is thesquare of the this is a second derivative dou square by dou x square dou square by dou y square dou square by dou z square. If you add all those three things you will get the the kinetic energy operator. And therefore, you need these three quantities it is just a long algebraic exercise and I usually tell my students who are doing it for the first time to allocate roughly between 2 and 3 hours to sit down and do thisin calm and very careful manner and not to make an algebraic mistake because tracing the algebraic mistake probably will take more time than re-deriving the whole thing. So, I usually give this as an exercise to the students in the class and at least about 5 to 10 students come back later saying that they have done it. It is a it is a worthwhile exercise for those of you who have not doneany serious mathematics, but who are also interested in understandingin following the algebraic details one should do that ok. So, let me just write down the expression for the derivatives I shall not do anything more than that here yeah. I do not remember these expressions. So, let me just copy them down dou by dou x is sin theta cos phi dou by dou r plus cos theta cos phi by r dou by dou theta minus sin phi by r sin theta dou by ok. Whenever you have an inverse with r make sure that every term has that because r usually represents length theta and phi r dimensionless. So, you have dou by dou r is important to have a 1 by r 1 by r things to remember ok dou by dou y if you do that it is sin theta sin phi dou by dou r r is already there in the denominator and therefore, when you have the next derivative cos theta sin phi by r dou by dou theta plus cos phi by r sin theta dou by dou theta ok. And the last expression that you need is dou by dou z which is cos theta dou by dou r minus sin theta by r dou by dou theta right. One can go through the algebra, but at the end of the day if you are going to write the derivative dou square by dou x square plus dou square by dou y square plus dou square by dou z square which is the Laplacian in three dimension in Cartesian coordinates. If you were to write that using the r and theta phi coordinates the some final expression is written in a compact form it is 1 by r square dou by dou r r square dou by dou r plus 1 by r square sin theta dou by dou theta sin theta dou by dou theta plus 1 by r square sin square theta dou square by dou phi square ok or phi square I use phi and phi as interchangeably. So, do not get confused if you here phi or if you here phi they are depending on where I am and what I am saying you know that they are the same in this particular case. Now all this means is that this is a compact way of writing the derivative the derivative is essentially this is equivalent to if you put the derivative dou by dou r on r square you will get 2 r therefore, 2 by r dou by dou r that is one term and the other derivative will be and the r square is here it is dou square by dou r square ok. This is written in that form and likewise here if you are looking at this term the analogy is 1 by r square dou square by dou theta square sin theta will cancel or when the sin theta is acted upon by the derivative you get cos theta by r square sin theta dou by dou theta ok that is the analogy that is that is that is the exact form which is written in a compact notation by mathematicians or I mean for years many many years ago and it is used in quantum mechanics ok. So, these are the derivative forms for the polar coordinates when you convert them from the Cartesian form and one of the reasons for the hydrogen atom being studied in the polar coordinates is that in a field free case the potential energy is centrally symmetric that is it is only a function of r it is not it is independent of theta and phi and therefore, the potential energy and for the part of kinetic energy that can be solved together for the total energy can be done much more quickly in the radial coordinate system and therefore, you have the radial functions that contain the total energy. The angular functions do not contain the total energy, but they contain the distribution of the function or the spread of the wave function on a sphere how it is done and for any arbitrary function on a sphere one would find out that the basis functions which are most suitable or the functions known as spherical harmonics and the solution of the Legendre the associated Legendre equation along with the differential equation for the hydrogen atom in the phi coordinates will give us the spherical harmonics as the natural solutions and therefore, the mathematics is in the case of hydrogen atom in terms of polar coordinates very nicely splits into a radial part and an angular part and the angular part is what you usually have seen in the form of the orbital structure the P orbitals the D orbitals and the F orbitals and so on. The radial parts tell you where how far away from the hydrogen atom the nucleus that the electron density is maximized and that depends on the quantum numbers associated with the radial coordinates and the energy is also a function of the radial coordinates. So, all these things come out when we use this particular form of the representation for the hydrogen atom. The last thing before we leave is to look at the integrals particularly when we do the integration of these functions in polar coordinates the equivalence between this minus infinity to infinity minus infinity to infinity that is over the entire 3 dimensional world d x d y d z of a function of x y and z its correspondence in the polar form when the function is replaced by the corresponding polar quantities and gives us a function of G in terms of r theta and phi what are these represented as same as what I had done in the 2 dimensional case the mathematics the simple partial differential calculus gives us the result that the volume element d x d y d z is related to the volume element of the r d theta and d phi in terms of the Jacobian which is the sinless determinant without the minus sign here it is dou x by dou r dou x by dou theta dou x by dou phi we had only 2 for the polar coordinates in 2 dimensions now we have all of them and likewise it is dou y by dou r dou y by dou theta and dou y by dou phi dou z by dou r dou z by dou theta and dou z by dou phi determinant 3 by 3 d r and it is easy these are all easily calculated because you know x is equal to r sin theta cos phi therefore it is easy to write down dou x by dou r is sin theta cos phi dou x by dou theta is cos theta r sin phi and dou x by dou phi is minus r sin theta sin phi so likewise you can get the other 3 derivatives in terms of y using y and using z when you do that substitution finally what you get out is that you get r square sin theta as the value of the determinant multiplied by d r d theta and d phi ok. So, this is the volume element transformation from the Cartesian to the spherical polar coordinate system. So, the last thing that remains is now the limits how do we transform this limits minus infinity to plus infinity minus infinity to plus infinity and so on again if you see the animation you will see immediately why the limits are what they are ok. So, let us look at the animation here which is to tell you the extension of the coordinates or the limits of the coordinates we start with the phi which is the coordinate the angle or it is called the azimuthal angle perpendicular to thez axis the polar axis phi is a full 360 degree extension from 0 to 360 degrees. So, it is a circular complete that. So, that is the phi now this is one value of theta theta is equal to 90 degrees and therefore, thechange of theta from 0 to 90 0 to 180 will give us the corresponding circular sections. So, if we take that arc theta is equal to 0 to 180 that is a semi circle and with the phi going from 0 to 360 it produces the surface area of the sphere ok. It is not necessary that you should do theta is equal to 0 to 2 phi it is only necessary that you do theta is equal to phi ok. So, you have the angle coordinates theta is equal to 0 to phi and phi is equal to 0 to 2 phi generate the surface area of the sphere this is the full surface section of the sphere and therefore, all that you need to do is to reduce the sphere from r is equal to 0 to r is equal to infinity. So, that you generate the entire three dimensional volume for each value of r and for every theta 0 to 2 0 to phi and phi is equal to 0 to 2 phi you are integrating over the volume of the sphere and as you vary the radius from 0 to infinity you are integrating over the entire universal three dimensional space. Therefore, r is equal to 0 to infinity theta is equal to 0 to phi and phi is equal to 0 to 2 phi is equivalent in the spherical polar coordinate system is the corresponding equivalent to the integration using the Cartesian coordinate system x y z d x d y z over all the three dimensional space ok. So, these analogies I mean these transformations are important. However, elementary or preliminary they may be one should not have any doubts on this one should not leave I mean there should not be any lack of clarity in these kinds of things. We will look at the hydrogen atom later and we shall also continue with the mathematics of solving the Schrodinger equation for specific cases later, but with this we can now go on and study the hydrogen atom split the hydrogen atom equations into the radial part and the angular part which I shall do in the next lecture and then tell you how the radial equation is associated with the Laguerre equation and the angular part gives rise to solutions known as the spherical harmonics. So, we will have one or two more lectures to dwell in the mathematics of the hydrogen atom with this as the preliminary and with the power series method you are only done with the Hermite polynomials and also for the second order differential equations as tools. Therefore, you see that these things keep on building from one to the next step and so on and the unfortunate thing is beyond hydrogen atom this kind of the so called mathematics cannot be carried analytically, but we will have approximation methods, but that is shooting too far away from now and let me conclude this lecture with the note that in the next lecture we shall start separating the hydrogen atom Schrodinger equation into radial and angular part. Until then thank you very much.