 We are on our way to seeing these beautiful pictures well this is slightly a little bit of a spoiler because we do not know yet what we are talking about but what we are showing you are depictions of hydrogen atom wave function. I think you are familiar with the shapes you know shapes of orbitals so we are going to say what orbitals actually are what you think orbitals are may not be the correct definition but more when we get there for now let us just say right now our quest is for wave functions for hydrogen atom and what you see here are spherical harmonics spherical harmonics means the solution of the angular part of Schrodinger equation what is angular part well this is where we were we could separate the Schrodinger equation for hydrogen atom into three different equations one in terms of the radial part the second one in terms of theta third one in terms of phi and in case you are rusty on what r theta and phi r I recommend that you please go back to I think lecture before the last and that is where we had discussed spherical polar coordinates in some detail but crux of the matter is now we are in a situation where the three the equations in the three variables r theta and phi are separated last one is simple let us try to solve that very simple differential equation in terms in phi right second order differential equation do not you know how to solve it I am sure you do how do we go about solving it well this is how we write it and we use a trial solution the trial solution we are going to use is a multiplied by e to the power plus minus i m phi in fact I do not even like to write e to the power plus minus i m phi that plus minus is also not required you might say that why are you using this it mathematically it might make more sense to write something like this that capital phi is equal to say a multiplied by e to the power i m phi plus b e to the power minus i m phi who will stop me if I write it like that nobody will stop me we can write it but I write I choose this solution itself is also correct this is a complete solution this is a partial solution I prefer to work with the partial solution because that gives me access to some property of the electron in the atom we will get there but is this solution incorrect no it is correct you can do that and in fact if a and b happen to be equal you know very well what that what the what the form of this thing is going to be well not happen to be equal if a equal to a and b equal to i a maybe you know what it is going to boil down to but we are going to work with a e to the power i m phi once again because you have the benefit of hindsight okay plug it in there as usual I hope everybody here is sitting with a pen and paper and I hope everybody is writing as you go along there is the only only way to understand if you just hear me speak nothing will sink in so please do keep on writing d 2 capital phi d phi 2 equal to minus m square minus m square capital phi so what you need to do is you need to differentiate this twice with respect to phi do it see what you get okay that is what you get this differentiated twice you get actually m square and this is why we had used m square in the separation of variables because we know that then I can write this conveniently in terms of e to the power i m phi and what is m y m and not q will come to that that also will take us to actually familiar territory now one thing to remember is that phi ranges from 0 to 2 pi by the way do we have quantization yet we do not but we are very close to it now see remember boundary condition so this is our wave function of course you can say that this is an imaginary wave function how are you drawing it like this bear with me for a moment what I am saying is whatever is the value of the wave function if you do not want to draw do not draw but it has to be continuous right and it has to be single valued actually I should have written single valued here rather than continuous this is again a persistent issue that is there with this slide I do not know I did not change it wave function has to be continuous it has to be single valued also so now see I start from a point and so this is some value of phi so the let us say this is my x y plane this is x this is y this is a point I start from a point go around in a full circle come back here this this point you can write as okay this is phi I can write it as phi I can write it as phi plus 2 pi also so will you agree with me if I say that capital phi at phi plus 2 pi is equal to capital phi at phi why because wave function has to be single valued do not have to draw it like this wave function has to be single valued so whatever is the value of capital phi at phi the same value must be obtained when you go around a full circle because it is you have reached the same point what is this it is a boundary condition and since this is boundary condition involves a periodicity of 2 pi it is called a periodic boundary condition okay even though this is a chemistry course this is nothing to do with power IOD or something periodic boundary condition alright so this is what it is a multiplied by e to the power im phi plus 2 pi I am not saying plus minus because I do not like it as you will see plus minus will come anyway later on a e to the power im phi plus 2 pi multiplied by a e to the power im phi what is the solution solution is very simple I can write like this I will write and then I will erase also I can write a e to the power im phi multiplied by e to the power im 2 pi is equal to a e to the power im phi up to here is fine and then these 2 cancel you are left with e to the power im 2 pi is equal to 1 so easiest thing to do here is to set m to be equal to 0 then of course it is equal to 1 okay but now we we if you just write this we actually miss out on some information so again this is a message for all of us how you write the wave function that is actually very important how you write the wave function might allow you to see some things or might actually hide some things from your view so to get the complete picture remembering that we are dealing with a situation where there is a periodicity of 2 pi periodicity of 2 pi means what what is it that has periodicity of 2 pi things like angles so what we will do is in order to extract the complete picture we will write this e to the power im 2 pi as your cos 2 pi m plus minus i sin 2 pi m that is equal to 1 all right we are going to write this in the trigonometric form now what do we see on the right hand side there is nothing in i so in the left hand side the sin 2 pi m must be equal to 0 then and only then does this vanish so cos 2 pi m equal to 1 holds when m is equal to 0 plus minus 1 plus minus 2 plus minus 3 so on and so forth we have got the possible values of m from here right now we have got quantization okay not quantization of energy something else to know what this something else is we have to think what kind of information does capital phi contain so what we can think is this this is what phi is right it is an angle angular displacement from x axis in x y plane so a circular motion in x y plane would that not involve an angular momentum along z axis yeah if you have circular motion in x y plane we are going to have an angular momentum along z axis depending on the direction of rotation it can point up can point down but it will be along z axis anyway so now we ask is it possible that the information contained in phi is that of z component of angular momentum makes sense but we have to verify it so this here is lz hat operator that we talked about earlier h cross by i del del phi this is a wave function make lz hat operate on capital phi what you get this is what you get do not you m h cross multiplied by capital phi an eigenvalue equation with a real eigenvalue m h cross and remembering that m is 0 1 2 3 rather 0 plus minus 1 plus minus 2 plus minus 3 now you see why I do not like to write plus minus here because in any case naturally the solution contains plus minus so this is your z component of angular momentum m is our familiar magnetic quantum number that is why we wrote m square and nothing else so what does it tell us it tells us that since m is equal to 0 plus minus 1 plus minus 2 so on and so forth we get back to the same kind of inference that we had from Bohr theory angular momentum can point in different directions the only thing is since we cannot really talk about the trajectory of an electron in a quantum mechanical system anymore we might as well shed these orbits but we should not shed the arrows so these arrows denote the direction of angular momentum what we learn is that the angular momentum vector can take up only specific orientations in space with respect to z what is z well to know what orientation it is you have to apply a magnetic field that is what defines the direction of z remember before making the measurement the system exists in an entangled state only upon making the measurement the wave function collapses into an eigenfunction which you see okay so we arrive at space quantization space quantization was a term that was actually from old quantum theory but we arrive at space quantization using quantum mechanics this is the solution to the magnetic quantum number then this is what we have learned so far m is equal to 0 plus minus 1 plus minus 2 plus minus 3 so on and so forth magnetic quantum number now it is restricted by another quantum number where does that come from that comes from solution of the theta dependent part of the equation and we will show you why magnitude of m has to be now here I better correct this mistake because in my regular class at least one student has been very worked up about this so let me correct it here less than equal to L all right so what do we do here now with the knowledge of m we go back and we have to work out this theta dependent part of the wave equation which is formidable we will not do it and fortunately the solution of this theta dependent part was already known by the time this Schrodinger equation was being done that is how science progresses you know to start with you are driven by curiosity you just work out as somebody had said why do you want to climb the mountain because it is there so you want to work out some mathematics because it is a challenging problem not because why doing that you will be able to develop a product or you will be able to sell it or you will be able to make money or you will be able to solve the energy crisis of the planet earth not necessary those are all big problems you should do it what I am saying is that curiosity driven research so called curiosity driven research is also a very very important thing to do for the progress of knowledge of mankind so when this kind of equation was solved there was no idea that this is going to be used in Schrodinger equation for hydrogen atom people just did it for the fun of it and it became useful because it was known that the solution is in terms of some polynomials in cos theta do not worry about these terms all I want to say is that capital theta I will write it because I am afraid that some of us might get scared looking at all this and some of us might stop thinking please do not this is all you need to remember capital theta of theta is equal to the some normalization constant I will just write n for now multiplied by a polynomial so I will write it like this and it is characterized by l as well as m two quantum numbers and this polynomial is in cos theta polynomial is in cos theta forget about this d d it is very scary will not go there please remember this capital theta is equal to a constant a normalization constant multiplied by a polynomial in cos theta what are polynomials in cos theta it can be one it can be something in cos theta something in cos square theta plus cos theta whatever but these are special kinds of polynomials they belong to a family of polynomials it is called associated legendary polynomials so in the series of polynomials each polynomial can be related to the previous one and the later one if you multiply it by cos theta we do not need to go into that they are called recursion relations but that is what gives the name associated legendary polynomials well what what is the meaning of legendary legendary was the name of a famous scientist or mathematician so capital theta is equal to n multiplied by a legendary polynomial and for the umpteenth time the polynomial is not in theta not in x y z this legendary polynomial is in cos theta please remember okay so that gives rise to this azimuthal quantum number l equal to 0 1 2 3 see from solution of theta dependent part you do not get that limit of l you know very well that l is less than or equal to n minus 1 you do not get that you only get the information that they can be 0 and positive integers the other thing that comes from out from here is m is less than equal to l how we will see all right one more thing that is very important remember beta beta is here in this equation also beta turns out to be l into l plus 1 where l is your secondary quantum number 0 1 2 in fact my animation is a little problematic here this beta expression should have come first l expression should have come later sorry about that but at least now everything is there in front of you okay so this is what it is we have got spherical harmonics spherical harmonics are the angular part of the solution of Schrodinger equation for hydrogen atom once again kindly forget this just think this is n multiplied by p of cos theta and this polynomial depends on what l is what m is multiplied by e to the power i m phi okay one more thing okay I should have animated this sorry about that but now I take you back to this l square operator see l square operator as we had shown a couple of lectures ago is minus h cross square multiplied by 1 by sin theta del l theta sin theta del l theta plus 1 by sin square theta del 2 del phi 2 if you remember this is the angular equation how do I go from angular equation to l square this multiplied by h cross square yeah y capital phi y capital theta that is how you get it so you do that this is what turns out so on the left hand side now you have the l square operator that operates on spherical harmonics to give you what beta remember what beta is beta is equal to root of l into l plus 1 so h cross square multiplied by I have made a little bit of mistake here the square root sign should not be there there is no square root sign so I should remove h cross square multiplied by l into l plus 1 that is what beta is remember remember beta is beta equal to l into l plus 1 so finally you get an eigenvalue equation for l square l square what is l square this square of total angular momentum operator that operates on the angular part to give you h cross square into l into l plus 1 multiplied by y of theta phi what is the value of the total angular momentum then total angular momentum then is equal to now I will write I can write like this l is equal to square root of l into l plus 1 multiplied by h cross now if I try to draw something let us see what I get this is my z axis and this is the angular momentum vector what is the length here root over l into l plus 1 multiplied by h cross remember what the z component is z component of angular momentum was we found out m h cross so can I say that m h cross has to be lesser than or equal to root over l into l plus 1 make sense z component of the angular momentum in the best case scenario would be equal to l into l plus 1 under square root isn't it because this z component here can never be more than the length of the arrow whose component it is this is theta so m h cross is lesser than equal to root over l into l plus 1 that is my first equation so m well I forgot z cross here m is lesser than equal to root over l into l plus 1 can I make it a little better remember what are the values of m m values are 0 plus minus 1 plus minus 2 so on and so forth so it has to be 0 or some positive or negative integers isn't it so square root of l into l plus 1 will it ever be like that can it be l it can be 0 but suppose l equal to 1 what is square root of l into l plus 1 root of 2 that is not an integer so the best case scenario actually turns out to be m can be lesser than equal to square root of l square I got to get rid of that 1 because m is 0 plus minus 1 plus minus 2 and so forth so what does that mean it means m is less than equal to l we have all learned this expression now we know how it comes it comes because m h cross is the z component of angular momentum root over l into l plus 1 h cross is the total angular momentum z component can never be more than the total angular momentum that is why m is less than equal to l so we have arrived at an expression that we have all learned while studying in 11 and 12 okay there is a more formal way of doing it using operators I think it should be elementary for you now it is just that it looks a little scary so I will not do it I encourage you to try and work this out yourself it says the same thing you it turns out that lz square minus l square minus lz square that operator is written as lx square plus ly square it turns out that eigenvalue of that operator is h cross square multiplied by l into l plus 1 minus m square now lx square plus ly square that has to be a positive quantity so it turns out that this l into l plus 1 minus m square has to be greater than equal to 0 and then the rest of it is very simple that is what we have learned today we have talked about the angular parts we have talked about the angular equation we have learned how to solve the phi dependent part we have shown you the solutions of the theta dependent part and we learn that from the angular part we can get two very important quantities of hydrogen atom well of electron and hydrogen atom total angular momentum which is determined by the secondary quantum number l and z component of angular momentum determined by the magnetic quantum number m okay where are those beautiful pictures that I drew at the beginning you have not got that yet we will let us wait a little bit before getting there let us think of the coordinate that we have neglected draw highly so far like poor good old smaller the radius let us see what the r dependent part of the wave function is then we will bring together the r dependent theta dependent phi dependent part and we will talk about the shapes of the wave functions and that time we will also formally say what an orbital is so homework for you now before you see the next videos is find out from whatever resources you have what is an orbital