 We have discussed rigid rotor and very soon we are going to discuss hydrogen atom. After that we will talk about multi-electron atoms where we will discuss something very strange enigmatic called spin. And in all these cases one quantity that keeps on coming at us is angular momentum. So what we will do in the next one or two modules is that we are going to discuss quantum mechanical description of this quantity angular momentum. The quantity as such is not very new to us. We are familiar with this kind of a diagram. This is a cartoon that I have downloaded from this website. But in physics we have studied that whenever there is circular motion where the direction of motion is given by the direction of fingers of right hand. Angular momentum is a vector that arises perpendicular to the plane of rotation and is along the thumb of this right hand whose fingers are curled to tell us the direction of circular motion. This is of utmost important because angular momentum is what defines rotational kinetic energy. And when we talk about rigid rotor for example it is kinetic energy all the way. Remember we had reduced the problem to a one body problem so there is no question of potential energy. It is only kinetic energy that we talk about and that kinetic energy is given by you might remember L square by 2i. So let us see what is the quantum mechanical way of talking about angular momentum. And also what we learn in this module and maybe the next is see we have been talking about this theta and phi part of the wave functions. Phi part I think we understand very well because we actually solved that equation. We did not solve the theta part of the equation. But we said we proved that Lz has this e to the power i m phi the phi part of the wave function as an eigenfunction with an eigenvalue of mh cross. So we said that the z component of the angular momentum of rigid rotor is mh cross. And then we said that the total angular momentum is h cross square multiplied by j into j plus 1. This is something we are going to use in hydrogen atom also. What we have not really explained is why is it that there is an upper limit of the values of m? Why is it that there are 2j plus 1 values of m in this module or maybe the next we will get to learn that. And initially I did not think I will go that far. But now I am tempted to also talk about ladder operators in angular momentum. Remember ladder operators in rigid rotor. Similarly we can talk about ladder operators here also. And I am tempted to go a little further and talk a little more about energies and wave functions. Let us see whether we get there but at least by the end of this discussion we will get to know why there is an upper cap of the values of m. And also we will get to encounter a very, very important phenomenon in quantum mechanics which involves operators that commute with each other. What is the meaning of commute? I guess you know. But in any case we are going to talk about it. So, here of course. Let us start at the very beginning. And since I have got a feedback from many of our potential students that they would like to have things done from scratch as far as possible. We will start at the very beginning here. We will talk about the classical description, the classical definition of angular momentum. As you know angular momentum is defined as the cross product of r and p. l is equal to r cross p. What is the meaning of cross product? Magnitude of, well first of all cross product is a vector product. Magnitude is given by r well magnitude of the 2 vectors multiplied by each other multiplied by sin theta. And what about direction? We have already talked about direction of the angular momentum vector. Here what we will do is we will work with the components. So, we can write like this that x the position we can write as i multiplied by x well r sorry r the position r is the position vector. So, length of the position vector is given by i multiplied by x plus j multiplied by y plus k multiplied by z. Perhaps I do not have to explain to you what ijk are but I still will. i is a unit vector along x axis, j is unit vector along y axis, k is unit vector along z axis and x, y, z are the magnitudes of the x, y, z coordinates of the point we are talking about in the first place. I hope that is clear. Similarly, p is defined as i into px plus j into py plus k into pz. Please remember these are vector sums i, j, k are vectors that is why I have written them in bold and italics. And px, py, pz are the x, y, z components of momentum, linear momentum x, y, z are x, y, z components of the position you can say. So, these are vector sums please do not forget. Now, when I take a cross product between r and p what I really have to do is I have to take a cross product between this and this ix plus jy plus kz and i px plus jpy plus kpz. While doing that how many terms will I get? Well 3 into 3, 9 terms let us take them in groups. First of all what you see here is the i cross i, j cross j, k cross k terms. The cross products of the same vectors, same unit vectors. So, ix cross ipx gives you i cross i xpx. Similarly, gy cross jpy gives you j cross jypy kz cross kpz would give you k cross k zpz. What i cross i, j cross j are we will come to that shortly. And I am hoping that most of you know anyway still in case you have forgotten or in case you have not come across this we will do it. What will the next term be? Next term what we can do is we can take an alphabetical order i cross j, j cross k, k cross i. So, i cross j term what will it be? ix cross p y plus jy, well j cross k will be multiplied by ypz. So, sorry I will say that again. ix cross jpy will be i cross j multiplied by x dot py. Remember x and py are just magnitudes. The direction is given by the direction of the unit vectors i, j and k. So, then kz cross kpz would well sorry i cross j, j cross k and k cross i. So, kz cross ipx would be k cross i multiplied by zpx. This is what it is. Sorry for the overlap here was not like this, but anyway sorry about that. i cross j multiplied by xpy plus j cross k multiplied by ypz plus k cross i multiplied by zpx. So, we have got 6 terms out of the 9 that we promised would be there. What would the last terms be? We can take the reverse order alphabetically. We have taken i cross j now let us do j cross i. We have taken k cross i now let us take k cross j and let us take i cross k. So, this is what we will get jy cross ipx that gives you j cross i ypx. Then kz cross jpy gives you k cross j multiplied by zpy. Why am I saying j again and again? kz cross jpy gives us k cross j zpy and finally ix cross kpz gives us i cross k xpz. These are our 9 terms. Fortunately not all 9 terms exist. I am sure most of you would know that cross product of same vector is actually 0 sin theta remember I said that magnitude contains sin theta and of course the angle between i and i is 0. So, sin 0 is 0. So, i cross i is equal to j cross j is equal to k cross k is equal to 0. So, this first 3 terms in i cross i j cross j and k cross k these are going to become 0 and therefore vanish. We are left with 6 terms i cross j xpy plus j cross k ypz plus k cross i sorry about this plus j cross i ypx plus k cross j zpy plus i cross k xpz. Maybe I will just correct this for you otherwise it is difficult to see that is much better. So, I have i cross j xpy plus j cross k ypz plus k cross y zpx plus j cross i ypx plus k cross j zpy plus i cross k xpz 6 terms. Now we need to know what is i cross j what is j cross k so on and so forth and that is also very well known these are unit vectors remember. So, unit vector means magnitude is 1 and i j k are perpendicular to each other. So, when you take cross products of say i and j you get the third one i cross j is k j cross k is i k cross i is j when you take them in the right correct cyclic order you take them in reverse cyclic order instead of i cross j you write j cross i it just becomes minus k xyz it becomes minus z something like that k cross j is minus i and i cross k is minus j these are very standard things that one learns while studying the chapter on vector in high secondary mathematics. Good. So, let us start substituting the first term we have is i cross j instead of i cross j we will write k second one is j cross k what is j cross k j cross k is obviously i is very simple I mean you just write the third one if it is in correct cyclic order you write plus sign if it is in reverse cyclic order you write minus sign. So, j cross k is going to be i k cross i correct cyclic order is going to be j now j cross i is reverse cyclic order right in alphabet i comes before j. So, you are going to get minus k k cross j similarly is minus i and i cross k is similarly minus j. So, now what we can do is we can collect the terms in i and j and k and we can get instead of 6 terms we can get 3 terms. So, if I collect the terms in i what do I get i multiplied by y p z minus z p y is not it j multiplied by z p x minus x p z and k multiplied by x p y minus y p x right. So, i into y p z minus z p y j into z p x minus x p z and k into x p y minus y p x what I really need you to do is please make sure you have a pen and paper in your hand when you attend these modules you should keep writing that way you understand better if you just look at the screen you may get it may not get it of course the advantages is that you can pause me and you can write do it no problem but please write when you are hearing otherwise it is then it will register better the more senses we use while studying better we grasp it that is a very well known phenomenon. So, remember this i multiplied by y p z minus z p y plus j multiplied by z p x minus x p z plus k multiplied by x p y minus y p x. So, now see what is this that right hand side is an expression for angular momentum. So, obviously whatever is multiplying i is the x component of angular momentum whatever is multiplying j is y component of angular momentum whatever is multiplying k is z component of angular momentum. So, we can write l x is equal to y p z minus z p y very nice right y p z minus z p y and what is that equal to l x. So, you can think like this if you think of three random variables I cannot use i j k and I cannot use x y z. So, maybe we will say p q r is p q r you know p is also there. So, we cannot use let me say 1 2 3 that is better. So, you can think like this when you talk about the components l 1 is equal to i into 2 p 3 minus 3 p 2 plus j into 3 p 1 well sorry l y. So, that is your x component right l 1 is equal to well that holds for everything actually l 1 equal to 2 multiplied by p 3 minus 3 p 2 this is a general way of writing your x y and z components of angular momentum. So, let us write that this is what we have one way in which we can express very nicely and we are going to come across this tool in some other context also is determinants. I hope we all know what determinants are if not please brush up it is just a way of writing sums like this. In a determinant what you have is if you have something like you just write once in case somebody needs to know how a determinant is written I will just write it once this is a determinant you write like this say a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 this can be anything we will just see an example of a determinant that is useful for us. But this is what a determinant is and what it means essentially is you take a 1 1 multiplied by this a 2 2 into a 3 3 minus a 3 2 into a 3 3 a 3 2 into a 2 3. So, this determinant will be a 1 1 into a 2 2 into a 3 3 minus a 3 2 into a 2 3 then next term is in a 1 2 but a 1 2 the second term has a minus sign. So, you can write a minus a 1 2 multiplied by again a 2 1 a 3. So, what you do is you are leaving out this column and this row whatever is left in that block you take cross product and subtract the other diagonal product. So, a 1 2 is in is in the first row and second column. So, leave out those multiplied by you can write a 2 1 multiplied by a 3 3 minus a 3 1 a 2 3 or if you want to write plus then all that will happen is that this will become plus and this will become minus then again plus a 1 3 a 1 3 is first row and third column. So, leave those out it will be a 2 1 into a 3 2 minus a 3 1 into a 2 2 a 2 1 a 3 2 minus a 3 1 a 2 2. So, this is how determinant is written and if you now compare this what I have written here by hand with what I have here you can correlate a 1 1 in this case is i a 2 2 is y a 3 3 is p z a 3 2 is z a 2 3 is p y a 1 2 is j a 2 1 is z p x a 3 3 sorry a 2 1 is z a 3 3 is p x sorry this is there is a minus sign here. So, this has to come here. So, a 2 1 is actually x and a 3 3 so this way you can just correlate with this and we will just write it in the determinant form before that let me erase this. Now, let us show you the determinant here this is what it is. So, i j k x y z p x p y p z very nice. The way we have written it is that the first row is just the unit vectors along x y and z the second row is the x y z coordinates position third one third row is p x p y and p z the components of angular momentum along x y and z. So, very nice systematic way if you go from left to right you go from x direction to y direction to z direction if you go from top to bottom you go from the unit vector to position to momentum it is very nice systematic way of writing it and this is how it expands y p z minus z p y j into z p x minus x p z remember j had a minus sign. So, that has gone in plus k into x p y minus y p x this is what the determinant is. Now, we know the x y and z components already these are the x y and z components the other quantity that is very important is l square square of angular momentum and you will appreciate it even more when we go into quantum mechanical description. That is given by l x square plus l y square plus l z square I hope I will not work this out explicitly because I have worked out the other one you can do it yourself but remember when you try to do l dot l that is a dot product first of all. So, it will be a scalar product no direction but then you have to take i into l x plus i j into l y plus k into l z take dot product of that with itself while doing that you have to remember that i dot i is equal to j dot j is equal to k dot k is equal to 1 and i dot j is equal to j dot k is equal to k dot i is equal to the same thing written in the other direction is equal to 0 because dot product has cos theta in the expression. So, if angle between the two vectors is 90 degrees then of course you are going to get 0. So, that is how we get l x square plus l y square plus l z square it is important to understand that this is a scalar quantity not a vector quantity and l x l y l z are the components of angular momentum which itself is a vector quantity. So, this is the classical description of angular momentum. Now, let us just build the quantum mechanical description and then we will end this module we will go to the next module for the next part. So, quantum mechanical description is as you know we start from the classical definition and what we do is for every variable every observable that is there we replace it by the corresponding operator. So, if you want to write this matrix what do we have to do we have to replace x by the position operator x which is just multiplying it by itself you have to replace p x by the momentum operator p x hat. So, these are the two rows in which this determinant is going to look different but mean the same when we try and build a quantum mechanical description while doing that x hat is very simple remember p q hat where q can be x or y or z is h cross by i del del q or you can write minus i h cross del del q in the later part we have actually use that minus i h cross it means the same it is no big deal do not forget. The way I remember it is that where if i is in the denominator then it is positive if i is in the numerator then it is negative h cross always is in the numerator of course. Now with that let us try to build try to write a similar determinant for the L hat operator the angular momentum operator what will it be it will be something like this first row remains the same i j k second row also remains the same in the third row you are going to have h cross by i del del x h cross by i del del y h cross by i del del z. So, you might as well take the constant h cross by i outside the determinant because the determinant multiplied by a constant is what you get and then the x component very similar to what you have there instead of p z you write p z hat that is all instead of I have made a mistake here sorry about that this is not z y hat it is z p y hat of course it is obvious but it is a typo let us see if I have repeated the typo later I have not L y is equal to L y hat is equal to z p x hat minus x p z hat please do not forget that this is actually p y hat this is a typo here maybe I will just write it with my pen so this one I will cut and I will write p y hat for you good L z hat L z hat is x p y hat minus y p x hat these are the operators for L x and L y and L z components and these operators are extremely useful you keep on using them in many situations later on. So, L x you can write like this instead of p z hat you write this del z and this is similar. Now what is what about L square operator L square operator for that you have to simply substitute this by L x square L y square L z square and that will come from these L x operators so I am not working it out explicitly we will do it whenever required but please remember that we have to build an operator for L x square L square also in fact that is the most useful operator I am not doing it because we are not going to use the Cartesian form we are going to use these spherical polar coordinate form remember your rigid rotor problem or any problem where rotation is involved angular momentum means rotation some sort of rotation would be there notionally yes even though for hydrogen atom we say that we cannot talk about rotation but still angular momentum is there same for spin but we will cross those bridges when we come to them but what I am saying is this L square operator is actually of utmost importance that is what we work with and we always use this spherical polar coordinate form which we do not have to remember we will give it to you but since we are talking about spherical polar coordinates form it makes perfect sense to show you what these L x hat L y hat and L z hat operators are for in these spherical polar coordinates this minus i h cross multiplied by minus sin phi del del theta minus cot theta cos phi del del phi this is L x hat operator sometimes I make mistakes in writing these expressions so please cross check with the book this part you can study from Piller's book but I have followed Macquarie not Macquarie and Simon physical chemistry Macquarie quantum chemistry book or Prasad or any book that you are comfortable with but please double check and make sure that whatever is there is correct it should not be that I make a mistake while typing here and then I do not notice it and you learn the wrong expression that should not be the case this is L x hat next we have L y hat very similar instead of sin we have cos instead of cos we have sin L z hat fortunately is very simple i h cross del del phi why is L z hat so simple because remember what is phi we have talked about this by talking about rigid rotor changing phi essentially means circular motion in the x y plane so the associated angular momentum has to be in z direction so that is L z hat and it makes perfect sense it you can correlate it you can correlate say L z hat the angular momentum operator in z direction with the linear momentum operator also they are very similar isn't it this h cross by i or minus i h cross del del q you can even write this as h cross by i del del q if you write q equal to phi very similar what linear momentum is in linear motion angular momentum is in circular motion it is very important to get these correlations right and sometimes students ask this question how do I know how do I know which is x axis which is y axis which is z axis and the honest answer to that is that we do not we define one direction as z if we have an external perturbation like if we apply an external field electric field or magnetic field then of course we define that to be z axis why because then we have to deal with an easier operator okay so it is of course in our hands but you will see that we will always work with the z direction because it is easier to handle okay maybe we will stop this module here and continue from here in the next module