 Greetings, we begin the afternoon session now and in some sense this gets to be more exciting and I hope will be more enjoyable than we had in the morning. The reason I hope it will be so, despite the fact that it's coming after a heavy lunch, the hope comes from the fact that physics gets to be more realistic as we get into the domain of quantum theory. So once again I will remind you of what we started out in the morning with Feynman's quote that it is the atomic hypothesis which is the one line sentence. You have any questions? Thanks. Yeah, let's have the technology in place. Yeah. No problem, you're quite welcome to do that. These people can edit the lecture and I have requested them to do the editing in such a way that they can at some point replace my lecture by Feynman's. I'm trying to persuade them to do that for the last five years, they have not agreed to do that as yet, but I know they have tremendous handle. They can change tracks, they can change the audio, they can change the video and if only they can put Amitabh Bachchan in place of me, his wife in place of me and Feynman's lecture in place of mine, I think they are in business. So are we ready? But now my director, I lost my director. So the atomic hypothesis is a very important one from Feynman's perspective. I gave my reasons speculating on why Feynman would choose this and let us review the development of atomic physics. We know the atomic spectrum from the times of Rydberg-Bahmer, these were like semi-empirical laws and this was known based on experimental parameters for which Rydberg and Bahmer were able to fit the formula that we call as the 1 over n square formula. And then in 1930, he became the model of Nies Bohr, which is the quantization of angular momentum in a classical orbit. That was the idea in Bohr's model. Now this model envisages atomic physics to be described in terms of electrons which are revolving around the nucleus in Kepler-like orbits. But an orbit is the trajectory of a particle in the phase space. So the position and momentum have to be defined at every instant of time and that is what defines an orbit. And this idea really breaks down because position and momentum cannot really be measured as such. And therefore this model really breaks down and I would like to begin with the same question that we had in the morning as to how do you describe the state of a mechanical state of the system and how does this state evolve with time? So this is the central question in mechanics. So normally we would solve this using equation to the motion and subject it to appropriate initial conditions. But then this is essentially a classical idea. Whether you use Newton's laws, Lagrange's equations or Hamilton's equations, in every case the state of the system is described by a point in the phase space, described by position and velocity or equivalently by position and momentum. So this is essentially a classical idea. You can also describe it like I said by a function of position and velocity or a function of position and momentum and thereby get the Lagrangian function or the Hamilton's function. Observables in classical mechanics are then functions of these variables. These are the dynamical variables. So an observable like energy, you write it in terms of mv square by 2 plus a function potential energy which is a function of position. So again, observables are functions of position and velocity, whether it is energy or angle of momentum or anything else. So observables are functions of the Lagrangian or functions of the position and momentum. And now the question is, if you make a measurement of position and momentum, what kind of an answer do you get? Because when you pose these problems in high school physics and you ask the students to solve a problem in high school physics and we tell them that this is a force which is acting on another particle, find where the particle will be after a certain instant of time or find the particle's trajectory. What he would do is to integrate the equation of motion, put in the initial conditions and typically he's dealing with a second-order differential equation. So there are two constants of integration or two first-order equations like Hamilton's equations. So again, there are two constants of integration and these would be the initial position and momentum and you provide him with that. Who is going to give these initial position and initial momentum? I mean, these are not given to us by nature. These physicists ask to determine by himself. You have to carry out an experiment because what is fundamental to physics is the process of measurement. And you cannot get the position and momentum. It's not something that you can take for granted. You have to carry out a measurement and get an answer. So now you carry out a measurement. You'll say that, okay, I need the position of an object. So I carry out a measurement, get the position. You do get an answer, nothing wrong with that. Again, you perform, you need the momentum also. Again, you perform a measurement of the momentum. You get the momentum, no problem with that. Now, suppose you want to check your answer. So you perform a measurement of the momentum again. Again, you get the same answer. You want to repeat it one more time. Two times, 10 times, 1037 times. You will always get the same answer. Every time you measure the momentum, you will always get the same answer, okay? But then you come back and make a measurement of the position. You had made this measurement earlier and you have recorded it in your notebook. And now you make a measurement of position. Now, the answer you get will be different. But you will get some answer. Now, you want to check this answer, so you repeat a measurement of position. Again, you will get the same answer. So as long as you're repeating a measurement of the same quantity again and again and again and again, you will keep getting the same answer again and again and again and again. You will get this answer absolutely accurately, okay? Assuming that there is no measurement uncertainty because the uncertainty we are talking about has nothing to do with the experimental limitation. It is not, uncertainty is not coming from the inaccuracy in the measurement process. So as long as you repeat a measurement of position, no matter how many times, 1037 times or 1038 times or any number of times, because what a measurement does to a system is the system collapses into an eigenstate of that measurement. It remains in that eigenstate. And if you keep performing that measurement again and again, you will always get the same answer. The problem comes when you measure some other quantity for which, again, you will get an answer. But then if you come back and repeat the first measurement, you will not recover that answer. So repeated measurements of position and momentum if performed in arbitrary sequence, okay, do not produce results which are consistent with each other. Because one measurement disturbs the value of another. Now this doesn't happen with everything. For example, if there is a free particle and you measure its kinetic energy and you measure its momentum and you can measure it alternately any number of times, 17 times momentum, 23 times the energy, again, seven times the momentum, doesn't matter whether the number of measurements are even numbers or odd numbers, prime numbers or not, doesn't matter because some measurements are compatible with each other. Some are not, okay? So those measurements which are not compatible with each other are the canonically conjugate ones in the classical sense, like position and momentum. So these are canonically conjugate variables and repeated measurements of these canonically conjugate variables do not reproduce results which are consistent with the previous measurement, with the result of the previous measurement. What is the upshot of this? Our whole formulation of mechanics, Newtonian mechanics which was based on causality principle, Lagrangian and Hamilton mechanics which was based on the principle of variation. These were based on the fundamental idea that the mechanical state of the system is described by position and momentum which is a point in the phase spectrum. And now we find that you cannot get this information which means that we have no choice but to abandon the point of view that the system is described by position and momentum. You cannot use it anymore. Then what do you do? Quit physics is one solution. The other is to find an alternate way of describing the mechanical state of the system that it's not position and momentum, okay? Come to terms with it. What is this alternative way? It can be described by something else which is a state vector, which is a vector in the Hilbert space. Now many students in particular feel that, okay, we are getting into the domain of mathematics. We are getting into the domain of abstraction. When you are talking about position and momentum, I knew what you were talking about. Now you say that, okay, the state of the system is described by a vector in the Hilbert space. Now what does this vector in the Hilbert space which in my mind, just a mathematical quantity and how is it related to the physics of the system? This is a question which every student would ask you. The answers are in fact very simple because yes, ultimately you do expect the physical quantities that you're using to describe the state of the system to be related to the physical, observable properties of the system, right? You expect that. And that is exactly how it is in quantum theory. The vectors and so on are related to physical observables which is why it is a viable and a valid theory in physics. But what the connections are, take us a little time to discover. You need to go through a course in quantum mechanics, maybe two courses, sometimes even three courses. And when you do that, and this is best described in the words of Dirac, who said that these connections are best established as you develop the algebra further. So now that you have introduced the vector in a Hilbert space to describe the state of the system, you're not stopping at that. You're going to describe certain other things in addition to that. You will introduce what quantization is, okay? And as you develop quantum mechanics further, you will then end up reaching a point from which you will be able to connect the results of quantum mechanics algebra directly to the observables just as much as you would do so in classical mechanics. And there is a certain correspondence between this, which is very nicely stated in Bohr's correspondence principle and so on. I won't get into those details. But the idea is essentially this, that whatever mathematics you're introducing is ultimately aimed at explaining observables so students must get it out of their mind that this is theoretical physics that has nothing to do with experiments because quantum mechanics is a theory which is designed to explain the observables. It is a theory of measurement. Measurement and experiments are central to quantum theory and as a matter of fact, they are far more closer to measurement than classical mechanics is because classical mechanics assumes that you know the position and momentum, initial position and momentum even without a measurement. Whereas in quantum mechanics, you will reconcile with the fact that you have to perform a measurement. And only as a result of this measurement, you can get information about the state of the system. Now, when you do that, all measurements are not compatible with each other. Some measurements are. And that is, these are the details that are involved in the development of quantum theory or in the development of atomic physics. And I like to develop the subject of atomic physics and quantum theory hand in hand that is how quantum theory developed in the last century, quantum mechanics and atomic physics. So it is in a certain sense, a course in atomic physics or also a course in quantum theory. And as such, since position and momentum cannot be determined and cannot represent the state of the system, the idea of an orbit fails. The trajectory, the ellipse, orbits and so on of the Bohr-Sommerfeld model, they fail. And as a matter of fact, in the book by Banish Hoffman, she describes Bohr's model in one chapter, calls it as the atom of Niels Bohr. And the next chapter title is the atom of Bohr-Niels. The reason is, orbits cannot exist in quantum theory. So the idea of Bohr's theory is not correct. That is something that we now call as the old quantum theory. And we have to abandon the system completely. So the new system and massages, so the state of the system is described by a vector in the Hilbert space. And what do you do with these vectors? You can operate upon these vectors by operators. Okay, what do operators do? The operators operate on the vectors. And it seems that we are now completely in the realm of mathematics. But then, all this mathematics is a lot closer to measurement and observations than classical mechanics was. Because this is realistic, okay? What are the connections? How does it become closer? That is something we will know as we develop the algebra further. Okay, so we need to have a little bit of patience, understand how this works. And that's what quantum theory is about. So dynamical variables, which were in classical mechanics described as functions of position and momentum, are now in classical mechanics described as operators. And this is what quantization is. When you say that you are now dealing with the system in quantum mechanics, you have quantized the system. The process of quantization of a system involves dispensing with classical dynamical variables and treating them as operators. So this is quantization. Mind you, I have not talked about discreteness of a spectrum anywhere. And there is a real reason for that. So quantization has been described in terms of using operators instead of dynamical variables, describing the state of the system as vectors in the space and getting physical observables out of it by developing the algebra further. So this is our contention. This is the premise on which we build quantum theory and atomic physics. And as a matter of fact, the entire domain of physics. So now that we have got a vector, we need to label it, right? We need to designate it. Means you have written a vector and we are asking, how is the state of the system to be described? So what can you describe? How can you describe the state of the system? You can describe anything only in terms of the observables, right? If you want to describe this water bottle, you will say that it is about so many milliliters. It has got so much mass. It has got this shake. These are the observables. You cannot describe it in any way other than the observables. So now, having said that the state of the system is described by a vector in the Hilbert space, you ask, what is it that you can observe about the system? And what you can observe about the system is what you can measure. And you carry out a measurement, which is fundamental to quantum mechanics. You carry out a measurement. You get an answer. What you get in return is the observable. So what the system does is it collapses into an eigenstate of the measurement and it returns the eigenvalue, which is a description of the system. So you have a system which you want to describe. You carry out a measurement for which the operator is A. It can be energy operator. It can be something else. So you perform a measurement and if the result of the measurement gives you a state which is proportional to the original vector, if the new vector is proportional to the old vector, you write an equation, which is an eigenvalue equation. The answer that you get is this proportionality, which is the little A. This is the result of the measurement and you can use this to label the system. So you insert this label in the state vector and that is what becomes the designation of the system. So this is an eigenvalue equation and notice how this eigenvalue equation is written. It is not written in one step from left to right. How did we do it? We began with an arbitrary state of the system. Then we asked what is it that I can measure? This is what I can measure, the property A. This is represented in quantum theory by an operator. The operator operates on this. So this is a measurement which is carried out. And you get a result of the measurement, which is this little A. And you use this A in this blank space to designate the state of the vector and then you use the same label over here on the left-hand side as well. So you have written this equation not really left to right, but you have written it in pieces. You first find what is the proportionality, which is the result of the measurement. So this is how an eigenvalue equation is developed and now you say that you've got some information about the system and this information has come from the measurement. So measurement is fundamental to quantum theory. It wasn't in the case of classical mechanics. As if it was taken for granted. But here you actually demand that you make that measurement and then use that measurement to describe the system. So then it comes necessary to reconcile with the uncertain difference. Because some measurements are compatible and some are not. Like in a free particle I said that kinetic energy and momentum measurements are compatible with each other. But that is not true for position and momentum. And the uncertainty principle, which I'm not going to discuss in great length, but I would like to flash the slide and urge you to go through Sakurai's modern quantum mechanics, where this is very nicely discussed, in which the uncertainty principle is stated as you'll find on the screen. The proof is based on the three lemmas which are given over here. And it is absolutely important to go through these three lemmas and then enunciate the uncertainty principle because delta X and delta P have to be understood exactly as they have been defined over here and in no other manner. Because students often write the uncertainty principle in so many different ways. I've sometimes we sit on these VAIWA committees and ask the student what is uncertainty principle. And he says that okay delta X delta P is of the order of H. The next student comes and says it is of the order of H cross. The third student comes and says it is about H over two. Four students says it is H cross over two. Then another says it is greater than or equal to. So you get like 10 different answers. And they are not correct. The only correct answer is what you have on the screen. And it has to be understood. Delta X has to be understood as a root mean square deviation as has been defined over here in terms of these expectation values and all. And it is absolutely important to understand the uncertainty principle in this manner. And the reason you have this minimum uncertainty is because these operators, the inequality is determined by this commutation of the two operators A and B. Now what happens is that certain measurements are compatible with each other. So that you perform a measurement and then you do some other measurement. And after having performed the second measurement you come back and repeat the first measurement, you get consistent answers. So those measurements are called as compatible measurements but some measurements are not compatible. And whether they are compatible or not determines whether the corresponding operators commute or not. So if the operators commute, they do so only when the measurements are compatible not otherwise. And the uncertainty principle is a consequence of this. Now uncertainty is an idea which many physicists resisted including Einstein because they thought that this is not how nature nature's laws would be determined. Einstein would say that this is not how we expect that God's scheme of things to be. And there is a famous code attributed to him that God does not play dice because uncertainty would mean as if you're depending on some kind of gambling kind of thing, right? But then that's how nature is, okay? And these things were hotly debated in the last century and most famous debates are between Nies Bohr and Einstein and every time Bohr won the argument against Einstein. So although Bohr's orbits failed, it is because of Bohr that the criticism on the uncertainty principle on the questions that were raised about quantum mechanics could be satisfactorily answered by Nies Bohr himself and Bohr would say that it is not your business to tell God what to do. Okay, so that's how the laws of nature are. That's how the laws of nature are and it's not because of Einstein's taste about what he would want God to do that nature would adjust its properties according to that. So anyhow, the question is back to this that how do you describe the state of the system and we agree that it can be described only in terms of its observables and you have performed a measurement, you get an observable, right? You use that as a label in the vector, right? And that becomes a description of the state of the system. But then a physicist will like to get maximum information about the system, okay? You want to know more about the system. Either I have a visitor who comes to my house and says that, okay, I want to meet Deshmukh and I tell him that, okay, when I was a child living with my parents, my brothers and sisters and everybody in the family was Deshmukh. Then he asked, okay, I want to know the Deshmukh who goes to the physics department or whose initials are PC and there were three of us. So you want to get maximum information about the system, not just one parameter because that may not be sufficient to characterize the system, right? So you perform more measurements and some measurements will be compatible, some will not be compatible. So what you do is you carry out a measurement, get an answer, so the system collapses into an eigenstate of that measurement. You get an eigenvalue equation and you use that to describe or to characterize the state of the system and then you carry out all measurements which are compatible with the first one. And when you have exhausted that set, you get which is what is called is a complete set of compatible observables for which there is a complete set of commuting operators and both can be written as CSCO and this is the complete set which will describe the state of the system. So the state of the system will be described not just by one label, but by as many labels as you can get, okay? And you look for all the labels which are compatible with each other, all the measurements which are compatible with each other to get a complete description of the state of the system. And then as you develop the algebra further, as I'm sure all of you know, you can get the expectation values, you can get the transition matrix elements related to observable spectral intensities and so on. So that is how you connect this to actual measurements. So that takes the development of quantum mechanics to a different level. So you need a complete set of commuting operators and everything that you can measure will depend on the symmetry properties which is what I was underlying in the morning session. That symmetry plays an extremely important role and what is the symmetry of the Hamiltonian? What are the things that the Hamiltonian commutes with? What is it that you can measure the energy along with? You can measure the energy as an eigenvalue of the Hamiltonian, fine, but then what else can you measure? And when you carry out all of those measurements which are possible, that will depend only on what observables are compatible with the measurement of energy and in quantum theory, this will be represented by those operators which commute with the Hamiltonian, okay? Because then you will have a complete set of commuting operators for the Hamiltonian. So we have reconciled with the idea that the state of the system is described by a vector in the Hilbert space that dynamical observables are represented by operators. So what is the operator for moment, for example? How do you get that? It is not a postulate of quantum mechanics, okay? That's the operator for momentum involves a gradient operator, is not accidental. It is not a postulate, it's not what you do because it works, okay? It is something because it has to be that, okay? The reason it has to be that is because the momentum is the generator of translations in classical mechanics, okay? And one can refer to Goldstein if you like, but the momentum is the generator of displacements and if you compose an operator which will affect the displacement of a vector and ask what properties it should have, you can expect that it should have such a property that it must be unitary because it must preserve the norm. So you can make some common sense expectations of these operators and when you apply these considerations, you are automatically led to the fact that the momentum has to be minus IH cross gradient operator. It's not to be introduced as a postulate or something which Bohr or Dirac or somebody told us how to do it. If you were to do it starting from first principle, you cannot escape this example, okay? Because all you have to reconcile is with the fact that okay, the state of the system has to be described by a vector in the Hilbert space. What would happen if you subject an arbitrary state vector to an infinitesimally displacement? What properties should that operator have? If you ask these questions to yourself, you will automatically arrive at this result. And if you want to see what steps you should go through to get this result, either you sit down and work it out yourself or as you will find a very easy answer in just half page in Sakurai's book, okay? So I will let you work that out. And then from the symmetry, you can get a complete set of compatible observable. So momentum has to be this operator and I will illustrate it for the angular momentum. Again, there are a few semi-technical articles which I have written with some of my colleagues and students which you might find useful in this context. They are available at my website. So let me illustrate to you what the operator for angular momentum should be. Okay, I already told you how to get it for the linear momentum. So for angular momentum, if you are to subject a certain point to a rotation, so you begin with a certain point here whose position vector is R, you take it to a new point which is here and you get to this point through a certain rotation about this axis, right? So all you are doing is a simple rotation and you write the new vector with the subscript R which is the rotated vector as the old vector plus this displacement vector which for infinite decimal displacement will be delta phi cross R as you can see directly from the cross product. You can write this as a matrix equation and get the elements of this matrix, right? And you can ask the question that if this is how a point responds to rotation, how would a state vector respond? Okay, a state vector and a wave function are pretty much the same things. A wave function is in the function space. It is nothing but the coordinate representation of a state vector. So it's just a matter of notation whether you use the de Broglie Schrodinger notation of wave functions or the Dirac notation of the ket vectors. It's pretty much the same thing. So the whole mathematics is completely equivalent. So I will use the term state vector and the wave function more or less equivalently because just the coordinate representation of a state vector is the wave function. So there's nothing new about it. So we are asking the question, how does a wave function respond to rotation? So wave function is some function of the position. The thing about it is just a function. Forget about quantum mechanics, forget de Broglie wavelength, forget Schrodinger everything. Just a function of position, just the way the potential function is, okay? Or if you look at that back, the contours of that back, okay? Describe a certain function. It has the value of a certain physical quantity at every coordinate x, y, z, okay? And if you describe this property correctly, you will generate the shape of the back, right? So it can be anything. It is some function of position. It is a scalar point function and you take a scalar point function psi alpha, it's subjected to rotation and you get a new function which I call as phi. The rotation that you have in mind is this rotation that from an old position vector r, you get a new one by adding delta phi cross r to it, okay? So now this function could be any shape function. You can subject it to rotations and you get a new function, right? And what the rotation has done is it keeps all the relative orientation the same, but it only subjects that object to a rotation as you see in this figure, okay? So what the operator U, this is the rotation operator, what the rotation operator has done is to change the contours of this function in such a way that the new function has got the same value or amplitude at the new point as the old function at the old points, right? Now that's exactly what a rotation operator has done. So now let us ask the question, how would you express the function phi in terms of psi? If you find this, you will get an explicit expression for the rotation operator U. So this is what you're looking for and obviously you can see that this rotation operator is, you can write it in terms of psi itself if you take this backward by operating on r inverse on r, and what our inverse of r will do is to give you a new point which is less than the old point through the same displacement, delta r, okay? So this is very straightforward. This is first order derivative calculus, differential first order differential calculus. So there is nothing very fancy about this. So you take an arbitrary state vector alpha subjected to rotations, you get a new vector. Over here, I have sandwiched a unit operator which is the resolution of the unity. You can write a unit operator in terms of these projection operators. And this tells you that you can move this operator U inside this integral so that the U would operate on r. So let's do that. So now you are subjecting the state vector r which is an eigenstate of the position vector to a rotation, okay? Now what will it do? It will change the coordinate r. It will change it to what? It will take it to r plus delta r, okay? So this is really extremely simple. We are using just the initial premise and extending it step by step. So you move this to r plus delta r and rewrite this integral now. Here the integrand is no longer in terms of the ket vector r but it is in terms of vector r plus delta r. How are you adding this up? This integration you know is a sum, okay? It is the limit of the sum. So what are you doing? You're taking all of these vectors, adding them but before you add them you are scaling them by this factor. This is a scalar factor, right? So you are scaling every vector. You're multiplying it by some complex number by some scalar and then adding all of these vectors. So it's a simple addition that you are carrying out and where are you, how many pieces are you adding up? You are carrying out the integration over whole space. So if you count the number of persons sitting in this hall I can begin counting with that person over there and count everybody and then get a complete number of persons or I can count, begin my counting with somebody in the middle, go around here and then pick up the rest on this side, okay? So it doesn't matter where you begin counting and whether I began counting at the vector r or at r plus delta r doesn't matter because I'm going to integrate over the whole space. So what I do is I, over here I was integrating all these vectors r scaled by alpha r. Now instead I'm integrating all of the vectors r plus delta r. So I've shifted where I begin my counting but the scaling factor is also different, okay? And then I carry out this summation because integration is nothing but a summation. So you carry out this over r plus delta r and just for simplicity if I relabel r plus delta r as r prime then I have instead of r plus delta r I have r prime over here the integration is now over r prime instead of r and the scalar multiplier is not r alpha it is r prime minus delta r alpha, okay? Because you've shifted the point at which you begin counting. So let's take this last result to the top of the next slide here and here this is as I have been saying a function is nothing but the coordinate representation of a vector. So you have got a vector alpha its coordinate representation with respect to r prime is the function psi alpha prime, psi alpha r prime. So this r prime minus delta r is nothing but psi alpha r prime minus delta, right? And how would you express this in terms of psi alpha r prime? Just the first order Taylor series, okay? The value of a function at a neighboring point is equal to the value of the function at the given point plus the derivative of the function multiplied by the displacement. And because we are dealing with displacements in three dimensions I've got the gradient instead of the first derivative d by dx, right? So this is what I get and I can represent this left-hand side by these two terms on the right-hand side. So now this integral over here has got this term replaced by two terms. Let us take it at the top of the next slide. Now delta r we already know is delta phi cross r. So I write this delta r into the terms of delta phi cross r. Now I've got the scalar triple product, one vector, second vector, and a third vector. So this is the box product of three vectors. What we call is the scalar triple product, right? And I can always interchange the position of a dot and a cross there, right? This high school students know. And if you interchange the cross and the dot over here you get the dot over here and you have a cross over here but the cross gives you something very interesting because the r cross del, del we know from our previous discussion is the momentum operator, right? Because the momentum is minus h cross gradient operator. So you write this gradient operator in terms of the momentum operator and now you have got r cross p which is the orbital angular momentum, okay? r cross p is orbital angular momentum which is j, which I've written as j. So now if you look at this you have got the angular momentum which has popped up over here. And if you just rearrange these terms a little bit and I will not go through the algebra in details you can look it up in the main lecture. But the algebra is very straightforward. It is just the scalar product of two vectors or a scalar triple product of three vectors that is all that is used over here. And essentially what you'll find by identifying the unit operator is that the rotation operator acts on an arbitrary vector alpha. And what is this alpha? Alpha is any arbitrary vector. It doesn't matter what it is. And if this relation holds good for an arbitrary vector there is an operator identity that you can extract. What is that identity? That this operator must represent the rotation operator because it operates on the same vector to give you the result of this operation a beautiful bracket operating on the same vector. So this gives you this rotation operator. See how the angular momentum has appeared in this. So just the way the linear momentum is a generator of translations. Mind you it is the linear momentum which is conserved in homogeneous space. Now you have got the angular momentum which is conserved in isotopic space space which has got the same properties in all directions, right? So the angular momentum is conserved in an isotropic space and it is a generator of rotations. So the operator for rotations or the operator for translations or generator of translation whether it is a linear momentum or the angular momentum these come automatically by considering these essential physical properties which are directly related to the process of measurement in quantum mechanics. So now we have got an operator for rotations and this is a very simple exercise which you can do or you can ask your students to do which is to subject to use this explicit form of the rotation operator and carry out rotations about orthogonal axis one after another, okay? Carry out a rotation about the x-axis then after the y-axis but the next time you do the same exercise but in the reverse direction first about the y-axis and then about the x-axis and when you do that you will find that if you subject the body to rotations to finite rotations but in the reverse order you will not get the same orientation at the end because finite rotations do not commute which is why finite rotations are not vectors. So rotations you can describe by magnitude 90 degrees rotation by a direction direction is rotation about the vertical axis rotation about a horizontal axis, right? So you can define finite rotations by magnitude and direction but they are not vectors. So as a result of that the rotation group is not an Abelian group is made up of those elements which commute with each other but these do not commute with each other. So this is a non-Abelian group whereas translations give you an Abelian group so px and py commute it is x which does not commute with px but x will commute with py. So this is now the definition of angular momentum in quantum mechanics, okay? That angular momentum is such a measurable property which is represented by operators which whose Cartesian components do not commute equivalently if you construct the cross product of angular momentum with itself it doesn't go to zero. Now this is unlike a classical vector if you take a classical vector and construct a cross product of that vector with any other vector which is collinear with it let alone itself, okay? Any vector which is collinear with it the cross product will vanish but not with the angular momentum. The angular momentum cross product will give you i cross j. If you construct the commutator of angular momentum with itself it will not give you zero. A classical operator would give you zero but this is quantum so this will not give you zero. And all of these are completely equivalent definitions of an angular momentum and it has got three attributes one is that it is a quantum creature it is a vector and it is an operator. So you must apply the operator algebra the vector algebra and quantum mechanics to interpret angular momentum. If you leave out any one of these you're not doing physics. Okay, it would be wrong. So these are the angular momentum operators all of these three attributes are important and that is the reason these classical models of angular momentum. When you talk about spin and sometimes books also some books also describe an electron spin in terms of they give the analogy of the earth spinning about itself. Now these models are absolutely wrong, okay? So they are wrong and that is the reason there is a big cross over it and not only that why is this ghost hiding? Let me bring it to the surface. Yeah, these ghosts should not hide. Yeah, because if you use these models these ghosts will come after you. One is Pali, the second is Dirac and the third I'm sure you recognize is me. Okay, so be careful. These models of spin are wrong they should not be given. Not only that even in as much as a spin angular momentum of an electron spin cannot be and should not be described in terms of such models the orbital angular momentum of an electron should not be described in terms of this orbit. We just discussed a little while ago that orbits do not exist in quantum mechanics and orbit requires position and momentum. There is no orbit that you can talk about no trajectory that you can talk about. That trajectory is essentially defined as the locus of a point in phase space it requires Q and P simultaneously. So orbits don't exist. Okay, so the orbital angular momentum cannot be described in terms of orbit. So what is it? What is an angular momentum? Whether it is an orbital angular momentum or spin angular momentum it is such a quantity which is described by these commutation rules. Okay, now come to terms. Now that's exactly what it is. Okay, so let's not describe it in terms of the Earth's rotation or orbits and so on. It has to be described in terms of these commutation relations and how do you get physics out of it? By developing the algebra further. Okay, when you develop the algebra further you relate it to actual measurements, actual spectroscopy and that's what Dr. Arvin's lecture will be about that when you use these ideas the angular momentum correctly you get the correct magnetic moment and then with that correct magnetic moment you're able to get the correct spectrum of an atom in a magnetic field like Z-manifact and so on. Okay, so that is where the observations come in. And unless you develop the algebra of quantum mechanics the tools of quantum mechanics to that extent you cannot relate these quantities to observables but they are designed essentially to produce the results of measurements. So quantum mechanics is not theoretical physics. It is physics, okay? So we recognize the angular momentum to be the generator of rotations and I think I may be taking more time than what I should. How much time do I have? I have until 3.15, is it? Three o'clock. Or just three o'clock, two more minutes? Thank you. I started about an hour late, I think. So symmetry plays a very important role because I mentioned that, okay, what is it that you can measure depends on what measurements are compatible with each other and this is where the symmetry of the Hamiltonian comes into picture. And there are different symmetries. We talked about the rotations which are continuous symmetries. There are these discrete symmetries like parity, reflections and so on. You can ask why the left goes to the right but why doesn't the top go to the bottom? And these are interesting questions. I'm not sure that all of your students really know the answer. And the answers are quite subtle. It is because the rotation operator and the reflection operator, they have different representations. One is represented by a determinant whose determinant is plus one whereas in case of reflection, the determinant is minus one. So they have different properties altogether. So when you take the set of all rotation operators, you have got an orthogonal group which is called as the SO3, okay? That is the rotation group. And that explains the quantum mechanics of the hydrogen atom because the operators which commute with each other which give you the set of compatible measurements. What is it that you can measure compatible with each other? So there are three measurements that you can carry out. And these are the energy, the L square and the LZ. And these give you the complete set of labels for the hydrogen atom which are the energy N, L and M. So the first thing to do is to recognize what is it that remains invariant with time and that is energy like Zendulkar's energy as he grew up in his career, right? And then what is it that is compatible with the measurement of energy? What are the observables which you can carry out without disturbing the result you got from the measurement of energy? So those are represented by operators which commute with the Hamiltonian and that is where the symmetry of the Hamiltonian comes in. So these operators H, L square and LZ, they commute with each other. They give you what we call as the good quantum numbers energy, L and M. And you can write a state of the system and describe in terms of what we call as good quantum numbers. Now this is not the whole story because there are just two points I would like to underscore before I take a break. One point I would like to emphasize is the fact that I introduce quantization in terms of representing the state of the system by a vector and observables by operator. I did not talk about energy spectrum being discrete and that was deliberate because if you look at the hydrogen atom spectrum, yes, it does have this discrete spectrum but above this energy equal to zero, you've got a continuum of energies and all of them are eigenstates of exactly the same Hamiltonian and that's also quantum mechanics, okay? All of them are eigenstates of the very same Hamiltonian, solutions of the same Schrodinger equation and you have a continuum of energies. So it is not at all correct to describe quantum mechanics in terms of discrete spectrum. You will have a discrete spectrum when you're dealing with bound states but when you have continuum states, you'll have oscillatory solutions. There's nothing surprising about it because after all you're dealing with a differential equation and the solutions depend on the boundary conditions. So do the boundary conditions, what are the boundary conditions for discrete part of the spectrum? The boundary conditions and the wave function must go to zero as in the asymptotic region as r goes to infinity, whereas in the continuum, they are oscillatory, they do not die out. So the boundary conditions being different, you'll get different kinds of solutions. So this is one point which I would like to highlight and the other point I would like to highlight goes back to what I stated earlier that if you look at the sodium spectrum, if you compare the difference between the sodium spectrum and the hydrogen spectrum, you could discover the columns. The reason is the following, that the actual symmetry group of the hydrogen atom is not SO3, which is this rotation, but there is an additional symmetry which is present for the hydrogen atom but which is not present for the sodium atom. So the sodium atom has got the possibility of transitions between 3P and 3S, but the hydrogen atom does not have this possibility because 3P and 3S are degenerate. And if you ask this question, why is 3P and 3S degenerate in the hydrogen atom, whereas it is not so for the sodium atom? The answer is that for the hydrogen atom, the potential is decided by one over R Coulomb's law all the way in the entire region of space from zero to infinity, whereas for the sodium atom, the one over R law would be valid only in the asymptotic region. When you take the 3S electron, take it out to the infinity and look at the rest of the ionic core. You have got 11 protons screened by 10 electrons. So effectively there is a single positive charge at the center. So in the asymptotic region, that's what this electron will see, but as it comes close to the core, it will see a departure from one over R. Whereas for the hydrogen atom, it will see one over R in the entire region of space. In other words, the Laplace vector is conserved for the hydrogen atom but not for the sodium atom. In other words, the Coulomb's law is sitting in the fact that sodium atom has got a transition from 3P to 3S but the hydrogen atom has the one over R potential. So you can really discover a lot out of doing elementary quantum mechanics and atomic physics. So I think I have overused my time, have I? So in fact, this degeneracy in the hydrogen atom is referred to in some books as an accidental degeneracy. But it is only an accident as an accident only as long as you do not explain why it happened. Okay? I mean, when you know that the accident took place because this particular actor was drunk and then he drove over them. Once you know that, you know you cannot call it as an accident, right? So it is called as an accidental, it is called as an accidental degeneracy only as long as you don't find the reason behind it. That the driver was drunk. Okay? Then it is no longer an accident and this degeneracy in the hydrogen atom is not an accident. It is because the interaction between them is the electrodynamic interaction, which is the Coulomb's law. So one could have actually discovered that. So anyway, I think I will stop here and you can do a lot of very exciting atomic physics from this platform but I will take a break. I will of course be happy to take questions later. But now it is time for Dr. Arvind's talk and he will talk about the Zeeman effect and the family of Zeeman spectroscopies. So thank you all very much. If there's any quick question, I will take it. Yes? And Dr. Kambu, can I ask? Yes. That is what I am asking for. Sure. Kambu, can you elaborate on this? Kambu is with each other. Arvind and Kambu, which is with each other. For example, the position of the X-cap and X-cap of the momentum, see, PX-cap is IH bar, which is equal to zero, PY-cap. What is the physical meaning of that? The physical meaning is that X component of the position of a particle can be measured, then you perform a measurement of the Y component of the momentum. You will get a result. Now you come back and repeat your measurement of the X component of the position of that particle. You will get exactly the same result as you had got earlier, because these are compatible with each other. But if you measure the X component and now you measure the momentum X component and now you come back and measure the X component again, this answer will not reproduce the result of the first measurement, because X and PX are canonically conjugate. PX is the generator of X, displacements along the X-axis. Momentum is the generator of translations. PX is not the generator of translation along the Y-axis. That is the reason those measurements are compatible. So thank you very much and we will, of course, continue the discussions, but I will...