Somewhere around 2018, that year we made the million. In September, we ended up pitching a shark. And that's when we knew we were on track to do a million, but we still weren't really sure. It was our first six-figure month. We were like, if this keeps going the way it's going, like, we're definitely gonna get a million in the year. But we weren't. We were still unsure. Was your valuation based on the growth rate? And so were you like, if we grow another 10X, we're going to be a $10 million company next year? No, we didn't even, I wish it wasn't in my right opinion. It wasn't even that high. We had basically won a million in sales for the year, and we had done like a, originally we had went in with like a $7 million valuation, and then the shark, not the shark, but the people who were advising us around the show suggested that we go lower. So it was okay, fine, we went lower. We didn't default on wanting it, but we just ended up just going a little bit lower. We did a $4 million valuation for Robert. And Robert's offer was a $2 million valuation, keeping in mind that we'd already made like almost a half a million for the year. And so I was like, this is not fair.  I was like, we're not doing this. And the night before we had made, we had our like walk away numbers, like we're not going to give up more than 15% of the company. It does not matter what they offer. It's a hard, fast no. And Robert would negotiate and he asked for 20%, and so we just, we ended up walking away. But later that year, I got offers from like Angels Syndicate for $7 million valuation, on the same data, you know? And so it was just like, I knew that I was right in that valuation. But we ended up getting an offer from the CEO of LinkedIn, Shirley Acton, Jeff Weiner. He's a former CEO. He just stepped down for a million, 1.2 million. And so I'm skipping Backstage Capital ago. I was just, I was going to go back to it. You could finish. Backstage Capital gave us $25,000 a year.  They first started with us back and we were doing the boxes. That's Arlan Hamilton, and she's been a mentor since then. And so they gave us 25K on a convertible note. The month that we made 30,000 off. So in the top of 2018, we cut off all the subscription box. We had $3,000 in sales in January. And then by March, we were doing $30,000 a month. And we were just like overwhelmed. We're like, we don't know what to do. We know we need help, we need money. We need to get out of our kitchen. And Arlan came through with the 25K that helped us move into a new manufacturing facility that allowed us to then scale to that million. Yeah. That's great. To go back to Shark Tank, we had Bad Birdie on recently who actually almost or did deal with Robert specifically. And he was telling us, apparently there's some, I think it's a girl he said that keeps a Google sheet of every single episode, every single person, the valuation and whether it won or not. And basically what he shared with me was, every shark cuts your valuation in half. Every single time. That's what they said. It's like, that's exactly what they do. They just cut it in half. And so when he went and he was prepared for that cut because statistically they were going to do it anyway.  But if you set it at something realistic, they're going to cut it in half no matter what, which is interesting. That's good to know. I wish we had known that at least initially. I'm glad we did it though, because we might have made a just and strong for it to get a deal. Versus value that what we personally value the best. And hopefully, Robert said he didn't know anything about the industry. He literally said, I don't know anything about it, but you guys sound like great entrepreneurs. The numbers don't lie. Everything looks good. I want to make it awful because I believe in you, but I can't really help you. And at that point, we needed more than just money. Like most new, like most new entrepreneurs or even even people who don't have like a big network of people. It's the information is almost just as important, if not more important than the money. Like knowing what opportunities are gonna be good or not or getting the right kind of people in our network. Those things have been just as valuable as any check that we've gotten over the years. So what happened after you guys aired? So you air what happens to the website? Does it blow up? You get three weeks notice, right?  So the month we aired, the month before we aired, we did about 300,000, 295, something like that in sales. And then the month that we aired, we did about 900,000. Hey everyone, thanks for checking out that clip. If you enjoyed it, be sure to hit the like button down below. And if you're interested in hearing the full episode, it's out right now on our YouTube channel. We've had a lot of great guests come on this show before and we've got a lot of great guests coming up in the future. So hit subscribe so that you don't miss a single episode. And one final note, we're always looking for new ideas and new companies feature on the show. So if you know of someone or know of a company, write us a comment down below letting us know who they are and what they do. We'd be happy to have them on the show. Until then, I'll just be here waiting for your comments. So see you later.


So today I'll be talking about this Zeeman spectroscopy. I'm just using the slides which are already there in the website. These are all slides from Procedesh Mook's course and I have not altered it. I just introduced a couple of labels for convenience. That's it. So it's all available. I hope I'll do justice to this for a lot of information and a lot of effort has gone in preparing this lecture. See, I had taught Zeeman spectroscopy a couple of times for students and both the times the emphasis was mostly on regenerate perturbation theory, treating Zeeman spectroscopy as an example. Okay, but what you will see in this lecture is Procedesh Mook has beautifully brought out the underlying fundamental concepts in quantum mechanics which are linked to Zeeman spectroscopy such as the transformation, symmetry transformations, the degeneracy due to symmetry operations, degeneracy due to symmetry in the Hamiltonian and how an external perturbation which will break the symmetry will lead to lifting of this degeneracy. And through this, he also introduces the couple of very important powerful tools such as the Fletch-Cordam Coehibitions and the Big Neckars theorem. And so we will see these. So mostly I have been using this, teaching Zeeman spectroscopy as I mentioned as an example for degenerate perturbation theory but there is so much of underlying concept that has to be, which are linked to this which we can convey to the students and make it more deeper. This we will realize in this lecture. So as I mentioned, there can be several transformations. You can have rotational transformation. You can have inversion which will inversion about the origin or you can have several combinations. The question is, what do these transformations do to your Hamiltonian? Do they change your Hamiltonian or not? There may be some transformations which leave the Hamiltonian unchanged. So we call that symmetry, those are symmetry operations. So those which leave the Hamiltonian invariant under those transformations. For example, if you take a simple harmonic oscillator with the simple harmonic potential in one dimension with the minimum of the origin, so then you know if you do an inversion, if you change x to minus x, what you have is there is no change in the potential energy. So there is this parity inversion transformation leads to this parity which is a symmetry operator in this case. It does not do anything to the Hamiltonian. The Hamiltonian which is the kinetic energy plus the potential energy remains unchanged. So that is what we call a symmetry operation and we will see what it leads to. There are fundamental properties that symmetry has. So transformations can be anything rotation inversion but only those which leave the Hamiltonian invariant is what we are interested in. They are the symmetry operators. If you have a large number of symmetry operations like which are R1, R2 and so on and if they all commute with the Hamiltonian, that means if they leave the Hamiltonian invariant then we can call this as a group of Schrodinger equations. These are all the groups of symmetry operators that system has, symmetry operations that the system possess. Why do we do this? If you identify the symmetry of the system, it actually makes your job much more easy. Just what remains is you are almost half the work is done when you want to find the eigenstate of the system. So half the work is done if you know the symmetry of the system. So if you have say Ri which is a symmetry operation which commutes with H, you know that both can be simultaneously diagonalized because when they both commute with each other and hence if you know the eigenstates of the symmetry operator then you know that there is a possibility to find out the eigenstate of the Hamiltonian system. If you remember that not all the eigenstates of the symmetry operator will be eigenstate of Hamiltonian but of course there is a scope to find the eigenstates of the Hamiltonian. So the group theory and the geometry of the system that is the properties of transformations is a lot of information about the quantum mechanical aspects of the system. So if you have a Hamiltonian expressed as a matrix in the eigenbasis of a symmetry operator say R then you know that you can show very easily that the Hamiltonian will not have the matrix element of the Hamiltonian will vanish if you have it evaluated between two eigenstates of a symmetry operator with different eigenvalue. This is very simple to show. This is what we see here. So if you have Ri which is a symmetry operator that commutes with the Hamiltonian and if you represent this is an operator. The commutation operator is an operator and this if you evaluate on the matrix the eigenbasis of as a matrix and the eigenbasis of R the symmetry operator then it is easy to show that this the Hij which is the matrix element of the Hamiltonian between Ri, eigencat of R, eigencat Ri and eigencat Rj this will be 0 if these two if Ri, Ri minus Rjj is non-zero. So what you need to do is the following you just have to search for the eigenstates of the symmetry operator with different eigenvalues. Once you do that you are very close to the problem that is of identifying the eigenstates of the Hamiltonian. We have already done this in the class. I just wanted to show you this. It is not due to us. We will take this one-dimensional problem simple harmonic oscillator. The time independent squatting equation for this problem is the size of the energy eigencase and when we try to we know that there are two ways to solve this to get the eigenstates of the simple harmonic oscillator one is the operator approach and the other one is the wave function approach. In the wave function approach what you do is you express this eigenstates obviously this and then you will substitute this in this equation I am not getting into details but I am just trying to get to the essentialities so you would substitute this in this equation and you will get the various coefficients. Once you get the various coefficients you have already solved for the wave function. When you do this what you would find is that the coefficients are related in the following fashion. The even coefficients are actually related to C0 where C0 is the coefficient of C0 is the constant of the x C0 is the coefficient of x power 0 in this series and you will realize the odd coefficients are related to the g and n or some functions of n are related to C1. Now you think that you have solved the problem because you have already got this solved but only two things are unknown once you know C0 you know all the even coefficients once you know C1 you know all the odd coefficients so you may be happy I have found the entire solution for this eigenstates but the answer is no here are two unknowns which are C0 and C1. Now the problem is if you had just one unknown you could get rid of it by just normalizing this wave function and then say that yes I know C0 because this has to be normalized so I can actually find out one of the unknown but we have two unknowns here how did we solve this problem? We use the symmetry of this problem that is if you do an inversion about origin that is if we consider v of minus x the potential energy is same for v of x and this tells you that the parity operator is an eigenstate of the Hamiltonian of this Hamiltonian and hence the parity the eigenstates of parity operator is going to be can also be are the good candidates for the eigenstate of the Hamiltonian and hence we know that what are the eigenstates of parity operator they are the even and odd states now I know that the states have to be either even or odd so if I choose C1 to be 0 then I get rid of all the odd components odd components of this expansion and when I get C0 to 0 I get rid of all the even components and I get the other parity equation so this is what we do when we solve this equation so we use the symmetry of the problem and that helps us to solve the problem very efficiently so this is what we need to appreciate here so if there is another the symmetry also leads to degeneracy this we can see here if we have a Hamiltonian operating on a wave function eigen wave function if this energy is given by E then if you consider a symmetry operator pr and so let pr be operated on this this is the squatting equation since we know that it is a symmetry operator it is supposed to leave the Hamiltonian invariant and hence you can actually bring this pr inside so because they both commute so what you see is not only psi n which is an eigenfunction of Hamiltonian with eigenvalue E even pr psi n is also an eigenfunction of the Hamiltonian with the same eigenvalue E so what we understand is that symmetry also leads to degeneracy okay so we had first the transformation then we went to symmetry and then symmetry the link between symmetry and degeneracy now if we consider a hydrogen atom problem without considering any non-relativistic hydrogen atom problem without any magnetic field then there is a lot of the column force is the central force there and you have the force depends only on R it is a non-theta and phi and you have degeneracy because of this symmetry because Hamiltonian is invariant under rotation in this case and hence because of this what you have is the 2s and 2p states are degenerate so suppose if you introduce an external perturbation that is you add a term perturbing term to the Hamiltonian which will break the symmetry then there is a possibility to lift this degeneracy this is what all about this about the basic thing of this Zeeman effect you are introducing a magnetic field say along z axis so there is no symmetry there is rotational variant is no more there there is symmetry about this z axis so you lost the symmetry and hence you lost this degeneracy so that is what this Zeeman splitting is going to be so you can either partially remove it or in some cases you can fully remove the what I mean is there are 5 states with same energy that is 5 fold degenerate the particular perturbation can leave one of them two of them at the same level and the other two can be raised or lower so it can be partial lifting of degeneracy and beauty is that how do we find this degeneracy being lifted after lifting after the line split we can still have transitions between the split levels so we can identify that initially I had just one line now I have two lines or three lines so that is a degeneracy being lifted so that is an indication of the degeneracy so today we will talk three important cases of Zeeman effect normal, anomalous and passion there is nothing abnormal about anomalous there is nothing normal about normal and everything is normal but therefore historical reason and I will also reserve the comment on this for why we call it as normal and why the anomalous will be really get into the physics but as of now these three different cases or the names for the three different cases are purely historical and the so called normal effects happen when you have strong fields and anomalous is when you have weak fields passion back effect is the intermediate field we will discuss these cases in detail now Zeeman found that the spectroscopic lines split in the presence of magnetic field so let's see how the splitting will happen now if you consider a magnetic field it need not be uniform it can be varying in atomic scale dimension of the in atomic scale you can consider it to be a uniform field though you have variation within the dimensions of the atom you can consider the applied magnetic field to be uniform and the vector potential can be represented as half B cross R where A is the vector potential of the magnetic field which is applied extra to the magnetic field which is the B vector now we know that in classical mechanics Hamiltonian is given by this expression where P minus E A by C is called the generalized momentum this is the Hamiltonian of a charged particle in a magnetic field we need to write down the Hamiltonian in this fashion so that we can get the Lorentz force correctly so that is this expression for the P is the momentum and P minus E A by C is the generalized momentum now what do we do is when you want to someone is asking you why don't you write the Hamiltonian for a charged particle in a magnetic field we simply borrow things from classical mechanics we just take this equation replace P with P operator A with A operator and see whether the results are consistent with the experiment so that is what we are going to do so this is going to be our Hamiltonian in quantum mechanics with P being replaced by the momentum operator with this minus AS bar del and A being a function of position operator now as you can see here we have to be very careful in quantum mechanics it is not just the square of this operator it is P dot this P minus E A by C operator dot P minus E A by C and we need to be careful because momentum and position do not commute with each other so we need to work out this very explicitly and we may not go into the details of this it is easy to see because momentum operator is anyway minus IH bar del that is where you get this del here and this term in Coulomb gauge vanishes so we can take divergence of A to be 0 and so this is what you get for this equation and similarly with this your Hamiltonian becomes the following that is this is your full time dependent scoring a equation for a charged particle in magnetic field so what you see here is the kinetic energy term the potential energy term and two terms which are linear and quadratic in magnetic field and these come up because of the external magnetic field and we are supposed to solve this problem now we can what we want to do now do is we have two terms on magnetic field which is really making us uncomfortable let us just look at these two things closely that is we see we want to evaluate the related strength of these two terms now if you consider this A dot del term A is the vector potential which is what is appearing here and we just do a slight modification here by bringing this minus ih bar into this del so that we can see this identify this to be the momentum operator so what you get is r cross p in classical mechanics r cross p is the angular momentum r vector operator cross vector operator r cross product with momentum operator is going to be the angular momentum operator as Professor Deschhoek stressed many times angular momentum we define angular for us in quantum mechanics angular momentum operators of rotations that's it they are not something about particle spinning or moving about spinning about themselves or something like that so if you compare this term this particular term is a dot b which r cross p is nothing but the angular momentum operator which is L operator so you have B dot L the angular momentum operator will be the order of angular momentum operator can will be the order of h bar which is which has units of angular momentum as well we may have to h bar is h bar is the order of magnitude so we are just considering the order of magnitude of this linear term when you do the same for the for evaluating the order of magnitude for the quadratic term and do a bit of mathematics and when you do this take a ratio of quadratic to linear term what you find is a very small number multiplied with the so you will see that quadratic terms become significant only in very large magnetic fields so which are not achievable in lab so we can very well ignore the quadratic term in these expressions you can just live with the linear term so only at very since we only at very large magnetic field these become significant so what I am going to do what we are going to do is just we are going to eliminate this term in the quadratic term and live the remaining because the lab fields are B so this is what we have here now what we do is we have identified the Hamiltonian to be this so we already have solved the hydrogen atom problem for kinetic energy with a Hamiltonian being the kinetic composed of kinetic energy and the potential energy now what we have in addition is this term due to the external magnet field so our Hamiltonian is now h naught which is the unperturbed Hamiltonian with a perturbation which is dependent on angular momentum and the external magnet field which is what we call it as HL prime now if you look at this this term the perturbing Hamiltonian we can do a rearrangement by identifying by introducing a magnetic moment orbital magnetic moment which we identify to be the both magneton times the angular momentum operated by H bar so and therefore magneton is given by this expression EH bar by 2 MC once you do this you can identify this perturbation perturbing Hamiltonian to be merely minus mu dot B this is not a surprise because we know that when you have a dipole kept in a magnetic field it gets an energy the potential energy is minus mu dot B where mu is the magnetic dipole okay so this is the energy of the magnetic dipole in an external magnetic field which is a perturbing term in this case so our escorting an equation now looks like this so you have this is a time-dependent escorting an equation with the hydrogen atom unperturbed Hamiltonian plus this perturbing Hamiltonian now we are supposed to solve this equation now what happened was when Ullenbeck and Samuel Gouds made when they did this experiment they found that it is not sufficient to deal with just mu L and there are other component sources for this magnetic moment of the atom which they couldn't really physically interpret but they just in order to in order to explain their experimental results they introduced another source of magnetic moment which they called as a spin magnetic moment in addition to this angular momentum magnetic moment but they didn't have any expression until we had the relativistic quantum mechanics now in the classical picture when you have an electron in motion you can see that you can identify this to be a current carrying loop and a magnetic moment can be associated with this but this is a classical picture with not exact analogy to quantum mechanics because you cannot talk about r and p together position the momentum together and so but this analogy with this analogy we can see that the magnet orbital magnetic moment is related to the orbital angular momentum and the Bohr magneton with GL given by 1 this is what we got in the previous slide now with the relativistic quantum mechanics we naturally also get the spin magnetic another source for magnetic moment which is the spin magnetic moment which is now related to the spin angular momentum it is no way related to physical spin of electron it is just an intrinsic quantity which naturally comes from the relativistic quantum mechanics so the magnetic moment due to spin is directly proportional to the spin angular momentum just like the magnetic moment due to dual is directly proportional to the orbital angular momentum but now here the value for GS turns out to be 2 these are discussed in length in the course on relativistic quantum in lectures in relativistic quantum mechanics so what you have is a total magnetic moment which is composed of the magnetic moment due to angular momentum and the spin angular momentum orbital and the spin angular momentum so total mu is given by L plus 2s because GS is 2 this is what is going to be our total mu so you are perturbing Hamilton is calling a equation needs to be replaced instead of mu L you need to introduce mu dot p where mu includes both the magnetic moment due to orbital angular momentum and spin angular momentum but you have one more term here what is that? once you have introduced spin when you agree to include spin you also have to agree to including the effects of spin in terms of coupling spin getting coupled with the orbital angular momentum so when you do relativistic quantum mechanics there is a spin orbit coupling a term which comes naturally in the Hamiltonian which is given by zeta of r times s dot L where s is the spin angular momentum L is the orbital angular momentum this term comes out in the relativistic quantum mechanics just as naturally as the magnetic moment now what we have before s is two terms which are in brown colour here minus mu dot b plus zeta of r s dot L these two terms are going to be the perturbing term perturbation term we need to handle this now in the relativistic quantum mechanics this spin orbit term is explicitly can be evaluated it is a function of r and that function of r is essentially a partial derivative of the Coulomb potential with respect to r and this so this is the perturbing Hamiltonian before us now we are in position now to discuss three different cases of magnetic field one such that magnetic field is such that the mu dot b term is much larger than this term second the spin orbit term so when you do that you can simply ignore this and do your calculation treating this as the unperturbed ket and this is the perturbing Hamiltonian this is the case when you have strong magnetic field so that mu dot b is much larger than this remember mu dot b is an external perturbation whereas the spin orbit coupling is an internal perturbation even if you do not have a magnetic field this perturbation is going to be there because it is just a consequence of spin and angular momentum which are intrinsic the second case is an intermediate case still mu dot b is larger than the spin orbit term but it is not as larger as in the first case it is an intermediate case the third one is such that the external magnetic field is so low that the spin orbit term becomes significant now the reason for discussing Zeeman spectroscopy is that it actually exposes you to reach the variety of concepts in quantum mechanics because in all these three cases we are essentially solving an eigenvalue problem but the choice of basis set is going to be different so you are going to learn a lot about wisely solving an eigenvalue problem as you go through these three cases now the Hamiltonian as we saw can could be the perturbing Hamilton could be internal as in the case of the spin orbit term and it can also be external because of external magnetic field so this is what you can there are other important internal perturbations such as the hyperfine interactions because there are others which are mentioned in the lecture now that we know the perturbing term the job is to find the eigenvalue that is the energy of the system under perturbations how do we do that? we can take recourse to the first order perturbation theory now because of the symmetry of the system we have degeneracy so we need to take recourse to the degenerate perturbation theory so what you do here is to the idea in the degenerate perturbation theory is to diagonalize the perturbing Hamiltonian in the basis of the unperturbed ket and in the basis of the unperturbed ket so if psi naught is the unperturbed ket we know that the Hamiltonian will lead to this is the escorting equation for psi naught and the energy shift is given by the matrix element of the perturbing Hamiltonian in between psi naught this is the shift in energy for a state psi naught this is from the first order perturbation theory now the question before us is okay, we can evaluate this energy shift by evaluating this matrix element but what is going to be the unperturbed ket in this case? how do we choose this unperturbed ket? there are two choices before us one with a ket which is an Eigen ket which has Eigen values of Hamiltonian L square L z and S z and the other set of corner numbers being N, L, J and M where L z and S z are the z component of the orbital and the spin angle moment respectively J is the total angular momentum and J z is the component of the total angular momentum along the z axis now we have two bases in front of us you cannot just go blindly and use any one of them you don't get a wrong answer if you choose any of these but if you choose one of them be the right one you get the answer quickly and without much of work so I mean Sakurai makes a comment if you have to choose a wrong you have to be a fool or a masochist to choose the wrong basis you know what a masochist is? I think Griffith mentions this a masochist is someone who says please beat me Sadist says I won't so there is a choice between there are two choices one is this Eigen basis which is having which are Eigen ket of L z and S z operators represented as M L M S bases the other one is J M bases which are Eigen values so J square and J z so we need to choose between them we will do that considering individual cases now if you consider these three different cases normal, bastion and anomalous Zeeman effect in the normal Zeeman effect we are going to the magnetic field is strong enough so that the spin orbit coupling is rather very weak so what you have and of course we are discarding the quadratic component of the magnetic field now it turns out that it is easy to see that the appropriate choice of the basis for the unperturbed ket has to be the M L M S Eigen basis why because if you if you if this is the only perturbing term if your magnetic field B is along the Z axis so what you have is this term is boils around boils down to mu Z B Z okay and mu Z is nothing but yet mu Z has L Z and L Z in it mu Z operator mu Z has L Z operator and S Z operator in it and naturally L Z basis is going to be a choice because they are Eigen kets of L Z and L Z operator so you can easily diagonalize there is nothing you just have to evaluate the Eigen value and the matrix element in this basis it's going to be easy to it's just it's going to be diagonal because they are Eigen kets of L Z operator so the choice is M L M S basis for the normal Zeeman effect and not the J M basis and hence we call it as good quantum numbers the other we call it good because they are they they favor us in evaluating quickly the others are bad in the sense they are not forbidden others are not forbidden basis but this is this makes life lot lot more easier so the what we need to do is to evaluate this splitting in energy that means to evaluate this perturbing Hamiltonian in M L M S basis so we need to evaluate this particular matrix element this is very simple so how do we do that so we know that mu is nothing but mu B L plus L operator plus 2 S operator by h bar now if the magnetic field which if the direction of the man if you take a uniform magnetic field along a particular direction which is which we can call us as the Z direction then what you have is this B has just one component along the Z direction so when you take a dot product what you get is mu B L Z plus 2 S Z by h bar and these are any way egging kits of L Z and S Z so finally what you get is this term so when you want to evaluate the matrix element when this is along the Z direction so you have L Z operating on this ML MS kit to give you ML H bar and 2 S minute twice the S Z operating on this will give you 2 H bar so that is 2 MS H bar so that is why you have this matrix element to be given by more magneton times ML plus 2 MS times B now there is look at this this splitting actually depends on this combination ML plus 2 MS so there may be two different states with different ML and MS but both for both of them ML plus 2 MS can be the same number so what does that mean that means both will have same energy both will have same shift in energy so look at this combination for example so if you take ML equal to 1 MS equal to minus half this gives you ML plus 2 MS is 0 that means this level undergoes no shift there is no shift for this term for this particular state which is having ML equal to 1 and MS equal to minus half if you look at ML equal to minus 1 and MS equal to half the ML plus 2 MS is again 0 that means for both these terms the delta E is 0 all other other terms will have different delta E so that means there are two terms who are still having same energy in spite of introducing this perturbation so you have not lifted the degeneracy completely you have partially lifted it how did you lift it because I introduced an external magnetic field which broke the symmetry earlier I had rotation Hamiltonian was rotationally invariant but now it is not so so there are two pairs of ML MS which will give this shift to be 0 and here we have a table of ML and MS so if you take a P state then with angular momentum equal to 1 you can take values 1 0 1 0 and minus 1 MS can take half and plus minus half what we have done here is you have 1 and half and the 1 can be you can have 1 and half and you also have 1 and minus half you have 0 half 0 minus half minus 1 half minus 1 minus half so you have listed all these six possibilities here the reason why you have this return in this fashion is that to show that these two combinations that is 1 and minus half or minus 1 and half gives you 0 shifted energy and hence ML plus MS is 0 for this the others have different shift so two levels rise up two levels come down so you have actually five different states now six different states with five different energies so equals the spacing so six degenerate levels splits now into five levels this degeneracy is not fully lifted okay fine that is not a problem but now look at this we are going to have now we are going to arrive at the reason why we are going to see why we call this is a normal Z1 aspect people considered it considered spin to be abnormal I mean everything if there is no spin everything is fine so that is what they meant by abnormal but you look at it but we have already included a spin here so there is no reason why they should be happy and they say that this is a normal Z1 effect we will see why they still call it as normal Z1 effect now if you have a transition from Nd to Np if you have D so that for D annual momentum is 2 so ML can take values 2 1 0 minus 1 and minus 2 and because of Z1 splitting which is given by ML plus 2 MS you can have five different states here when you consider only MS equal to plus half if you consider MS equal to minus half you have additional five more terms then we have already seen that P state will have six states but if you consider MS equal to plus half only so then you have three states now here now when you have transitions between these you can actually have nine lines which are allowed by electric dipole transition you can have nine lines but do we see that don't we see that though we have nine transitions only three there are three different frequencies involved all these three bunches there is lines here one this this this all these are same energy you can see roughly from the length of these lines and these are this bunch have a different frequency line so at the end you get only three lines nine lines is that fine but now if you consider here we have considered only MS equal to plus half of both these states if you consider MS equal to minus half of both these states then you will see that again you will get the same three different frequencies because these this will be shifted downwards and these three states will also be shifted downwards the energy difference will lead to same frequencies so eventually you will see that you get only three frequencies as if there is no difference if it was MS equal to half there is a transition between MS equal to half to MS equal to half or MS equal to minus half MS equal to minus half it is very it is very much the same as the situation where there was no spin at all only the spin comes into play when it when it tries to give you a shift in the energy but otherwise there's no way to actually glean the influence of the presence of spin and win or not one more thing there is a question that we may want to ask a student what about transition from MS equal to half MS equal to minus half well the electric dipole transition that's not all over there is nothing in the electric dipole interaction term that will involve that that can mix MS equal to half and minus half there is no way there is no reason why the spin should flip okay so this is called Lorentz triplet because you get only three terms three frequencies and this very three very same three frequencies up here for MS equal to minus half to minus half transitions as well so that is why this is called normal Seaman effect so historically this is normal called normal Seaman effect only for this reason now you can see that like just like we got for D to P transition we got nine lines for three frequency you can also work out the same thing for P to S you will see that six lines but you have only three different frequencies for this and these are normal now we let us go into the intermediate case this is a Passian back effect wherein B is strong but not strong enough so that we have to we can ignore the second this term also unfortunately okay so then we can still use ML MS basis ML MS basis in this case and still they are good quantum numbers now the perturbing Hamiltonian is going to be combination of these two so let us work out so the energy shift in Passian back Seaman effect is given by two terms one is the term having my external magnetic field and which is this and the other shift is due to the internal spin orbit coupling which we represent is to spin orbit coupling so there are two components to the energy shift it is very trivial to evaluate it so because if you evaluate mu dot s then magnetic field is in along the z direction again you get the same result which is ML plus 2M 2MS and for this you need to play you need to see this the second term more closely the other term is just what we already see so what you have is the shift in the Passian back Seaman effect due to external magnetic field is going to be given by the served same term mu B B times ML plus 2MS but the shift due to the spin orbit term needs to be evaluated it is not difficult you see that you have a radial component of the Hamiltonian which actually bothers only the radial the you can actually this is actually a direct product of NL ket and ML MS ket so if you want to evaluate the expectation value of this two components it makes sense in psi of r is no way related to ML MS ket it is only related to NL okay so you can just split it into two terms so this is product of this term into this term this we can easily evaluate because this is just a function radial function which needs to be evaluated in hydrogen atom wave functions this we can do and what we need to bother about is this term what is L dot s L dot s is nothing but Lx Sx this is just a dot product of these two operators now it is easy to see that Lx and Ly and Sx and Ly can be rewritten in terms of ladder operators so ladder operator is similarly you will have the same for spin operator as well now the ladder operator has an interest so all the terms with Lx and Sx Ly and Ly you can do something and write it down in terms of ladder operators only when you do that you can easily see the following the ladder operators which are going to be here instead of these two terms will rise and lower ML and MS once it rises or lowers it is going to be when you take an inner product with ML so if it is going to rise ML to ML plus 1 then ML plus 1 MS ket will not go with ML MS because of auto morality so those terms will vanish all remain that remains is just this term finally so simply we need to bother about only this term and that is trivial to evaluate because the LZ will give you ML H bar this will give you MS H bar so we have ML MS H bar squared so this is what you have here and this particular term one can evaluate easily by once you are given the radial wave functions ok so and this is what is appearing here this Zeta of R is basically this you can work out this we are not going to do this I am just going to give you the final result here wherein these are discussed in detail in the in the video lectures as well so when you evaluate the after you evaluate the radial integral this is what you finally have you do not have to worry about L this term blowing up when L goes to 0 because when L is 0 there is no spin orbit term at all the spin orbit term is only for non-zero L so this particular perturbation shift is only for L not equal to 0 so and hence this is the split shift in energy due to only the spin orbit term so when you put them together finally that is if you put them all together what you do here is you do a rearrangement here to get to write down the energy of the against states of a hydrogen atom so then you can get it in this fashion so that the shift is dependent on N and L and ML and MS this is just an algebra you can easily work it out and let us not go into that so the perturbation energy corresponds correction depends on L, L, N and also ML, MS okay so finally putting both the shift due to magnetic external magnetic perturbation and the internal perturbation so you need to just add these two terms to get the total perturbation so once you have this total perturbation then you can for any state N N prime you can write down the shift this is the unpert energy when you do not have any perturbation this is the correction tool first order correction tool and similarly this is the same for another state N and you can always look for transitions between these once you when you look for transition between these then it exposes the presence of spin so this is why this also exposes the presence of spin because you will have delta MS equal to 0 and delta ML is 0 plus minus 1 so this is delta E is going to be transition frequencies so this we can work it out so now what we are left out with is the anomalous Zeeman effect which is anomalous because the presence of spin is evident here now it will be evident in this in the spectra so again we now we are going to consider the last two terms what happens is the spin orbit magnetic field is so low so that the spin orbit term is dominant than the mu dot beta so what we do is we can actually to begin with we can ignore this particular term and solve this identity problem by considering this to be the unperturbed Hamiltonian we are not going to do that but I am just telling you what I mean is B is in this regime B is much smaller than S dot L so we can first consider an Hamiltonian which is composed of this hydrogen unperturbed Hamiltonian plus this and we can solve this problem and we can get the Eigen function so what happens is if you have say for example when you take only these two when you have Hamiltonian composed of when you have this then you can solve this problem and you can treat this as a perturbation so what you have is say you will have one mistake and you will have unchanged because L is 0 and you will have a 2S state as 2S state also unchanged but you will have a lift in 2P because in 2P you have L equal to 1 and S equal to half they are together then you will have 2P 3 half and 2P half term you have a change in the energy you can always work out these and now what we do is consider the Hamiltonian unperturbed state and we will treat this mu dot B term which you have here as your additional perturbation okay when you try to add it I mean you can write because you already know the shift due to this term you can add the shift due to mu dot B term that's what we are going to see here now so in the anomalous Zeeman effect you have B very very weak it is important to use J square S Z term because this is we know that L dot S is we know that J's egging ket of J square Z is also egging ket of L dot S because you can write down L dot S in terms of J square L square S square now so the basis is going to be different it is not no more J M basis but it's going to be sorry it's no more M L M S basis now we have to work on J M basis now let us see how to do that now what we have is the full Hamiltonian is this we already have evaluated the split due to the spin orbit coupling now what we need to do is to evaluate treat this as the perturbation when you come when compared to this because this is much larger term and compared to this so the perturbing Hamiltonian here is mu minus mu dot B which is mu B again the same term L plus 2 S by H bar dot B but we need to be careful here because when you consider magnitude along Z direction this perturbing Hamiltonian takes this form but J M egging ket is not an egging ket of L Z and S Z operator that's when the problem comes problem is also leads to opportunities to learn more things like that's what we are going to do now we try to evaluate this so ML and MS are not good quantum numbers so that is why we are going to J M egging ket but J M egging ket is not an egging ket of L Z and S Z operator so when you try to evaluate this particular perturbation term in J M basis this is what you have because L plus 2 S is nothing but J plus S and since you have magnetic field only along Z direction you have this J plus S is just J Z S Z then you have B because B is just only along the Z direction it's a dot product of this term operator vector operator with this so now it is easy to see that J M J is anyway an egging ket of J Z so when J Z acts on this you get M J H bar so that is simple so you have mu B by H bar times B M J H bar so that H bar goes away so you have mu B by B M J for this term and what remains is this really troublesome matrix element that is we need to evaluate this S Z operator between M J M J it's looking very humble but it is going to take a lot of effort from us to grab get a grip of this so now this IV say that this particular term this split in energy is proportional to M J so we call this we put a subscript on to indicate that with J superscript and this term is dependent on S operator so we have in superscript S so now this is evaluated all what is required to do with that is to evaluate this particular term okay now we can evaluate this particular term by two different ways one is to use the vector operator identity and the other one is the beginner Eckhardt's theorem now this S Z operator both are very powerful tools if you look at this particular matrix element S Z is a vector operator Professor Deshmukh in the previous talk he explained what a vector operator is vector operator has to satisfy a certain commutation relationship with angular momentum operator if it does then that is what is defined as a vector operator so if you want to call an operator A as a vector operator it has to satisfy these commutation relationship with the total angular momentum operator J so these are if it satisfies then it is so you can easily verify that S Z is a vector operator so we can do that ourselves later so we can take as of now what we have is this mid vector operator S Z to be evaluated and we know that S Z is a vector operator so the task before us is to evaluate the matrix element of this vector operator in JMJ basis now following these definitions there are also these two identities which as of now you can take it but it is very simple to evaluate these check these identities based on this based on this commutation relationship once this is given then these two follow so let us for the moment we can just take these two identities and work ahead among these two I am going to take the second one this is what we have here look at the way I mean this is very interesting though the mathematics is abstract once you have your paper and pen doing this it is really exciting to work with this now what I am going to do is remember we are we want to evaluate the matrix element of S Z that is now we are taking a small digression but very relevant digression now what we are doing now is we have this identity before us we have vectors on both vector operators on both sides what I am going to do is I am going to sandwich these operators in JMJ base JMJ again between JMJ again gets so that is what I have here so I have bring in a JMJ bra here JMJ kit here and do the same thing in the right hand side also now look at it the left hand side is really it can be seen to be zero because JMJ kit is an eigen kit of J square operator and what we have is a commutator here so it will give two terms subtraction of two terms which are equal so this term will make LHS will vanish it is easy to really see that the left hand side is zero what you have is this term which is now equal to zero okay now this is zero but so we take them on two sides we can always do that so this is what we have before us and you can easily see that again J square being again JMJ being eigen kit of J square values J J plus J squared it is yeah J into J plus one H bar square so you can bring this J square out evaluating it so what you have is this J square acting on this just J square acting on this eventually will give you this H bar J J plus one and what remains is B plus V term and similarly here we just don't do anything here we just bring this term now look at it this is to twice the V operator so we have a two term here to here we just cancel them finally we have this simple expression here now just to remind us we are doing something but we want something else and that is the following we wanted to evaluate the SZ operator matrix element of the SZ operator between these two states but I assure you that once you know this you can evaluate it right now please be assured of that but I assure you that with knowing this so I just write this right down here so once you know this particular term you can evaluate this matrix element very quickly so I just write it down here J in J I am just dropping the J1 and J2 terms okay so once you know once you arrive at this we can do this but we will take one more digression and that is though we know that once we know this term in box we know we can evaluate the shift we will try to evaluate it using a different approach as well that is to use the Wigner-Eckart's theorem okay so you look at it you have a scope to learn so many different tools when you talk about Zeman spectroscopy so this is what we need to try to do and see this so this is the Wigner-Eckart's theorem which we know from the theory of angular momentum this is also dealt in detail in process of lectures so what we have here let us just look at this Wigner-Eckart's theorem here what we have here is an irreducible tensor tensor operator of rank K sandwiched between the Jm Jm eigencates and this is equated to two terms okay and one of this term is actually we call it as a geometric part because it involves M1 and M2 because it involves the geometry the way the system is oriented in space with respect to the external magnetic field or whatever so this is the geometric part and this particular part does not depend on the orientation of the system so this is physical part and this is the geometrical part and this physical part also has this vector operator in it so this part has just the Klebsch-Gordan coefficient what is Klebsch-Gordan coefficient there when you expand the Jm ket in an ML basis then the coefficients of the expansions are nothing but the Klebsch-Gordan coefficients now what you have we need to go through do we have times I think we have till 450 but I can take 5 more because going to the Wigner-Eckart's theorem we need to be more patient and pay a lot of attention to this now look at this what you have here if the geometric part is identified this is the Klebsch-Gordan coefficient because it is nothing but the projection of Jm eigen ket on ML M1 into eigen ket now this is the Klebsch-Gordan coefficient when you add J1 and J2 that is J and K to get J' so two angular momentum are added which are J1 equal to J, J2 equal to K to get this total angular momentum J' now it is the same equation that you have now what we do is when you have K equal to 1 that represents a vector operator okay so when you have K equal to 1 we represent the vector operator and now what we do is we replace this vector operator T with the angular momentum total angular momentum operator and this is what you get that is the Wigner-Eckart's theorem for the total angular momentum operator now what we do here is we if you replace Q if you make Q to be 0 then you have if you why do we do that because we are interested we know that this JQ when K is 1 it represents vector operator and Q can take three values minus 1, 0 and plus 1 when Q is equal to 0 this is the element of the vector operator and remember we have been behind this Z operator evaluating the matrix element of Z operator and hence we need to look at the case of interest so we just put Q equal to 0 here so when you put Q equal to 0 here this is what you get and simply this is nothing but JZ for the total angular momentum operator because K is 1 so it is a vector operator so what we have is this particular form for matrix element of JZ operator so this is what we are going to evaluate so if you look at this JZ is a JM egg and ket is an egg and ket of JZ operator so when JZ operates on this it will give you an MH bar and this MH but then you will have JM taking in a product with J prime M prime which will vanish unless J and J prime are same and M and M prime are same so that gives you a chronicle data and of course alpha also have to be same because otherwise in that space you will have orthogonality will also give you a vanish in term so you have MH bar and 3 chronicle data to ensure that this is 0 or non 0 accordingly now so this is the equation before us this is the Wigner-Kahr's theorem now this particular term is the clefcord and coefficient when two different angular momentum are added where in J1 two different angular momentum J1 equal to J and J2 equal to 1 are added how do we evaluate this so the task before us is to evaluate the clefcord and coefficient there are tables available for this clefcord and coefficient you can download it from the internet it is also in the standard books so the task before us is to find out this particular term now look at it in this particular term you have J1 equal to J J2 equal to 1 and that is what is here J1 and J2 are the two different angular momentum that are added so here J1 is J and J2 equal to 1 M1 is M and M2 is 0 now this is the clefcord and coefficient for clefcord and coefficient table for clefcord and of this similar type here so what we need to do is as we just discussed M1 is M M2 is 0 here so you just have to look at this particular term column because M2 is 0 and you look at it here you have J and J prime have to be same okay so J and J prime have to be same other way this term will vanish there is no meaning for this so J and J prime have to be same that means J should be equal to J1 which means that I need to take this column and hence the value of interest in the clefcord and coefficient table is M by root of J1 into J1 plus J so this is going to be the value for this clefcord and coefficient table and take them okay one can also evaluate this rigorously individual terms on this column but since the tables are available you can just look at them and pick the clefcord and coefficient so what you have is this particular term is nothing but this term so we have just merely write it down here so we have identified this to be this so your Wignacar theorem finally becomes this particular clefcord and coefficient from table to be M prime by root of J into J plus 1 now if you look at this term you have two terms under square root one is 2J prime plus 1 the other one is J into J plus 1 and if you yeah so this is what by rearranging this you get this particular term but there is a discrepancy there is a difference in two different books which Praveeshmu has mentioned also talks about it in this lecture that is in one of the book which is Branson and Joe Shane you don't see this 2 into J plus 1 term but that is fine because this particular Wignacar theorem is nothing but a definition of tensor operator okay so one can actually observe this 2J plus 1 term in this term so let us not bother about it right now so what we need to worry about is the blue color item and the blue color box so let us do that so this is what we have before us now what I am going to do is I have I write down the same Wignacar theorem for a different operator which is tensor operator which is V and I take this ratio once I take this ratio this square root term will vanish so it doesn't come into picture now what we can do is we can rearrange this we can take this particular matrix element to the right hand side and what we see here is two terms one term which is not dependent on M1 or M2 the other term has one term depends only on the physics of the operator the other term is there are two terms in this both yeah so what we see here is this particular ratio okay can be is a constant and you have this particular term here so by taking these two ratios we can arrive at this particular result which is what we have here now what we do is now that we have this result remember we wanted to we had terms like v.j so we will evaluate this so this v.j matrix element of v.j can be evaluated by introducing identity operators in between which is a outer product of jm with itself so the summation over all the basis will give you one so when you introduce that you will have v.j expressed in this term now you know that this particular term is expressed as c times this and we take the advantage of this because we have this term here so we replace this with which is nothing but this term as c into c into this particular matrix element so what you have before you is the v.j matrix element is equal to j square term because you have j.j which is just the j square and the eigenvalue of j square which is j plus one so finally you can arrive at the projection theorem I think you can work it out if you have more time you can work it out it is not difficult to get this to this projection theorem so the projection theorem is what you have here now we are very close to the result now once we have this projection theorem which relates any look at this projection real theorem relates matrix element of any vector operator to the matrix element of the angular momentum operator so if you substitute m prime equal to m so you have this equation now what you do is look at it so what you have is if you take since this particular term is valid for any value of q okay you can this is same for any value of q so you can write it in general for v vector operator it can stand if you have here q equal to one then you have j q where q equal to one so you can write it down simply like this then we do a rearrangement of terms so we have this term into this is j into j plus one h square which is equal to this now what did we do now we have arrived at this result and I promised you that I showed you that once you have this we will be able to evaluate the matrix element all we did by taking the second migration is that to arrive at this using Wigner-Eckard's theorem okay so for some time we can just get relaxed for the take yourself away from Wigner-Eckard's theorem because we have already used Wigner-Eckard's theorem to arrive at this result now just to remind us we wanted this how are we going to use this result to get that is how are we going to use this result to evaluate this it's very simple look at it we have v operator and we have j operator here what about the z-comparate of the v operator you replace v as s operator and this is what you get then if you are interested in the z-comparate of this you will also be interested in the z-comparate of this so this is the equation that you are having before you and this is what this is the matrix element that you were behind all this while and this is very this is very simple this is equal to this term by this and what is this j z j mj ket is an eigen ket of jz and that will give you mj h bar okay so what you have is mj h bar times this this is the LHS sorry RHS then you have this and we can write down j dot s in terms of j square L square and L z s square and all these three operators or eigen operators were j mj kets so you can when j square operates on this you get j into j plus 1 h bar square when this operates on this you have L into L plus 1 h bar square this operates on this you get s into s plus 1 h bar square so this is very clearly comes to this so finally you are evaluated the matrix element for s z operator in j mj basis and that is proportional to mj and terms involving j L and s one and numbers now so this is just the what is this term this is just the term this is just the perturbation due to the magnetic field we also have a perturbation due to this pin orbit coupling I know I am sorry this is the this is also due to the magnetic field but it is just one component one part of it so so when you add both these together so this is the shift due to Zeman anomalous Zeman effect because of this term the other term which we have already derived previously so putting these together you get a complete Zeman shift and where this particular term is we have already evaluated so you can simplify this to get the shift to be proportional to mj and the proportionality factor includes the Bohr magneton and this particular term the square bracket which we call which is nothing but the Landau's g factor which depends on j L and s so you can see that the shift in the anomalous Zeman effect is proportional to the magnetic field and mj for so if you are given for a given state with j and L is known all you have is shift for various mj the shift is proportional to mj so this is the simple expression that we get for the anomalous Zeman effect and just to remind us the source of magnetic moment was both from orbital angular momentum and the spin angular momentum and we can rearrange this again because j is nothing but addition of orbital angular momentum and the spin angular momentum you get two terms L plus r and L minus r so you can replace j for that so this is the final so when you consider coming back to this np half state why did we have p state split into two because of spin orbit coupling the spin orbit coupling gave us split and we had two regions we lifted and we had p half and p three half state having two different energy but p three half three half is nothing but the total angular momentum value and corresponding to that you have four mj values three half half minus half minus three half and this for this total j value you have again mj values half and minus half and we know that this splitting is proportional to mj and hence you have these the three half state the degeneracy in three half state which was initially degenerate in mj value so that every different state states with different mj value will have different energy now you can look at transitions between them so now this was really anomalous because this cannot be reconciled unless you unless you take spin into consideration so that is so then and hence if you consider the d line which is from three half you know sodium line has two you have two lines one is from the three half p half to s half the other line is from p three half to s half but this two lines the presence of magnetic field was split into ten lines and that is what you have because of the anomalous human effect this is it can be explained only when you consider spin into consideration do we go for the intensity you know now just a comment here we do not have to go into this we to be frankly we though we use dignity theorem just as a tool we we did not exploit its advantage completely but one can do it suppose if someone ask you a question you get d ten lines in this spectra when you have external magnetic field and someone is asking you what is the relative intensity of these ten lines then it really depends on evaluating it depends when you consider say line five and line ten suppose someone is asking you these two inter intensities then it depends on the transition this transition probability the transition which leads to line five actually depends on this particular matrix element which is initial state interaction Hamiltonian sandwich between initial state and the final state so for this state this which you can evaluate for both these states and depending on the value of this the relative strength of this particular term for these two states the intensity will vary accordingly and it is really one has to be a mass searcher to discard Wigner-Kart's theorem and go ahead and calculate this particular term and evaluate the relative intensity but one can see from the lecture there are a few more slides on this you can see that with Wigner-Kart's theorem you can simply see in a moment that the intensities have to be so all you need to do is look at the class code of coefficient and take the ratio of them this will require some amount of pen and paper work and I definitely would tell you to encourage your students to do this and we will stop here if you have any questions prior to this we were saying that you represent the state of the system by a vector but space and all that seems so abstract but this is what immediately connects the state vectors and operators to observations because spectroscopy is all about measurements and experiments and the only way you can interpret your observations is by doing quantum theory so it's not something which belongs to the realm of abstract mathematics but it relates directly to measurements and the family of spectroscopies which come under the demon spectroscopy or starship act or anomalies you can actually get the frequencies of the sector lines you can get their intensities using these theorem so this is what we learn from Dirac that as you develop the algebra of quantum mechanics further you connect them to physical observations which is what quantum theory is for what do we normally observe normally the magnetic field well it depends on the conditions of the moment the conditions of the experiment as he pointed out because it is the strength of the magnetic field if it is very strong then you will observe what is typically called as the normal moment but that is just a matter of terminology nothing else anything else is abnormal so this is normal because that's what it was considered to be normal when they did not know about electron spectra