 Okay, this is where we stopped in the last module, we are talking about second order nonlinear phenomena and we have arrived at an expression for the second order polarization and we see that it consists of two terms, one is a constant polarization which is called DC effect and the second one is the term that is responsible for second harmonic generation and the reason why we say that this is the term that is responsible for second harmonic generation is that it is associated with an angular frequency 2 omega 1 which is double the frequency of what we had started with, okay. Now let us go ahead and see how to actually get the second harmonic generation but before that we have unfinished business from the last module, we have to talk a little bit about this k vector, now to do that I want to show you the expression of something that I think we have discussed earlier phase velocity, phase velocity V photon is given by omega by k which is equal to C by N, what is N, number of photons of frequency omega, no, what is N, we hint N also goes by the symbol mu, refractive index, refractive index nothing else, why do I write refractive index and then I write omega n bracket because refractive index is color dependent, frequency dependent, remember this, this is something that we are going to build upon in the subsequent discussion, okay. So this is the expression for phase velocity and it is going to come handy in a minute. Now the reason why we are even talking about the k vector again without going into too much of discussion of optics as such is that the k vector tells you what the momentum of photon is, momentum of photon is P equal to k h cross, see h cross is the unit of everything in quantum mechanics is not it, what is the angular momentum, if you ask that question hydrogen atom let us say, angular momentum is root over L into L plus 1 into h cross, what is the z component of angular momentum, M into h cross, right even when you talk about NMR spectroscopy you get this h cross everywhere, this is a unit. So what we see is that this k then is a number that gives you an idea of momentum of photon, okay, right. Now comes the important discussion again more or less axiomatically without going into the math, it is important that we know this that phase matching condition, condition that has to be satisfied and we will see why it has to be satisfied for getting second harmonic generation is that the momentum well here this expression is written in the in a little more general term where we are saying omega 1 and omega 2, two different kinds of photons are combining to give you some frequency generation, okay. So for some frequency generation the phase matching condition is k 3 which is the k vector of the some frequency photon has to be equal to k 1 plus k 2 where k 1 and k 2 are the characteristic k vectors for photons of angular frequency omega 1 and omega 2, what is the meaning of this, what is the meaning of this, to understand the meaning of this first of all let us say we are working with collinear beams, do we use collinear beams for some frequency generation or second harmonic generation, for second harmonic generation in any case it is collinear, right it is one beam that goes in for some frequency you can have a collinear geometry, you can have a non collinear geometry. So for now let us only talk about collinear geometry, so if it is collinear geometry then things become a little easier, why does it become easier because these k vectors all have the same direction, we do not really have to worry about what is the angle, what is the direction so on and so forth, collinear means all k vectors are aligned. So for a collinear beam what will happen, what is k from this phase velocity definition k multiplied by c is equal to omega multiplied by n, yeah omega multiplied by n of omega and I am not worried about the direction because it is collinear, so what can I write instead of k3, instead of k3 I can write omega 3 multiplied by n for omega 3 that is the left hand side, what will I have on the right hand side, n1 multiplied by omega 1 plus n2 multiplied by omega 2, okay, okay, but even well before going further what is the implication of phase matching, k vector of the sum frequency, sum frequency light is equal to sum of the k vectors of the combining frequencies, what is the implication of that, once again the answer is there in front of you in the projection, what does k stand for, what does k define which property, momentum, so are we not saying that momentum has to be conserved in some frequency generation that is the physical meaning, all we are saying is that when 2 photons combine they should combine in such a way that the momentum well momentum is conserved, momentum of the sum frequency photon should be equal to sum of momentum of the combining photons that is all we are saying and then it boils down to for collinear beams it boils down to something very simple n3 omega 3 is equal to n1 omega 1 plus n2 omega 2 and it becomes even more simple if you talk about second harmonic generation, in second harmonic generation omega 1 equal to omega 2, yeah, so n1 equal to n2 what does the right hand side become second harmonic generation omega 1 equal to omega 2 equal to say omega and n1 equal to n2 equal to say n1, n1 omega 1 right, what is the right hand side then, 2 multiplied by n1 omega 1 what is the left hand side what is omega 3 second harmonic generation, second harmonic generation 2 omega 1 right, so left hand side is 2 n3 omega 1, right hand side is 2 n1 omega 1, so 2 and 2 cancel and you are left with this condition that diffractive index for 2 omega 1 sum frequency is equal to diffractive index of the incident frequency, okay, but a little while ago Sharma sister told us that diffractive index is a function of omega, so do we expect in a crystal solid medium do we expect the diffractive index to be the same for omega 1 and 2 omega 1, so see where we have got, we are saying that you can get second harmonic generation only when diffractive index for omega 1 is equal to that for 2 omega 1 on one hand. On the other hand we know for ourselves that in a medium usually the diffractive index is a function of frequency, so diffractive index of omega 1 for omega 1 should not be equal to diffractive index for 2 omega 1 right, which means that second harmonic generation cannot be done and we can go home to get the point, what we have shown here is that an essential condition for second harmonic generation is that the diffractive index indices have to be the same, diffractive index has to be the same for incident frequency and second harmonic and on the other hand from our conventional knowledge we are saying that it is not possible, which means it is not possible to get second harmonic generation unless we have material in which somehow your diffractive index is the same for omega 1 and 2 omega 1 under some conditions ok, that is why not everything can give you second harmonic generation. So, as you are going to discuss it is possible to achieve this condition only for birefringent crystals, what is the meaning of birefringent crystals, what is the meaning of birefringent, y is 2 then birefringent yes, so it has a simpler sounding name actually birefringent goes by another name, which you actually just said almost double refraction ok, we will come back to birefringence shortly, in fact we might have forgotten but all chemists have read about birefringence, if not in 11 and 12 definitely in first year BSc, that too not in physical chemistry, organic chemistry yeah, tartaric acid, what does tartaric acid do or you might have even done some experiment, polarization remember rotation of polarization and all that, polarimeter what do you use, what is there inside a polarimeter other than lenses and all, don't you have some birefringent crystal, what is it calcite calcite cube cut diagonally and joined by Canada balsam yeah, so that is where we have all encountered birefringence, we are not saying anything new, it is as if we might not have used that term, so we might have forgotten ok, so we will come back to see how elegantly birefringence can be used to get second harmonic generation and we will come back to see why it is that we take that birefringent crystal, what is the magic that we turn it like this and all of a sudden we get second harmonic, we will get to get to that, before that for the sake of completion of this discussion, let me show you the expression, we will not derive this, if you are interested there is plenty of resources are available, two things since this is for the sake of completeness, the psi of psi which is a function of x and t equal to a cos kx minus omega t plus phi is just a little more elaborate way of writing the expression we have written already, instead of e we have written psi here, this is another way of writing, so if you read some literature where suddenly you see psi do not get scared, sometimes we get scared the moment we seek the greek alphabet, there is no reason ok, second thing is what is phi, yes we did not write it earlier, we assume that phi 1 equal to phi 2 that is not very difficult to achieve, but look at the intensity expression for intensity of the second harmonic, intensity of second harmonic is equal to some constant k multiplied by square of the second order non-linear susceptibility multiplied by L square where L is the length of the crystal multiplied by our old friend, sin delta k L by 2 divided by delta k L by 2 square, where have we encountered this, this function does it ring a bell sin square theta by theta square, where have we encountered this everywhere, we have encountered it when we talked about not in this course, when we talked about interaction of radiation with matter, this is what appears in the probability of transition from one state to the other and in this course we encountered this kind of expression, sin square theta by theta square when we talked about the shape of the pulse, modulocking remember, modulocking we encountered sin square theta by theta square, so what does this function look like, sin square theta by theta square where does it have a maximum at 0, but then there is a problem right, because at 0 theta equal to 0 of course theta equal to 0 and sin theta is also 0, so you have 0 by 0 which is not defined, then why are you saying that it has a maximum at theta equal to 0, because the shape is as you approach 0 from both sides this is the shape, yeah so you can close as you can as close as you want to 0, just for that one point where theta equal to 0 it will be undefined, but you can keep going close close close close as much as you want and the limiting value will be 1 sin square theta by theta square, so this is called a removable discontinuity and you can take the limit of 1 for that one point where theta equal to 0, okay, now remember what it looks like, it starts from 0 maximum goes down and then it has some ripples that slowly die off, remember that shape something like this, so where will this be maximum, okay we will come back to that, then that is multiplied by I0 square of omega, square of intensity of the incident light, so let us take factor by factor, so first of all in order to get I2 omega you should have a high value of second order non-linear susceptibility, point number 1 and then it is well known that if there is centrosymmetry then that second order non-linear susceptibility is actually equal to 0, so first thing you look for is that you must have a crystal which does not have centrosymmetry, point number 1, point number 2 and then of course it has to be high also, whether it will be high or not that I cannot really discuss so qualitatively, second thing is what is the second factor, L, so L should be large, but then it cannot be too large also, I think it is better to use a big crystal, so I try to use a one foot long crystal, that is not a good idea, first of all if you are using pulse laser it is going to get broadened, secondly it is not so easy to grow a one foot long crystal anyway, millimeter long crystal is extremely difficult to grow that is why they cost something like 50,000 rupees, 1 lakh rupees, 2 lakh rupees something like that and thirdly there is a problem, there is something called a coherence length, coherence length roughly means for which length omega and 2 omega are in constructive interference, see problems you have 2 kinds of wavelengths, omega and 2 omega, initially they are in phase, as they go out then you will have destructive interference, so if you go beyond a certain length and in fact there is something odd multiple of coherence length or even multiple of coherence length it becomes 0, destructive interference is there, so one has to take care of coherence length as well, then we are going to discuss later that we can do angle tuning to get phase matching, so delta K has to be equal to 0, this is what we have discussed already, we said that condition for second harmonic generation is that delta K must be equal to 0, momentum must be conserved, this is why momentum has to be conserved, because when delta K equal to 0 that is when the sin square theta by theta square function has a maximum value of 1, that is why delta K must be equal to 0, what about this I0, if I0 is large will intensity of second harmonic be large or small, large definitely but then like everything else in laser spectroscopy too good is no good, if your intensity is too large then your optics will burn, so it has to be large but not so large that the crystal will burn anyway, okay and then one more thing that we are going to see why this happens, polarization of the second harmonic is going to be perpendicular to that of fundamental, this is not really included in this expression but we will see shortly in the next 10-12 minutes why that is going to happen, in the next module we are going to take it from there, great. Now let us go a little bit slower and see why we need birefringent crystals and how you can get phase matching using a birefringent crystal, remember it is not enough to be birefringent right, it should be birefringent and it should have a high value of second order non-linear susceptibility then only it will work, so there are very few materials that fulfill all the conditions, earlier it used to be KTP, KDP all these things, now it is BVO all the way by and large, okay great. So let us say that this rectangle is the crystal and let us say this arrow here is the optic axis Z, what is the optic axis? So let us hold off our angle of incidence is actually correct, well I would say direction of incidence rather than angle at the moment but it has something to do with direction of incidence but let us wait a little bit we will come back to this and then let us say this is the incident ray, omega 1 okay, now see this is a birefringent crystal, what is the meaning of birefringent? The moment this incident light enters the crystal it breaks into 2 parts, 2 rays, does anybody remember what these rays are called? Lantian Organic Chemistry or in 11-12 physics, physical optics, ore is an E-rays does that ring a bell, ore and E-ray, I am not calling anybody these are names, ore means ordinary ray and ordinary ray is the component of the incident light that continues in the same direction okay, if there is an angle of course it will be refracted that is a different issue, E-ray is the part of the light that goes in a different direction, this is called extraordinary ray, why ordinary, why extraordinary we will see in a moment but for now who did this experiment, does it look like, so I will tell you hygens and somebody else I forgot the second name, hygens, heard of hygens, hygens was one of the pioneers of physical optics long ago he did all these experiments that have to do with wave nature of light and all, so hygens did this experiment, so I do not exactly remember which century it was but it was a long long time ago, definitely not 20th century, much before that I think 19th century maybe okay, so hygens and scope workers are actually shown that what happens is this ordinary ray if you use polarized light, ordinary ray has the same polarization as the incident ray, however the extraordinary ray has perpendicular polarization, also the refractive index of the ordinary ray and now comes why it is ordinary, why it is extraordinary and this is the most crucial part of the discussion, refractive index of the ordinary ray is the same no matter what the angle of incidence is, of course refraction will be there because the light is passing from air to a denser medium but it is the same so refractive index of the ore is independent of the incident angle, however refractive index of the e-ray is dependent on the incident angle okay, I am just stating facts we are not proving anything, this is something that has been known for 150 years or more that if you are birefringent crystal then incident light divides into two parts in the crystal, one has polarization same as the same as that of the incident light, the other one has perpendicular polarization, first one is ordinary ray, second one is extraordinary ray and refractive index of ordinary ray does not depend on the angle of incidence, refractive index of the extraordinary ray does depend on the angle of incidence, we take a break here and complete the discussion in hopefully significantly shorter module that is to follow.