 We are discussing particle in a 1D box and we have succeeded in so far learning about the wave functions of a quantum particle in a 1 dimensional box. The wave functions have turned out to be sine waves and we have seen that only some will be allowed. The wave function that we have established so far are psi of x is equal to root over 2 by L, sine n pi x by L where n equal to 1, 2, 3, 4 so on and so forth positive integer. And we have established that this set of wave functions well we have normalized that is how we got root 2 by L. We have also established that they are orthogonal to each other that is where we stopped in the last module. And we have also shown that the first derivative is not continuous at x equal to 0 and x equal to L. So, that really is not such a stringent condition on a wave function. The next step is to understand what are the energies. Since only certain wavelengths are only certain wave functions are allowed only certain energies will be allowed as well. So, let us see how we get to that. This here is the expression for energy for a free particle if you remember. And this is the energy that our particle would have as long as it is inside the box and it can be only inside the box equal to h cross square k square by 2 m. But then we have also learned from the boundary condition that psi must vanish at x equal to L that k L is equal to n pi and n is a positive integer. So, the obvious next step is to take this k L in equal to n pi expression and plug it in to the energy expression of a free particle. The moment we do that the energy expression becomes E of n is equal to n square h square by 8 ml square where n is a positive integer. What just happened? Well, quantization of energy happened. Now, we are saying that only certain values are allowed and now we should say something that we deliberately did not say while discussing the wave functions in the last module. What is n? n turns out to be a quantum number for the particle in a box. And this is the beauty of Schrodinger's treatment. Unlike Bohr's treatment where the quantum numbers fell from the sky, quantum numbers arise naturally simply from the probabilistic interpretation of the wave function provided by Max Born. Here we see an example of energy like energy, other quantities like momentum, angular momentum, these are also quantized and everything arises from the imposition of some boundary condition or the other. So, now we learn a very, very important lesson in quantum mechanics. We learn that quantization finds its origin in the boundary conditions which we know arise out of Born interpretation. This is very important to understand. I have seen many students who have got MSc in chemistry, good marks and all but and they can do the math associated with quantum mechanics. But this very fundamental concept has sometimes not sunk in. Let us all be very clear about this. This is one of the founding tenets of founding principles of quantum mechanics. That quantization really arises out of boundary conditions. The moment that happens we can breathe easy. Now our conscience is clear. We have not imported a quantum number from anywhere just to satisfy some experimental observation somewhere. It has arisen naturally out of boundary condition. This is a great revelation but that is not all. What we see in particular in a box is that look at the energy ladder. The separation in energies increases as you go higher and higher up. That is one thing. The second thing is another important observation, zero energy is not possible because the smallest value of n the quantum number is 1. And for that energy has a value of h square divided by 8 ml square which is non-zero. This brings us to the concept of zero point energy and this is something that becomes a governing factor when we discuss simple harmonic oscillator. We find that a free particle, a quantum free particle can never be at rest and it can never be at rest because if it is at rest it is going to violate uncertainty principle. Let us see it comes to rest. It would have come to rest at some position. So, X is defined precisely. Delta X is 0 and since it has come to rest completely the momentum is also 0 plus minus 0. So, delta X into delta P X is 0 that violates uncertainty principle. Uncertainty principle we are going to talk a little more about uncertainty principle later on when we take a break from this quantum mechanical systems and discuss the operators and demonstrate uncertainty principle nicely using the position and momentum operators. But then the thing is uncertainty principle is a law of nature. It is a boundary beyond which one cannot probe nature. So, it is not a question of making a better equipment that will let us see something you cannot see today. We can never do it as far as the current understanding of quantum mechanics is you cannot do better than the best. You cannot violate uncertainty principle and that is why a free particle can never be at rest because that would violate uncertainty principle since it would be associated with 0 uncertainties of conjugate properties. That is another important revelation from particle in a box. Now, let us talk about the energy gaps between successive levels. What about that? If you take any energy gap well right now it is E f minus E i. So, n f square minus n i square let us say I take n f square to be n i square plus 1. We can work out what the gap will be. Let us keep it simple. Let us say n equal to 1, n equal to 2. 2 square is 4, 1 square is 1. Whatever it is no matter which pair of water numbers you take you get a constant. For that constant what is the value of h square by 8 ml square if I change L. If L is increased since L square is in the denominator energy gap between the two chosen energy levels is going to decrease. Larger the box smaller is the energy gap. Actually should not have written h 2 year that would should come after another piece of discussion that we have coming up but well bear with me until then we are really talking about the energy gaps. So, if L is small then energy gaps are larger. If L is large then energy gaps are smaller. What does that mean? If you keep on increasing the size of the box the energy levels keep coming closer and closer and closer. So, beyond a certain value the energy levels will come so close to each other that you cannot tell between this energy level and the next one. So, you do not have discrete energy levels anymore rather you have a band that would be the classical limit. So, this is another beautiful thing that comes out of our discussion. Just by changing the dimension of the box one can go from quantum world to classical world and back and it makes perfect sense in both the worlds. This is another point of strength another reason why we gain confidence in this treatment. Now comes a discussion that should have come before I could write h nu. Let us talk a little bit about the spectroscopy. So, what is spectroscopy? Spectroscopy is the interaction of radiation with matter and essentially it involves transition from one energy level to the other and to be honest we are not spectroscopy is something that is not at all new to us because after all quantum mechanics is something that has sort of arisen out of spectroscopy. I am deeply indebted to a colleague who many many years ago told me that spectroscopy is just quantum mechanics in action. All these calculations that we do in quantum mechanics is manifested in interaction of radiation with matter because that is the only way in which you can probe the energetics of a system experimentally and quantum mechanics is all about the energy levels and wave functions and so on and so forth. Now see just because there are two levels it does not mean that a transition will necessarily take place. In spectroscopy something that is very important is selection rule and selection rule which tells us which transitions are allowed and which transitions are not arises from something called transition moment integral. Transition moment integral is integral psi 2 mu psi 1 and the condition is that this transition moment integral must be non-zero. Now I think we might have introduced this nomenclature earlier as well but just to make sure in case we have not let me just do it once again what we have written here is this. In Dirac's notation this is called ket vector sorry bra vector and when we write a wave functions inside it let us say I write psi 2 in a ket vector bra vector bra vector it essentially means psi 2 star complex conjugate of psi 2. Of course in our case we do not really have to worry about complex conjugates because our wave functions of particle in a box are all real this is bra vector. If you write like this so this is psi 1 in what is called ket vector and what it essentially means is you can just write psi 1 itself. A lot of complicated discussion becomes simpler if one uses bracket notation in our case it is almost trivial we might as well write the integral but it at least saves us a hassle of writing d tau and all that all the time so we will use bracket notation but I should tell you what this means what we have written here. When you write like that so when you write bra psi 2 ket psi 1 then it essentially means integral over all space psi 2 star psi 1 d tau in our case d tau can be should be replaced as dx because you are working in one dimensional space. Now one more thing it is possible that I might want to multiply this psi 1 by something or make some operator operate on something what we have done is that we have used the operator dipole moment operator which means basically multiplication by dipole moment so this is what it means what you might notice that there is an additional vertical line after mu like this that is just to make things look good it still means bra psi 2 ket psi 1 ket mu psi 1 this is what it means this is your transition moment integral and the condition for transition moment integral is that for transition moment integral to be non-zero is what we have to determine if you want to know which transitions take place and which transitions do not take place. For that purpose we are going to use here the property of symmetry of wave functions one can of course plug in the expressions for wave functions and do it the hard way work out the integral and that has to be done many times but many times you do not have to go through all that hassle symmetry is a fundamental property of systems which determines many of its aspects. So here we will see how nicely one can use symmetry arguments to determine whether this integral is 0 or not remember we do not really need to know at this point whether what the value of the integral is if that is what we require then of course we have to work it out but if you only want to know whether it is 0 or non-zero we will see how we can do it nicely with symmetry to do that let us first recognize that all these wave functions where n is odd this one or this one n equal to 1 n equal to 3 these are symmetric with respect to inversion what is the meaning of symmetric with respect to inversion if I say this is Psi 1 right so I write something like this Psi 1 and then I invert it so I interchange 0 and L in this case let us say then upon inversion it will remove Psi 1 it will remain Psi 1 that is the definition of being symmetric or what is called an even function however look at this n equal to 2 or this n equal to 4 what happens if I interchange 0 and L the function is going to change shape right now this is plus Psi 2 if I interchange 0 and L you are going to have a function that essentially will look like this so upon inversion what happens for n equal to 2 4 etc where n is even we get Psi n becomes minus Psi n and that makes it antisymmetric with respect to inversion and such functions are called odd functions let me digress a little bit and point out that the number of nodes remember what nodes means nodes mean nodes node means a point where the wave function go through 0 and change sign so this is not a node this is so number of nodes is n minus 1 when n equal to 1 there is no node number of nodes is 0 when n equal to 2 number of nodes is 1 when n equal to 3 number of nodes is 1 here and 1 here so that is something that is just there we should know it now coming back to our original discussion what how do we know whether this integral is going to be 0 or not well let me write on this side so that I do not have to erase it later on see an integral is nonzero when let us call let me call this I the integrand this one is I so what happens if I change a sign upon inversion let us say upon inversion I becomes minus I what happens to the integral the integral is I d tau overall space that becomes minus I d tau but the issue is we are talking about a transition between two levels that cannot change depending upon which one we decide to be which point we decide to be 0 and which point we decide to be L just upon inversion the sign cannot change so when will this happen in any case if this is equal to this we are saying that the integral cannot change upon changing sign so that means integral I d tau must be equal to minus integral integral I d tau must be equal to minus integral I d tau when will that happen when both are equal to 0 is not it say q equal to minus q only when q is equal to 0 otherwise it is impossible so what we learn is that this I has to be symmetric and I here is a triple product two wave functions and your dipole moment is dipole moment symmetric or is it anti symmetric let us not forget that we can write dipole moment as mu equal to e into x where e is electronic charge x is displacement x is it symmetric or is it anti symmetric well we have said this already that a symmetric integrand is required for the integral to be nonzero we are now asking the question what about x is it symmetric or is it anti symmetric what happens when we invert x becomes minus x naturally so x is definitely anti symmetric so now when will this triple product be symmetric so we are doing inversion right we take this and do an inversion when we do an inversion x changes sign x becomes minus x let us say psi 2 is symmetric so psi 2 will remain psi 2 now if psi 1 is also symmetric what will happen then this triple product changes sign which means the integral changes sign which means the integral is 0 how can one avoid it only when psi is anti symmetric so if psi 2 is symmetric psi 1 would better be anti symmetric sorry for my horrible handwriting then what will happen is that due to x you have a minus sign due to psi 1 also you have another minus sign and the whole thing becomes plus so what we are saying is that if one wave function is symmetric the origin or the destination then the other the destination or the origin respectively has to be anti symmetric that is the condition for the transition moment integral to be nonzero that is the condition for the transition to be what we call in the language of spectroscopy allowed so the selection rule then is delta n equal to 1 3 5 so on and so forth why because if you remember the wave functions alternate wave functions are symmetric and anti symmetric n equal to 1 symmetric n equal to 2 anti symmetric n equal to 3 symmetric again n equal to 4 anti symmetric once again so delta n equal to 2 can never happen what happens if I want to go from say psi 1 to psi 2 psi 1 is symmetric sorry psi 1 to psi 3 psi 1 is symmetric x is anti symmetric psi 3 is symmetric so psi 1 x psi 3 the whole thing is anti symmetric so that integral vanishes and that transition is not allowed now if you want to know where this transition moment integral comes from that requires a little more of quantum mechanics in fact we have discussed it in some detail in an earlier course that we had floated in on NPTEL course on molecular spectroscopy of physical chemist perspective so whoever is interested in knowing more about where transition moment integrals come from can go through those videos of the course of that course which are now freely available on YouTube I believe but for the purpose of this course at least for now we are going to take this transition moment integral business axiomatically and we are going to proceed from here so doing that what we have learned is that for a particle in a box delta n has to be 135 odd to even even to odd transitions are only allowed which means that you can have this transition 1 to 2 you can have this transition 1 to 4 but you cannot have this transition 1 2 3 why not because as we said several times 1 and 3 are both symmetric making the triple product which is the integrand of transition moment integral anti-symmetric so this is the discussion we wanted to have on spectroscopy of particle in a box and this is what tells us which transitions take place and which transitions do not with this background we are now prepared to go ahead and talk about color and we are ready to talk about whether or not there can be some application of the simple particle in a box in real chemical systems.