 we actually looked at stationary states of energy. So, when I say stationary states of course, it is stationary states of energy which means solution of H psi equal to E psi for a time independent problem. So, all the problems that we have done are time independent problems. So, the problems that we actually discussed are one was particle in a box. Remember I am not going to go through them that is reminding you what all we did and all of you know the results of the particle in a box you get the quantum numbers which are and the energy is a n square h square by 8 mL square where L is the box length. So, this is for 1D box if you have 2D and 3D box similarly you will have more and more quantum numbers and energy will be some of the x, y and z. So, that you have done and of course n is equal to n starts from 1, 2, 3, etc. And the wave function is for the 1D box again wave function is some constant times sin n pi x over n. So, for 1D box you can write down the 2D and 3D box trivially by invoking the non-interacting theorem that I stated yesterday. What is the non-interacting theorem? There if you have Hamiltonian is sum of two operators or three operators wave function is a product of course for electrons they will become anti-symmetrize product that is a different matter but right now there is only single particle. So, it is just a product of x, y and z direction energy is a sum. So, basically invoke the same non-interacting theorem and you can generate the total wave function. I hope all of you remember y n cannot be y n negative is not written and y n equal to 0 is not written. I hope all of you know this. Somebody told me y n equal to 0 is not allowed in the derivation of the particle in a box. Wave function would not be a number, wave function will be 0. No, I did not understand why it is a number means? Number is a proper wave function. Oh, 0 is not just a number, 0 means the wave function does not exist. There is no probability of getting the particle, there is a serious consequence. Number is different if psi is just constant it is okay at least if it is not constant over entire space then it is not okay because then it is what is called not normalizable. So, otherwise it is okay but it is 0. So, that is not number is a really incomplete way of saying giving the answer please understand. It is 0 because of course you have a probability finite probability why it is not negative? Because if I take n is a negative this is nothing but negative of psi n. So, any psi n if psi n is a Eigen function minus psi n is automatically an Eigen function. So, that we do not have to take. So, the point is that if H psi equal to S E psi then any H psi is also an Eigen function with the same Eigen value but that is not a linearly independent Eigen functions because it is just a multiplication of a psi. So, any of these will be an Eigen function. So, we do not take this into account. So, that is the reason negative n is not allowed. Is it clear to everybody? In the solution of this particle in a box we encountered one important goodness condition and that was called the continuity of the wave function. So, I want to tell you that there are certain important points that the wave function must satisfy. One was a continuity of wave function that the wave function must be continuous over entire space. So, that is how we put a two boundary condition at the x equal to 0 and x equal to L. Remember of course the particle in a box I have not written the problem that V was infinity outside the box. The potential is infinity outside the box and 0 inside. Of course if it is not 0 if it is finite it does not matter there will be only a scaling. You can easily calculate as long as it is some constant value. If it is not a constant and that could be an interesting problem for example it is infinity here but inside it is not 0 it is not constant but let us say it is improved increasing. So, let us say V equal to x. You understand the problem. I hope you will be able to solve this problem. I may be entitled at least for this course to ask such problems now without solving them in the quiz at least. So, you should be able to solve such problems. You actually solve V equal to 0. Remember particle in a box but now if I put V equal to x there has to be still a continuity here. Remember so at this point whatever happens I must go back to 0. So, you think over it how to solve such problems? It can be a quadratic function x square for example. I had also discussed several nuances for this particle in a box that my coordinate need not be 0 to L. Coordinate may be minus L by 2 to plus L by 2. In which case eigenfunctions will be either symmetric or anti-symmetric which was a consequence of the fact that the Hamiltonian is a invariant to parity operator, parity symmetric which is exactly what I will come to the harmonic oscillator. In fact, now harmonic oscillator has the same problem. So, there are several nuances that we talked and we also looked at 2D and 2D, 3D boxes and their degeneracy. Remember so I just thought I will at least mention you the topic because I am not again going to cover this but those who have problem please revise these topics that what are the 2D and 3D boxes, they can be a square box, they can be a cubic box then need not be a square box. I can have a 2D box with L1 and L2 length then the degeneracies will become completely different. Do you remember? Then we have to see N1 square divided by L1 square plus N2 square divided by L2 square must constant. So, find out N1 and N2. So, those are very interesting problems but it is a only problem of arithmetic that you should be able to find out. So, I think we discussed many of these problems. Then we also discussed among these stationary states particular in a box remember was discussed very, very elaborately. It is a very, very famous problem. So, then the second was harmonic oscillator, simple harmonic oscillator where the potential was half of kx square and here in particular before we solve the problem we noted that the Hamiltonian itself is even. So, minus h of minus x is equal to h of plus x. Why? Because kinetic energy is a second derivative operator d2 dx2. So, if you second derivative operator if you change x to minus x it remains in variance. If it is d first derivative it changes sign. I hope that is clear. So, d of d minus x is minus d of dx. I hope that everybody understands. So, same way if I do d2 d of minus x square to it is same as d2 dx2. In this case it is plus exactly like minus minus becomes plus. So, it is exactly the same reason. So, the kinetic energy is always an even function even operator. The potential being an even operator makes ensures that the Hamiltonian is even and this is exactly what is called the parity invariant and we showed we actually introduce a term called the parity operator that parity operator p acts on any function phi of x to give you phi of minus x. So, that is a definition of the parity operator and then we showed that this Hamiltonian commutes with the parity operator and a proof that hence the Eigen functions are either odd or even. I remind you again this proof was something that we did quite elaborately. A very similar proof we will do for the permutation symmetry in many electron problem. But both on that I will actually do. So, I am not going to do it again but it is very easy to show that if h commutes with p the Eigen functions of h will be Eigen functions of p then we took the p square psi is equal to lambda square psi. So, it is very similar thing that the Eigen function value of p is either 0 or 1 very similar thing that we did to show that the psi must be either odd or even. I hope all of you again can prove this but if you remember the proof was actually absolutely trivial that p psi was equal to a psi then you have p square psi equal to a square psi and of course p square psi by definition is nothing but psi because p psi is psi of minus x p psi is again psi of plus x. So, we can say that a square equal to 1 which means a equal to plus minus 1 very similar if a equal to plus minus 1 then p psi is nothing but plus minus psi and p psi is nothing but psi of minus x. So, psi of minus x is plus minus psi of x that was the very three line proof that we did just again recall that. So, the Eigen functions are either odd or even very similar proof will come. I am again reminding you because it is very easy to forget. So, then eventually of course we solve the actual harmonic oscillator problem by two means by ladder operator remember and then by actual differential equation solution and we found that the energy quantum number is n plus half h cross omega where omega was the oscillator frequency and n of course in this case starts from 0 etc. 0 1 2 3 etc. Importantly you have to find that since E is n square dependent here E is linear dependent on n the spacing between the two are quite different in this case they keep on increasing in this case it is a constant alright. Further in trying to solve these differential equation we brought in an important concept and that is called the normalizability. If you remember that the wave function must be such that it must be normalizable just like we said continuity of a function we said that the normalizability of a function so the wave function must be normalizable. So, that was another important consideration if you remember that is the reason for this we had pre-multiplied by a factor exponential minus alpha x square by 2 because that ensures that at x equal to plus infinity and minus infinity it will make it 0 remember any wave function if it does not die at large x the function is not normalizable I hope all of you understand this. So, let us let us take a psi of x one dimensional where the limit psi of x at x tends to infinity is finite which is non-zero in such a case mod psi square dx if I integrate between minus infinity to plus infinity either plus or minus does not matter will become infinity and if it becomes infinity it is not normalizable I hope you understand this for example even if psi of x is just a constant across entire minus infinity to plus infinity some constant k then do the integration k square dx minus infinity to plus infinity it will become infinity k square will come out integral dx is infinity right. So, whenever you have a Cartesian axis which goes up to infinity that is very important psi must die this is not allowed so in the limit x tends to 0 the psi must die which means which essentially means that actually it should become 0 and that is a very important part that you have to so limit psi x as x tends to plus or minus infinity must become 0 note that you may ask the question why did I not use it for particular in a box I did not because particular in a box my potential was infinity so that automatically ensured that psi is 0 remember so that was not required but if I come here even if I am talking of this potential it is not required because the potential here is infinity it is only up to L there is no problem but in the case of the harmonic oscillator I have to ensure because the potential is not infinity potential is just half kx square so of course at some point it will become infinity but it is it is not automatically infinity so I have to ensure in the process that this has to be 0 and that is the reason the harmonic oscillator psi of x if you remember always starts with a pre factor e to the minus alpha x square by 2 and then the polynomial some polynomial which also actually has to be cut for the same reason and that is how the quantization comes this ensured that x equal to plus infinity minus infinity the wave function is 0 because the polynomial is only a finite power of x this is a infinite power of x x square actually so this will die much faster than its increase so either way that will make it 0 in the limit so that is really the calculus that whenever you have such a term it will be 0 x square because I have to make it 0 on either side x minus infinity plus infinity so it must be an even function this part must be an even function this polynomial will be either even or odd so that will make the entire function either even or odd right and that is what we have actually seen so this is something again remember we have done it for 1D box 1D harmonic oscillator sorry you can do the same thing for 2D and 3D which are uncoupled and then your energies will be some wave function will be product if it is uncoupled harmonic oscillator that is x and y there is no coupling we have just have a potential which is half k x square plus half k y square plus half k z square it is trivial to do by using my non-interacting theorem so again that is something that we did and then the third example that we did was hydrogen atom which is formally two particle but we said that we can actually convert this into a single particle problem by center of mass and the relative coordinate so we transform this into a center of mass and a relative coordinate right center of mass is just the kinetic energy part the relative coordinate is what we solve what we call the normally R or R theta phi three dimension electron with respect to the nuclei and we actually solve this problem it is a very detailed calculation detailed solution of the differential equation because it is a 3D problem by definition which is not coupled of course so it is not there x plus y plus z you have three quantum numbers to start with like here you have only one quantum number only if you go to 2D and 3D you have two and three quantum numbers here you automatically get three quantum numbers and they have been called n l n m so all of you know this is a principal quantum number it starts from 1 l goes from 0 to n minus 1 m goes from minus l to plus l in steps of 1 I think all of you hydrogen atom is very well discussed once again remember that the continuity condition for phi and theta and for R so you actually solve this problem in spherical polar coordinate R theta phi and note for R which can go to infinity again you have to use very similar condition like this so that is the reason for the radial part for the radial part which depends on two quantum numbers n l you have a pre-factor exponential minus alpha R and then something else note that everywhere this pre-factor is again there here you do not need R square because R by definition is only 0 to infinity spherical polar coordinate not minus infinity to plus infinity I hope you remember in spherical polar coordinate phi goes from 0 to 2 pi but R goes from 0 to infinity okay so I do not need R square R is sufficient so this is again the same effect the normalize normalizability wave function so this is basically sometimes these are these particular thing is called square integrable that is another name that the wave function must be square integrable what is square integrable that the square of the wave function modular square of the wave function if you integrate you must get a number which is finite and not infinity I showed you that if this is this was not there if I integrate with a minus infinity to plus infinity the number would have become infinity so that is not allowed so square integrable means square of the function must be integrated in the entire volume to give a result which should be finite so in general this should be square integrable so we learn two important conditions of the wave function which are called the goodness criteria one is the continuity of the wave function another is the square integrable of the wave function we also later in certain cases we have done the problem that not only the wave function must be continuous but it must be also derivative continuous which means the derivative of the wave function must be continuous in certain types of problem this is required in the particular in a box as it is defined it is not required but we have done the step functions in fact there is one of the problems that we did again in 4 to 5 I just thought I would remind you these are all called the goodness criteria of the wave function no it is not necessary condition but these are these conditions are required only when see I mean here here if you if you calculate the derivative here for a sin n n pi x it is a cos x so you will get a cos x so at cos l I do I am not yeah so I am not I am not I am not enforcing that as a condition I cannot enforce I cannot enforce because because this is a this is a higher level of criteria so that is the reason I cannot level if the wave function is not continue there is no point talking of derivative continuity okay so it is not required in that sense and this is not required because there are no other parameters to obtain okay so the derivative continuity is often given when there are more parameters in the wave function and you have less equation so that also comes up as a conditions to get those parameters that is correct I am saying the wave function is 0 that part is right right so I am saying that since if you want to make a continuity wave function then if I cannot impose the derivative continuity so this is a higher criteria okay so that is the point if I have already satisfied this then I will see remember these are all goodness criteria which is called the desirable part of the wave function even you can have wave function which are not square integral nothing wrong in that except that we do not like to work with that because I cannot define my probability so most of this criteria are not necessary conditions they are goodness can goodness criteria so they are very often called the goodness criteria of the wave function so the wave function is well behaved there is another term which is called it well good behavior of the wave function okay so in this case I cannot do for a simple reason because that will clash with my continuity and continuity wave function is a very important conceptual criteria because that means you are saying that the density abruptly changes probability density which is non-intuitive counterintuitive I would say so that is the reason it is very important to have this