 We have been introduced to perturbation theory and we have discussed why I think this cat represents perturbation theory nicely. Today we will complete our discussion we will get an expression for the delta E the first order correction to energy and we will do a very very simple example of an application of perturbation theory. Then for the next few modules we are going to talk about applications in progressively more chemical chemistry dated systems. But before we do that let us quickly recap what we have discussed already we have said that the scope of perturbation theory is that it works when the deviation from the unperturbed system is very very small. We can write the Hamiltonian of the perturbed system as 0th order Hamiltonian plus a first order correction we can write the wave function as 0th order wave function plus first order correction delta C we can write E as 0th order energy plus first order correction to energy delta E. Before going further it may make sense to say that this expression is of course incomplete. I might want to do I might want to invoke a second order third order perturbations sometimes it might be necessary. In the next module we are going to discuss a case where first order perturbation with a certain kind of expression of a quantity just does not work you either have to change that quantity or you have to invoke second order perturbation. So it is not necessary that first order correction is be all and end all of it but for most of our purposes we are going to stick to first order correction. Then what we did is we set up Schrodinger equation in terms of H hat psi and E this is what it is and since delta psi and delta E are small changes we figured that this H1 hat the first order correction to Hamiltonian operating on delta psi and delta E multiplied by delta psi remember it is an operator so it cannot be multiplied by delta psi it is operating on delta psi. So these two are products of small quantities so they are 0. We also said that H0 operating on psi 0 is the same as E0 multiplied by psi 0 because that is your Schrodinger equation for the unperturbed system. So they cancel each other we are left with H0 minus E0 operating on delta psi plus first order correction to Hamiltonian operating on psi 0 is equal to delta E multiplied by the unperturbed wave function. Here we said that let us not forget that this unperturbed energy subtracted from the unperturbed Hamiltonian is in itself an operator in this case it is operating on delta psi. The problem with that is that we do not know what happens when it operates on delta psi we perfectly know what happens when it operates on 0th order wave function. Because H0 is equal to E0 multiplied by psi 0 so H0 minus E0 operating on psi 0 is equal to 0. So somehow we can replace delta psi by psi 0 then we know how to go about this. That is what we are essentially trying to do. So to do that first of all we will multiply by psi 0th star and integrate over all space this is the expression we get on the right hand side we get delta E multiplied by the integral of the 0th order wave function unperturbed wave function and its complex conjugate this integral of course is equal to 1. So right hand side gets cleaned up we get only delta E and the left hand side we are now trying to find out what would be this expression here. And to do that this is where we have got so far we have remembered that quantum mechanical operators must necessarily have real eigenvalues because the one of the central themes of quantum mechanics is that the eigenvalues of quantum mechanical operators are the values of the observables these operators stand for. So you cannot have a physical observable with an imaginary value if I if you ask me what is the position and I say the position is root 5 into i what does that mean it means it is not in real space. So that cannot be the case if I ask what is the momentum and you say 0.32 multiplied by omega or 0.32 minus 5.6 i what kind of momentum is that momentum position these have to be real quantities. So eigenvalues must be real for quantum mechanical operators using this we have got this expression here integral f star a hat f d tau is equal to integral f multiplied by a hat f star d tau. So what we are seeing is what we are trying to do is we are trying to interchange the bases here at least we have been able to interchange this complex conjugate and a function by itself. Left multiplication here is by complex conjugate left multiplication here is by the actual function but our job is not done because here the 2 functions are different is something like integral f star a hat g d tau here we have integral f star a hat f d tau. So we have to somehow bring in a g here then we know what this is going to be and that is what we proceed to do from here. So since we need a g let us simply write this same expression in terms of g without any loss of generality and talking about a function how does it matter if you call it f or if you call it g or if you call it phi or whatever you want to call it arose called by any other name smells just as sweet. So we write this but then there is a reason why we have written it it is not just we have we want to bring in a different function here and to do that what we do is we take a linear combination of f and g and we take psi is equal to c 1 multiplied by f plus c 2 multiplied by g why because we want an integral f star a hat g d tau this linear combination will enable us to do exactly that we are going to jump steps there because it is lengthy if you want you can do it yourself you will not have to remember anything as long as we understand the logic we are good. So now see this f is any general function right. So instead of f I might as well write psi so I can write integral psi star a hat psi d tau alright let me write what happens when I take psi and put it here in this equation I get integral psi star a hat psi d tau is equal to integral psi a hat psi we take complex conjugate of the whole thing d tau it is a simple as that but what is psi psi is a linear combination of f and g yeah. So what I do is I just write that instead of this I will write integral c 1 f plus c 2 g star a hat operating on c 1 f plus c 2 g d tau is equal to integral c 1 f plus c 2 g multiplied by a hat operating on c 1 f plus c 2 g whole star d tau d tau okay and then what we do is we expand this that is the step that I want to skip it is not all the difficult also does a hat operating on this linear sum what will it be it will be c 1 multiplied by a hat f star a hat f plus c 2 multiplied by a hat g it is a simple as a linear operator is not it and then you just open this up. So this way you just do it using the values of f and g and you put in these expressions also it is a little tedious but not impossible to come to this expression c 1 star c 2 multiplied by integral of f star a hat g minus g into a star f star d tau is equal to c 1 c 2 star multiplied by integral f a hat star g star minus g star a hat f integrated over all space okay. Now see we are almost there look at this integral instead of psi 0 I will write f instead of delta psi I will write g integral f star a hat g f star a hat g. So we are getting to the functional form that we require but before we get there it is important to now develop an understanding of the situation and that is where we are going to embark upon or where we are going to reach a very important property of quantum mechanical operators let us see. See first of all this left hand side here is the complex conjugate of the right hand side is not it? See c 1 star c 1 c 2 c 2 star f star a hat g f a hat star g star g a hat star f star g star a hat f. So right hand side and left hand side are complex conjugates of each other point number 1. Point number 2 is well it is an equation so of course LHS equal to RHS so what are we saying here we are saying that LHS and RHS are complex conjugates of themselves that is going to happen only when LHS and RHS are real otherwise it is impossible. Something similar also comes when you talk about what are called Hermitian matrices matrix algebra. So LHS and RHS have to be real it is as simple as that. Now see the c 1 star c 1 and c 2 they are perfectly arbitrary constants I have not told you that this is 5 and this is 3i or other way around or anything. So naturally since these are arbitrary complex quantities we should expect that c 1 star c 2 in the general case is going to be complex because we are multiplying two complex numbers either c 1 star and c 2 both have to be real or c 2 has to be the complex conjugate of c 1 star then only this c 1 star c 2 c 1 c 2 star these are going to be real but that is not the case that would be a very very specific case we are we have written a general expression. So these are arbitrary complex quantities these are complex quantities. So what about the rest remember left hand side let us work with the left hand side first left hand side is real left hand side is made up of two factors the first one is c 1 star c 2 second one is this f star integral the second one is the integral. Similarly right hand side is c 1 c 2 star multiplied by another integral now each is real in the left hand side focusing there c 1 star c 2 we said in the general case has to be complex which means that the integral either has to be its complex conjugate because the product of the integral and c 1 star c 2 is real. So either the integral has to be complex conjugate of c 1 star c 2 but if that is the case once again generality is lost that is a specific case we do not want to talk about specific cases we want to see if there is something that is absolutely general similarly on the right hand side also the same thing will happen. So what is the only other situation in which we have a complex number multiplied by a quantity and the product is real unless the second quantity is a complex conjugate of the complex number the only other option you have is that the second quantity is 0 anything multiplied by 0 is 0 why it is not 0 1 by 0 anything real or imaginary does not matter multiplied by 0 you are going to get 0. So the general solution is that these integrals on the 2 sides must be equal to 0 integral f star a hat g minus g a hat star f star d tau must be equal to 0 similarly integral f multiplied by a hat star g star minus g star a hat f d tau is also equal to 0 well there are too many f's and g's and stars here please do not get scared please do not get confused stop the video here if you have not understood replay replay as many times as you need and it will sink in it will sink in faster if you do it by yourself so this is something that we are saying in almost every module please make sure that in addition to hearing what I am saying you write down the same thing so that you practice okay so this is 0 so let us clean up things a little bit and before doing that this is actually the definition of Hermitian operators okay an operator that satisfies this condition that this integral is 0 is called a Hermitian operator and Hermitian operators as we see have real eigenvalues this is the definition do not say that Hermitian operators definition is that they have real eigenvalues that is a coronary this is the really the definition we have sort of back calculated starting from real eigen functions we have shown that the operator has to be Hermitian this integral has to be equal to 0 okay so let us clean up things now and write this integral f star a hat g minus g multiplied by a hat star f star d tau equal to 0 okay so what we can do is we can expand the left hand side and get two terms and since one has minus sign we can take it to the right hand side that way we get integral f star a hat g d tau is equal to g multiplied by a hat star f star d tau and that is a happy situation why because look at this psi 0 star let us say that is f h 0 h 0 minus e 0 that is your let us say a hat and this is g the problem we have here is that g that is delta psi is not an eigenfunction of h 0 minus e 0th but psi 0 is an eigenfunction so if we can somehow take this delta 0 here and bring this psi 0 here then that is a happy situation I mean it is okay if you bring psi 0 star also but that will not happen okay and this equation is what allows me to do that I have an operator operating on a function left multiplied by another function what it says is that within the integral sign you can just interchange the two functions right instead of f star now f star will come here g will go there so this is what we get so using this expression we get delta psi star multiplied by h 0th minus e 0 h 0 complex is okay e 0th complex is complex h 0th complex does not make a difference so h 0th minus e 0th operating on psi 0 theta so this expression here gets transformed this integral here gets transformed to another integral where we have this problematic delta psi star factor multiplying an operator operating on its wave function on its eigenfunction now tell me what is the eigenvalue of this h 0th minus e 0th operator for psi 0th I can find eigenvalue and I can do it fairly easily remember Schrodinger equation for the unperturbed system unperturbed Hamiltonian operating on unperturbed wave function gives us energy of the unperturbed system multiplied by unperturbed wave function okay so let us bring this to the left hand side we get h hat 0th psi 0th minus e 0th psi 0th equal to 0 or we get I can write like this h hat 0th minus e 0th that is the operator we are working with operating on psi 0th is equal to 0 okay so first thing we have done is we have transformed this integral to another one in which we have h 0th minus e 0th operating on psi 0th and we have found that this h 0th minus e 0th operating on psi 0th is 0 so inside the integral we have delta psi 0th multiplied by 0 integrated over all space what will it be definitely it is going to be 0 okay so now after all this exercise we are happy to see that even this problematic term is actually 0 you do not have to worry about it so finally we get an expression for the first order correction to energy and that expression is delta e equals integral unperturbed wave function multiplied by the first order correction term to Hamiltonian operating on the unperturbed wave function itself integrated over all space okay another way of writing it indirect notation that is often done as we have discussed a little bit is we can write this same thing as psi 0th h hat h hat psi 0th integrated over all space this is your delta e I think we are now familiar with this expression there should be no problem so we have got an expression for the first order correction to energy now to conclude our discussion let us think of applications of this simple perturbation theory in some systems now before that this further records this is called time independent perturbation theory because we are working with time independent wave functions we are not working with time dependent wave functions time dependent because we are working with stationary states here right when we work with when we talk about spectroscopy at that time it will be imperative that we use time dependent perturbation theory but if you understand time independent perturbation theory time dependent perturbation theory is actually child's play not difficult at all so we are going to see its application today we are going to talk about this kind of a system a particle in a box with a slanting bottom and in the next module we are going to discuss what is called an an harmonic oscillator and that is where we will see what kind of problems we face if we do not define our perturbation properly or if we try to stop at too low level of correction so let us start so what we are doing is we want to talk about particle in a box with a slanting bottom which means a particle in a box with an x dependent potential okay so to start with what will be the unperturbed system our good old particle in a box where v equal to 0 for all values of x v equal well all values of x within 0 and a where 0 and a are the limits of the box and outside it is infinity so for such a system for a particle in a box with infinite height of the walls the wave function as we have discussed is root over 2 by a sin n pi x divided by a and energy is n square h square divided by 8 m a square simple now we want to bring in the perturbation to the system to arrive at our box with a slanting bottom so this is the perturbation we bring okay so let us say at x equal to a the perturbation potential is v remember what we are trying to do is we are trying to get an expression for the first order correction to the energy brought in by this slanting bottom to do that we will need to know what the first order correction to Hamiltonian is first order correction to Hamiltonian will be of course the potential energy what is the potential energy it is equal to 0 for x equal to 0 it is equal to capital V let us say for x equal to a so what you have here is a straight line going through the origin so very simple right h first order for any value of x is going to be x divided by a multiplied by v it is quite simple not difficult at all okay x divided by a multiplied by v v is the value at a so we take this expression for the first order correction to Hamiltonian and we plug it into the expression for delta e this is what we get and of course we have to use the expression for psi 0th as well so delta e turns out to be integral 0 to a psi 0th star which is the same as psi 0th because psi 0th is real wave function multiplied by x into v divided by a multiplied by psi 0th integrated over all values of x from 0 to a simple as that now here this v is a constant is not it a is also a constant so v and a can come outside the integral and what do we get here psi 0th is root over 2 by a sin n pi x by a and there are 2 of them so I get 2 by a from there as well so what comes outside the integral is 2 by a multiplied by v by a something like that so 2 v divided by a square that comes out inside your left with integral 0 to a x into sin square n pi x by a dx of course you can work this out yourself it is an integration not a very difficult one but it is a standard integral we know the solutions already we are going to just put the value from the compendium the value of the standard integral turns out to be a square by 4 so the value for delta e that we have is 2 v divided by a square multiplied by a square by 4 that turns out to be v by 2 so you see how simply how easily we have been able to find out the value for the first order correction to energy of the system even without working out what the wave function actually is this is the beauty of approximation methods you do not have to know the wave function to get the energy you can get the energy first then worry about the wave function so let me just write this remember energy in particular in a box is quantized we said earlier that e 0th order energy for nth level is n square h square by 8 m square now in the perturbed system you need this correction term of v by 2 so this is the expression for the energy of the nth energy level en is equal to n square h square by 8 m square plus v by 2 so what has happened to the energy level of particle in a box because of this slanting thing what has happened is that every energy level has simply got offset by v by 2 what is v v is the perturbation potential at x equal to a so this energy here is not x dependent potential energy is x dependent fine energy is not it would have been a little unhappy situation if it was okay but it is not so it is a good thing that we see that n square h square by 8 m square plus v by 2 this is what we get now see v would better be small right and you can think of very large values of v what about n square h square by 8 m square we have talked about what kind of energies they can be if v is very large then of course this treatment is not even valid yeah because that delta e into delta psi that you cannot equate to 0 anymore if delta is large so you have to work with these that are small enough if v itself is close to infinity of course the statement is not valid otherwise it is and we see an elegant and simple example of application of time independent perturbation theory to a simple model system next day next module you go on and discuss application of the same time independent perturbation theory to your anharmonic oscillator remember harmonic oscillator equi-space energy levels and so on and so forth we will discuss why it cannot really be the case for the diatomic molecules and it has to be a it has to look different so you have to bring in anharmonicity we are going to bring in anharmonicity as a perturbation and we will get a correction term to the energy while we do that we will come upon something interesting after anharmonic oscillator we want to see how one can use perturbation theory to talk about helium atom remember we were talking about helium atom two modules ago from there we have taken a sort of a break and we are learning perturbation theory only because we will can go back we have to go back to the same system and use it this is not a digression we are developing the tool so that we can use it in our atomic system please do not lose the focus of what we really want to do after this but to get there we must learn the workings of the theory and the best way of learning it is to make the theory work on model systems and while doing that we also get some interesting results that are relevant to molecules also right even this kind of a slanting box can be there in conjugated molecules in which all atoms are not carbon and anharmonic oscillators are ubiquitous in molecular vibrators so that is the reason why we are doing this but do not forget that all this is really a part of our discussion of many electron atoms to understand which we are going to use this perturbation theory.