 As we know quantum mechanics is all about orthonormal functions. The way functions that we use are almost always when we talk about exact solutions they are either they either form an orthonormal basis or they are made up by linear combinations of functions that form a complete orthonormal set. So it makes sense at this stage of our discussion to see what happens when we express trial functions as a linear combination of orthonormal functions. And we know about the properties of orthonormal functions that we have studied in the past. Great. So what we will do is we will do that and we will actually come back to particle in a box with a little bit of twist when we talk about orthonormal functions. So let us quickly recap what we learned in the previous module about variational calculation that we perform for particle in a box. There we had represented the wave function as C1 f1 plus C2 f2 where f1 was x into 1 minus x, f2 was x square into 1 minus x whole square. Are they orthonormal? Okay, you figure out. But they are definitely not the functions that we get by solving the Schrodinger equation exactly that we have done towards the beginning of this module of this course. So and this is the Hamiltonian minus h cross square by 2m d2 dx2. So what we did was that we wrote this secular equation and then we found out the expressions for all of these matrix elements h11, h12, h21, h22, s11, s12, s21, s22. And then knowing these expressions and substituting e dashed for em by h cross square this secular determinant simplified into this form. 1 by 6 minus e dashed by 30, 1 by 30 minus e dashed by 140, 1 by 30 minus e dashed by 140, 1 by 105 minus e dashed by 630 that determinant is equal to 0. This expands as a quadratic equation which has 2 roots 51.065 and 4.93487 of course out of these this is a lower value that this is ground state energy. And then what we did is we wrote this in terms of h cross square by m and we compared the mean this minimum value that we got of energy with the exact value of energy that we got earlier and we see the agreement is really, really good. If you only if you go to the 6th place of decimal there is a difference between the calculated minimum value of energy and the calculated exact value of energy that we got earlier. So we are good here but then just because it has worked nicely for particle in a box does not mean that it will work if we use arbitrary wave functions for more complicated systems. We want to simplify it that is why we want to talk about orthonormal basis. So in this linear combination that we have we want to write this f n's that are members of a complete orthonormal set of course even before going there if we simply write this expression then no matter whether f n's are orthonormal or not you will get this kind of a secular determinant. Last time we had just 2 by 2 determinant this will be n by n determinant that is all and we will get n roots we have to evaluate the integrals and the one with lower energy will be the ground state. Now when we use orthonormal basis and we just touched upon this before closing the previous discussion this the expressions become simpler and expressions become simpler for the simple reason that S ij integral of f i star f j over all space or the function space is equal to delta ij we know that when i is equal to j then you get 1 provided f i and f j are normal normalized and of course we are saying that they belong to an orthonormal basis set so they are normalized and if i is not equal to j then well once again remember they belong to an orthonormal set so if i is not equal to j then this integral becomes 0. So what happens to this determinant in that case h 11 minus e s 11 this s 11 becomes 1 so this becomes h 11 minus e this 1 2 position s 12 becomes 0 so the second term vanishes are completely or left with only h 12 and that happens for all the other elements all the way until h 1 n this s 1 n also becomes 0 when you go from top to bottom in the first column all these second terms vanish because they are of s ij type the 2 2 term what happens there s 2 2 is equal to 1 so you get h 2 2 minus e so what we see is that the diagonal elements are all like hi i minus e of diagonal elements are all h ij h ji that kind of thing okay so this is what you get h 11 minus e h 12 h 13 so on and so forth until h 1 n then when you go down h 11 minus e h 21 h 21 so on and so forth until h n 1 diagonal elements hi i minus e that is what you get and then already the expression has become simpler what about h 12 and h 21 can we say that they are 0 yeah actually we cannot okay if you think that they are 0 it is not necessarily correct we are going to encounter a situation where they are 0 but it is not that is not the general case why because if you write this h ij any i any j you write h ij that is equal to integral f i h f j the problem is see f j usually would be the eigenfunction of the Hamiltonian of the system that is solvable exactly generally that is what we choose so this Hamiltonian is different right for example well we have not taken that example yet when we take it we will see this Hamiltonian is not usually the Hamiltonian of the system for which we can solve Schrodinger equation exactly right even then we can write the Hamiltonian so it is not necessary that f j is an eigenfunction of this Hamiltonian right so we will not get and we will not get a constant coming out of the integral always sometimes we can have special cases we will see so what we will do is we will keep h ij for now and we will set this s ij for i not equal to j to 0 and we get this kind of a secular determinant now we take this example we go back to our particle in a box but we include in it a twist or rather a bend if you want to call it that what is that bend or twist we say that here potential energy is not 0 however we see that the potential energy is x dependent for x lying between 0 and l by 2 half of the box the potential energy is v 1 multiplied by x where v 1 is a constant for the other half for x ranging from l by 2 to l it is v 1 into l minus x okay what happens at x equal to l by 2 if you take the first expression you get v 1 into l by 2 if you take the second expression then also you get v 1 into l by 2 right l minus l by 2 that is l by 2 so there is no discontinuity in the function as such there is a discontinuity in slope what does the potential energy look like then let us say this is my particle in a box for the first half from the left your potential energy will be a straight line with a positive slope right v 1 into x for the second half it will be a straight line but with a slope that is negative equal in magnitude but negative now in this system let us use this trial function phi of x is equal to c 1 psi of psi 1 x plus c 2 psi 2 of x what are psi 1 and psi 2 remember what we want to do we want to use an orthonormal function and while using orthonormal function it makes sense to use orthonormal eigen functions of the Hamiltonian of the system for which we can solve Schrodinger equation exactly then things will start making sense you want to talk about particle in a box the orthonormal set that you use at least to start with it makes sense to take the wave functions of the Hamiltonian for a particle in a box in which the potential energy is 0 as long as the particle is inside box and we know that those solutions are something like this psi n of x is equal to root over 2 by L sin n pi x by L n as we know ranges from 1 2 3 4 all the way up to infinity that is the quant those are the quantum numbers okay so remember when I say psi 1 of x I essentially mean n equal to 1 put in this expression when I say psi 2 of x I mean n equal to 2 put in this expression okay so this is the formulation of our problem as usual I am not going to solve every step in fact in this case you also do not need to solve every step you need to understand the logic if you can solve it then it is even better but as long as you understand the logic you do not have to remember the final expression please remember that these are not things where you need to know the formula or anything okay you need to know only the very basic formula that we already know by now please do not try to remember the results it makes no sense at all okay try to understand the logic okay so what will we do now we know what the potential is we know what the trial function is and we have expressed it as a linear combination of the orthonormal eigenfunctions of the Hamiltonian of particle in a box for which v equal to 0 as long as the particle is within the box okay now we want to evaluate the matrix elements one by one I will show you some steps of h 11 how do we write h h 11 integral psi 1 x star in this case of course star is redundant just psi 1 x is fine because psi 1 of x is fine because it is a real function that left multiplying Hamiltonian operating on psi 1 of x let me for the record we will refer to it shortly but let me even now write it once what is this Hamiltonian what is the Hamiltonian for your particle in a box for which v equal to 0 inside and v equal to infinite outside you just have the kinetic energy term is it minus h cross square by 2 m or 1 d box so I might as well write d 2 dx 2 that is for the particle in a box that we have studied earlier in this case particle in a box with a twist we have an additional potential energy term which itself changes its expression at x equal to l by 2 so to write the general expression I will write that Hamiltonian is minus h cross square by 2 m d 2 dx 2 plus let me just write u of x depending on what you x is excuse me depending on what x is we are going to use the appropriate expression for u of x the potential okay so this is what we will use depending on the value of x okay of course we are going to get two terms isn't it when we expand this integral first of all there are two terms here and secondly even the second term is different for different ranges of x so I have to write two integrals one with limits from 0 to l by 2 the other with limits from l by 2 to l so that we can write this analytical form of u of x so this is what I will write 2 by l where does 2 by l come from well psi 1 psi 2 remember psi well in this case both are psi 1 so root 2 by l root 2 by l so I take root 2 by l outside then integral between limits 0 to l by 2 sin pi x by l multiplying the Hamiltonian that we have written here in this case u of x is substituted by v 1 x because I am integrating only up to l by 2 that operating on sin pi x by l dx okay what will the second integral be same thing except well similar except the limits would be from l by 2 to l and instead of v 1 x here I will write v 1 into l minus x okay this is what it is so I have written down the expression for h 11 all right what do you get what do you get if you try to say expand the first integral that is a final answer let me write the first integral so here if I take this sin pi x by l minus h cross by 2 m d 2 dx to sin pi x by l and what I will do is I will bring this here also root 2 by l here root 2 by l here so this first one I can write it as I 1 here will be equal to integral essentially psi 1 of x well do not forget the limit 0 to l by 2 this one is Hamiltonian okay since we know perturbation theory already I will use the language of perturbation theory here I will write 0th order Hamiltonian 0th order Hamiltonian means the Hamiltonian of particle in a box in which potential energy inside the box is 0 okay so you can think that this is the 0th order Hamiltonian and this is the first order correction okay so I will just write like this operating on psi 1 of x dx already know what h 0 h 0th operating on psi 1 of x is that is basically the left hand side of Schrodinger equation for n equal to 1 so it will be your e 1 multiplied by psi 1 n and for particle in a box where potential energy is 0 what is e 1 h square by 8 pi square m is it not yeah if l equal to 1 if l equal to l then it will be l square okay so this turns out to be the just the energy and then you integrate so this is equal to something like I will write epsilon 1 integral 0 to l by 2 psi 1 of x whole square dx so now tell me what is integral of psi 1 x whole square between limit 0 to l by 2 if I had integrated from 0 to l what would it have been it would have been 1 total probability so 0 to l by 2 for the symmetric wave function here that is going to be half isn't it so this is how you expand what about the second one in well second term arising out of the first part here it will be v 1 will come out and then you will get integral 0 to l by 2 we will write I will erase all this and I will write just erase everything what I am saying is I take root over 2 by l sin pi x by l and I am multiplying it by v 1 x then again sin pi by x by l will you agree that okay I will call it I 2 I 2 will be equal to integral 0 to l by 2 I can write again psi 1 as we discussed earlier multiplied by x multiplied by psi 1 dx and v 1 will come out isn't it v 1 is a constant remember so this will be v 1 integral 0 to l by 2 x into sin square pi x divided by l dx does that ring of l or does this ring of l if I only change the limit from 0 to l if I only change the limit from 0 to l by 2 to 0 to l then this integral would be your average value of position here I am just integrating from 0 to l by 2 so we know how to solve this integral using integration by parts so that is what one has to do so I am skipping all the steps here but I told you how to go about it and well I encourage you to do this by yourself when you do that and then when you do when you perform a similar treatment to the second term that is there then the final result that I will show you is this h square by 8 ml square remember what h square by 8 ml square is h square by 8 ml square is essentially the energy of the particle in a box when potential energy inside is 0 potential energy outside is infinity okay alright so that is what it is and the second term arises out of well I will call it the second term because I have written it as v 1 into l multiplied by something that arises from the second terms of these integrals okay if you do it is a little long but it is definitely not undoable okay so this is the result we get so this here is our h 1 1 h 1 2 and h 2 1 turn out to be 0 and we will discuss little later why it is that they are equal to 0 remember they are not necessarily equal to 0 but we will prove that they are equal to 0 h 2 2 turns out to be h square by 2 ml square multiplied by v 1 l by 4 this h square by 8 ml square that is energy of n equal to 1 right so I think I have written it later but in case I have not I will write it like this this is epsilon 1 do you agree with me that this is epsilon 2 what is epsilon n I will write here is equal to n square h square by 8 ml square right so if n equal to 2 then this becomes 4 4 by 8 is half that is how you get the 2 in the denominator so this is very nice isn't it the first energy that we get h 1 well the first matrix element that we get that is essentially the energy of particle in a box for v equal to 0 plus now again remembering our remembering what we learnt in perturbation theory plus we can say a correction term provided v 1 is very small h 2 2 is energy of the second energy level plus a correction term provided v 1 is small and the corrections are not exactly the same here it is v 1 l well for h 2 2 it is v 1 l by 4 for h 1 1 it is v 1 l by 4 plus 1 by pi square we will see how this is important but let us go ahead and finish our discussion the secular determinant we get is this okay and it is very nice because it is a block factorize determinant since h 1 2 into h o since h 1 2 is equal to h 2 1 is equal to 0 we get a block factorize determinant and block factorize matrices and determinants are lovely because now see the solution becomes so easy so what happens then is this this is h 1 1 this is h 2 2 ideally you have to write h 1 1 into h 2 2 is equal to 0 minus well h 1 1 into h 2 2 equal to 0 that is what we get if we did not have non-zero of diagonal elements then it would be h 1 1 into h 2 2 minus h 2 1 into h 1 2 and you would have to solve whatever equation you got now h 1 1 into h 2 2 equal to 0 directly gives you h 1 1 equal to 0 or h 2 2 equal to 0 right so you do not have to even go into a quadratic equation when h 1 1 equal to 0 I simply get e equal to h square by 8 ml square plus v 1 l into 1 by 4 plus 1 by pi square so this thing that we get here is really one of the roots for energy so what we see is that one of the solutions is the energy for h equal to for n equal to 1 plus a correction term that is one of the roots the second root you get by equating h 2 to 0 is e equal to h square by 2 ml square plus v 1 l by 4 so once again energy of n equal to 2 plus a correction term that is very nice so we know that these two are the energies of n equal to 1 and n equal to 2 for the regular particle in a box that we have studied earlier right and that is really very nice so whatever energy we got earlier plus some correction term so that is advantage of having block factorized determinant we have got the two roots which one is lower in energy which one is higher in energy most likely this is lower in energy unless this v 1 l into 1 by 4 plus 1 by pi square this is really really large if that happens then well but then that will add here also so this is the lower energy I think we can say that which with sufficient confidence now if you did this same treatment from perturbation theory which we are not doing explicitly you would get the same result and we should be willing to believe that also because the result is definitely in line with what one gets from perturbation theory the zeroth order energy plus first order correction right that is what you get so these two theories are in agreement with this the important thing here is that there is no contribution of Psi 2 in ground state energy ground state energy is the uh unperturbed energy now if I can say plus a correction term okay if this is small it is nowhere close to the energy of the second level and that happens because these h 1 2 and h 2 1 are 0 so now we better discuss why is it that h 1 2 or h 2 1 must be equal to 0 in the case that we have discussed so let us write it out h 1 2 is integral 1 to l and now I can write 1 to l because I am writing the general expression for u of x I am not writing uh v 1 x or v 1 into l minus x depending on the range I choose I I can substitute u of x as v 1 x or v 1 into l minus x so this is the general expression we get of course you can break it down into two terms and when we do that the in the first term again what do I have I have Psi 1 then Hamiltonian unperturbed Hamiltonian operating on Psi 1 so of course I will get uh sorry the unperturbed Hamiltonian operating on Psi 2 not Psi 1 I am not talking about h 1 1 anymore I am talking about h 1 2 so the eigenfunction that we have of the Hamiltonian gives me an eigenvalue of e 2 when the unperturbed Hamiltonian operates on Psi 2 I get e 2 which is constant comes out of the bracket integrate from 0 to l Psi 1 of x multiplied by Psi 2 of x dx what is that that is definitely equal to 0 because Psi 1 and Psi 2 are orthogonal to each other what about the second integral integral 0 to l Psi 1 of x u of x Psi 2 of x uh actually to evaluate it there is no need for me to write the explicit form of u of x because one thing I know for sure is that it goes up and goes down remember it is like a triangle so it is symmetric with respect to the midpoint so this Psi 1 of x is symmetric that we know already it is assigned function that does not go through a node Psi 2 of x is anti-symmetric like this u of x is symmetric and now we have studied this when we talked about particle in a box allowed transition disallowed transition what happens when the integrand is a triple product of 2 symmetric and 1 anti-symmetric functions the integrand is anti-symmetric and we know by now that an integral of an anti-symmetric integrand is equal to 0 is necessarily equal to 0 so both the terms of h 1 2 or h 2 1 for that matter would be equal to 0 that is why h 1 2 and h 2 1 is 0 in this case and that is what makes life very simple for us this leads to this interesting observation that Psi 2 of x makes no contribution to ground state energy only Psi 1 of x makes the contribution to it this is a general this is sort of a subset of a more general phenomenon that when we express our trial wave function as a linear combination of orthonormal functions then the functions with higher energy have lower contribution to the ground state so now let us write again using the notation we learnt in perturbation theory Hamiltonian is 0th order Hamiltonian plus first order correction phi this is what we have written in our earlier treatment of variation variational treatment phi equal to C 1 Psi 1 plus C 2 Psi 2 now we are saying that the size are eigen functions of the 0th order Hamiltonian with eigenvalues E j 0th so the secular equations we get in this case what would they be they would be h 1 1 minus e h 1 2 h 2 1 h 2 2 minus e what we have learnt earlier what is h i i instead of h 1 1 h 2 2 separately I will just h i i this is what h i i is two terms the first term I hope you can see without much hassle we have done it just now first term is E j 0th multiplied by integral Psi i star Psi i over all space which will be equal to 1 plus integral Psi 1 will Psi i first order correction to Hamiltonian operating on Psi i Psi E i so what is this the second term the first term of course is E i 0th uncorrected energy second term as we have learnt in our perturbation well perturbation theoretical treatment this is the first order correction to energy so we get E i 0th plus first order correction to energy so that is what we will use and this is what we get and then if you solve the quadratic equation I am jumping steps here and if you use a power series expansion then this is the approximate result you get of you will get there will be more terms but they will make progressively lower contributions we get E 1 equal to E 1 0th plus first order correction to E 1 plus this h 1 2 square divided by E 1 0th plus E 1 first minus 1 and so forth if you now remember the earlier problem of particle in a box with that triangular potential you will remember that h 1 2 is equal to 0 there it automatically follows that E is equal to the 0th order energy plus the first order correction for other general solutions if you work in the regime that the energy levels are sufficiently the spacing between the energy level successive energy levels is sufficiently large if you work in that regime which is a reasonable approximation reasonable thing to imagine then this difference is going to be large if this is large then this term will become small because h 1 2 is never very large so then approximately you get E equal to E 1 0th plus E 1 first so the higher terms do not even contribute to the ground state energy so that is what we have and now to close this discussion let us just introduce the general case we are talking about linear combination of orthonormal wave functions and what we have said so far is that it is the coefficient that is the variational parameter that is not necessarily the case I can also have variational parameters within these orthonormal functions themselves the F's themselves. For example for hydrogen atom we can write the function like this sum over j c j e to the power minus alpha j r square so this is this function here is similar to the 1 s wave function with the exception that this coefficient in the exponent is also a variational parameter so 2 variational parameters gives me greater flexibility and of course you can play around with number of terms so once again remember hydrogen atom of course is exactly solvable but if you do a variational calculation using this kind of parameter this kind of a this kind of a linear combination and if you keep increasing the number of n's what we see is this if you use n equal to 1 this is the energy that you get minus 0.424413 if you increase the numbers 1 go from 1 to 2 it becomes lower to 5 even lower 16 minus 0.49980 and now the exact calculation let me remind you in terms of what we have given here mu e to the power 4 by 16 pi square epsilon 0 square h square the exact calculation that we got earlier is minus 0.5 so now when you increase the number of terms to 16 you get very close to the exact solution so what we learn from here is that in the general case if you increase the number of terms you get closer to the energy and that is by using the power of variation theorem upper limit theorem one thing to understand here if we include the variational parameter in the functions themselves is that the secular determinant is complicated because the variational parameter is also there in the functions so in fact we will not be able to get an analytical solution like what we have been getting so far you have to solve this numerically putting number sever solution that you get change the number sever solution that you get and you need different kinds of algorithms and this opens up the field of computational chemistry I think I said this opens up field of computational chemistry once earlier well this really takes us into the domain of computational chemistry but let us talk a little more about this little later the other thing that happens here is that now see we are we have slowly started going away from the exact solutions in we have changed the exact solution 1s orbital by this introducing this variational parameter a little more significantly than what we have been doing so far so eventually we will reach a situation where you perhaps even do not have to start from the exact solutions any kind of orthonormal set will do okay but let us wait for that for now we close this discussion and finally we are now ready to resume our discussion of many electron atom the discussion will perform will be on helia