 Welcome back to the lectures on the principles of quantum mechanics from the department of chemistry IIT Madras. This is part of the course that I have been offering to our master students. The course contains some mathematical preliminaries which we are currently looking at and today and possibly the next one or two lectures we will look at an important area of mathematics that has been employed in quantum mechanics very thoroughly in the beginning namely differential equations. The Schrodinger equation was proposed initially as a second order differential equation in coordinates and a first order differential equation in time. Differential equations have been well known the theory of functions and solutions of mathematical equations have been studied for more than 200 years earlier and therefore there is a rich history you will not look into any of that and nor are we going to look at even formal aspects of the differential equations that people look at in quantum mechanics namely this term Louisville theory etcetera. I mean it will be nice to do those things but in this course approaching this from the point of view of the chemistry students who want to learn the principles of quantum mechanics to apply them to molecular systems. We shall look at the differential equations that the students have seen in their physical chemistry courses and then give some general rule of thumb rules of thumb or whatever you call it as directives to solving the simple as basically directional aids to solving the differential equations ok. So, I shall start with this with a few simple differential equations and their solutions I am going to state them and if you are not familiar we I expect that you would have studied them earlier and probably not do not remember it now. So, please go back and look at some of these things from the earlier textbooks but solutions of these equations are important. So, this lecture will contain some descriptions of the following four equations d y of x where y is a function of x is equal to a constant times that function ok. Now in all respects it is also a nice starting point because if you remember d by dx as an operator acting on a function giving you the function back multiplied by a constant that immediately this is an eigenvalue equation ok that is a relation of this to what we talked about so far. Anyway I will see these things later on. The next one is d by dx of y of x plus a constant times y of x is equal to another function d of x. So, the first one is the second one third equation that we will look at is d y of x by dx plus p of x instead of a constant now it is a function of x which multiplies y of x is equal to a different function d prime of x or there is the danger of g prime being a derivative. So, let us call it a h of x some other some other function and the last equation that we shall see is d y by dx is equal to f of x y homogeneous and first order that is all we will see. Some of these things you would have seen in the course on thermodynamics or will be seeing them in the course on thermodynamics when you look at the partial derivatives and the Maxwell's equations. In fact, if I have time I shall just end my lecture with a brief introduction to one of the four Maxwell's equations that we can write down very quickly using these types of methods and it is important for you to remember some of the results while being able to understand the others for you to derive them when you need them ok. So, the first part of this lecture will deal with simple first order differential equations d ok. It is easy when you write d y by dx is equal to a constant times y of x that the answer has to be an exponential because that is the only one which when you take the derivative gives itself with the constant. So, immediately you should see that y of x is a constant times exponential of a x this constant is defined for some arbitrary value of x that is the function of y at a particular value of x is the one that determines the c usually at x is equal to 0 the exponential goes to 1 therefore, y of 0 is equal to c is a an initial condition it is a general solution for that equation. And there are a number of such equations particularly first order chemical kinetics first order rate law in chemical kinetics has a as a negative constant d of the concentration of a substance as a function of time by d t. So, this is the rate of change of the concentration of a of t and when you put in a minus it is a rate of decrease of the concentration of a of t as a function of time because a reacts with some other species or with itself this the first order law is it is given by the fact that it is k times the constant a therefore, you can see that minus k is equal to this a and y is a of t and x is t here. So, the solution for this is obviously a of t is equal to a of 0 which is c here and it is e to the minus k of t ok. So, this is for the kinetics of the reactions a giving products and the rate of change of a being proportional to linearly to the concentration a of t. Nuclear reactions you have seen the decay of particles of the nuclei and with the decay constant lambda often written as lambda. So, you can see that d x the concentration of the radioactive a nucleus d t as lambda times x of t and you see that this also has the solution that x of t is equal to x of 0 times e to the minus lambda t ok. Now, the exponential law is very specific law often found in nature that you would have seen the half life or what is called the rate of change in the concentration to become one half of the initial concentration then one fourth and one eighth and so on. So, if you think of that then x naught of t becoming x naught by 2 at t is equal to t the exponential law implies that x naught by 2 will become x naught by 4 at t is equal to 2 t that is if you put the time let us be careful. So, what we have is x naught of t goes over to x naught by 2 at t plus e and this goes over to x naught by 4 at t plus 2 of t 2 times t and so on. So, it becomes x naught by 2 raised to n at t plus n t ok. There is nothing special about 2 about 2 suppose you have x naught goes over to x naught of 3 t x naught of t divided by 3 at t plus t and x naught t of 3 at that instant of time goes over to x naught t one third of that value at t plus 2 t this is x naught by 9 and so x naught by 9 goes over to x naught by 27 same fraction equal fractions decay in equal amounts of time is the phrase with which you are familiar with the exponential law that is what this is. The fraction need not have to be half the fraction can be any value, but it is a same fraction that disappears at any given interval of time and that homogeneity ensures that it is a first order rate law. So, in fact, the best way to construct to the rate law is to look at that fraction constantly. So, if you say at 1 0 at half at time t at one fourth at time 2 t and one eighth at time 3 t and so on and if you start from here and if you connect to that that is the exponential law ok. Again there is nothing special about the 1 by 2 1 by 3 etcetera any if it is 1 by 3 the curve will be steeper if it is 1 by n it is even steeper, but it is still exponential ok. So, this is the first law and a similar equation that you see in the growth of nuclear magnetization when you put a substance in the NMR cube and then you want to study the nuclear magnetic resonance spectroscopy the growth of the nuclear magnetization in a magnetic field d m z the z component of the magnetic field as a function of time which is 0 when the substance is outside there is no magnetic moment, but the moment you put it in the magnetic field the magnetization grows to a certain value and if that value is m naught the differential equation that this satisfies is by t ok m z ah m naught minus m z of t at a relaxation time constant t is called the relaxation time and in this case of course, it is called the spin lattice relaxation time, but that is not what is important the law that governs the growth of magnetization in a magnetic field is subject to a time constant t and the growth is an exponential growth to a saturation it does not grow forever it is grows to be a saturation value and the saturation value is m naught which is 0 initially at t is equal to 0, but it is in principle m naught only at t is equal to infinity, but you know you can hardly discern the difference between the m naught and m z of t for very long times. So, this is the exponential growth with saturation of course, the solution of this equation is ah an exponential minus t by t now simple equations ah the other chemical kinetics equation which are linear you can think about a plus b giving products and you know that the differential equation for second order is the minus d a by d t is equal to minus d v by d t and if it is proportional to the concentrations of a and b then it is called the second order equation, but it is linear in a linear in b, but this equation is by linear and the solution can be obtained by a simple first simple differentiation process involving separation of variables ok. Also you know that minus d a by d t d t is equal to k of a of t square of course, this does not come into the linear equation that we started with. So, we look at that later ok these are one set of differential equations. The next that we can look at is the first order d with constant coefficients that is typically d y by d x plus a y of x this is the constant coefficient is equal to some g of x. The solution is immediate in the sense that you write the left hand side as a simple function whose derivative gives you that and is equal to g of x then you can immediately integrate the both sides and write the function is equal to integral of g of x ok. But what is this function from looking at this form what you have is d by d x plus a acting on y gives you some other function p of x ok. So, in order to get this as the differential of a function all that we need to do is to multiply the left hand side by multiply both sides by exponential a x where this a is the same as this a same a. So, if you do that then you have e to the a x d by d x of y plus a e to the a x y of x y of x is equal to e to the a x g of x. Now, you can see immediately that this is nothing but d by d x of the product of two functions namely e to the a x y of x. So, this is the function that we talked about this is the d by d x of the function that we have here if you remember is equal to it is not g of x as you see it it is e to the a x d of x ok. Then you can write the integral as essentially integral of e to the a x y of x d x is equal to d by d x sorry integral d by d x of this. If you integrate this you get e to the a x y of x and on this side you have e to the a x prime g of x prime d x prime integrated to the upper limit x and if you do not know the lower limit you can write this as a constant. So, the solution is therefore, y of x is given by e to the minus a x inside the integral x e to the a x prime g of x prime d x prime plus c ok e to the minus x. So, this is a simple general solution for a linearfirst order differential equation with constant coefficients. I have some examples in the some problems in the lecture notes that you will find on the website http colon double slash nptl dot iitm dot ac dot in and these this lecture notes on this contains some of those problems and all I would do is request you to solve the following equations namely d by d x of y of y minus 2 y is equal to 2 x e to the 2 x y of 0 is equal to 1. Then I have another example d y by d x plus y is equal to 5 sin 3 x with no initial conditions these are two examples. Just to tell you that to give you a hint this is nothing, but d by d x times e to the minus 2 x y of x is equal to 2 x. The solution of this is when you integrate both sides you have e to the minus 2 x y of x is equal to the integral of 2 x prime d x prime to x plus a constant and then you can see immediately that y of x is equal to e to the 2 x you bring it to the side times x square plus e to the 2 x times s e ok. So, this is the solution that you have to and this you can verify that it gives you when you say d y by d x minus 2 y it should give you the result namely the original equation 2 x e to the 2 x ok. So, this is easy to verify. So, let me stop here and then start with the next set of first order differential equations. So, the next is general first order linear differential equations. So, the equation has the general form d y by d x plus p of x of y of x is equal to g of x ok. So, to do this exactly we do what we did before namely we want to write this d by d x of a function is equal to a g of x times some function some other quantity function 2 this is function 1 you call it function 1 ok. But, since p of x is itself a function of x we cannot use the e to the a formula, but we have to remember the following identity among the differentials the differential of an exponential or not even that let us just look at this. The differential of an integral which depends on x only as a limit and a function inside g of t d t if you do that the result is g of x. This is only if the limit alone depends on x and the function inside does not depend on x and also only the upper limit depends on upper or lower upper limit depends on x ok. Now therefore, if we have an integral in the exponential with the x p of t d t and if we differentiate this we will get the differential of the exponential will anyway give the exponential back. So, you will get p of t d t ok times the differential of what is there in the exponent and the exponent will give you p of x. So, we have to remember this in order to solve this equation d by dx of d by dx of y plus p of x y is equal to g of x. So, what we do is multiply both sides with the integral with the exponential of integral x to p of t d t you multiply that d by dx y plus p of x y is equal to exponential integral of x p of t d t g of x ok. Now, you know that this whole quantity is d by dx of exponential integral x p of t d t times y of x. So, product u v u v is u times d v plus v times d u. So, the differential of this will give you this particular quantity and the differential of y will give you this quantity and therefore, that is equal to exponential of integral x p of t d t g of x. Now it is easy you integrate both sides with respect to x and the answer that you will get is e to the integral x p of t d t y of x is equal to integral with respect to the limit with the upper limit x e to change all the x's to x prime some other dummy variable x prime p of t d t g of x prime d x prime. So, that all quantities inside do not have any x dependence the x is defined by the limit of the integration upper limit plus a constant when youintegrate this equation you get an arbitrary constant. Therefore, now the solution is y of x as you take this to the other side you have e to the minus integral of x p of t d t multiplying this whole quantity namely integral of x e to the integral x prime p of t d t g of x prime d x prime plus c this is the general solution to linear first order equation general solution integral of x e to the integral of x prime yes g x prime d x prime that is correct ok. So, this is the solution slightly more complicated, but these are allstill simple enough anyway and as examples I would want you to check with one or two ofsimple applications namely do these problems like d y by d x plus 1 by x of y is equal to 3 cos 4 x and the condition is that x has all values greater than 0 that is it is defined only for those. Another example is d y by d x minuslet us see or sorry plus y cos x is equal to 1 by 2 sin 2 x ok. This is easy to solve because you know that 1 by x is the p function remember integral of 1 by x d x to x x prime d x prime to x is log x plus c ok. So, if we have to do we will not worry about the c here. So, if we have to worry about e to the integral of x p of t d t then what we have is e to the integral x 1 by x prime d x prime we have e to the integral sorry e to log x, but of course, exponential and log or inverse of each other. So, it is equal to x. So, you can see immediately that x times d y by d x plus 1 by x y which is given as d sorry yeah d of x y. So, this is equal to d by d y by d x plus y that is that this equation is this one 3 cos 4 x. So, you have it is 3 cos 4 x. So, the solution is of course, x y is equal to integral x 3 cos 4 x prime d x prime plus c and therefore, y of x is equal to 1 by x times integral x 3 cos 4 x prime d x prime plus c ok. So, this is the solution. So, if you do the differential of this you will see that this equation is satisfied namely d y by d x plus 1 by x of y if you do that if you calculate the left the left hand side with the differential and then multiply y with 1 by x you will see that the sum gives you 3 cos 4 x. So, that is a verification. So, this is another form of first order differential equation that we have to remember ok. Then the next one that I will talk about in the next part is the separable equation, but we will stop now and we will continue that in the next part of today's lecture separable first order. So, this section we look at both the separable first order d and homogeneous first order shall do very quickly look at what is relevant to us a lot more can be studied, butlet us look at the simple equation d y by d x is equal to f of x y which can be written as n m of x y plus n of x y d y by d x is equal to 0 ok. So, what f of x y is nothing, but the ratio m by n with the negative sign that is how it is this is minus m x comma y by n x comma y that is what we have written ok. So, specific form now if m of x y is only a function of x and m of x only and n of x y is only a function of y only. Then it is easy to separate this equation because it is m x plus n y d y by d x is equal to 0 and you can immediately solve this as m of x d x is equal to minus n of y d y and. So, you can integrate this into m of x d x prime x is equal to y minus n of y prime d y prime plus of course, a constant that you add as the constant of integration. This is an easier way of solving such equations and a typical example of that is for example, for as d y by d x is equal to x square by 1 minus y square. You can see immediately that this x square is like the m of x and 1 minus y square is like the n of y and. So, the solution for this is immediate namely d y times 1 minus y square minus x square d x is equal to 0 and therefore, when you integrate this and you integrate this you get y minus y cube by 3 minus x cube by 3 is equal to 0 or the equation is x cube minus 3 y plus y cube sorry is equal to a constant the integration constant x cube minus 3 y minus plus y cube is equal to a constant ok. Easy to separate the equations if they have a special form that is called a separable form. Another example that I can give you is the solution of this which I will not do, but I will give you this for you to figure out x square by a square times 3 y square solve. It is easy to solve by this separable first order equation and whenever you have an equation like this d y by d x is equal to a x plus b by c y plus d it is a separable equation ok. We will see them when we come across the second order differential equations in the particle in a 2 dimensional box particle in a 3 dimensional box in the hydrogen atom. In fact, the whole body of the model problems in quantum mechanics uses this separability of the wave functions into products of functions of lesser number of variables. Here we have x and y in real world we may have x y z or we may have polar coordinates like r theta and phi and various things. Therefore, separable equations both first order and second order are extremely important to remember and some manipulatory skills are needed for you to appreciate the method of solution for some of the differential equations in quantum mechanics ok. Now, homogeneous equation first order d there again of the form d y by d x is equal to f of x comma y except that the function x f will not depend on either x or y, but the ratio x by y or y by x as the only clearly dependent quantity ok. So, it is d y by d x is either f of x by y or f of you call it f 1 or f 2 of y by x ok. An example is for is this one d y by d x is equal to y square plus 2 x y by x square. Now, it is clear that if you divide the whole thing by x square on the right hand side you have y square by x square plus 2 y by x you see immediately that this is what the function depends on it is y by x whole square. So, it is not dependent on y or x, but the ratio y by x. Therefore, y by x can be treated as a variable and then the differentiation can be transferred to the differentiation of that variable and then once you solve that you can come back to the regular equation because these are forms which will be eventually reduced to what are called the linear first order differential equations the general form constant coefficients or this arbitrary functions that we saw in the first part of this lecture. So, the idea is to reduce somewhat more complicated equation, but specific forms this is homogeneous simply because of this ratio that y by x square y by x etcetera that happen. There is a general way of defining that namely a function f of x y is homogeneous in degree or if if x is replaced by a x a y then this function will be exactly the same as a raise to r times f of x y ok. Such functions are known as functions of homogeneous nature and with the specific degree r ok. If r is 1 it is simply that you have a x or a y and if it is x by y of course, a will cancel out and you can see that it simplifies ok. So, this is the definition of a homogeneous function that we often use in in physical in situations in physical chemistry and an example here and example here is for example for dy by dx is equal to x raise to n plus y raise to n by x raise to n minus y raise to n ok. Now, you can see that if we divide the whole thing by either x or y the numerator and the denominator you will get the ratio y by x or x by y. So, if we divide this by x raise to n you have 1 plus y by x raise to n by 1 minus y by x raise to n. So, the general equation that you write dy by dx is equal to f of x y homogeneous is essentially a function of f y by x ok. Now, how do we solve that? It is very simple. Let u be y by x then y of x is equal to x u of x this is a function of x therefore, dy by dx is equal to x du by dx plus u of x and the this is a function of u ok because we have replaced y by x by u. So, this is the quantity that goes in here ok. Then it is easily separable because you can write this x du by dx is equal to f of u minus u or du by f of u minus u is equal to dx by x ok. So, now, you can integrate both sides this with respect to u and this with respect to x with with appropriate limits as given by the problem and then solve for u x solve for u and once you know the answer u, you know that y of x is nothing but x times u of x ok and you get this once you get this. This is called the homogeneous equation and these are easily separable. Now, as a last example let me just give you the differential equation from thermodynamics and let us look at what are known asexact equations and equations that can be made exact can be made exact with an integrating factor. So, let us solve the equation that we started with namely m x y plus n x y d y by dx is equal to 0. Then suppose there is a psi for this such that m of x y such that m of x y is equal to dou psi by dou x and n of x y is equal to dou psi by dou y. Then you can see that the same equation can be written as dou psi by dou x plus dou psi by dou y which is n times dy by dx is equal to 0 or dou psi by dou x dx plus dou psi by dou y dy is equal to 0 and you recognize that this is nothing but the perfect differential d of psi x comma y in two variables. Therefore, the fact that a psi exists such that it satisfies these two relations these m is this and n is the derivative with respect to y and then you can write this as a perfect differential and so if this is 0 then you have the solution psi of x y is a constant. Is the solution also please remember we wrote dou m by sorry we wrote the following dou psi by dou x is equal to m dou psi by dou y is equal to n. Therefore, dou m by dou y partial derivative will be dou by dou y of dou psi by dou x which is dou square psi by dou y dou x and likewise dou psi by dou y is equal to n gives dou n by dou x is equal to dou by dou x of n which is dou psi by dou y and that is dou square psi by dou x dou y. The function is continuous then both these quantities are the same dou square psi by dou x dou y and dou square psi by dou y dou x both of these are the same and therefore, we have in addition a relation namely dou m by dou y is equal to dou n by dou x ok. So, these are all under the assumption that the m and n which are there in the differential equation can be written as partial derivatives with respect to one of the variables only. If we do that then we have a perfect differential and we can solve that and the m and n will also satisfy the additional property that their derivatives with respect to the other variable or equal ok. Now, if they are not what will happen if the condition d psi by d x is equal to 0 that is a perfect when it is called the perfect differential. However, if this is not satisfied that is dou m by dou y is not equal to dou n by dou x ok. Then we have a problem namely m d x plus n d y is equal to 0 is not exact it cannot be written as a derivative of a function with respect to x d x a derivative of a function with respect to y d y is equal to 0. Therefore, the same function if it cannot be written that way if m and n cannot be written that way then it is not exact often what you can do is you can multiply this by another function mu such that mu m and mu n satisfy these conditions that the d by d x of mu m the d by d x of mu m is the new function and the d by d y of the mu n is the new function their sum with with the d x and d y is equal to 0 and then the multiplication by mu to make this an exact quantity an exact differential the factor mu is known as integrating factor integrating factor. Let us see a little bit of this in the thermodynamics as an example and then we will go back to the second order and other differential equations in one lecture before we move on to the formal solution of the Schrodinger equations and model problems. So, let us take the 4 examples that 4 equations that you are familiar with in thermodynamics for the internal energy d u which is T d u which is T d s minus e d v infinitesimally small but what is it called infinitesimally small and reversible processes this is nothing but the delta q and this is the delta w the heat content and the heat change and the work done on the system the two together gives you the total derivative d u, but this itself is not a derivative they mean this is not an exact differential, but d u s. Now, since u as we have written is a function of s and v we can write d u as dou u by dou s times d s plus dou u by dou v times d v of course, please remember we are writing partial derivatives therefore, we keep the other variable constant when we take the derivative and you can see that this is T d s and this is minus p d v giving you the equation dou u by dou s is equal to T and dou u by dou v is equal to minus p ok. Now, it is important to put the constancy of the other variable here of course, keep this volume constant and here we keep the entropy constant therefore, you can see that dou squared u by dou v dou s will be dou of T dou v and that is at constant s dou square u for this part by dou s dou v will be minus dou of p dou s at constant v, but u is analytic or it is continuous mathematics tells us that derivative of u with respect to v and s the second derivative it should not matter which order we do the second derivative. If these two are equal then essentially what we have is the relation between the derivatives namely dou T dou v of s is equal to minus dou p of dou s at constant v this is one of the celebrated Maxwell's equations with the statement that d u is a perfect differential. So, you can see that T is expressed as a derivative with respect to one variable and p is expressed as a derivative with respect to another variable and it leads to one Maxwell equation and it is immediate that when you write the other three such quantities namely for enthalpy d h is equal to T d s plus v d p or h as a function of s a functional of s and p you want to write that and d a the Helmholtz free energy as minus p d v minus s d t that is a is a function of v and t and lastly the Gibbs free energy as d g is equal to v d p minus s d t that is g is a function of g p of t you can see that all of these are perfect differentials and therefore, it is possible to take the derivative of t with respect to p and the derivative of v with respect to s and make use of and equate them. Likewise for the derivative of p with respect to t and the derivative of s with respect to v can be equal to the derivative of v with respect to t and the derivative of s with respect to p can be made equal leading to the other three Maxwell's equations and these are all based on the fact that we have an exact differential and therefore, it is possible to express these differentials which are functions of two variables in the form of the celebrated Maxwell's equation. So, a lot of these things are taught in other courses the idea of bringing those things here is a to recall them and see that these are employed in physical chemistry and we will see similar things in the dealings of quantum mechanics at some point or the other perhaps not all of them, but these are related ideas and the use of differential equations in quantum mechanics has been of course, very very I mean it goes back to more than 80 years right from the Deyser-Schrodinger equation a lot of these differential equations have been solved. So, it is important to go back and refresh a bit of this mathematics the lecture notes on the website gives you more problems and assignments for you to solve and some of the solutions to these are also given on in the website. So, I would invite you to go visit the website look at those problems and solutions and if you find something strange or you find something interesting feel free to write to me on the by email or by your comments on the website. We will continue with the differential equations in the next lecture.