 Let us start with recapitulating what we did in today morning's lecture on quantum mechanics. We discussed that in Young's double sheet experiment we cannot determine from which slit the electron goes and simultaneously observe the interference pattern. Whenever we try to observe electrons it will perturb the experiment enough to destroy the pattern. So this is what we discussed in the morning lecture. If we see the interference pattern we may not be able to say how electrons reach this screen. Particles tend to behave mysteriously when they are as tiny as fundamental particles. Then we started looking at new equations which can describe the behavior of a particle which has an observable wave character. And we started trying this particular equation which is generally known as a wave equation. Electromagnetic waves satisfy this particular equation. So we said let us just try to look at this particular equation and see whether this can work for us. We will soon realize that it does not work because it requires certain modifications and that is what I am going to tell you. What we did? We initially just put various type of waveforms in this particular equation and just showed that those happen to be solution of this thing. Now let us go a little ahead in this particular thing and let us talk of something which is very important concept when we try to describe our things. That is about the superposition of waves. See whenever we are talking of interference pattern you remember you are talking of two waves which come and superimpose on each other and that is why there is an interference and therefore you get either destructive or constructive interference. Essentially when we are trying to make a wave packet we are trying to take a large number of ideal waves and then we are superimposing on them so that eventually that leads to a wave packet. What I want to say that superposition is actually a very inherent part, very important part of these wave theories that we have been talking. It means there is a particular wave and there is another wave. These two waves if they want to interfere then actually our psi should be equal to psi 1 plus psi 2 they must be able to superimpose on each other. If they are not able to superimpose then in principle if one wave goes one particular way another wave goes another way they will not interfere they will just keep on going exactly in the same direction. Now this superposition of waves should be inherent in your equation because once we say that the wave equation governs the waves let us say electromagnetic waves then superposition principle should be built in that particular way. So this is something which I want to describe now. So let us just read we know that electromagnetic waves even of different wavelengths superimpose I mean we get beats when the wavelengths are different. If we say that the wave equation is the wave governing the electromagnetic waves it implies that superposition should also be allowed by the wave equation. What it means that if there is a solution psi 1 of wave equation which we have seen that psi 1 of the form of A cos kx minus omega t or sin kx minus omega t is indeed a part a solution of this particular wave equation then and psi 2 is another equation which let us say slightly different value of k or slightly different value of omega or slightly different value of amplitude or whatever it is then psi 1 plus psi 2 must also be a solution of this equation. If that does not happen to be solution of this particular equation it means this particular wave equation is not allowing superposition. If it is not allowing superposition it means I will never get interference I will never get diffraction because all these things are based on the fact that if there is a disturbance which is given by psi 1 another disturbance which is given by psi 2 then psi 1 plus psi 2 is also a disturbance therefore I can add these terms and once I can add these terms then I can find out where there will be interference constructive interference where there will be destructive interference where the amplitude will be large whether the amplitude will be smaller that is inherent in this equation that psi 1 plus psi 2 or in general to be say a psi 1 plus b psi 2 is also a solution of this particular equation should also be a solution of this particular wave equation. So, let us first show that this is really true for this particular wave equation and the electromagnetic waves. So, I am going to the next transparency if psi 1 and psi 2 are solutions of the wave equation psi 1 plus psi 2 should also be a solution of the wave equation. Let us verify because this is actually a key of interference this is key to formation of wave packet because wave packet can form only when I super I am able to super impose these two disturbances or multiple disturbances for that matter. So, let us take let us assume that psi 1 and psi 2 are solutions of the same equation of the wave equation it means psi 1 satisfies this equation psi 2 also satisfies this equation it means d 2 psi 1 d x square is equal to 1 upon c square d 2 psi 1 d t square and because psi 2 is also a solution. So, we must have d 2 psi 2 upon d x square is equal to 1 upon c square d 2 psi 2 d t square. Now, what I will do I will just add these two equations and show that psi 1 plus psi 2 is also a solution to it. So, I have just added these equations remember this was my original equations. So, I have just added and because derivative of sum is sum of derivatives. So, therefore, I can write these equations in this particular form d 2 psi 1 plus d x square plus d 2 psi 2 plus d x square this I have just added. See remember c is constant. So, I have taken the c is 1 upon c square outside. So, it becomes d 2 psi 1 upon d t square plus d 2 psi 2 plus d t square and this I can write as d 2 d x square of psi 1 plus psi 2 is equal to 1 upon c square d 2 psi 1 plus psi 2 d t square. Now, you can very clearly see that psi 1 plus psi 2 is also a solution of this equation. Now, you can also see that instead of psi 1 plus psi 2 if I would have written a psi 1 plus a psi 2 this would also mean 2. Then what I should have done I would have written a here and I would have written b then there will be a here and there will be b here. This I can write as a and I can write as b here I can write a and plus b here. So, a psi 1 plus b psi 2 will also be a solution. In general if I have 3 waves or 4 waves or 5 waves or n waves or infinite waves they can all superimpose and they will all be solution by of this particular equation. So, this particular equation allows superposition and therefore we can really see superposition. See once we say that this is the equation which governs the wave nature of electromagnetic waves then superposition should be built in in this particular equation then only superposition is possible which we know experimentally that superposition does happen we do get constructive and destructive interferences. Now, let us see suppose the same equation was to represent a particle wave where the superposition will be allowed. See remember in particle waves also we have seen interference I mean remember when we are talking about the de Broglie wavelength at that time we did talk that electrons can deflect when electrons can deflect essentially means the electron waves can also superimpose. When we are talking about a wave packet being associated with a particular particle at that time also we talked that their infinite number of waves which are superposing to create a wave packet. So, obviously superposition has to be present in particle wave but in this case whether superposition will be allowed or not. What I want to show that if this particular equation is governing the wave nature of the particle then superposition principle will not be allowed. How? Let us just look at this before we start reading this particular thing. Remember inherent in this particular wave that this particular wave equation is this 1 upon c square which happens to be constant. In the case of electromagnetic waves c is speed of light and this c is constant for both the waves even though the two waves may have different wavelength may have different frequencies but they will have the same velocity. Therefore, 1 upon c square can be taken out and therefore, it has become possible that psi 1 plus psi 2 also becomes a solution of this particular equation. But if you imagine a particle wave, particles need not travel with the speed of light. In fact, they do not travel with the speed of light, they travel with a speed which is lesser than the speed of light. And in that particular case and of course the wavelength if I say that there are two particle waves which have different wavelengths. Obviously, different wavelengths means different momentum by de Broglie relationship. It means their velocities would be different. So, one particular wave may have c 1 here or rather v 1 here. Another wave will have 1 upon v 2 here. So, then in this particular thing I will not be able to take 1 upon v square out. I will be having 1 upon v 1 square plus 1 upon v 2 square. Then I will not be able to write this equation in this particular form. And we will see that the superposition would not be possible. So, if psi 1 is a particular wave which represents a particle of a particular wavelength psi 2 is another wave which is representing a particle of slightly different wavelength. Then psi 1 plus psi 2 will not be a solution to this particular equation. So, if this equation is to govern the wave nature of the particle, this equation will not be good enough. Remember as I have told in the morning we are quite a bit in dark. I have no idea we are trying to assemble our ideas to come to an equation which is likely to represent the particle nature rather wave nature of a particle. So, that is what I have written. If we take two particle waves corresponding to the same type of particle, for example, I am talking only one type of particles, let us say electrons. Of course, we have never seen electrons interfering with neutrons, let us say. So, we are talking of always a wave of a particular particle, of a particular type of particle. So, if we have the same type of particles, for example, electron with different wavelengths, they would not superimpose if this standard wave equation was governing it. So, how can we form a wave packet? We cannot work with this particular equation, we definitely need another equation. So, this is basically what I am trying to say that if they have been traveling with different speeds, then it will not be possible for us to have superposition principle valid. And remember superposition is inherent, is built in, it has to be built in because otherwise no interference is possible, no diffraction is possible, no wave packet formation is possible. So, we do need an equation where it is possible to superimpose different waves. So, how do we solve the problem? Now, you realize the basic reason why we were not able to get this particular superposition is this particular thing C1, which depends on the speed of the particle. Now, whatever is this particular constant which is coming in this particular equation, see remember here, you have d2 psi dx square, here you have d2 psi dt square. So, you must have a constant in between just if nothing else at least for dimensional matching. So, whatever is the constant appearing here, if that constant, it depends on any dynamical parameter, for example, momentum or energy or velocity, then we will always land up into this particular problem. So, what I must do? I must have some constant here, if at all possible, which does not depend on any of the dynamical properties of this particular particle. So, if this particular constant would have become independent of velocity or momentum or for that matter energy, then like here C is independent for electromagnetic waves irrespective of its wavelength C is same, then only I am able to superimpose. Similarly, in the case of particle, if this particular constant would have been independent of speed or momentum or energy, then this particular constant could have been taken out and superposition would have been possible, which as I say is very very important for us to have. So, this way say if the constant appearing in the wave equation was not dependent on dynamical properties like speed, momentum, energy, then superposition position could have been made valid. Now, let us try to think of a equation where let this particular superposition is possible and whether we can have a somehow a constant, which is now independent of these particular dynamical quantities. Let us just think a little bit about this. See, remember the equation that we had the wave equation was relating to del 2 psi del x square to del 2 psi del t square. There was equality sign and there was a constant. This constant I want to make independent of the dynamical properties. Now, this particular psi has to be given by a raise for E i kx or let us say the sign does not matter minus omega t, we can even put sign, it does not matter. But what I want to essentially point it out that or you can take, let us say A sin kx minus omega t does not matter. If I take the derivative once with respect to x, I will get k. If I take derivative twice with respect to x, I will get k square because once I take derivative with respect to x, I will get k first time, second time again it will get multiplied by k. So, I will get k square. Once I take derivative with respect to time, I will get an omega term. If I take derivative twice, I will get omega square term. So, in a normal wave equation, here I will be getting k square and here I will be getting omega square and that series omega square upon k square turned out to be the speed of light and that is what appeared in this particular equation which happened to be constant as I have said for case of electromagnetic waves. But now I want this particular parameter not to depend on the dynamical properties, but it should depend purely on the, in the case of, I mean purely it should be purely a constant. Now remember I am talking now a massive particle, when I am talking of this particular thing as a massive particle, it has to obey a condition and let us assume that it is a free particle, there is no force on it. So, there is no potential energy involved. So, what I will realize that k has to be equal to p square by 2m, kinetic energy of this particular particle has to be equal to p square by 2m. Now if kinetic energy has to be equal to p square by 2m, if something appears as a square of momentum and something which appears as just a single power of k, then p square and k, this will cancel it out on the two sides. What I am trying to mention is that remember k depends on the momentum, h cross k is momentum and we know that h cross omega is supposed to be the energy, which in this case I am taking as kinetic energy. Once I take this particular thing, I differentiate twice with respect to time, I will get k square. If I differentiate twice with respect to x, I will get p square, that does not help me. But on the other hand, if I take second derivative here, I will get p square, if I take only the first derivative, then I will get k. So, what I am suggesting is that why not try an equation which is something like this, del 2 psi del x square and put some constant let us say gamma and relate it to del psi del t. Now what will happen when I take the second derivative with respect to x, then in that case here I will get k square. When I take only one time derivative with respect to psi, I will get here omega. This omega is related to energy, this is related to momentum. So, I will get p square here, I will get k here and because I know that k is equal to p square by 2 m, this k will cancel, this k square will cancel with omega and then this particular gamma will become independent of the dynamical properties. So, why not try this particular type of equation and let us try to understand whether I can reach into a better way of looking into this equation. Initially, consider a non-relativistic free particle, there is no force, there is no kinetic energy. For this particle, we have the following relationship, k is equal to p square by 2 m. As I said momentum can be related to wave vector k while energy to omega, both involved involving the Planck's constant. If the second derivative with x was related to the first derivative with respect to time, things could be different. Why not try this equation? Why not try this equation where I am relating the second derivative of psi with the first derivative of p. As I have mentioned, then I will get a k square term here, so which will be p square term and when I take derivative, I will get omega which will be of the type of k and then p and k square will cancel giving you just mass. So, let us try this type of equation but there is a slight problem. Our problem is that if I take the relationship as let us say a sin kx minus omega t, let us take this as psi. If I take second derivative with respect to x, then first time sin will become a cos, second time again cos will become sin. You will have sin kx minus omega t here. If I take derivative with respect to time, then sin will become cos because I am relating the first derivative to the second derivative. Here there will be cos term, here there will be a sin term. Now cos and sin term will not cancel. So, we have solved one problem but we have added one problem. I could make this particular gamma independent of dynamical properties but then this particular solution is not at all a solution of this type of equation because here I will be getting sin kx, here I will be getting cos kx and they will not cancel it out. Advantage in the original dynamical equation, original wave equation was that here also there was second derivative, here also there was second derivative. So, sin first becomes cos then cos again becomes sin when we take the second derivative. Here also sin becomes cos and then cos becomes sin, then sin, sin term we are cancelling it out but now they do not cancel. But remember instead of this term had I taken psi is equal to A e raised power i kx minus omega t then I do not face this particular problem. Then what will happen? Once I take the derivatives twice with respect to that I will still get k square but this still remains in the exponential form. When I take the derivative with respect to time for the first time it will still remain e raised power i kx and I will get just omega. So, an equation of this type is a solution while this equation is not a solution. Problem is that this is complex but as I said earlier that we do not mind things to become complex. So, long the things which are actually observable they must be real. So, long they remain real I do not mind this particular type of no psi to become a complex quantity. So, I will be able to get a desired equation provided I accept the fact that this particular psi the wave that we have been assigning into a particle is complex. It is a very very important point for us to realize and that is what makes quantum mechanics much more abstract and much more no not easily comprehensible in that in a standard way especially for the undervalued students. So, I said traditionally when we take the Schrodinger equation we always take the equation of this particular form. So, equation has been obtained by using this type of functional form of psi. So, psi is equal to a e raised power i kx minus omega t advantages that in this particular case I could still cancel this particular term which is the exponential term and we could relate the second derivative of x with the first derivative of time. So, let us now let us take psi is equal to a e raised power i kx minus omega t I take the first derivative with respect to time. So, I will get minus i omega high out from here once I differentiate it will be differential of this exponential term which will give me the original exponential term multiplied by differential of this particular function i kx minus omega t this being constant because this is a partial derivative I will just get minus i omega. So, this minus i omega term has come once you take derivative with respect to x you will get again similar thing but you will get now i k. So, instead of i omega you get i k you take the second derivative you will get minus a k square in this power i kx minus omega t. So, on the left hand side you have got k square on the right hand side you have got omega whenever I am trying to write an equation which involves the second derivative of x with the first derivative of time. I substitute in this particular equation the original trial equation which was this equation del 2 psi del x square is equal to gamma times del psi del t. So, this is del 2 psi del x square this is del psi del t if I substitute this particular thing gamma turns out to be equal to 2 m upon i h cross of course this gamma turns out to be a complex number i is of course imaginary number also it depends on mass but as I have said that we have never seen let us say electron beam interfering with a neutron beam electron interferes with another electron beam and then the mass of the electron is same. Similarly, a neutron beam interferes with another neutron beam then the mass of m mass of the neutron it remains same m is mass of neutron which is remaining identical. So, therefore I do not mind that this constant is it contains m but it definitely does not contain velocity it does not contain momentum does not contain energy because if they would have been contained then I would not have been able to interfere even the electron wave one electron wave with a different electron wave of different wavelength. Now, we go back to our original equation and I substitute this particular equation and this equation can be written I for gamma I have substituted 2 m by i h cross which I have just now obtained this equation is generally written in a slightly different form. So, I write this as minus I multiplied by I h square which is minus h square from the both the sides. So, you get minus h square 2 m comes on this side. So, this becomes minus h square by 2 m and this instead of 1 upon i h cross becomes i h cross. So, this equation becomes minus h square by 2 m del 2 psi del x square is equal to i h cross del psi del t important thing to realize that we are relating the second derivative of psi with the first derivative of time. This in fact some people had asked the question in the morning or yesterday saying that the wave function has to be complex and that is the reason this wave function has to be complex. But remember they weigh I mean see whatever we have done till the in the quantum ideas they were all vague ideas they were not based on a proper theory. Now, this is the time when we are actually developing the proper theory we are developing our new concepts we are developing our new mechanics. Now, all the ideas that we have talked earlier should fit into this particular idea this becomes our basic equation from which we are going to derive not only the original ideas but more complicated problems that we are going to face in nature. Now, just a few comments on this particular thing the value of gamma indeed turned out to be independent of dynamical variables but depends on m which I said I do not care. Remember while obtaining this equation I have assumed the psi to be of the form A e raise power plus i k x minus omega t. Now, that is you remember we have earlier mentioned that this would represent a wave travelling in plus x direction because the direction of the motion of a wave always depends on the relative sign of the x term and the omega term. Now, this is another wave which is representing a wave which is travelling in plus x direction but in this case I have slightly different sign here I have a minus sign. Now, if I would have started the wave equation taking the general equation of this particular form my gamma would have changed sign it does not matter nothing would have changed in physics. But all I want to say that we always take a standard psi value in which the time dependent term is always assumed to be negative signs. Remember here the time dependent term minus i minus omega. So, time the coefficient of time is positive while normally which I take the standard value of psi this is always of this particular term where the time term is always negative this is important to realize because when we apply certain boundary condition in actual quantum mechanical problem we have to remember that time dependent term is always taken of the form of minus i omega t not plus i omega t otherwise my equation that I have obtained would have been different this equation has been obtained with a specific type of disturbance in mind where the time dependent term is of the form of e raised to the power minus i omega t. In that case the value of gamma would have been changed sign no physics would have changed conventionally we always use this type of equation. So, which has time dependent term of the form of minus i omega t the wave function turns out to be complex but that shall not pose any problem so long the observables turn out to be real. We shall later see the physical interpretation of wave function as we will be seeing it later which again I mean from now onwards we will realize that quantum mechanics is becoming more and more abstract and things are not so easily comprehensible in the sense of our normal traditional classical concepts but there is no other way of doing it that is the way we have to do and that is what actually makes quantum mechanics in a sense more interesting because it essentially stretches our imagination you know some something to something which is fairly abstract. Now, we define certain quantities which are called operative operators operators are essentially very very abstract concept and they are defined in order to maintain the consistency with classical mechanics. I will not be able to give more details as far as this particular course is concerned why a particular operator is defined in a particular fashion. But one can see later I mean if one does a course on advanced quantum mechanics one can see that these particular operators have been defined so that eventually we have a consistency with classical mechanics because we know classical mechanics does work for bigger particles. If I have to have a motion of a cricket ball of a motion of aeroplane or a motion of let us say train or a car or a bicycle I do not require quantum mechanics classical mechanics is good enough it gives me excellent results. So eventually this particular quantum mechanics should lead to classical mechanics in the limit of bigger particles so if we want to maintain this consistency these are the ways the operators have to be defined. Now the concept of operator is fairly simple though only the name appears to be somewhat big for example if I define let us say a sine operator it essentially means that if it operates on let us say a value x then my result is sine x. So I will call sine as an operator which when it operates on a quantity called x results into a different quantity which is now sine of x. Now let me define some another funny operator let call o operator which is let us suppose a square operator. Now we define this operator let us say such that when it operates on any particular quantity x my resultant is x square. So this operator o will now be a square operator so basically an operator means it operates on certain things so we write something on the right of it so it operates on that particular quantity and results into something which is different from the original one. Similarly I could define a cube operator which if operates on x then my result will be x cube. So we can think of infinite type of operators we can talk of a d dx operator when it operates on any function then for example d dx is also an operator if it operates on let us say any function of x my result is df of dx. So all these things this is a differential operator this is sine operator this is square operator we can define multiple type of operators and these operators are inherent in the quantum mechanics they are essential parts of the quantum mechanics and we have once we sort of formalize quantum mechanics to a particular way okay then we will see that these operators are very very essential part of the quantum mechanics okay. So now without giving any reasoning I will define certain operators first thing that I will define is what I call as momentum operator and to be more precise I am calling at px operator okay it is x component of the momentum operators and it is defined when I have written three lines here okay it means this is the definition of this operator this definition is minus ih cross del del x it means anything which we write on the side of px and for operator I have put this cap on this thing to realize that this is an operator and this is not anything which is simple because for px we have also used for the x component of the momentum okay. So there should not be any confusion so I have put a cap on this particular thing just to mention that this is just to sort of remind us that this is an operator and not just simple x component of momentum. So whatever is on the right hand side this d dx operator will actually differentiate this take a partial differentiation with respect to x then whatever is the result it will multiply it by minus ih cross then I will get px okay I will get the result of applying px operator okay what it essentially means that if px has to apply on let us say psi okay which could be any function of position in time the outcome of application of this operator will be that my result will be minus ih cross del psi del x. Now similarly I can define an energy operator an energy operator is defined as ih cross del del t okay remember there is no negative sign in the case of px there was a negative sign okay here there is no negative sign so this is ih cross del del t so it is taken as a derivative with respect to time like before it implies that if E operates on psi xt my result will be ih cross del psi del t. Now having defined these operators what I will do the equation that I have just now obtained okay by giving some sort of logic okay I will rewrite that equation in the form of an operator equation this is the equation that I obtained earlier which I have written on the top line okay by taking the value of gamma and taking the second derivative of x with respect to the first derivative of time now I am writing this I take 1 upon 2 m out here then I am writing minus ih cross del del x minus ih cross del del x remember square of an operator always means that you operate it twice it does not mean that you operate and then take a square. So if I am writing p square operator it means you first apply px then whatever is the outcome on that you again apply px on that okay so I can write this is minus ih cross del del x because these are anyway normal multiplication constants so minus ih cross minus ih cross minus minus will give you plus and h square there is h square and this i square will give you minus one so this is minus h square there is a 1 upon 2 m here and once you take the derivative with respect to x you will get del psi del x when you take second time derivative you get del 2 psi del x square so this is the term which has been just written in this particular form here it is comparatively simpler all I have written is ih cross del del t I have taken out and this is operating on psi xt now as we have seen that this operator is px operator this operator is also px operator so this equation in the form of an operator can be written as 1 upon 2 m px operating again on px psi xt and this also write I can write as an energy operator so eventually this equation which I have written earlier can be written in the form of an operator equation which is px square by 2 m this is the entire operator operating on psi xt is equal to energy operator applied on psi xt so this equation has been looked has been rewritten in terms of an operator equation now see remember I want to go to a step further this equation I have written assuming a free particle on which there is no force but that probably is not the problem which I am going to only solve it eventually I have to solve problem like hydrogen atom problem or any other complicated problem in which the particle will be under the influence of a force in that case the energy equation has to also include a potential energy now I want to write a general equation which will also take care of those type of particles which are under the influence of a force or they are having a potential energy function of course force and potential energy as you know are related to each other so now we come back to what we call as a Schrodinger equation now we move our consideration from a free particle to a particle under force for such a particle which we must assume that this force are conservative forces that are of course of interest for such a particle the following equation would be varied e is equal to p square by 2 m v in the case of free particle this v was 0 and how I have obtained that particular equation was multiplying on the right hand side or writing psi on the right hand side of both the equations then replacing energy by its operator and replacing p square by its operator I will do the same thing here because in the presence of a force this becomes my general equation relation of energy to p square by 2 m and kinetic energy and the potential energy so what I will do I will multiply by psi here I will multiply by psi here then replace e p square and v by its by their operators and it so happens that the operator of v happens just to be a simple product operator it means when v operates on psi the resultant is just v multiplied by psi so that makes it makes things very very simple so this is what I am doing multiply each side by psi and replace and remember psi has to be written on the right hand side so probably I should not have said multiply I should have said just write psi on the right hand side of the each of the equation and replace their you know dynamical quantities by their corresponding operators of course at the moment I am assuming only one dimensional case okay we can generalize this equation to three dimension later so I have just multiplied this by psi replace e by i h cross del del t p square by minus h square by 2 m del 2 del x square and v as I have said is a simple multiplication product so operator so this becomes v psi now in three dimensions you can write this equation in terms of a general operator which is generally called a del square operator these equation is called this is a the general equation which is called time dependent Schrodinger equation this is our basic equation which we believe represents the dynamics of a particular particle represents the behavior of a particular particle with significant wave character in that or observable wave character in that particular thing which is now our new equation and we expect everything to follow this particular equation which is the Schrodinger equation now most of the time at least not always most of the time but at least some of the interesting type of times at least as for this particular part is concerned most of the time you will find that this potential energy is not an explicit function of time it just depends on x y but it does not depend on time it means if you are at the same position at all the time this particular potential energy will always remain constant of course if the particle changes the position the potential energy will become different but at the same position it will always remain always remain same if it so happens when potential energy is not a specific function of time then this particular equation can be written in a slightly different form and that is what we call as a time independent Schrodinger equation and as far as this particular course is concerned we will be using time independent Schrodinger equation more than time dependent equation. So, let us now come to what we call as a time independent Schrodinger equation if the potential Vx is independent of time it is possible to separate out the special and time part of the Schrodinger equation and this is a method which we call as a method of separation of variables this is a very very standard method which we will be using also later but very very commonly used in mathematics. So, if we have to solve a particular equation which is consisting of let us say multiple variables in this case two variables we can separate this particular equation into two separate equation for example if it contains three variables it may be possible to separate the equation into three different equations this is what we call as a method of separation of variables. So, what I assume in this particular thing let us start with the assumption that I am able to write psi xt as a product of two functions one let me call as phi another I call as f and let us assume that this phi is purely a function of x it means it does not contain any time dependence. Similarly, this f a function is purely a time a function of time it does not contain any x term. So, this is purely a function of x this is purely a function of time and I assume that psi for a given problem can be expressed in this particular form. If it cannot be expressed then we will find that this is I mean when I substitute in this particular equation I will find that this does not happen to be a solution I will not be able to separate the variable out. If this happens to be a solution then I will be able to get two different equations one corresponding to phi another corresponding to f and therefore I can solve two independent equation one of the equation will not contain time another equation will contain only time. So, let us try to do this particular trick of applying method of separation of variables and try to see whether I can get a simpler equation. Now because I have assumed psi to a function of phi and f I substitute this in the Schrodinger equation and I am again going back to the first to one dimension because that just makes life simple we could have done the same thing for a three dimensional case it would not have made any difference. So, we write I mean psi as product of two functions this was my original Schrodinger equation I substitute this particular thing here. So, for psi I write phi x into f t here also I write f x into f t here also write phi x into f t. So, all I have done that this psi which was coming appearing in the original equation I have written a product of the two functions in place of that psi. Now let us look at this particular equation when I take the left hand side of this equation this is partial derivative with respect to time it means anything which contains x has to be treated constant this phi x contains only x it does not contain time. So, this particular thing has to be treated as a constant when I take a partial derivative with respect to time. So, this phi x will come out of this particular thing then I have to take derivative of f with respect to time but this term contains only f it does not contain any x. So, instead of writing partial derivative I can write as full derivative I can write d dt. So, this is what I have done here this ih cross is anyway here phi x has to be treated constant because it does not depend on time. So, this I can take out here. So, I have taken it out here then d del del t operates on f but f is purely a function of time. So, I can write is a complete derivative df dt. Similarly, I do here here I have del 2 psi del x square therefore f t has to be treated as constant. So, I take f t out then I have to take second derivative of phi but this contains only x. So, I can write as a full derivative d 2 phi x v x square plus v x of course this term remains exactly identical. Now, what I do I divide this particular thing by phi x into f t divide I this term by on both the sides and of course we can say that the particle exists the wave function cannot be 0. So, phi x multiplied by f t will not be equal to 0. So, I can divide both the sides by this particular term if I divide this thing on the both the sides let us see what happens divide by phi x into f t if I divide by phi x into f t. So, what will happen this phi x will cancel and you will get 1 upon f t here. So, this is what is happening here I am getting 1 upon f t here. Similarly, if I divide this here with phi x then this phi x will cancel it out where I divide by phi x by f t then f t will cancel it out and you will get 1 upon phi x here. And when I get phi x divide by phi x f t here I will just get v of x this is what is happening here. So, here get I am getting h square by 2 m 1 upon phi x and then this v just remains because I have already divided by phi x and f t. So, just remains v x. Now, we come into a very interesting situation this is an equation here are the left hand side of this particular equation everything is only a function of time there is nothing which depends on x on the other hand on the right hand side everything is a function of x there is nothing which depends on time. Now, for example, if time changes for the same value of x then this term may change but this term will not change or for the same time I am looking at a different value of x then this term will change but this term may not change this term will definitely not change because it is not a function of x. And if I want both these things to be always valid then the only possibility is that this also a constant and this also a constant if both the sides are equal to constant then only it is possible to maintain this equality forever irrespective of x irrespective so I can write this particular equation as this particular thing has to be constant which I am writing as e and this particular thing also obviously has to be equal to the same constant because this has to be maintained equal to this. So, this equation has now been separated out into two different equation one of this equation contains only time another equation contains only x. So, I can solve this equation I can solve this equation separately. Now remember this has been possible only because x did not contain v potential energy did not contain the time part if it had contained time part explicitly then I could not have separated this equation because this particular part on the right hand side would have contained a time dependent part nor I could have put this v here because then it would have contained a x part then it would not have been possible for us to separate out this equation this separation has been possible because we have assumed that v is only a function of x and is not a function of time. In that case I have been able to solve these two equations. Now one thing which is very interesting that this particular solution of this particular equation this equation there is no v term when I go from one problem to another problem what is likely to change is only the potential energy. The force under which the particle is moving that is the one which is likely to change and therefore v of x is likely to change. What is interesting in this thing that as far as the solution of the first equation is concerned this will always remain same it is not dependent on v x it is only this particular equation which is dependent on v x and this will be different the solution of this particular part phi will be different for different problem but so long v x is only a function of x its time dependent part will always be identical which will be just a solution of this particular equation. Now let us try to first solve this equation which is a very very simple equation to solve as all of you know this is standard differential equation. So, let us solve this particular equation let us write this equation in this particular form and when I write this particular equation in this particular form I can just solve this equation this is an exponential solution. So, f t will be of the form a e raise power minus i upon h cross times time p and this is an always an equation I mean whenever I mean this particular thing will always be constant irrespective of whatever is the potential energy. Now I want to make sort of a strong statement here I hope all of you would agree that if this equation is a general equation then this should also be varied for a free particle and for a free particle I know that the solution is to be written on the form of psi k x minus omega t. Now as we have said that omega is going to be h cross omega is going to be e. So, this omega has to be written as e upon h cross where e is energy all right. So, now I know that time dependent part is of this particular thing is e raise power minus i e by h cross time and as I said this particular time dependent part does not depends on phi. Therefore, what appears here as e or what appeared as a constant in that particular equation can be identified as the energy of the particle. So, this energy of the particle because this is what comes in the case of free particle. So, when I compare this particular thing for the free particle I can identify this particular constant e as the energy. So, that is what I have done in that particular equation. Comparing with the standard wave equation we interpret e as the energy of the particle. Now what remains is the second equation which is what I have written here and this is what you call as time independent Schrodinger equation. Remember the solution of this particular equation can be real because phi here there is no i here. So, the solution of phi in this equation may yield you a value of phi which is always real but you must always remember that actual wave function is this multiplied by e raise power minus i e t by h cross. There are many problems in which this particular term will not matter which I will be discussing later. But if you want to write a general wave function, general solution of this particular equation the general solution is always of the form phi x into e raise power minus i e t by h cross which always will make the wave function complex. Though solution of time independent Schrodinger equation can be real but overall wave function will always be complex. So, this is my time dependent Schrodinger equation which I am writing in this particular form. This is a three dimensional form. So, when you are writing in this particular form where this is a del square operator which is given by del 2 psi del x square plus del 2 psi del y square plus del 2 psi del z square. Now of course, there is an advantage this if you are not using r theta phi I mean if you are not using Cartesian coordinates x, y, z coordinates but you are using r theta phi coordinates in that case we can correspondingly change del square operator by its value in any other coordinate system. This is my time independent Schrodinger equation and remember when v is purely a function of x psi x t is phi x into e raise power minus i e t by h cross. So, we have obtained the equation. Now let us try to see what can we do with these equations. See unlike what I am going to do in relativity here I am not going to give the postulates of quantum mechanics immediately because they are as I said fairly abstract complex. So, what I will be keeping on doing will be introducing these concepts one by one slowly and slowly I mean this is what I have found it is quite helpful when I am dealing with the students because you do not give them a big shock right in the beginning you start slowly and slowly exposing them to more and more complex ideas. So, then students are able to take these concepts somewhat easily. So, now we will start talking about this what is this psi. Once I have got an equation suppose I know what is the value of v of x and I have solved this equation and I have obtained this value of phi and hence I have obtained the value of psi what I am going to do with this particular thing. How do I physically interpret? Remember this psi is what we are saying is the wave which is associated with this particular particle. Now we have also said that this wave is not a realistic I mean that sense is a sort of a complex wave. So, in that sense it is not very easy I have seen it is so different from the normal waves that we are used to it the normal waves like you know let us say water waves or for that matter sound waves or electromagnetic waves where we can really think of something which is a real which is being associated with these particular things. While in this case whatever we are talking is a complex space this wave function or wave which is representing the particle is having some sort of complex behavior but the outcome the wave I am going to use this equation must provide me something which has to be real so that it has a physical meaning. Now, let us first try to give an interpretation of what is this psi what does this psi represent this particular wave what is eventually it represents and as some of you had already pointed out in the morning this does represent the probability. So, let us try to understand this wave function which is actually not again all that very simply understood. So, I am trying to give some example to make things somewhat clearer. I have written in a form of sort of a postulate but you know as I said all the postulates I will sort of I mean not all the postulates as quantum mechanics we keep on developing you find that many more postulates are coming out they are not as clean as for example Newton's laws when you see Newton's first law second law third law everyone understand what is first law what is second law. Similarly, when you talk of the postulates of special theory of creativity which will talk first postulate second postulate everyone understand it but here in the case of quantum mechanics they are sort of mixed up. So, we are talking of one of the postulates quantum mechanics which says that a microscopic particle is described by a wave function which you can think of a wave associated with the particle and we assume that this contains all the information that we can have about the physical properties of the particle. So, all that you can obtain about a particular particle is its wave function that is going to contain all the information all the information that you want about the dynamical properties of this particular particle is contained in that particular wave function. This is all that you can obtain only from this particular wave function you have to extract appropriate things. How do we abstract that we are going to discuss in our subsequent lectures. The Schrodinger equation gives the position and the time dependence of the wave function. So, how do I obtain the wave function? Solve Schrodinger equation first we say this postulate says that there is something called wave function which contains all the information about the particle and this Schrodinger equation will has to be solved in order to obtain this particular wave function. So, second statement says that how to obtain this wave function you obtain it by solving Schrodinger equation. Then, next thing that we say that probability of finding the again switching back to one dimension because one dimension is rather easy for us to understand especially it is much easier to make it understand to students. As I say we want to introduce complexity slowly rather than talking of all the complexity right in the first go. So, the probability of finding the particle between x and x plus dx is given by psi x t square it means psi star psi dx. See I hope this is clear that when we are talking this essentially means psi star x t multiplied by psi x t this is what I mean by psi square. So, we take the complex conjugate of psi multiplied by it obviously you will get something which is real but because any complex number if you take complex conjugate and multiply you will always get real number. So, this particular thing will always be real this will tell me what is the probability of obtaining the particle between x and x plus dx. What does this probability mean? Let us first understand what is this probability mean? So, let us define this probability is exceedingly important to understand this thing very properly. Imagine a very large number of separate identical particles described by the same wave function. So, you assume that they are very large number of particles described by the wave functions psi x t. Remember somebody had asked in the morning also that can we talk of a single particle of course we talk of the single particle wave function but if I want to look at the probabilistic character I have to have a large number of systems. All right. So, let us assume that we have very large number of separate identical particles and all of them are described by exactly the same wave functions psi x t. If a measurement of position is done on all of them at the same time t this particular probability means that the result will not be identical that is what is important. For example, if we are sure that they will all be at one point okay then the probability is one. For example, if we are toss a coin and we are sure that the I am always going to get head then the probability of obtaining head is one. But if I do let us say 1000 measurements then sometimes I will get head sometimes I will get tail okay this is just to draw a parallelism similarly if there is a probability associated with the particular of particle being formed at a particular value of x then very very large number of particles I try to locate the position of that particular particle at a given time then I will find that the results are not identical. If the probability of finding the particle let us say between x naught and x 1 is 0.1 it means then in 10 percent of measurement the position of the particle would be formed between x naught and x 1 in 90 percent of the measurements on the average you will not find the particles to be lying between x naught and x 1. So you have a particle you have multiple particles okay and multiple particles you make the measurement exactly at the same time find out what is the location of the particle and when you find out the location of the particle you find that in 10 percent of the measurement the particle will be formed between x naught and x 1 if the probability of finding the particle is 0.1 between x naught and x 1. See what is important to realize that this probability does not talk of making measurement on the same particle at a different time this is very very important to realize the probability is not defined over a measurement on a single particle. Suppose a measurement was made at a time and the particle was found to be very very close to x naught okay if we make immediately a second measurement remember the word important is immediately it will still be found there okay it is not that that if I make a measurement of particle and find the particle here okay let us say I find the particular particle here then immediately after that I make the measurement I will find particle somewhere here then there is no meaning of experimental reproducibility it means what is the experiment sometimes I make a measurement here and immediately I make the next measurement particle is somewhere else remember probability does not talk about it when we are talking about the quantum mechanic probability it does not talk about measurements being done on the same particle okay they are a very large number of identical particles and these measurements are being done on different particles described by the same wave function that is what is most important aspect of this particular probability I will give you more examples in our later classes to make this particular thing very very explicit so let me read it again defined over a measurement on a single particle the probability is not defined over a measurement on a single particle suppose a measurement was made at a time t and the particle was found to be in the close vicinity of x naught if we make a measurement immediately afterward it will still be found there experimental reproducibility is necessary of course at a much later time the wave function may again evolve under the Schrodinger equation and then of course it may change its position okay so the word what was important was immediately okay okay at a later time if you leave the particle for another 10 seconds or 10 minutes okay the particle may again change its position and may again evolve under the Schrodinger equation okay but if you make the measurement immediately after that I must find the particle there itself measurement on different particles at the same time how may our may not yield the same result when I am making measurements on different particles then I do not get the same result that is what is important and that is what is inherent in this particular probability so I think I will stop here I mean our time is over 430 if we have some questions I can sort of answer now you can post your questions okay of course the centers we have where there is no question if they want they can leave it there is no problem we will take for another 5 10 minutes some questions and then yes please tell your question good afternoon sir sir I have question I have a question sir first question is yeah in intensity will increase or decrease yes if it will decrease is it true sir no see thing is that you know intensity as I said I mean this psi square I mean I I presume you are talking of the wave function then of course psi square represents the intensity and it is this intensity which is proportional to the probability all right so a larger intensity means a larger probability of finding the particle there I am not sure whether I have understood your question and whether I have answered your question it is able to find the intensity losses during interference see I do not mean what do you mean by intensity losses see when you are talking of interference and you are say finding out at some place intensity to be low it means the probability of finding particle there is low if there is certain point where you are finding the intensity to be large it means the probability of finding particle there is large I mean this is what I have been mentioning also when we are talking about the interference experiments okay sir thank you PSG college in the case of electromagnetic wave yes is there an interference in dispersive media well I mean most of the media that we are trying to see are you know partly dispersive okay I mean it is not really not I mean we are not never performing experiments in non-dispersive media but of course depends on what is the dilatry constant things like that I think professor Ghosh is going to talk about electromagnetic waves a little bit more in general okay at that time you can raise this question about you know what happens when there is a media okay in the case of matter waves yes do you think the wavelength or whatever changes with the medium is there a does the medium matter for a matter being see as far as the Schrodinger equation is concerned you know there is no I mean effect of media in that sense because see this does not depend you know the on the dilatry constant of the medium or the permittivity or permeability of the medium so at least I do not see any way how does it matter see the media may change alter its momentum therefore the de Broglie wavelength may change okay let me put it like that if you want to really draw a parallelism of media which we normally draw in the case of particle wave it is only in terms of potential energy okay if you talk of a medium in it is electromagnetic wave the closest parallel to that particular thing in the case of matter wave is when a particular particle moves from one potential to another potential so like a step potential you know example of some of which we will give later okay that probably mimics the media more securely in that case if you talk of that type of media where potential energies are different okay of course the wave equation will get affected the kinetic energy the energy will get affected the probabilities will get affected okay so they so long we talk take of that parallelism of medium I think okay it does depend on media but if you are talking about medium like let us say air or water okay that has a totally different connotation as far as quantum mechanics is concerned it doesn't matter so so I mean that I am saying so as far as kinetic energy is concerned it depends on what are the forces that are experiencing see it's almost the media matters almost exactly the same way for example if you fire a bullet inside the water okay what will happen to the bullet okay this is basically interacting with whatever are the forces there so whatever are the forces as far as this particular particle is concerned they have to be mimicked all right so media will matter so long we can actually find out what in what way this particular media changes the potential energy of the particle all right yeah so if you take a complex representation yes for a wave function yes right so but we don't find any analogous in a real picture right you can find any if it's only for a mathematical it's yes that's right in here so it is only for a mathematical convenience in that sense that's right no I don't lose any generality but still we could get some physics out of it that's the thing is that you know if you want to really talk about the wave function it does represent a probability wave in a sort of a complex medium okay but this particular method can be used to derive something which is realistic about this thing as probably all of you would be knowing because I am sure you at some level you would have done an advanced quantum mechanics course it's not really necessary to talk about the wave mechanics we can also approach quantum mechanics is a totally different fashion by using totally operators okay but to introduce them to first year students is not all that simple so I am sort of avoiding because I am sort of trying to talk at a level which we can actually explain to the first year students in that case you know we are not talking about that aspect we are still talking about the wave aspect so this particular complex thing which we have introduced as a complex wave pattern okay which yields us some realistic results okay need not be present at least in some problems you can do totally in a different fashion and obtain the same results of the you know whatever dynamical quantities we are interested in okay but there is no approximation in such case when we take only the real part right no no you see we are not taking the real part so see for every quantity that we are measuring we have to find a method how I evolve that particular information using this particular wave function which we will describe later when we are talking about these things okay I am not taking real part it is unlike that you know see for example when I am taking probability I am not taking the real part I am taking complex conjugate and multiplying by it okay so it is unlike the circuit analysis where we are used to taking only the real part it is not that circuit analysis type we are for once we have obtained wave function how do I obtain realistic information for that we have to have a postulate and all that I have told at this particular moment is the postulate of finding the probability okay and that is not taking by real by taking real part but by taking a complex conjugate and multiplying by that that is okay but the system the real system is represented by a real part alone right but what is the role of the complex part in see that that's the only way I can describe that that's the only way I can describe the wave function I mean I that's the only way I can talk about the wave nature of the particle okay there is no other way of talking about it you know I I cannot draw a parallelism with a real system because you know the system the realistic system is a particle which has a significant wave character what happens if the forces are not conservative well that's a little more difficult problem you know because I'm not very sure whether this can be done in a very simple fashion of the quantum mechanics I'm not very familiar with you know if you're talking I mean the thing is that the level at which the quantum mechanics is applicable okay see we are not I mean we are not going to solve a problem of a block moving on a inclined plane and introduce things like friction which I which I've done on conservative forces okay I'm going to talk at a much more atomic level where I mean I don't see immediately a sort of a non-conservative force in that sense thank you sir welcome probably we'll go to two more questions sir what are canonically conjugate variables sir what are which is used in canonically conjugate variables see I'm not very sure actually mean I mean when we are talking conjugate variables we talk only those variables amongst which there is a uncertainty principle exists for example x and px okay I hope that that's what you mean okay or let's say energy and time or angle and angular momentum so these are sort of conjugate variables okay amongst which a uncertainty principle exists the same quantities are evolved from classical mechanics from pysand brackets that's correct that's right in fact you know when you're talking about the quantum mechanical aspects many of these things they concept to Hamiltonian etc all those things have actually come from the traditional kinetic classical mechanics which have now been a sort of expanded to be to include in the in the quantum mechanics yes you're right so I just want to know what is the special about only energy and time position and momentum and as you said here as you use your use you have it see things that eventually if you take a commutation relation it does not they do not commute with respect to each other that that's how you know that we eventually come to the uncertainty principle but you know I don't want to introduce these things right at this particular level because you know that that's much beyond the scope of this particular course sir you can we say that the light photon is less localized compared to x-ray photon since it's well in this shatter no no see what is it's localized depends on the conditions whatever the condition that's going to be put in you know it does not depend on what is the wavelength okay what type of condition that you are letting it go through okay if the localization depends basically on that particular aspect it essentially depends I mean the eventually the wave function localization also depends on the boundary conditions because the particle nature of microwaves radios the long wavelength radiations are never a any practical examples are not there that's why I'm asking no no no see things that you're right the wavelength is very large there okay but still as I say I do not immediately see a role of the wavelength being there I mean see what it could mean that the probability may be constant or whatever it is but you know eventually the the determination of the wave packet or the length of the wave packet will depend on the type of the boundary condition that you're applying so eventually if you're looking at why waveguide okay you are putting them some putting their microwaves inside a waveguide it's the waveguide which is sort of localizing the electromagnetic waves okay thank you Gandhi institute sir do you get any idea of quantization of mass from quantum mechanics well I at least I'm not aware of it because definitely not aware of any quantization of mass type of because we're talking about in a sense you can always talk that there are certain fundamental particles and you do not know I mean how to divide those particular particles so in that sense you can talk if you want about quantization of mass but I do not think that's a part of any quantum mechanical aspects at least I'm not aware of course the the massive particle you know probably a result of particle physics about which I'm not aware personally that's not my area so how the masses of these particular particles come in and things like that that I'm not very very I'm definitely not familiar with that particular thing okay thank you is that one more question why do you get the application of time dependent score in your equations see for example when we were talking of transitions etc you know at different levels that we talk of the time dependent perturbation theory so we do talk of time dependence okay for example you know whenever there are things which depends on time for example a typical example which I have given is transition if a particular atom has gone to an excited state and you know it sort of emits a photon and comes comes back to the ground state or a photon get absorbed and it goes to excited state all this becomes a part of the time dependent shooting equation thank you thank you sir okay thank you so we are closing now and I if you have any questions please do post it and we'll try to answer and all I want to say that you know because we are getting large number of questions so we'll be able to answer only those questions which are really pertinent to the type of lectures that we are giving here thank you