 We have established the tenets of quantum mechanics. Now we are going to slowly get into systems in which quantum mechanics is applicable and this how we when we will see how we can build a description of these systems. The first system that we are going to talk about is a particle that is as free as this colorful little bird here we call it a free particle. What is the meaning of a free particle? We will come to that. But before we do let us remind ourselves that so far we have discussed Schrodinger equation. We have said that Schrodinger equation yields wave functions and this wave functions sorry about the mistake in spelling here. These wave functions are interpreted by Max Born to be associated with what we call probability waves. So mod psi psi mod psi square is equal to the probability density that is what we have said. So far we know that psi has to be a solution of Schrodinger equation. It must be normalizable. Why? Because if psi psi star is probability density then we remember that psi psi star or the way it is conventionally written psi star psi d tau is the probability of finding the particle in a small volume element d tau at a particular position. So if it is Cartesian coordinate d tau will be dx dy dz. So probability of finding a particle somewhere in space at this point is given by the probability of finding it in this small little box here which is called the volume element and we are saying the volume of this volume element is d tau this is probability and psi star psi is probability density. Naturally you have to find the particle somewhere in space. So we write integral psi star psi d tau over all space I will just write all space here. I think in the last class I had written minus infinity to plus infinity I will not do that because you see what infinity might mean here it is not necessarily infinity infinity. So integrated over all space that has to be equal to 1 that is your normalization condition. Next we said psi has to be continuous and we will see how this is very useful in our discussion maybe not in this module in the next module. Now d psi d q this has to be continuous in q we are going to see that this is not a very rigid condition this does not really arise out of bond interpretation it arises out of the requirement that we are writing a second derivative that is why psi must be single value this of course comes from bond interpretation you cannot have more than one probability density for a given point and it has to be quadratically integrable that these are in a nutshell a the conditions that are wave functions must satisfy and as we are going to see in the next couple of modules quantization arises out of application of these conditions imposition of these conditions and we will do that and we will study these three systems. Today we will talk about free particle then we are going to engage that particle put it in what is called a square wave potential that is called the particle in a box problem and finally we will go on to discuss hydrogen atom but before that I want to talk a little bit about Eigen functions Eigen values and expectation values and I have been a little lazy that is why I have not really written these things here I will write it in real time. So, as we said the postulates of quantum mechanics tell us that for every observable we should be able to construct an operator here I am writing a and when this operator operates on the wave function psi we get an Eigen value equation where the same psi is multiplied by a number that number is called the Eigen value. It is not necessary that always a system will be described by a wave function that is an Eigen function of the operator that we desire and that is what we will need to understand. You might remember that in the postulates we have said that the wave function has to be a sum of mutually orthogonal Eigen functions. So, let us say that I have a wave function psi for some system and we are actually going to encounter system like this very soon which is a linear combination of say a 1, phi 1 remember I do not necessarily have to write psi for wave function I can write anything. So, a 1, phi 1 plus a 2, phi 2 plus a 3, phi 3 and these phi's are such that they are Eigen functions. So, let us say a hat operates on the wave function phi 1 to give us I have written a already. So, let us say c 1, phi 1 let us say a hat operates on phi 2 to give us c 2, phi 2 and a hat operates on phi 3 to give us c 3, phi 3 is psi an Eigen function of a in that case. Let us see remember these are all linear operators. So, when a hat operates on psi what do I get? I get a 1 c 1, phi 1 plus a 2 c 2, phi 2 where is the c 1, phi 1 c 2, phi 2 coming from here. Plus a 3 c 3, phi 3 now see is there any way in which I can take this a 1, phi 1 plus a 2, phi 2 plus a 3, phi 3 out this will happen only if c 1 equal to c 2 equal to c 3 is not it. If c 1 equal to c 2 equal to c 3 equal to some c then only then and then only I am going to get a hat psi is equal to now c will come out and inside you get a 1, phi 1 plus a 2, phi 2 plus a 3, phi 3 which is essentially psi. So, now I get c multiplied by psi. So, the problem now is that that is the only condition in which it will be an Eigen value equation. So, a linear sum of Eigen functions is an Eigen function if the constituent Eigen functions phi 1, phi 2, phi 3 here have the same Eigen value. This is something that we are going to use when we talk about p orbitals later on, but this is a very specific case. What is the general case? The general case is that this is not an Eigen function if c 1, c 2 and c 3 are not equal to each other then what happens? Then what are the values that I will get if I perform some experiment? Remember we have talked about wave function collapse. We had said that the system before measurement exists in an entangled state. When you perform a measurement it collapses into a particular wave function depending on the experiment you perform and that is what you see. So, when you perform an experiment on the system I am going to see either phi 1 or phi 2 or phi 3. For a given experiment for rather let me say for a given observation the system is going to reveal itself to us either as phi 1 or phi 2 or phi 3. And one thing I forgot to say is that phi 1, phi 2, phi 3 let us say these are all normalized wave functions. So, you are going to see either phi 1 or phi 2 or phi 3. So, the variable a that you measure you are going to for some experiments you are going to see a value of a 1 for some experiments sorry for some experiments you are going to see a value of c 1 for some experiments you are going to get a value of c 2 for some you are going to see a value of c 3. What is the average value that you will get? As we said you are going to see either c 1 or c 2 or c 3 when you perform a particular observation or particular experiment. What is the average value that you will see? Average value of let us say c or I could have written a hat also that is going to be actually I am using a bad convention I should have written c as coefficient and a as eigenvalue but it is fine. So, that is going to be integral and I hope you remember this direct notation a 1, phi 1 plus a 2, phi 2 plus a 3, phi 3 then a hat operating on this second vertical line is absolutely not necessary it is just written so that it looks a little better a 1, phi 1 plus a 2, phi 2 plus a 3, phi 3. What will I do? First of all I will work with the ket vector I will give the bra vector as it is and write a 1, phi 1 plus a 2, phi 2 plus a 3, phi 3 and now what happens when a hat operates on phi 1 I get c 1 and a 1 is already there. So, in the ket vector I get a 1, c 1, phi 1 plus a 2, c 2, phi 2 plus a 3, c 3, phi 3 I hope you understand what I am doing when a hat operates on phi 1 I get c 1, phi 1. So, when a hat operates on a 1, phi 1 I get a 1 multiplied by a hat phi 1 that is equal to a 1 multiplied by c 1, phi 1. These are linear operators remember it is phi 1 and similarly we can proceed with phi 2 and phi 3 also. Now, I will open the bracket and write what I get this is what I get integral a 1 square will come out c 1 I say I have missed something though in the denominator I will get something denominator is also there and I am not writing it in the first time second line at least I should write. In the denominator I get integral a 1, phi 1 plus a 2, phi 2 plus a 3, phi 3 multiplied by a 1, phi 1 plus a 2, phi 2 plus a 3, phi 3 because you know there is no guarantee that this psi is normalized. So, in order to normalize it I have to divide it by this integral psi star psi also all right. So, let us go ahead and open this now what we get in the numerator first term phi 1. So, remember this phi 1 star and this is phi 1. So, I get well something like this a 1 star a 1 because the coefficients can also be imaginary multiplied by c 1 integral phi 1 star phi 1 that is the first term I get. Similarly, second term I get will be from say phi 2 and phi 2 let us just take the phi i phi i terms first what will I get there a 2 star a 2 c 2 1 integral phi 2 star phi 2 over all space plus a 3 star a 3 c 3 integral phi 3 star phi 3 over all space and then I just write one more term and not write anything else because what will be this one for example, if I take a 1 phi 1 and a 2 phi 2 then I will get a 1 star remember anything in bra vector is actually complex conjugate a 1 star a 2 then I will get c 2 is a constant it comes out integral phi 1 phi 2 over all space and I will get many more terms like this the problem is integral of phi 1 phi 2 over all space is actually 0 why because they are the wave functions that we said we are defined that phi 1 phi 2 phi 3 are mutually orthogonal. So, the only terms that will survive are the ones on the top what about the denominator sorry the three terms that I have written out this is going to be 0 what is the denominator denominator is going to be a 1 star a 1 this time I will just get phi 1 star phi 1 plus a 2 star a 2 integral phi 2 star phi 2 plus a 3 star a 3 a square better correct it integral phi 3 star phi 3. Now, see these phi 1 phi 2 phi 3 they are all normalized. So, I can write like this now I can just write I can be 1 or 2 or 3 integral phi i phi i over all space is actually equal to 1 as they are all normalized. So, what does the average value turn out to be it turns out to be a 1 star a 1 c 1 plus a 2 star a 2 c 2 plus a 3 star a 3 c 3 divided by a 1 star a 1 plus a 2 star a 2 plus a 3 star a 3 all right this is your average value. So, it is important to understand what we have just seen this is a very important aspect of quantum mechanics. Now, what we have learned is that when you perform a particular experiment you are going to experience one of these Eigen functions. So, you will get one of their Eigen values fine and then you are going to when you perform an average measurement what will be the number of times you observe c 1 or c 2 or c 3 that depends on a i star a i. So, mod square of the coefficient of the this wave functions gives you the fraction of times you are going to see a particular Eigen value all right this is an important thing in quantum mechanics that we needed to know before we can go ahead further and we have also learned in the process that if all the Eigen values are the same then the linear combination is going to be an Eigen function as well all right. That being said let us now go ahead and talk about our free particle the meaning of free particle is that if you write the Schrodinger equation this v of x is going to be 0 free particle means we have created a situation where the particle is not under any field of anything in the universe. So, you can think it is like you have an atom and you have provided ionization energy to the electron. So, the electron has just come out of the potential of the atom or you can think of a satellite that has just come out of the gravitational potential of earth that kind of a particle is a free particle. In fact, you can create free electrons and stuff in equipment like what are called synchrotrons ok. So, v of x is going to be equal to 0. So, the Schrodinger equation then for free particle boils down to this minus h cross square by 2 m multiplied by d 2 dx 2 psi of x. Now we are working in one dimension to start with if you are working in three dimensions if we said the particle is free in all directions then I just have to add d 2 dy 2 and d 2 dz 2 and psi would have been a function of x and y and z right not very difficult to extrapolate from here. So, this is Schrodinger equation for a free particle in one dimensional space and this here minus h cross square by 2 m d 2 dx 2 this is the kinetic energy operator. It is often called the T operator and here since it is along x direction I will write T hat of x T hat x this is the kinetic energy operator. So, this E of course, is going to be only kinetic energy for a free particle as we said there is no potential energy anyway. Now this is a differential equation and we are familiar with these differential equations I hope. So, the trial solution that we can think of is psi of x equal to a sin kx plus b cos kx we can think of another kind of trial function as solution as well we will come to that. Now here is this a good trial solution well to know that all we have to do is differentiate differentiate once this is what you get differentiate twice this is what you get. I am not going through the steps because it is very simple this is the solution I encourage you to just do it once yourself and then you will be convinced. Now the question I want to ask is what is this k I have got d 2 dx 2 of psi of x equal to minus k square psi x what is k then if I plug k back into the original Schrodinger equation we see that if I replace d 2 dx 2 psi of x by minus k square psi of x then I get h cross by 2m from here k square psi of x from here that is equal to e multiplied by psi of x or in other words I can write this energy as h cross square k square by 2m k is a measure of energy or rather k is a measure of energy and remember that this energy is entirely kinetic energy. So I can write another expression here I know that if it is kinetic energy then what is the relationship between kinetic energy and say linear momentum I will write px here because the motion is only along x so e is equal to px square by 2m we know that. So if I compare this expression and this expression then what will I get 2m in the denominator so px square turns out to be h cross square k square and so well what is px then that will be equal to plus minus k h cross. So what we see is that it appears that we are going to have a momentum of k h cross and the particle move can move in this direction or in that direction. So that is what seems to come out and we are going to arrive at it at in another way also. So k then is equal to root 2me divided by h cross but do not forget the plus minus sign. Plug it back into the trial solution you get an expression for psi of x as a sign root over 2me by h cross into x plus b cos root over 2me by h cross into x and if I plot it this is the kind of function that I get. This looks like a cos function or a sin function it is not I actually multiplied sin x by some 3 or something and added it to cos x by 2 or something so this is what you get. Now is this a good wave function actually it is not because you can see clearly that it goes from plus infinity to minus infinity it does not become 0 anywhere it is not normalizable so there is a little bit of a problem. So to work with a free particle what one does is one uses the technique of box normalization which means that you say that you set some limits you set long limits set 10 angstrom 15 angstrom 20 angstrom something like that and you say that the particle exists within this limit and you normalize within that of course it is not a very very very rigid way of doing it but that is the best you can do. So be aware that the wave function for free particle is not really a perfect wave function all right that being said we can actually write the wave function in different manner also instead of writing the sin and cosine functions I can write this kind of an exponential function e to the power i k x plus d e to the power minus i k x I leave it to you to differentiate and convince yourself that this function also satisfies the Schrodinger equation for a free particle you can arrive at it in another way I think we all know that e to the power i k x is equal to cos k x plus i sin k x e to the power minus i k x is equal to cos k x minus i sin k x. Now if you multiply e to the power i k x plus by c multiply e to the power minus i k x by d add them up you are going to get something that will be a sum of a sin and cosine functions all right but the reason the good thing about writing the function in this particular form is that it starts making sense why does it start making sense well the momentum operator what is momentum operator I will come to that before that let me just say this I differentiate e to the power i k x let us say with respect to x what do I get I get i k multiplied by e to the power i k x so this is actually an eigenvalue equation the only problem is that the eigenvalue is imaginary but it is not really a problem because if you think of the linear momentum operator linear momentum operator is actually h cross by i multiplied by d dx so you multiply the previous result i k e to the power i k x by h cross by i you end up getting h cross k what is h cross k h cross k then is the value of the linear momentum that you get if the linear momentum operates on the first term in the wave function what would you get if the linear momentum operates on the second term of the wave function you are going to get minus h cross k this is absolutely in line with the argument that we had proposed a little while ago is not it we had also obtained the values of plus h cross k and minus h cross k for the momentum even before using the operator so using the operator it makes sense that is point number one so we see that the particle can move in this direction or in that direction what is the probability probability is equal there is no bias for any direction and probability being equal is given well ensures that the coefficients also have to be such that their mod squares have to be same okay this the discussion that we just performed I hope it rings a bell about the discussion that we performed maybe 5 10 minutes ago writing on the surface remember what happened we perform a measurement we are going to experience one of the eigen functions of that particular operator that is what we see here so linear momentum can be either plus h cross k or minus h cross k right that is what we learn average value of course has to be equal to 0 again I leave it to you to plug in this wave function into the expression for average value of course you will have to use this second form and you can work it out yourself but even without doing it we can see from simple logic that probability of moving in either direction should be the same so this average value must be equal to 0 average value of momentum is equal to 0 average value of energy is not equal to 0 because energy depends on k square and not k okay so what have we got so far is there any restriction of k no so there is no restriction of E as well so for a free particle the first quantum mechanical system we have studied there is no quantization where will quantization arise from if we confine the particle if you put it in a box we will see in the next part in the next class that quantization arises nicely