Okay, let us start now, there are some questions which I received day before yesterday. So I will try to quickly answer some of them. One question is about a particle under motion behaves like a wave which is quite obvious. However, I do not know accurately what is psi that I have anyway discussed day before yesterday where I said psi is to be treated as somewhat abstract way, it is a probability wave which is happening in a complex system unfortunately that is the best that we can do. So in that sense like a normal wave about which you can sort of say what is getting displaced is only the probability which is sort of oscillating if you want to call it like that. Electromagnetic waves travel this velocity of light, my question is that there is any possibility of superposition of two waves having two different wavelengths, of course there is a possibility of superposition I mean the phenomenon of beats is exactly the same thing where two wave superimpose which have slightly different wavelength. So things that the superposition principle essentially means that if you have n number of waves let us say y1, y2, y3 up to yn the total displacement is just sum of y1 plus y2 plus y3 plus yn. So see whatever number of the waves are there you have to just simply go and keep on adding mathematically that is what you are going to get. So there is a superposition is definitely valid for them. Now for quantization of energy level only boundary conditions are required is normalization condition required for quantization of energy levels. So normalization is just to normalize the probability in fact even if you would not have normalized the wave function the energy levels would have been same okay the only thing is that when you are actually looking at the probabilities in that case if I have not normalized they will only give you the relative probability they will not give you the absolute probability. If you want to find out the absolute probability of finding particle let us say between x and x plus dx then you have to use normalized wave function that is the only difference. How did you obtain the general solution of a particle and rigid body box of this little bit I will discuss today when this becomes more of a sort of a differential equation problem but you know they are sort of two standard type of differential equation which we always use in physics and most of the time just by looking at those equations I will be able to find out what is the solution of it. So this I will explain little bit in today's lecture when I am doing some of the problems. When I have two questions instead of using identical system what would happen if we take repeated readings on the same system provided we ensure that the system returns to the original state see that is what is the tricky question and how do we ensure that the system returns to the original state. Of course if it returns to the original state whatever I have said earlier will be true but what I am trying to say that once we have made the measurement it is no longer in the original state okay so things get changed and after that you know repeated measurements will give values according to the new wave function and not according to the old wave function. If at all it is possible to generate, regenerate the system to the old wave function then whatever we are saying earlier would be true. If n increases to infinity what will happen to behavior of wave function of course you know when n tends to infinity and you will I mean it depends again on the system but you know for a very large value of energy you can show that this will approximately be equal to the classical system I know when you have very large energies in principle it should be behaving like classical system. Then there is a question if there are 5 electrons in the ground state what will be their energy of course you will not have 5 electrons in the ground state because electrons obey poly exclusion principle I think when Professor Suresh is going to talk about solid state physics he is going to talk about these things okay in principle if whatever the states that we are getting as a result of solution of Schrodinger equation okay in principle not more than 2 electrons can occupy that state and 2 we say because you know there is spin which we have not accommodated in the normal quantum mechanics if we accommodate spin also then there are 2 possible states of spin spin up and spin down so therefore each state which we evaluate from Schrodinger equation can in principle be occupied maximum by 2 electrons one with spin up and another with spin down so there is no question that there are 5 electrons which can occupy the states of course if these happens to be bosons which again I think probably Dr. Suresh may discuss a little bit only in that case it is possible that 5 of the particles will occupy all the ground state in that case this problem is of a statistics I mean we have discussed Maxwell Boltzmann statistics you have to go to more refined statistics when you are talking about quantum systems which we call as a Fermi Dirac statistics when particles obey Pauli solution principle and Bose Einstein statistics when particles do not obey Pauli solution principle so that becomes more of a part of a statistical mechanics then another question is that in a double slit experiment using electrons we consider electron wave moving with velocity of light or with the velocity less than C no no I mean this I think we have discussed number of times that the electron wave has to travel with the same speed as it electron and the electron will travel with whatever energy we have given so when I have accelerated the electrons the electrons are emitted they would have a particular amount of kinetic energy and whatever is the kinetic energy corresponding to that the electrons would have a speed and that is the speed which will remain and the wave group also has to travel with the same speed so there is no question of course you know a particular particle wave can never travel and that yesterday I had all I am not yesterday but day before yesterday I had also explained that what can travel with larger than C is only the phase speed that also if we define frequency in a particular way but you know that is of purely mathematical consequence you know there is nothing physical about that okay then there is another question which is interesting question instead of one particle one dimensional potential well if you deduce two particles like electron, electron or proton, proton or newton, newton what are the energy eigenvalues and eigenfunctions of course this is interesting question in principle when you are taking you I mean if you are taking for example charged particles you have to also accommodate for the forces which are there between the charged particles the potential energy becomes much more complicated but you know as we normally do in the case of solids which professor Suresh will do okay you always I mean as a first approximation you assume that even because even though the particles are interacting they are charged and there is force between them we neglect that and we assume that the electrons or the particles are experiencing only the force because of the coulomb interaction of the solid and assuming that you know the energy states do not change and we can then talk about statistics and find out how they are going to be occupied but if you want to solve the problem exactly when you have more than one particular particles you have to also take inter particle forces into consideration when you are talking to the total potential energy I mean the another question is that in a single state experiment is not the wall stopping the electrons of course you know there is a wall that will stop the electron so there is nothing very surprising about that is it necessary that along every wave function should there be a constant that I have seen that I have told you that the nature of the Schrodinger equation is like that if psi is a solution a psi will always be a solution you can substitute there and see because a eventually will cancel out when I substitute a psi so there is always an a which will always be present which as I said has to be found from normalization condition the next question is that what you mean by dynamical variable by dynamical variable which I mean any variable which defines anything related to dynamics for example it could be just displacement it could be velocity it could be acceleration it could be angular momentum it could be whatever you want to solve the talk of x square talk of x cube all those things does e mean the total energy or the kinetic energy in this case Schrodinger equation e always represents the total non relativistic energy because when we come to relativity we also talk that energy is also defined in a slightly different fashion by Einstein which is for some of the standard well-known relationship e is equal to mc square but that energy e does not have any classical analog in Schrodinger equation when we talk about e that e is always the classical energy which can always be taken as potential energy plus kinetic energy only thing in the case of particle in a box because the potential energy is zero so e happens to be only the kinetic energy the next question is that we can calculate the energies of the a by using one-dimensional works can we compare these energies with more model of energy levels so more or model is totally different because in more model the potential energy is not just constant but it is given by your Coulombian interaction e square about 4 pi epsilon not r that particular problem has to be solved in a totally different fashion. So the solution of hydrogen atom in its completeness I mean is quite long I am sure all of you would have done mc in physics you know would have undergone that particular thing which is fairly long and that is of course beyond the scope of this particular course but if you look at any book on quantum mechanics they will always solve the energy corresponding to a hydrogen atom and those energies do turn out to be equal to the Bohr s energy that was actually the success of quantum mechanics that you know when quantum mechanics was being developed that see we know that the energies that were found by Bohr okay can are able to explain the atomic spectra and therefore people had faith in Bohr s theory and then we realize that Bohr s theory does not always work and then you know but the energy seemed to be working and what was the success of quantum mechanics originally that one could get exactly the same energies as Bohr had obtained Bohr had obtained by some sort of phenomenology while this is actually by a solution of Schrodinger equation okay those you know sort of derivations of those energies are much more complex and obviously has nothing to do with the particle in a box. What is the effect observed on wave function energy the particle in a box with different boundary values in fact it is very interesting problem one can always work it out from x is equal to minus l to x is equal to plus l or minus l by 2 plus l by 2 you will always get the same values of energy provided you take the appropriate length of the box okay the boundary conditions become somewhat different if it is a good exercise to work out this particular problem assuming that the origin is not at 0 but origin as at half of the box okay then you have to apply boundary conditions which are not which are much more symmetrical and still you will get exactly the same eigen I mean same values of same wave function and the same values of energy that does not cause a problem there are a lot of questions which ask for what is the justification and what are the physical significance these are very very difficult questions okay I mean we have discussed that in quantum mechanics we cannot treat everything in a classical way in classical mechanics we can ask a lot of questions about you know how the particle moves okay how it goes from one point particular point to another point while in quantum mechanics many of these things are you know somewhat hidden in uncertain difference pool many of the questions we cannot answer exactly the same way that we can sort of ask in a classical mechanics problem so these cannot be justified in a simple fashion I mean let me put it like that I mean I would like to say that quantum mechanics itself is a theory and whatever you obtain from quantum mechanics is essentially an outcome of the solution of that particular equation so see like for example when we try to explain the normal motion we always take this stand okay that we obtain this for Newton's law of motion because Newton's law of motion is my original theory and whatever I got it from I can explain a particular behavior on the basis of Newton's laws of motion similarly we can argue in this particular case that okay my Schrodinger equation or basic quantum mechanics is the original theory and whatever is the justification that I have to give has to be given only in the basis of Schrodinger equation I do not have any independent justification okay similarly you know just like for a classical particle our justification lies in Newton's law of motion I am not giving any other justification so that becomes my original theory original model from which I derive all those things similarly here Schrodinger equation becomes my original equation and the only way to justify all these things is using Schrodinger equation then what is the justification to derive Schrodinger equation time dependent equation using time independent equation in fact it is the other way you obtain time independent Schrodinger equation from time dependent Schrodinger equation and that is why just mathematically separating the variable which is the very standard technique in differential equations so I do not think we can give any specific justification for that particular thing then what is the significance of finding the expectation value that is what I have said there is no significance other that you are interested in finding out the average value of certain quantity because of whatever reason you want for example I used a certain principle for that I obtained the value of x square which is expected value of square I obtained the expected value of x okay so there cannot be any specific significance other than that because of some reason I need that particular value I may like to ask a question what is the average value of the position when I make you know let us say very large number of measurements on the position of a particle so if I am interested in that then I will obtain this is the way to obtain it so there is no other justification that we can give or there is no significance that we can attach it is the significance as for example if I have to calculate the average mark of the entire class okay so that is the justification that I am interested in finding out the average mark of the class just to get how has been the performance of the student similarly I am I may be interested in finding out what is the average position of the particle if I make a very large number of particles if I am interested that is what is the significance there is nothing else that we can say about that the last question is that zero rest mass is called massive in fact that is not correct in fact zero rest mass we have not been calling massive when I say massive particles it means a particle which does not have a zero rest mass a particle with this mass non-zero that is what I have called as a massive particle zero rest mass particle is called is it because so no so I am never calling zero a smart particle a particle as a massive particle so I mean whether I mean when we do relativity we will see that massive particle and a particle with zero rest mass are treated exactly by the same equations so other than that the fact that m naught is equal to zero in that one has to substitute in a in certain expressions otherwise we use exactly the same basic equations as we use for a particle with m naught is equal to zero and for a particle with m naught not equal to zero okay this sort of gives us the idea of the questions which I had got it now let us come back to the lectures so we recapitulate of what we did day before yesterday we talked about the orthogonality of wave functions and then we talked about the non-stationary states we introduced many new concepts we introduced the concept of collapse of a function we introduced the concept of eigenvalues and eigenfunctions and using all those things we eventually gave some of the postulates of quantum mechanics some of these things we will be using as we keep on I mean we have another I am today and then in the afternoon we have two lectures in which I will be covering basically the formal part of quantum mechanics we will have one more session which will be more on problems and quantum mechanics and then we will move to crystallography and then eventually to special theory fluidity okay now let us go a little bit ahead and try to make my problems slightly more realistic earlier I have solved a particular problem in which the particle was contained in what I call as infinite potential wave or many times it is called infinite square well potential now I will call it finite square well potential so the difference between the earlier problem and the problem that I am going to discuss now is that in the earlier problem v was infinite when I am going between for values of x less than 0 and for values of x greater than l now the potential is not infinite but the potential is finite v is equal to v not here and we have v is equal to v not here and so only within this particular box between x is equal to 0 and x is equal to l that potential energy v is 0 so this is what I have said is the condition of potential energy vx is equal to 0 for values of x between 0 and l and is equal to v not whenever x is less than 0 and whenever x greater than l you will always say that this is somewhat more realistic okay but we will see that just by making it little more realistic the problem has become much more difficult okay and in fact we will not be able to find out the energy values in this particular case in a very very straightforward in a sort of close manner we will not be able to find out an expression the way we have obtained for a particle in a infinite square potential well okay but nevertheless this particular problem is very very illustrative because it gives you the way we apply boundary conditions in a more regenerate fashion okay and also introduces some new ideas now first thing that I would like to mention here is that that when I am solving this particular problem I am assuming that the energy of the particular particle is less than v not it means the particle has energy somewhere here because that is the only way this particular particle will be confined in this particular potential well see whenever we have situation like this we call these particular problems as a bound state problem so this is a bound state problem because the particle is supposed to be bound in this particular region even though as we will be seeing that there is a small amount of probability that the particle could be found outside this particular box that is what happens in the case of quantum mechanics this particular aspect we will discuss little bit more in this particular case so this problem is little more interesting than the earlier problem which as I say was the simplest problem in quantum mechanics that we had worked it out okay so this is the thing and whenever we have discontinuity of potentials which we are still maintaining here so I have said that is not really realistic in a real situation you generally do not have discontinuity of potential energies but let us assume that as a value of x we have discontinuity of potential energy we always divide this particular problem into various regions this is what I have done region 1 this I am calling as r 1 this is region 2 which I am calling as r 2 and this is region 3 which I am calling as r 3 so my r 1 my potential energy is v is equal to v not and r 2 my potential energy is v is equal to 0 and in r 3 my potential energy is equal to v is equal to v not again again I repeat my e is less than v not so my particle is bound in this particular case if my e happens to be larger than v not then this becomes what we call as a free state problem and immediately after this problem I will start discussing free state problems and that are a slightly of different type as you will see that the way we apply boundary conditions are in a slightly different fashion in those cases and the way we interpret wave functions is also somewhat different fashions at this moment let us just consider this particular problem now what happens because there is a other discontinuity of potential energies here we always try to solve independently the Schrodinger equation in this region in this region and in the third region so we first solve the Schrodinger equation in this particular case and then eventually write the general equations general solutions of the equation in all the three region region 1 2 and 3 and then eventually we apply the boundary conditions that is what is the most important aspect in quantum mechanics the boundary conditions that we imply that keeping physics in mind we have to or keeping postulates of quantum mechanics in mind we apply appropriate boundary conditions and then we will be able to see how do we eventually are could obtain the energy of the particles which is confined in this particular box. Okay now let us write these equations which I am writing in the next transparency in these three regions separately so this is my region 1 so this is my d 2 5 dx square remember this is the Schrodinger equation plus 2 m upon h square e minus v naught psi 1 x all right so this is exactly the Schrodinger equation the only difference here is that this I am writing as phi 1 so I am saying that this is a solution which is valid only for r 1 region 1 in the second region which is actually the region of the box potential energy is 0 so this v naught has been put 0 but this particular wave function or this particular equation is valid only for region 2 so I am writing psi 2 phi 2 here so this is phi 2 this was region 1 this was region 2 so phi 1 phi 2 now in region 3 again my potential energy goes back to v naught so again we write exactly the same equation which is here but here I am writing phi 3 because this is the wave function which will be valid only in region 3 okay so at the moment I have written only the equations I have to solve these equations only thing when I solve I have to realize this particular fact that in this particular case my e is supposed to be less than v naught because as I have said earlier that I am assuming that my particle is bound here so energy has to be lower than v naught if it is larger than v naught then this particular particle this potential energy I mean this particular particle is essentially what we call as a free particle which is not a bound particle here the particle is bound so essentially if or in the other words if I look classically if I look classically only when the energy of the particle is less than v naught the particle will be confined within the box okay otherwise if e is larger than v naught this particle is free to move outside the box so if this particular particle has to be confined in the box in a classical sense then e has to be less than v naught okay so if I compare for example with the classical mechanics I can get an idea that this particular particle will actually be called a bound particle because this is not existing I mean it is not possible for it to exist I mean in the entire region of this space okay this particular part of being free state and bound state I will explain little later so what will happen here this particular quantity will be negative of particular thing so actually what I will prefer to write I will rather prefer to write this as v naught minus e and put a negative sign here when I put a negative sign here because normally I would like to always handle with real numbers and numbers which are sort of positive so I will rewrite this equation as v naught minus e and then substitute it here now that as I have said that people have been asking this question so these are the two type of questions that we generally write here see if we have write written this equation let us write d 2 general equations differential equation d 2 by dx square it could be let us say k square y or d 2 y dx square is equal to minus k square y so I am putting k square just to say that this particular quantity is actually positive so if we have equation of the type d 2 y dx square is equal to k square y and we have equation d 2 y dx square is equal to minus k square y generally by look we can write the solution in this case remember this is second order differential equation so they have to be two unknowns whenever I have negative sign generally I prefer to write this equation as a sin kx plus b cos kx and you can see that if I substitute this equation in this particular thing this will solve if I differentiate y with respect to x I will get a k cos kx and this I will get minus b k sin kx if I write d 2 y dx square I have to differentiate it further once more so this cosine will become now sin but with the negative sign so this will become minus k square sin kx and there is already minus sign but when you differentiate sin you get only plus so you get minus b k square sin kx this you get as minus k square then a sin kx plus b sin kx I can write as y so you can see that this equation is a solution of this equation on the other hand if I take k square y then I write this general solution in the form of y dx square is equal to a e raise power kx plus b e raise power minus kx sorry what I am doing I am writing y like this so I differentiate d y dx I differentiate this with respect to x so I will get a k e raise power a kx minus b k e raise power minus kx I differentiate once more with respect to x I will get another k out here so I will get a k square e raise power kx differentiate again so I will get b k square e raise power minus kx so you will get k square into y so as we see that if this particular term happens to be positive now we write the general solution in the form of exponential on the other hand if this is negative then we write generally in terms of the sinusoidal functions and you should realize that the nature of sinusoidal function and the exponential function is different sinusoidal functions are so sine functions and cosine functions are oscillatory function while these functions which we are talking about exponential functions here these are with respect to x there is no oscillation either it will go to plus infinity or it will go to minus infinity depending upon the sine here so the nature of the two relationships are different of course you can always say that I can always imagine k to be imaginary then cannot I write the general solution also in that particular form of course you can also write in this case if you want the solution to be in the form of y is equal to a e raise power i kx plus b e raise power minus i kx this is what I will be using whenever I am doing for the free state problem here in this case it does not matter for a particle in a box even if you have tried a wave function of this particular type which is correct you would have got exactly the same answer nothing would have changed okay it is just because we are a little more familiar with the sine term and as far as possible we may not like to have use I mean it is not always possible so but you know if we can avoid use of imaginary numbers somewhat simpler I mean I will let me put it like that it is a matter of choice which you want to do it in this particular case when your equation is of this type whether you write phi like this or whether you write like this it does not make a difference like this of course you can also say in this case also cannot I write this solution in terms of sine but then it will be sine of imaginary number then it will become hyperbolic functions so in that case again the equations have to be looked into different fashion again we are not very familiar with hyperbolic functions at least the students at the first year level are not so much familiar so generally what I have told that that either when I am having equation of this particular type either I use sine kx plus cos kx or as this okay whichever is more convenient okay in the case of bound state problem many times I write like this this being more convenient but in the case of free state problem I am obliged to write in this particular fashion as I will be discussing little later on the other hand when I have a equation of this particular type I always prefer to write in terms of kx and minus kx and also because these are not oscillatory functions many times I avoid using k in fact this instead of k I use alpha square so I will always write this as alpha square and here I will write as alpha and here also I will write as alpha so this is just a matter of how you sort of represent so there is I mean depends again as I said in a matter of choice this is the way when I look at this equations I can find out sort of solution so as you can see if I look at this particular equation because E minus V naught is negative so this should be negative sine so this equation will be of the form d2y dx square is equal to alpha square phi 1 while this equation because this particular quantity is positive so when I take it to the right hand side this equation will be of the form d2 phi dx square is equal to minus k square phi while this equation will be again of this particular form so therefore now I am in a position to write the general equations in this particular form that is what I have written note the sine of the quantity in square bracket for different regions the functional behavior would depend on this for phi 1 I have written a e raise power alpha x and remember when I have defined alpha square I have defined this as 2m V naught minus e because remember it is this quantity which is going to be positive because V naught is going to be greater than e so this alpha has to be positive so this alpha number here which is coming here is going to be a real number and a e raise power alpha x with real x is an exponentially growing function while this is an exponentially decaying function now in region 2 because my equation nature was slightly different so I wrote general solution as C sine kx plus d cos kx and my k square was defined as 2m e upon h square okay again if you have some doubts you can substitute these solutions back into the equation which I have written of the last transparencies and you can convince yourself that this is going to be the value of alpha square this is going to be the value of k square now similarly in region 3 I can write the solution only in this particular form so this is what I have written alpha turns out to be exactly same because the potential energy in region 1 and region 3 are same so alphas are same remember I am taking the constants to be different because these constants in general have to be found out the boundary conditions and in this case in principle their values could be very different so I am writing the most general solution in the three regions as you can very clearly appreciate there are going to be six constants a b c d f and g okay now let us start looking into the boundary conditions go back to this particular figure of this thing see remember when we had particle in a box problem infinite box problem which we worked out earlier in that case the only condition that I had applied was that the wave function should be continuous at that time we had argued that the wave function here is going to be 0 the wave function here is going to be 0 so all that you wanted that the wave function should disappear at x is equal to 0 and should disappear at x is equal to l the wave function should be 0 at the two ends alright but we have mentioned when we are talking about the postulates of quantum mechanics that matching wave function is always true but in general derivative of the wave function should also match and we said that unless v is equal to there is infinite jump in potential energy we need not we should always match the derivatives in that particular problem that we did earlier v was equal to infinite so at x is equal to 0 and at x is equal to l there was an infinite jump in the potential so we did not match the derivative of the wave function we only match phi we did not match d phi dx but in this case because this jump is only finite this v naught is assumed to be finite in this case then I have to match their derivatives also so I have to match not only phi in this particular end from solution of this particular end solution of this particular end at this particular point at x is equal to 0 not only their individual value should match but their derivatives should also match so this will give me two equations similarly when I match the wave function between region two and region three and their derivatives then I will get two more equations then I am getting four more equations now what is happening I have six constants I get only four equations of course there is one particular constant which I can never get rid of that is has to be found from normalization condition okay how do I eliminate other constants in order to find out the general solution that is what is interesting and that is why we have to apply the boundary conditions okay the condition that now I am going to apply in order to get rid of some constants is again that wave function must be well behaved when we had talked about the wave function being well behaved we had said that wave function should be continuous it should be single valued and it should always be finite this is what we have said we also applied the condition on d phi dx to be continuous alright now if you look at the solution of region one remember region one extends from minus infinity to x is equal to 0 so it extends only for negative values of x this solution of region one is not valid for x is equal to x greater than 0 or for positive values of x between x is equal to 0 and l phi 2 is a valid solution for x greater than l phi 3 is the valid solution so let us first look at the solution r 1 which is valid only for negative values of x the solution in r 3 is valid only for values of x greater than l so this was my solution remember just now I said that this relationship is valid only for negative values of x if I put x is equal to minus infinity remember I cannot put plus infinity here because this is not valid for positive values of x so if I put x is equal to minus infinity this term will become 0 okay no problem this will remain finite for all negative values of x this term will always remain finite but what happens about this particular term if I put x is equal to minus infinity if I keep on increasing negative values of x you will find that this particular term will keep on blowing up and as x tends to infinity this particular x tends to minus infinity this term will actually blow up to infinite and as I expect that wave function should always be finite this particular term cannot be allowed into the wave function so I force this particular term to become 0 or in other words I put b is equal to 0 so if I put b is equal to 0 then only this particular term is present so you are able to get rid of this particular term purely from the condition that wave function must always be finite now as far as region 2 is concerned no issues because region 2 you know is only between 0 and l and between 0 and l this value is always going to remain finite moreover sine and cosine are always oscillatory function they will never become infinite even if this particular solution would have been valid at x is equal to plus infinity or minus infinity which it is not even then sine and cosine will always be oscillating between plus 1 and minus 1 they would never blow to infinity same thing is not true for exponential function. Now let us come to region 3 now region 3 is valid for all values of x greater than l it means it is also valid as x tends to plus infinity where x tends to plus infinity this term will tend to 0 because when you put infinity you know this is minus infinity so this term will tend to 0 so for all values of x greater than l this term will be finite but as far as this term is concerned as x tends to infinity this term will blow up and blow up and therefore I do not want this term because I want the wave function to be always finite so what I will get I will put this g equal to 0 I will not be able to allow this particular thing so what we have seen that I would have got 4 equations from matching the wave functions and their derivatives now I am also able to get rid of 2 other constants I mean I have got 6 equations essentially okay and there is a another equation which is the normalization condition we have always one equation in excess of how many number of constants we have in this case the energy will turn out to be quantized that initial condition will always try to normal to sort of will be responsible for quantization of the energy so this is what I have said we put b is equal to 0 and g is equal to 0 so I have put b is equal to 0 and g is equal to 0 here this is what I have done so my equations now become this my solutions become only this this is valid for region 1 this is valid for region 2 this is valid for region 3 so I have got rid of the other term here also I have got rid of other term ensuring that phi 1 always remains finite and phi 3 always remains finite now I will match their values I will put first boundary is between region 1 and region 2 which occurs at x is equal to 0 so phi 1 at x is equal to 0 must be equal to phi 2 at x is equal to 0 similarly I will match another boundary condition at boundary between region 2 and region 3 which occurs at x is equal to l so I will put at x is equal to l what the wave function should be continuous then I also want their derivatives should be constant so first I will take differentiate this particular thing with respect to x remember first we have to differentiate then put the value of x so we differentiate and then put x is equal to 0 and then match their derivatives similarly between here and here I will differentiate this with respect to x differentiate this with respect to x put x is equal to l here then equate that so this is what I am doing in the next transparency so this is a boundary between region 1 and region 2 so I have put phi 1 at x is equal to 0 to phi 2 is equal to 0 there was only e raise power alpha x e raise power alpha x so I am putting x is equal to 0 so this becomes just a so e raise power 0 so this becomes a sine 0 is anyway 0 cos 0 is 1 so this gives you a is equal to d then I take the derivative first I differentiate if I differentiate this term I will get a alpha e raise power alpha x put x is equal to 0 so I will get only a times alpha because e raise power 0 will be 1 so a alpha e raise power 0 then I differentiate when I differentiate sine becomes cosine and a k term will come out so it will be ck cos kx and put x is equal to 0 so it will become ck cos 0 when I differentiate cos I will get a negative sine so I will get minus dk sine kx I put x is equal to 0 so this will come sine is equal to 0 this term now will become 0 so I will get a alpha is equal to ck so this is the boundary condition which has been applied between region 1 and region 2 now I will rather go quickly because pure mathematics between region and 2 and region 3 I will apply the boundary conditions between phi 2 and phi 3 because this is the wave function which is valid in region 2 this is the wave function which is valid in region 3 I just put the values of l here I just put the value of l after differentiating here so these are the boundary conditions that I have applied now what I will try to do is to solve them and normally what I try to do try to solve in with respect to a I take one constant because I know my equation is such that I will never be able to get rid of one particular constant so let us choose that particular constant to be a so express all other constants in terms of a and let this a be found out by normalization at a later point so these are my conditions four conditions a is equal to d a alpha is equal to ck and these are little more complicated condition so what I will do I will find out f d have already found in terms of a c is also very clearly c is equal to alpha upon k alpha upon k times a so what I will do I will put c and d in terms of a then I will get two equations and both the equations will contain only a then in all that the equations are consistent I will divide the two equations and then a will cancel out then I will get one particular equation which is independent of any constant and that is the equation which must get satisfied if all the boundary conditions should be satisfied and the wave function should be well behaved this is what I am doing express all constants in terms of a so I put d is equal to a c is equal to alpha by k a then other equation I have put the c as alpha upon k d I have put as a there also there was a c which I have put as alpha upon k the d I have put as minus a these are the equations there is f here see you look at this equation on the left hand side there is a the right hand side there is a here there is a there is a f if I divide these two equations in that particular case what you will find the a will cancel out remember a cannot be 0 and f will cancel out when both cancels out you will get a purely an equation which does not have any constants this equation is the one equation which must get satisfied if all boundary conditions are satisfied and if wave function is well behaved this equation is the one which is going to give rise to energies so this is what I am calling as energy eigenvalues I think now we can use the term eigenvalues that because these are eigenvalues of the Hamiltonian operator so divide the last two equations this is what you are going to get remember alpha had defined as h cross alpha under root 2 m v not minus e and k have defined as h cross k under root 2 m e all right so basically k and alpha both depend on energy alpha also depends on what is the potential energy in for negative values of x and for values of x greater than l so depending upon what is the value of v not this happens to be a complicated transcendental equation if you can solve this equation but these days when the computer sets quite easy to solve it otherwise people were using earlier graphical method to solve these equations okay you can find out what will be the values of energy for which this particular condition is going to be satisfied it will get satisfied only for certain specific values of energy and those will be the only values of energy that will be allowed in the system so this is what we will be calling that energies again quantized but I am not able to find out this energy in a close form as I was able to find out for a particle in infinite potential well here these to find out the energy values you must solve this equation all I am trying to say that this equation is given a very particular value of l this equation is going to be satisfied only for some specific values of energies and those will be the values of energy that will be allowed for this particular particle energy will turn out to be quantized now there is another interesting question here which is what we call as a classically forbidden region when I call this particular bond particle I say classically this particular particle which has a energy e less than v not can never leak out to this particular area in region 1 and region 3 because if at all it leaks out its potential energy is larger than its total energy it means its kinetic energy will turn out to be negative and in classical mechanics negative kinetic energy does not mean anything a particle cannot have a negative kinetic energy it means velocity is imaginary a velocity imaginary velocity does not mean anything a quantity which is a real a physical quantity which is velocity velocity cannot take a imaginary value therefore this is what I call as a classically forbidden region classically forbidden region are these regions where classically the particle could not have been found because in that case kinetic energy would turn out to be negative but now we see that the wave function exists here here is an exponentially decaying function here also it is exponentially decaying function but on the other hand it has its presence it means if it has its presence okay there is a finite probability of finding the particle in what we call as a classically forbidden region so in principle it is possible for the particle to be present in what we call as a classically forbidden region okay now you can always ask the question if it is I mean okay let me first say that had this been a wave then probably you can understand because there is always a skin depth through which the very traits but in a pure classical particle sense okay it is very surprising I mean that a particle could exist in what is a classically forbidden region but you should understand that in quantum mechanics okay particle is not really the classical particle this particular particle has a wave character okay in fact we have discussed earlier quite quite clearly that if you have a picture of a particle which is a localized entity that is not going to work when you are talking of quantum mechanical particles in these type of systems okay and we have said that this sort of somewhat mysterious on the other hand if you really want to insist on particle picture and say that I would like to localize this particular particle and really try to measure it is almost as we are trying to determine from which particular particle from which particular slit the electron went in the double slit experiment whenever I perform this experiment I want to localize this particular particle into the classically forbidden region and measure its kinetic energy to find out whether it is really turning out to be negative okay I will create enough perturbation into the system that this particular particle will no longer be in a classically forbidden region this can of course be shown through a particular problem which I am not sure we have time it will probably do it but basically that is what is the idea that if we take a pure wave picture then you know that a wave even if there is a particular let us say sheet of paper or a sheet of a material which is completely opaque to light light nevertheless penetrates to some extent inside this particular medium okay which we call as a skin depth so if we look at the wave nature then I can understand that this particular particle wave has a small amount of penetration okay but if you look at purely classical nature of the particle then I will not be able to understand it again it is not always possible to mix up classical ideas of particle with wave mechanical picture so we say that there is a non-zero probability of finding the particle in the classically forbidden region this probability decreases exponentially as we move away from the box remember this we still call the localized particle because probability of the particle decays as we go out of the box so this particular particle is still called a localized particle however the uncertainty principle prohibits one to localize the particle in that region and measure its kinetic energy and really show that this is negative and then what will happen to its velocity you will not be able to perform this experiment from wave nature we can understand the finite penetration in the form of skin depth as just now I have mentioned now if at all I want to normalize I have expressed all the constants in terms of A so this becomes my equation this is what is my wave function in the region 1 this was my wave function region 2 I have to square it I have to square it this was my wave function region 3 then I have to integrate take psi star psi integrate and put this equal to 1 that will give me the value of A finally which is the final value of A that one has to obtain to make the wave function complete okay if at all we want to normalize it and as you can see that this particular solution itself is going to be fairly complex if I want to find out the value of A just by making the problem little more realistic my mathematics has become much more complex remember quantum mechanics is a highly mathematical subject I mean a lot of people have been asking question what is the need of doing quantum mechanics the thing is that there are so many things so many wonderful things in the nature or so many I mean even the form of devices that you cannot understand for example you just simply cannot understand how a semiconductor device will work I mean semiconductor devices so much of spread of some semiconductor devices everywhere you know you are using semiconductor devices okay now they you cannot understand the way they work unless you know quantum mechanics so many parts of the sciences where quantum mechanics is inherent remember whenever we have to describe for anything where I want to talk of a particle at a microscopic level I need quantum mechanics if you are talking about recording magnetic recording people are using magnetic recording they are the whole area of spin planets which has come I mean in fact people call that this is for the first time that we have been able to find out quantum mechanical heads which are used for reading information from let us say hard disk of your magnetic computers quantum mechanics is necessary to understand you will not be able to understand how these devices do are forget about you know further research and further advancement unless you know quantum mechanics okay house over abstract is the quantum mechanics okay it is necessary to understand if you want to understand the working of the devices and want to make those advancements now let us come to a free state problem which is a slightly different type of problem these are also very very interesting type of problems so let us just discuss you will see that the language has become somewhat different these problems are often used when we call of a scattering so many times let us say a beam of particle is coming and there they encounter a potential energy and you know a step potential as we call it and because of that you find the particle change their path so we call them free state because generally you will find that the region where particle can exist with finite probability that is infinite so such type of problems are generally called free state problems and as I have told earlier that these problems in general it is not possible to normalize the wave function because the probability the a will turn out to be 0 but the normalization is not that great amount of interest other important thing that we will discuss in this particular case that in free state problem we will find that energies do not turn out to be quantized all possible values of energy are possible there is a continuum of energy which are possible only when the particular particle is bound that the energy turns out to be quantized I think I have mentioned even this earlier that for a bound particle there are two things which are very very important the first thing is that the ground state of the particle that is the lowest energy state will not be 0 and the second thing is that the energy levels will be quantized when you are talking of free state problems none of these conditions occupy okay all possible values of energies are allowed now let us look at this particular system now we say that there is a particular particle of course we always try to work out with the single particle potential wave functions okay there is a particle which is coming from the left hand side and let us suppose it comes with the energy E okay now in all these region right from minus infinity to x is equal to 0 the potential energy was 0 it means this particular particle the energy that it has E was purely kinetic because this potential energy turns out to be 0 so you have only kinetic energy a particle which is coming with the energy E which is approaching here now let us look at this particular problem classically first now at x is equal to 0 suddenly the particle encounters a potential I have not got help it is step potential potential hill okay potential suddenly changes from v is equal to 0 to v is equal to v0 all right I mean in classical mechanics whenever there is a change of potential energy there will always be a force okay but let us not bother about this particular thing as of now because in quantum mechanics we are talking purely from the point of view of potential energies okay now classically if I look at this particular problem because I am assuming that this E is greater than v0 as far as this particular case is concerned if E is greater than v0 you will find out that in this region when the particle goes beyond x is equal to 0 or it goes in positive values of x its kinetic energy will go down energy has to be conserved so whatever was the value of energy then E minus v0 will be its kinetic energy so the particle which is coming from the left hand side will definitely will be able to go and cross over the x is equal to 0 reason and eventually go to x is equal to plus infinity but now beyond x is equal to 0 its kinetic energy and hence its velocity will go down this is what I would have looked at classical way now when I try to solve the problem quantum mechanically we will find something interesting which says that there is also a probability there is small amount of probability that this particle which is going this way can come back okay again this I can understand if I am doing a wave nature because in wave nature whenever there is a change of medium there is always going to be some reflection okay but on the other hand purely from particle nature I will not be able to understand this particular issue so I mean remember in this case as I have been always talking that these particular particles do have a wave character so I cannot just think this particular particle is a pure 100% classical particle okay so let us try to solve the problem we have only two regions here region 1 and region 2 I will write Schrodinger equation region 1 and region 2 let me do my mathematics little faster because we have to cover quite a bit of portion and I think this is pure mathematics the wherever there is a physics concept that I am trying to describe a little bit more in detail so I write general solution now at this particular point see remember in both these cases if I write my Schrodinger equation this was my Schrodinger equation both in region 1 and 2 this is positive now I had just now mentioned in the morning that I could have written the solution in terms of phi is equal to a cos kx plus b sin kx or in the form a e raise power i kx plus b e raise power minus i kx this I had just mentioned in the morning now in the case of bound state problem it made no difference whether I choose this particular type of solution or whether I choose this type of solution okay I would have got exactly the same result but in the case of free state problem it is almost essential to write the solution in this particular part because one of the boundary condition that I am going to use is going to depend on this particular type of writing the wave function so the basic thumb rule is that whenever we have a free state problem it is better to write the solutions in this particular part I mean you have to write the solution in this particular form let me do it like that okay in the case at least the type of problems that I am going to describe where there is a step potential involved okay and when I do the boundary condition you will realize why it is important to write the solution in this particular form while in the case of bound state problem you could have written in either of the two ways it made no difference okay so what I am doing I am writing the solutions in this particular form and remember because here v was equal to 0 and here v was equal to v naught so there will be two different values of k so for phi 1 the value of k I am calling as k 1 of course the energy is same okay which will be given by this and for region 2 there will be different value of k because of potential energy has become different and this k value will be given by this equation again if you are not happy and you are feeling that and I am going fast please substitute this equation into your original Schrodinger equation and you will get automatically the solutions to be this type please do it yourself that will also give you confidence in whatever I am trying to say alright now we apply boundary conditions and I have only two equations continuity wave functions and continuity of its derivatives can I get rid of any other constants can I put my condition that wave function should be finite at plus x x is equal to plus infinity and x is equal to minus infinity unfortunately I cannot this is something which you must understand see when I had put that condition that wave function should be always finite my solution was of the form a e raised power alpha x and e raised power alpha x or minus alpha x are exponential functions which are only growing functions or decaying functions okay while e raised power i k 1 x is actually a sum of sin and cosine term if you remember e raised power i k x can be written cos k x 1 x plus i sin k x or whatever it is alright this particular exponential function with an imaginary coefficient gives you a sin or cosine term okay and a sin term and a cosine term or oscillatory terms they will never become infinite they will always become remain finite so even if you go to x is equal to plus infinity or x is equal to minus infinity I mean this is valid for x is equal to minus infinity if I put x is equal to minus infinity this term will not become infinity this term will still oscillate between plus 1 and minus 1 or whatever it is because sin term will oscillate between plus 1 and minus 1 cosine term will also oscillate between plus 1 and minus 1 so irrespective of whatever value of x you choose these equations are never going to become infinite similarly they are not going to become infinite. Therefore, it will not be possible for me to get rid of these constants by the logic which I used earlier in a particle in a finite box. Is there any other way of doing it? There is another way of doing it if I give a physical meaning of these two terms. This is what I want to emphasize. What I want to insist that we first make a statement then I will explain to you. I insist that this particular part of the wave function gives me a travelling wave which moves in plus x direction. This particular part of the wave function gives me a travelling wave which is moving in minus x direction. This particular part of the wave function gives me a wave which is moving in plus x direction. This particular part of the wave function gives me a wave which is travelling in minus x direction. How? Remember just by looking x term I cannot say in which direction does it move. But remember it has to be multiplied by a time dependent term and a time dependent term is always of the form e raise to the power minus i e t by h cross. Remember I have been insisting this thing right from the time when we are talking about the Schrodinger equation. Implied in this Schrodinger equation is that there is a time dependent term which is negative in sign e raise to the power minus i e t by h cross. And I have said earlier that the direction of the motion of the wave depends on the relative sign of x and t. Because here this particular thing is having a positive sign and time dependent term has a negative time. Therefore, in order that even time increases x has to increase in order to make the phase or the displacement same. Therefore, this represents a wave moving in plus x direction. This represents a wave moving in minus x direction. Of course, as a thumb rule the advantage of taking negative time dependent term was that just by looking at the sign, if this is a plus sign you can always say that this wave moves in plus x direction, this wave moves in minus x direction. Similarly, using exactly the same logic I can say this wave moves in plus x direction, this wave moves in minus x direction. So, if I look at this particular thing, I write a solution of this equation which consists of two terms. One represents a wave which is going in this direction, another represents a wave which is going in this direction. Similarly, when I write the solution in this region too, one particular solution represents a wave moving in this particular direction, another solution represents a wave which is moving in minus x direction. Take the analogy with normal waves, whenever there is a change in medium I can expect some reflection. So, though my original particle is coming in this particular direction, because there is a change in medium, there is a change in potential energy, there is a possibility that this electron may go back or this particular particle, I have not said this electron, it could be any particle, this particular particle goes back. But once the particle has crossed, if at all the particle has crossed to this region, has gone to positive values of x, my problem does not state that this particular particle encounters any further change of potential or any other step potential. So, after that this particular particle can never come back, because now there is no change in the medium, this particular particle has moved from one medium to another medium. In this particular medium, this particular particle has to keep on traveling to plus infinity direction. There is no way that this particular particle will be able to revert its direction. It could have reverted its direction had it seen a change in the medium or for that matter a change in potential energy, which does not happen. Therefore, though it is possible for a wave to go and come back in this particular direction, so if you are in negative values of x, you may find electron going this particular direction, which might have resulted as a result of reflection from here. But when particle has crossed this particular region, there is no way that this particular particle is going to come back. So, for positive values of x, it means in region 2 there is no way that I will be able to get a particle traveling in minus x direction. So, that particular part of the wave function which represents a wave traveling in my minus x direction cannot exist in region 2. Therefore, this particular part thing I will make 0, it means I will put d is equal to 0. That is what is important thing that I am going to mention in this particular case. So, my wave function becomes this way, physical interpretation of different components yield d is equal to 0, then I apply boundary condition, you apply the wave functions to be same at x is equal to 0, you apply the derivatives to be same, this is the wave functions to be same, this is derivative to be same, you just get these things. Remember, we have only two equations, we have three constants, one constant we got rid of, one constant I can never get rid of. We have just enough number of constants and therefore, there is no further condition which is being put here to quantize the energy, the particle could have all possible values of energies. Now, you just solve these equations, this is a plus b is equal to c, a minus b is equal to this thing. So, you can write, take some add them and subtract them, you will get the values of a and b, you can find out what is c upon a and what is a b upon a, that will give you what is the probability that the particle gets reflected or what is the probability that particle gets transmitted. And eventually, as I say these particular problems are more interesting, when you do not have one particle, but you have number of particles which are coming, it is a b more particle is coming. So, you can find out what is the reflection coefficient and what is the transmission coefficient. Come in, so we have one less number of equations than bound state. Remember, when we did a bound state problem, we could get rid of two constants, one on this side, negative side, one on the positive side, here we could get rid of only one constant. So, you have one less number of equations, has no quantization of energy, all energies are allowed. Strictly speaking, non-normalizable wave function that I have mentioned earlier. These are interesting from the point of view scattering of particles, we can evaluate transmission coefficient and reflection coefficient. Here, there is also one of the important point which I must just mention, then I will go to another case, another type of the problem. We have to look at the relative probability of finding the particle different parts of the beam. Probability is different from the number of particles crossing which I must mention. Let me just sort of explain this particular part a little bit. So, let us suppose I am interested in finding what is the probability of finding the particle or let me put it like that, let me top cost in terms of beam because that is little more easily understandable. So, you have let us say beam of particle, you do not have just one particle. Of course, we solve one particle equation. I assume that there is only one particle and then find out the wave function and assume that these particles do not interact like the answer which I was trying to give you in today's morning. And therefore, assume that all the particles have the same wave function. Now, let us suppose this is a small region, let us say x and x plus dx and within a given time I want to find out how many number of particles are there. So, I can calculate what is the probability of finding the particle. Now, for example, if the particle is entering from here and going with a very large velocity, then this particle enters and remain spends very little time in this particular region and goes out. Now, let us suppose their same number of particles which are coming but they are coming with smaller speed. If they come out, come with smaller speed, they will spend more time here before they come out. If they are coming out with smaller speed, you will find out that the probability of finding the particle gets enhanced because you are seeing a larger number of particles which are being there. Now, when I am defining transmission coefficient and reflection coefficient, I define them in terms of how many number of particles cross a particular region. And that particular crossing will also depend on what are their velocities. It will not just depend on probability of finding it. Let us say B upon A square gives this particular part of phi 1 had two parts which was A raised to the power ikx k1x plus B raised to the power minus ik1x. This represents a wave moving in plus x direction. This represents a wave which is moving in minus x direction. That is what we have discussed. Now, if I take complex conjugate of this and multiply by this, I will get A square. Now, this A square is going to be proportional. Remember, we have not normalized. It is going to be proportional to the probability of finding the particle beam which is going in plus x direction. While B square, I take complex conjugate of this. So, it becomes plus ikx multiplied by this. This exponential term goes away. So, this becomes B square. B square is going to be proportional to the probability of finding a particle in the reflected beam. This represents a reflected beam. This represents the incident beam. So, B upon A square ratio gives me the relative ratio of the probability of reflection. Similarly, C upon A square ratio gives me the probability of finding a particle reflected beam, relative probability of finding particle reflected beam. Here, when the particle gets reflected, it moves with the same velocity. But on the other hand, when particle goes into region 2, its velocity goes down. So, for the given region, a larger number of particles could exist. While C upon A square gives me only the probability, if I have to calculate the transmission coefficient, I must multiply by their individual velocities to find how many of them actually cross a given region. So, as far as reflection coefficient is concerned, it will give you B upon A square, which you just put those numbers. When put transmission coefficient, it will be multiplied by V2 by V1 because this is only a relative probability and this particular thing has to take care in order to find out how many particles actually cross this particular thing. Whenever we are comparing things in two different regions, then you have to, where the velocities are different, then velocities have to be taken care. Here also, it should have been V1 upon V1, but because V1s are same, so they get cancelled out. Now, you can show very easily that reflection coefficient plus transmission coefficient is equal to 1. Now, that this particular thing can also be interpreted from the value of the momentum eigenvalues. We have talked about eigenvalues. If I apply momentum operator to A sin kx, you find out that this particular sin kx is not if eigenfunction. That is the reason I did not want to represent these terms of sin kx, because in that case, I will not be able to interpret this in terms of wave moving plus x direction and minus x direction, which is important because we are interested in reflection coefficient and also, we are interested in using this to get rid of one constant. Therefore, I decided not to use sin kx term because sin kx does not turn out to be a momentum eigenvalues of momentum operator. On the other hand, if I apply momentum operator to this particular thing, you can see that this is just this thing which can be written as h cross k times this. You can say px when I apply on this particular function, I get a constant times the same function. So, this a raise power ikx is an eigenfunction of momentum operator with an eigenvalue of h cross k. Hence, this is an eigenfunction of the momentum operator with eigenvalue of h cross k. Now, if I write the solution in this particular form a raise power kx plus b raise power minus ikx, you realize this is also an eigenfunction with an eigenvalue of minus h cross k. This is also an eigenfunction which is with eigenvalue h cross k. When I write these two things, this is a linear combination of two eigenfunctions and therefore, probability that you will get a particular particle with an eigenvalue h cross k is proportional to a square coefficient square. Probability that you find the particle in a with eigenvalue of minus h cross k will be proportional to b square which gives me exactly the same thing that I have discussed earlier. Now, let us just quickly go through the second step potential. Now, we assume that e is less than v naught. So, a particle approaches again from the left-hand side, but now it is approaching with a very energy less than v naught. Now, classically in this particular case, the particle would have always gone back because as I said a particle classically could not have crossed two regions x greater than 0 because then its kinetic energy will be negative. We still call it a free state wave function even though there are certain values of x to which this particular particle could not exist classically. But nevertheless, the physical extent of this particle is infinite. It exists from minus infinity to 0 which is infinite extent. So, this is still a free state problem. This is not a bound state problem. Now, we try to work out exactly the same thing and just let us see. I will just highlight the difference in the solutions that I have used from the earlier case. Now, thing is that in region 2, I will write the solution in this particular form not in terms of i kx and minus i kx because my v naught is larger than e just like in particle in a finite box. Once I do this thing, now these are no longer sinusoidally varying function. I can apply that particular boundary condition that wave function must remain finite. When I say wave function remains finite, it means it cannot blow at x is equal to infinity to infinite value which this term would have done. So, I force this particular term to be 0 and therefore put c equal to 0. But remember, I could get rid of only one constant still. But the physical argument that I have used is different here. In the earlier case, I had used the physical argument that there cannot be any wave travelling in minus x direction. Here, I have used the condition that my wave function has to remain finite. Again, apply boundary conditions, continuity of wave functions, continuity of derivatives. Just take reflection coefficient and in this case, you will find reflection coefficient to be equal to 1. So, you will really realize that all the particles actually come back. They all get reflected back. But nevertheless, there is a finite probability of finding particle in the classically forbidden region just like skin depth. Particle may penetrate to some extent in the classically forbidden region and may then eventually get reflected back. Reflection coefficient is 1. So, all particles eventually get reflected. Still, there is a finite probability for finding particle in classically forbidden region. Now, this gives you one very interesting possibility. See, I had just now said that there is a finite probability of finding the particle in the classically forbidden region. Now, suppose I make this thing very, very small and I have something like this. In this particular region, this is classically forbidden region. But there is a finite probability of finding the particle here. Is it possible that the particle crosses this region? Once it crosses this particular region, then in principle, again the particle can exist because now its energy is greater than V0 and this is no longer classically forbidden region. Now, in principle, there should be a possibility which is something like this. And this is what I want to just quickly go through. I will not give the complete mathematics. I would like to say that this possibility exists. That in principle, this particular particle could eventually be formed in this particular region which we call a phenomenon of tunneling as probably all of you would have heard. Now, this essentially, if you look from the pure nature, you can think that if you make a region very thin enough, if you take a particular opaque material and make it extremely thin, it will always transmit some light. It is almost of this type that you have made this particular thing thin enough so that you can find out always some particles which could, which you would be able to penetrate through. This is what is called tunneling. And you know that tunnel diodes, etc. There are many tunneling phenomenon. I think Professor Ghosh may talk about, you know, sub-superconductive tunneling. They are all based on this particular idea. This I said phenomenon can be understood on the basis of wave theory. If the classically forbidden region is of small width can particle pass through to the other side. This is what I said is a tunneling phenomena. So, I have considered this particular thing. It is exactly similar type of problem. Remember, this is exactly the well that we had talked about earlier, except that now it is on the other side. So, you see the problem which I solved just the first problem today. Between x is equal to 0 and l, the well was on the other side. Now it has gone to v is equal to v, v0. Except that, you know, also the origin of potential energy was somewhat different. But otherwise exactly the same problem. But this is a free state problem that I am looking. A particle is coming from here with a finite amount of energy. The shape is somewhat like the earlier problem. I will divide region 1, region 2, region 3 as before. I will write the wave function and I will go rather quickly. But I do not want to go into the particle approaches from left with e less than v0. General solution, we agree this has to be of this form. In region 2, it has to be this form because this is a classically forbidden region. Region 3 has to be again in this particular form. Here, of course, k has to be same because v0 is same in region 1 and region 3. Is there any coefficient 0? This coefficient cannot be 0, c or d because this is valid only between 0 and l. And between 0 and l, neither this term nor that term becomes infinite. This term could have become infinite. This term could have become infinite had this been valid at x is equal to plus infinity. This term would have become infinite had this term would have been valid at x is equal to minus infinity. None of the 2 is the case. This term is valid only between 0 and l. Therefore, these terms always remain finite. Therefore, I cannot get rid of these constants. Only thing that we can get rid of is this constant because once the particle has cross to region 3, in that case you cannot find a particle which is moving in minus x direction. Therefore, this particular particle, this coefficient becomes 0. I hope the logic is becoming clear of how we are applying different boundary conditions. These are my boundary conditions which I will get. Now, I will match one constant is 0. Then match the two regions now, two boundaries. So, I match their wave functions. I match their derivatives. I get four terms. I express everything in terms of A. Then actually you can do some calculation which is not trivial calculation. You can find out 1 upon t which is the inverse of transmission coefficient A upon F is square. Remember, this represents the incident beam. This represents the reflected beam in region 1. This represents the transmitted beam. So, F upon A square, remember velocities are same in region 1 and region 3. So, F upon A square will represent the transmission coefficient from region 1 to region 3 or represent tunneling current. So, what I have calculated or rather what I have given here is A upon F and that value turns out to be equal to this. So, this is a standard tunneling phenomenon. I will stop here. We can take a few questions. M.E.S. Pillay is. In finite potential, well, if the particle is considered as wave. Yes. Then every time the particle hits the wall, there should be loss of energy due to transmission and reflection. No, no, see, see, see. Because wave, when incident on a surface, there will be three phenomena transmission, reflection and absorption. Basically, I would say that the entire wave function represents that particular particle. My impression is the way when you start talking like this, you are still insisting that your particle is a localized entity which has to be very, very tiny, which is going to go and hit the wall. I say that this picture, we have to get rid of that picture. Particle is no longer localized. The entire wave packet, entire wave which I am writing, entire wave functions represents that particle. And its physical interpretation is only that there is a probability of finding the particle. We are not talking of anything else other than that. I mean, that is what is difficulty at at least at definitely at first year level. Because we are so used to thinking of a particle is a tiny particle which has to go and hit the wall. And therefore, this particular particle has to be here.  I mean, things that those concepts are not, do not exist for a quantum mechanical particle. This wave represents the particle itself. So, entire particle is represented by this particular wave. And this particular wave has to be interpreted as how, what is the probability of localizing a particular particle in a given region. That is all I can say. Nothing more. Thank you sir. Welcome. PSC College of Technology. Sir, we are dealing with optical materials. We always have by theory the sum of absorption, reflection and the transmission is equal to 1. That is right. You are 100 percent right that this particular thing should be always true. But remember when we are talking of reflection coefficient, transmission coefficient, we are talking of free state problem. See, remember there is a difference between free state problem and the bound state problem. The question that you have asked is perfectly valid when we are talking of free state problem. The type of the problem which I did in the later part. Not the first part. Now, there when I am seeing a particular beam of particle is coming, if I take reflection coefficient plus transmission coefficient, that does turn out to be equal to 1. Similarly, in the case of tunneling also, if you find out what is the reflection coefficient and what is the transmission coefficient, that has to turn out to be equal to 1. See, there cannot be a loss of particles. Is there a BO particle? The number of particles have to be conserved. So, transmission coefficient plus reflection coefficient always have to be turned out to be same and to be equal to 1. There is absolute ambiguity. What you have said is a perfectly correct statement. Sir, does it mean that the absorption is also inclusive in this transmission, sir? No. See, in this case, we are not considering any absorption because we are talking of quantum mechanical particles and these wells are somewhere where particles cannot be absorbed. So, it is not like that optical wave where you know you can some photons can be absorbed and can give rise to some sort of excitation. So, I am assuming that none of those processes are possible in which these particles could be absorbed. See, these particles are sort of a massive particle. I mean, it is not so. I mean, you can imagine always a system, you know, in the relativistic system when the particles could get absorbed, can disappear. But I do not expect these things to happen in these cases. At least, these equations are not valid for those relativistic cases. Okay. Thank you, sir. Thank you. Goa College of Engineering. My question is related to the tutorial session that we had yesterday on quantum mechanics. Yes. And there we discussed the application of this wave particle duality as applicable to Bose magnetization condition. That is right. Yeah. So, my question is since an electron is said to be in stationary orbit because it is not supposed to lose energy. Yeah. My question is when a stationary wave is formed, we require an incident and a reflected wave. So, that the stationary wave pattern is formed. How do we interpret it? Will you please elaborate in that context, sir? The basic issue is something like this. See, that particular problem which was given was the way it appears, the Broglie thought of the issue. Okay. It is very, very difficult to imagine a stationary wave in a circular string. All right. So, in fact, normally in the normal lectures, I always want to avoid this particular thing. But nevertheless, I do want to mention because this particular thing has a historical impact. Let me put it like something like this. Generally, for example, you take a vibration of a string. You find only those type of frequencies possible in that particular string, which satisfies certain boundary condition and which gives you a stationary wave. All other frequencies will not sort of exist in that particular string. It is like a musical instrument. If you are trying to play a violin or playing, want to play sitar, okay. So, there is only certain frequencies which satisfies certain boundary conditions. Now, remember, let me complete it. Now, Bohr's model is strictly spring not correct. We cannot talk about Bohr's orbit these days with quantum mechanical idea. But at that time of D Broglie, Bohr's model was in fashion. And what he was thinking, what is so surprising that only these particular orbits are possible and no other orbits are possible. Okay. So, he thought that why not imply a condition of a stationary wave maybe some stationary wave is propagating inside. And that's the reason these are the only orbits which are possible. All right. So, using that particular concept, he obtained this D Broglie relationship and then applied it. So, it is more of historical importance rather than talking of more physical aspect. Because concept of Bohr orbit itself is not correct. Yes. Please go ahead. But how to reach out to the students while explaining this? They will ask the question that how stationary waves are formed there. Exactly the same way this way I have described. This is the way I would like to answer to the question. See, this is only of historical impact. This is the way D Broglie obtained this particular thing. I know that this particular thing I cannot talk. I cannot talk of a well defined orbit. I can only talk about wave function in today's world. But historically, if you look, that is the way D Broglie found it out. Because that's the way he convinced himself that this is the way probably the particle should have a wavelength. So, that's the way I will take and I will stop here. Yes, you have another question. Sir, can we go back to the slide where we had this finite potential well solution to the three regions. While selecting the solutions for the three different regions. That's right. Is there a specific condition that we have to adopt or we can generally take any solution for the three regions? See, the thing is that as I have mentioned in the morning today, whenever you are getting the equation of the form d2y dx square is equal to a constant times y, the solution can be obtained by just looking into it. Of course, I am not very sure whether mathematicians have a rigorous way of deriving it. And so long I have my two constants and it is a second order equation. This is a general solution. So, all these equations of the type d2y dx square is equal to a constant times y. I can write the solution depending upon whether this constant k is positive or negative. If it is positive, the solution is in the exponential form. If it is negative, then it is in the form of sine and cosine. So, just by looking at that equation, I can find out the solution. So, there is no specific rules by which these solutions have to be taken. At least I do not know. I mean probably mathematicians would have a way of doing it. I am not sure about it. And for the barrier potential, if we look at the tunneling phenomena of an electron crossing over into the forbidden region, to go into a third region, the electron not only has to have the v0, it also has to travel the distance L. So, what would that term v0 into L give me? No, the thing is that again when we are talking of a particle going from one region to another region and passing through a classically forbidden region, again I am insisting on a localized picture of the particle. It means again I am thinking of my ideas are classical. Now, what is happening? Let me put that not v0, I think v0 square L has some kind of significance because this will actually determine how much is the tunneling etc. That is a different issue. But if your question is about that how the electron passes from that region, if at all I want to perform an experiment and want to be very sure that my particle is in the classically forbidden region and also want to measure its kinetic energy to find out that it is negative, I will never be able to perform this experiment. My uncertainty principle protects, that is what I was trying to tell you that uncertainty principle protects wave particle duality. Whenever I have this uncomfortable question, I can always raise my hands uncertainty principle, I cannot answer this question. Whenever I try to do this thing, I will land into a problem, I will always perturb the experiment enough. Then I will either not be sure that my particle is in the classical forbidden region or I will not be sure about the measurement of kinetic energy. Just like I will never be able to perform the experiment, whether particle goes from this slit or it goes from this slit. So is that why we take an i term in the first in the third region e raise to i? No, we take i because I want to assign a direction to the wave function. That is only possible when we have solution of e raise power ikx because I want to say that there is no wave which is travelling in minus x direction. So reason why I am putting e raise power ikx and minus ikx, because I want to interpret them in the form of a travelling wave which is moving plus x direction and minus x direction. And this is absolutely essential whenever I am talking of free state solutions. In a bound state problem, it does not matter. All right, I think I will just close the discussion.