 Let us start now this will be my last lecture on the quantum mechanics portion. Let us just discuss what we have sort of done in the last lecture which was a early morning, today morning. We worked out for a particle in a finite potential box, then we discussed some of the free state problems. One of the most important result that we obtained is that there is a certain probability for particle to get reflected even though classically the particle could not have crossed the barrier. This is what we have said, then we also said there is a finite probability of finding the particle in the classically forbidden region and eventually we discussed rather briefly a tunneling phenomenon very very important phenomenon which does not have a classical analog where you can have a small separation of a classically forbidden region and because particle have a finite probability of being found there if this is thin enough you can have a reasonable amount of probability of the particle crossing to another area and once the particle crosses to that area because now that is no longer classically forbidden the particle can keep on propagating. So this is a phenomenon which is very well known as you know there is a lot of semiconductor devices which are based there are many other tunneling phenomena which are pretty well known. I mean in fact even these days in magnetism tunneling magneto resistance there is so many things tunneling phenomenon has found a lot of applications and this can be understood on the basis of quantum mechanics. So that is why I am always surprised when people say you know what is the use of quantum mechanics because once you see of all these things all the recent developments which have taken place in varieties of areas you realize that how quantum mechanics has played a role in actually pushing many of these things many of these devices many of these applications. Okay now we will try to work out one or two problem I will take one particular specific problem which is a sort of interesting problem which is half the way between a particle in a finite box and in finite box and there is only one particular aspect which I want to discuss I want to use this particular thing to discuss a very very specific aspect actually I could have discussed this particular thing also using particle in a finite potential well but I thought you know let me do one more problem and try to sort of understand it try to discuss it. After that you know what I will take one or two more examples before I come to particle in a three-dimensional box. So let us look at this particular problem. So this is a particular problem which as I say is a sort of a mixture of particle in an infinite box and a particle in a finite box. So again you have a box let us say between x is equal to 0 and x is equal to L and like particle in an infinite box for all values of x less than 0 the potential energy is infinite while as far as x greater than L is concerned the potential energy is V0 but is finite alright. So in that sense this is not really a symmetrical problem unlike the problem that we have solved so far and let us discuss the bound state solutions of this of course you can always think of a free state solution when a particle comes here but that problem is not very interesting because if you know if a beam of particle comes from this side okay it will has to because there is a what we call as a perfect reflector here where the potential energy is infinite all of them have to go back and therefore in principle reflection coefficient will always be equal to 1. So that problem is not of that much interest but let us try to find out the solution of corresponding to the bound state and for bound state it means I am looking only for those solutions for which energy E is less than V0 then only this particular particle is bound between x is equal to 0 and x is equal to L. So let us look at the bound state solution of this particular problem where potential energy is infinite for all negative values of x is 0 between x is equal to 0 and x is equal to L and for L values of x greater than L it is positive and it is equal to V0. Like before I will divide into region 1 and region 2 there is no point in devising in this take as a fact of this particular region because as we have discussed earlier the wave function has to be 0 of course if you are not very satisfied by wave function should be 0 you can always take potential a finite potential well and take the limit of V tending to infinite then you can say those C that the only solution possible will be that wave function is 0 in this particular case. So I am dividing only in 2 region region 1 and region 2 and in this region 1 the solutions will be of the sinusoidal form while when we are talking of this particular solution this region the solution will be exponentially decaying because this is a classically forbidden region. So in region 1 I have written the general solution as A sin kx plus B cos kx and because V is 0 in this particular case k square will just turn out to be 2 Me upon h bar square. Again I repeat that if you are not satisfied how I put k square is equal to this thing you can just substitute this equation this particular expression back into the Schrodinger equation and then you will get k square to be equal to this. For region 2 I have as I have said this is going to be an exponential type of functions. So I will write this particular thing as C e raise power alpha x plus D e raise power minus alpha x and remember alpha when I am writing this is V naught minus e because this is V naught minus e which is positive and therefore alpha is real. Region 2 for bound state solutions V naught minus e is going to be positive and alpha is going to be real. Now again I putting my boundary conditions which again I am saying as I am jumping steps. One condition is that this solution is valid for x is equal to plus infinity at x is equal to plus infinity this term will blow to infinity I do not want this term. So I remove this particular term by putting C is equal to 0. So the only solution which is possible only part of this solution which is possible is D raise power minus alpha x. Similarly I will put the condition that phi 1 of at x is equal to 0 should be equal to 0 remember because at this particular point there is a infinite jump in potential energy therefore I do not have to match d phi dx I have to match only phi 1 and for values of x less than 0 here the wave function is 0. So wave function must be 0 at x is equal to 0. So you put x is equal to 0 then you get sine 0 and you will get b cos 0 is 1. So a sine 0 is 0 so you will get just b it means b should be equal to 0. So you will get b is equal to 0 and c is equal to 0. So remember if the solution has to start from here at x is equal to 0 it is only a sine term which can be present this is just from pure thinking because cos term at x is equal to 0 will not be equal to 0. So cos term will not be allowed because at x is equal to 0 you expect only the sine term will be 0 therefore you will always have this particular b coefficient b to be equal to 0. Now you put the boundary condition at x is equal to l which is the wave function should match and remember at x is equal to l the potential jump is finite so I have to match derivative is also at x is equal to 0 I did not match because at x is equal to 0 the potential jump was infinite here the potential jump is only finite here the potential jump is only finite. So at this particular edge I have to match not only the wave function but also the derivative of the wave function. So just like before remember this was my sine kx so I put x is equal to l so you will get a sine kl this term was 0 so you will get d e raise power minus alpha l. So this is what I have written then you take the derivative of that and put x is equal to l so this is this remember this dash which I have written means derivative with respect to x you get this particular condition. So you must have a sine k l is equal to d e raise power minus alpha l and a k cos k l is equal to minus alpha d l I just divide the two d gets cancelled and the solution turns out to be cot k l is equal to minus alpha by k like what I have discussed earlier in a particular finite box alpha and k both are dependent on energy this is alpha and this is k both are dependent on energy therefore solving this equation cot k l is equal to minus alpha by k you will be able to get the values of energy only for certain values specific values of energy will this equation be satisfied and those will be the only values of energy which will be allowed for this particular particle. As you can see that in this particular case you find that the equation which is governing the energy is comparatively simpler in comparison to the particle in a finite box. Now the next question which I wanted to talk about it that if I have to solve a general solution for this particular equation and find energy in the closed equation in the form of an equation as I have said this is not very easy because this cannot be done in principle and definitely for a particle in a finite dimensional box one dimensional box it becomes much more difficult. But on the other hand if I want to solve the question at what should be the value of v naught or for that matter what should be the value of length of this particular box for which one of the allowed energy is a unique value a particular value that can be done comparatively easily. So this is what I have tried and tried to answer this particular question I want to find out that what should be the value of v naught or l corresponding to which I will get one particular energy just at e is equal to 0. So what I am looking that this is sorry at e is equal to v naught not e is equal to 0 because this is what so I want one particular energy level which is what I call is just bound. So remember if energy was greater than v naught this will become a free state. If energy is less than v naught then this becomes a bound state. Now I am want to find out a condition that what should be the value of for a given l what should be the value of v or for a given v what should be the value of l corresponding to which I will get just a state which is just bound. This particular thing has a specific meaning as we will be discussing little later. Let me also tell you I will give you the problem what will be the value for a given value of l what will be the condition on v naught so that the value of energy energy value of the bound state is just equal to v naught by 2 that also I can do that is much simpler. But to find out given an arbitrary value of v naught and given an arbitrary value of l what will be the value of energy of the bound state that is little more difficult question. Now I am saying that I want to look specifically at solution corresponding to e is equal to v naught let me just go back to this equation. I want a solution corresponding to which I will get a state which is just bound just bound okay it means it is just v is equal to v naught okay let us try to find out the condition for that particular case. If I do that then remember because I am looking specifically for solution v naught is equal to e so under root 2 m v naught minus e will this particular term will become 0 therefore alpha will become equal to 0 and because e is equal to v naught so therefore k will be under root 2 m v naught by h cross. Now I can relook at what will be my boundary conditions my boundary condition will just become a sin k l is equal to d remember this was my a sin k l is equal to d raise power minus alpha l because alpha has become equal to 0 so this condition becomes a sin k l is equal to d. Similarly if I look at this condition is a k cos k l is equal to 0 which is also obvious from this particular thing because this alpha is 0 so this entire term is going to be 0 if this entire term is going to be 0 this is going to be a k or cos k l should be equal to 0. Of course you could have obtained also by substituting it here but I thought of doing it right from basics so if at all we have a bound state just at e is equal to v naught in that case this condition must be satisfied. See for example if a k cos k l is equal to 0 then we said a cannot be 0 because if az is 0 the wave function itself can become 0 so only cos k l equal to 0 would be allowed. So what happens here that a cannot be 0 so only cos k l can be equal to 0 if cos k l is equal to 0 then it means k l must be equal to 2 n plus 1 pi by 2 only when k l takes one of these values where n is an integer it is it is possible for cos k l to become 0 it means if potential energy I just substitute the value of course we know k l which I have k I have already substituted under root 2 m v naught minus h cross it means for a given value of l if I have this potential energy only then I will have a bound state just at e is equal to v naught as you can see there are multiple values of v naught for which this condition will be satisfied. So if n is equal to let us say 0 then in that case this particular value of v naught will correspond to h square pi square by 2 square upon 2 m n square okay there will be one particular bound state at e is equal to v naught okay then if I put n is equal to 1 then I will get another value of v naught corresponding to which there will be another bound state. Now let us just sort of try to interpret this particular thing little more clearly in fact this is this particular aspect is a little difficult aspect which in fact I have found the students also find it little difficult to understand so let us just go little slowly about this particular thing. Now first statement which I make that there will be no bound state if v naught is less than equal to h square pi square by 8 m l square how would I obtain this particular expression I put n is equal to 0 if I put n is equal to 0 you get pi square by 4 square okay let me first write then I will display. So just you see because n was turning out to be equal to 0 so this is h square pi square by 4 this divided by 2 m l square so this becomes h square pi square by 8 m l square. Now what I am trying to say is that this is the least value of v naught which is required for which you will get a bound state corresponding to this energy what happens if v naught was smaller than this okay because v naught is something depending upon problem I mean I am just looking at a theoretical ways let us suppose there is a way by which I can continually vary v naught if I keep on reducing v naught and let us suppose there is a problem where v naught is less than this particular value okay what I insist that this particular problem will not have any bound state there is no energy allowed state allowed in which the particle will be bound only free states will be allowed the reason for that is little subtle so let us try to understand it. Let us suppose initially we have a condition when this particular well well was not existing at all okay so you have this particular thing and I reduce this so much that the well is only like this almost like this okay then I slowly increase like this. Now what I want to say that the least value of v naught for which you have a bound state here now if I keep on continually decreasing I will not find a bound state coming here until my v naught is dug to this particular value which is equal to h square pi square by 8 m l square for all values of v naught smaller than there is no bound state now it is not possible for you to suddenly create a bound state if I am reducing here then in that particular case whenever a new bound state comes that will always start appearing from is equal to v naught that is one particular thing which I want to insist on this particular thing let us see whether people understand this particular issue or not very correctly because all our variables are well behaved variables okay and I require a certain minimum value of v naught to create a bound state here let us suppose if we had just a situation like this this was x is equal to 0 and x is equal to l it means essentially this particular potential well was not existent at all in that case you agree that there is cannot be any bound state now I start digging my bound states I am digging my well and this is being slightly less but I know I have just dug it slightly but I know my bound state cannot appear here x is equal to 0 because it is just very very close if at all at bound state has to come I am starting from a continuous state and as I am slowly digging my well okay a bound state has to start appearing and a bound state cannot suddenly start coming half the way it has to start always from this particular case this is something which is very very important a bound state a new bound state will always start appearing from the boundary of a free state and a bound state that is what I am trying to essentially assess on this particular thing if it was very very close to this bound state I know does not exist because it is very very close to e is equal to v0 okay and for a bound state to appear at e is equal to v0 I must at least have this much v0 if I have any value of this if I slowly keep on going from this condition to this condition whenever a bound state comes it will first start from here then start going down so initially a bound state will be here then as I dig more this particular bound state will go down and keep on going down then unless I get another value of v0 corresponding to which a energy e is equal to v0 is allowed then a second bound state will start coming in think that I am trying to insist is that whenever a new bound state starts coming in okay that will always I if I let me put it like that if I continuously vary v0 okay start from a situation if I go from a situation where this is essentially v0 is equal to 0 so this point is here and continuously making v0 little larger little larger then whenever the first bound state will come it will always appear from this particular point then as I dig it more this particular bound state will start coming down and down and down then I of course given a particular arbitrary value of v0 and l I will not be able to find out for an arbitrary value where is this particular bound state but one thing is definite whenever if I have dug it enough so that a new bound state starts coming in that will first appear at this particular point and then only it will start going down in principle by looking at the value of v0 and l I can find out how many number of bound states will be there in such type of situation you cannot have infinite bound states unlike particle in infinite box where you can have infinite bound state here you will have depending upon value of v0 and l you will have only a finite number of bound states so what I am insisting is that there will be no bound state if v0 is less than h square pi square by 8 ml square because this is the least value of v0 which is required to create a bound state at is equal to v0. Now we will have only one bound state unless a second bound state starts appearing at is equal to v0 second bound states where we will start appearing at is equal to v0 when n becomes equal to 1 when n becomes equal to 1 so n is 1 so 2 plus 1 so this will become 9 square and 9 upon 8 ml square until this reaches a value of 9 h square pi square by 8 ml square once your v0 becomes equal to this then again I will get a bound state exactly at is equal to v0 so between this value and this value I will get only one particular bound state as I keep on increasing v0 further then the second bound state which was created which will now keep on going to lower value it will no longer be at that particular boundary it will keep on going down okay until I hit a particular value of v0 which corresponds to n is equal to 2 when it hits to n is equal to 2 a third bound state will start appearing at is equal to v0 and then at that particular state of time you will have three bound states so depending upon what is the value of v0 you can actually check you can actually found out without solving that equation how many number of bound states will appear in this particular problem this is what is very very interesting about this particular thing that you can actually create you can always find out the total number of the bound states in this particular case so between this and this you will have just one bound state then I can always find out what will be the condition when you have just two bound states then we can find out the condition when we will have just three bound state of course this question can also be answered in terms of l I could have fixed my v0 and said that what will be the values of l corresponding to which I will get a particular bound state because see what appears in these equations basically v0 multiplied by l square in fact morning I had to say v0 square into l that is not correct v0 into l square now let us have some comments on this first of all is that a finite v0 is needed for a bound state to exist in this particular problem you definitely require a finite value of v0 for a bound state to create if v0 is less than that you will not get a bound state this problem has an alternate allowed energies for a particle in one dimensional box of length 2l with the same potential this is also an interesting point this is actually invoking symmetry if you have a situation like this the problem which I solved in the morning actually if you look at the first bound state the wave function would look something like that because this is a sinusoidal term so it will be something like this and this will exponentially decay here exponentially decay here this will be the ground state solution the first excited solution for this particular thing see actually there is a theorem in quantum mechanics which says that Cartesian coordinate system when you are going to the next excited state the first excited state or the ground state will have no crossing with x axis when you are going to the first excited state you will have just one crossing okay let me not bother about this particular thing let us go to the second bound state so as a ground state if I look at the second excited state it will become like this because this has to have one crossing and because of symmetry this curve must be symmetric about this particular point okay and this will decay like this and this will decay exponentially like this this has to be symmetrical about this particular point now what you will notice that if I divide this particular thing half okay this will not satisfy see my problem which I am solving trying to solve now is that there is a perfect reflector at this particular point now what is very interesting to realize that this particular solution will just satisfy this particular condition also when there is a perfect boundary provided we take only this half of the wave function ignore this particular half of the wave function so what will happen whatever energy that I would have got corresponding to this value of v naught and a length 2l okay this will obviously will not be allowed solution because this actually at x is equal to l I mean in the present problem we will have to put it like 0 let me put plus l and minus l 0 plus l and minus l the problem that we had here was 0 and l all right now here in this particular problem I wanted that the wave function should be 0 at x is equal to 0 this wave function will not be 0 at x is equal to 0 but this wave function is 0 at x is equal to 0 now this particular thing will have a wavelength exactly of whatever is the this particular wavelength and it satisfies this boundary condition as well as this particular boundary condition so if I just cut this particular thing into half this will turn out to be a solution of this now if I look at the third one it will have two boundaries two crossings and then the wave function will be looking something like this and in that case again at x is equal to 0 you will not find that the condition of wave function being 0 gets satisfied okay only when I go to the fourth one then again because we will have two things again at x is equal to 0 you will find the wave function to be 0 so what you will find whatever the solutions corresponding to this you have obtained if you ignore the first one take the second one ignore third one take the fourth one then alternate of these solutions will turn out to be solution of this particular problem so this is very very interesting which can be argued purely from the symmetry point of view without being in fact symmetries and quantum minis are sometimes very very interesting so this problem has alternate allowed energies of a particle in one dimensional box of length 2L with the same potential purely from symmetrical orbits okay in this particular case I want to insist that you will always have a bound state irrespective of the value of V0 so if I go back to this particular picture see remember in this particular case this particular state is also there which is not present here now you can show this I leave this as an exercise for you to show that even if you reduce this particular thing to almost zero thing you will always get a bound state here this particular state will always be present but in this case because this is not the allowed one this state will require certain type of potential so in a finite potential the bell of like this you will always get one bound state irrespective of house over small is your V0 but in this case you will get only a bound state when V0 acquires a certain value depending upon the value of V0 and L so this is what is very very interesting a bound state always exist in a letter case which is the case of one dimensional box of length 2L finite bell okay obviously it should be finite now let me try to work out a little more realistic problem which is very very well known problem which is called the problem harmonic oscillator so let us assume that there is a particular particle which is experiencing a force of the type of Hooke's law which is minus kx and you know corresponding to that force the potential energy is given by half kx square so if I have to write a Schrodinger equation for an harmonic oscillator a particle which is under the influence of force which is called f is equal to minus kx Hooke's law type of force then this is the equation Schrodinger equation that one has to solve and let us assume that this is a one dimensional harmonic oscillator now if I want to start from basics and I want to really solve this particular equation in its most general fashion this is going to be much more difficult and definitely is beyond the scope of this particular course okay one has to look at the book of a proper quantum mechanics book the first level book and let us say like shift for any standard book there are so many books which deal with this then they will give you a general solution what I am trying to do is to do the other way I am giving you a solution which I know somehow okay so this solution suppose I know I want to show that this particular solution is going to be of the form which I have described here phi x is equal to a e raised power minus alpha x square and using this particular solution I can tell what will be the energy corresponding to this particular solution so I am doing this in a much more simple fashion let us say somehow I have hit upon a value of phi x let me just repeat this is not the way that you would like to solve harmonic oscillator problem you will actually start with this equation and try to work it out this particular thing in a more general fashion okay the solutions are much more complex which are called hermit polynomial which as I say is beyond the scope of this course but I want to just show it basically I want to demonstrate to first year students that this is a really a realistic problem harmonic oscillator okay and we know what are the energy levels for this particular thing okay so let us look at this particular question I am telling you that somehow I know this particular solution I want to only convince that this is actually a solution and then find out using this solution what will be the value of energy for which this will be the solution in fact this is the solution corresponding to the ground state of harmonics oscillator which is corresponding to the lowest energy state okay now what I will do because I want to show that this is a solution I will substitute this particular solution into this particular equation so let us go to the next one this is phi x so in order to substitute first I have to find out the I have to differentiate with respect to x if I differentiate okay this is exponential function so it will be present as it is e raise power minus alpha x square multiplied by differential of minus alpha x square okay alpha is a constant so it will be minus 2 alpha x so this is the differential coefficient of phi with respect to x now I take the second derivative second difference second order differential coefficient so I have to now this is a product of two functions of x okay so I have to differentiate this multiplied by this plus differential of this multiplied by this so this is the first thing which I have taken I have taken minus 2 alpha x into differential coefficient of this when I take differential coefficient of this I will get again this particular term okay so this is what is there and there is already minus 2 alpha x so now it becomes minus 2 alpha x square then I take this as a constant and differentiate this particular thing so I will get only minus 2 alpha so this is what it is so this is d 2 phi dx square I substitute this into Schrodinger equation this is the term which I have written this is what I have just now found out d 2 phi dx square plus 2 m upon h square e minus v v is half k x square and phi I know which I have started with the solution a e raise power minus alpha x square all right then I sort of start reorganizing the term so there is minus in fact a e raise power minus alpha x square but cancel everywhere the wave function cannot be zero so I can always cancel it out so I will get minus 2 alpha x square here then I will get minus 2 alpha which is the second term here and this third term will remain as plus 2 m upon h square e raise power half k x square because this term goes away now what I will do here I will try to reorganize this term if I expand this particular thing I will get x x square there will be x square there are certain terms which are constant for example this term does not contain a alpha this term will not contain x similarly this term will also not contain x so there will be 4 terms 1 2 3 4 2 of which will contain x square and 2 of which will be constant so all I am trying to do is to reorganize these terms so I am just reorganizing these terms okay x square this is the same equation which I have written earlier now first I assemble all the terms which are x square so there will be x square term which is here which is I take the square of this thing so I get 4 alpha square so this is 4 alpha square there is another x square term which is this half will cancel it out so m k upon h square you get m k upon h square both these terms contain x square there are 2 constant terms one is minus 2 alpha which is here minus 2 alpha there is another constant term plus 2 m upon h square so you get an equation of the type a x square plus b is equal to 0 so you have an equation of this type now this you can say is a quadratic equation so can I find out the value of f x okay for which this equation is satisfied no I do not want to do this I want that this particular equation should be a general solution it means for all values of x this should be correct okay I do not want a solution which has to be valid only for a specific value of x see this particular solution which I have obtained or which should be there from solution of Schrodinger equation should be valid for all values of x so this particular equation must get satisfied for all values of x not for some specific values of x and that is only possible when this a is also equal to 0 and b is also equal to 0 only when these 2 terms are 0 then this particular equation will always be satisfied irrespective of the value of x so for all values of x this equation has to be satisfied only when this coefficient of x square and this constant terms both become individually 0 so that is what I have shown last time in the next transparency so this is there so I will put this term to be 0 and I will put this term to be 0 this is the equation first thing I have put this equal to 0 so I get a value of alpha which turn out to be under root m k upon 2 h cross second I put this particular term to be equal to 0 so in fact from this I have found out the value of alpha in terms of the force constant and the mass of the oscillator 2 m upon h square this gives me energy in terms of alpha h square by m I substitute this particular value of alpha and I can find out what will be the energy of this particular particle this turns out to be I just substitute it here and this energy will turn out to be equal to half h cross under root k by m and as generally under root k by m is written as omega this is half h cross omega so I can see that this is the value of energy which will get corresponding to this wave function so in fact if you know about the harmonic oscillator energy energy levels harmonic oscillator energy levels are given by En is equal to n plus half h cross omega or h nu remember plank originally took En is equal to n h cross omega remember this was a hypothesis he did not have Schrodinger equation to derive these things okay but when we take Schrodinger equation we get exactly this form except for this additional term of half which I said that the ground state energy is lowest energy cannot be 0 here for n is equal to 0 the energy will be equal to 0 here what you actually get that there is a what we call as a zero point energy energy corresponding to the lowest energy state this being a bound state remember this is an exponential solution okay but still treated as a bound solution because the particle physical extent is still limited to a definite region of the space so for a bound state problem the energy lowest energy has to be a non-zero and this turns out to be equal to half h cross omega what I would suggest that you can try another type of form which is the first excited state which the solution will be of the form x into e raise power minus alpha x square and there you will find out that this particular thing you know the second energy that you will be getting 3 by 2 h cross omega exactly in the same fashion so it will just if you put if I would have given you this particular thing not as a into this but a x into e raise power minus alpha x square try this solution yourself try to work out exactly in the same fashion it requires exactly the same steps except that no second derivative will have a slightly more number of terms and you will get that now this equation will get satisfied with an energy 3 by 2 h cross omega so you can actually see that this is the way the energy spectrum will be looking like that I would also just like to mention something in support of the the zero point energy I am not sure whether you are familiar with this particular thing that helium gas which is the lightest inert gas okay see because the inert gas the bonding forces are very very small they are only the Wonderwall forces which are very very weak forces in fact that is why it is rather difficult to liquefy the inert gases and for helium it is believed that you cannot solidify even if you go to zero degree Kelvin of course people do not go to zero degree Kelvin you cannot reach zero degree Kelvin but looking at this particular helium they say that because of the ground zero point energy it is impossible to solidify helium at very even if you go to very very low temperatures only when you put pressure on liquid helium that you can get a solid helium okay this is actually treated as an evidence of actually possession of zero point energy now I come to the last problem of my lecture which is a particle in a three-dimensional cubical box this particular problem is probably the one of the most interesting problems from the point of view of solid state physics and when professor Suresh pj solid state physics he would be using this particular result see as we have said that one-dimensional boxes are generally only from the point of view of you know theoretical interest okay of course there are always certain cases from which you can draw the parallel of the one-dimensional box but three-dimensional cubical box is a more interesting problem okay you can always imagine a particular particle to be sort of contained within a particular box of three-dimension okay in a room or of course the room is too big but I mean you can imagine room at a microscopic level when there is a particular electron which is confined to a three-dimensional box there are certain specific things I want to introduce from here the concept of degeneracy okay which we have not yet introduced okay then essentially this particular aspect of the quantum mechanics course and then we will close it down. So now just like we had one-dimensional box we have three-dimensional box we find that the potential energy is zero only within this box it means only when x is less than zero x is between zero and l and similarly y is also between zero and l and also similarly z is also between zero and l okay in all other cases the potential energy is infinite so potential is zero only when this particular the by x, y, z values lie somewhere within this cubical box for easiness I have taken this to be cubical even it is rectangular I mean not really cubical you could have worked it out but it is more interesting from the point of view or rather more simple from the point of view cubical box. Now what I have to do because this is three-dimensional problem I have to solve a three-dimensional Schrodinger equation and this equation I have given you the general equation has been given to you earlier. So this becomes my three-dimensional Schrodinger equation which I have given earlier if you look at my earlier notes this is what we have written and then this del square operator in three-dimension is written as del 2 psi del x square del 2 psi del y square del 2 psi del z square remember generally because this is a three-dimensional space so wave function would consist of all these three coordinates x, y and z and therefore when I am taking partial derivative with respect to x it means y and z have to be treated as constant if I am taking partial derivative with respect to y x and z have to be treated constant and so on okay because potential energy is 0 within the box so I have put v is equal to 0 so the equation just becomes 2 upon h square e times phi is equal to 0. Now we use this particular method separation of variable to solve this equation. So remember we have used this particular method of separation of variable when I obtained time independent Schrodinger equation from time dependent Schrodinger equation. From time dependent Schrodinger equation I separated the time variable and x variable and then we got two differential equations okay one kind of containing time term and that particular time term dependent term we had because that did not contain potential energy so we solved it separately then whatever was remaining was time independent Schrodinger equation. Now we use the same procedure of separation of variables in fact one once you go to the problem of hydrogen atom that is what you exactly do except that in that particular problem you have a spherical symmetry so you express this del square using r theta phi coordinates you do not use x, y, z coordinates but here we use x, y and z coordinates they are also we separate out variable and here also we separate out the variables okay. So what I will do I will assume that this phi is product of three functions so this is an assumption okay if using this assumption I am able to solve separate out the variables I will I am successful it is fine if not it means this particular method of solving the Schrodinger equation would not work. My logic are exactly identical I have to take derivative first partial derivative with respect to x okay and I am assuming that this x is only a function of x coordinate this capital y is only a function of y coordinate this capital z is only a function of small z coordinate alright. So if I take partial derivative with respect to x capital y and capital z will be treated as constant even if I take the second derivative is still capital y and capital z will be treated as constant so it will become capital y capital z into del 2x del x square but as we have said that this particular term contains only x okay so there is no point in writing partial derivatives because this has only x no other variable so I just write this as d 2x dx square. So this equation this particular first term becomes y z d 2x dz square remember this is my equation this is what I have found out. Similarly I will now find out d 2 phi dy square in that case x and z will be treated as constant and only capital y will be differentiated with respect to y here capital x and capital y will be treated as constant and only capital z will be differentiated with respect to small z this what has happened capital y capital z d 2x dx square capital x capital z d 2y dy square and capital x capital y d 2z dz square plus of course in this equation place of y I have written capital x capital y capital z as before so this is what has been written here alright now what I will do exactly the same thing I will divide by xyz when I divide this by xyz I will get here 1 upon x d 2x dx square remember xyz cannot be 0 so I can always divide it okay then when I divide this by capital x capital y capital z I will get 1 upon capital y similarly I will get capital z and this particular term will become 2m upon h square 2m upon h square this will become 1 xyz divided by xyz so this particular thing I can take on the right hand side and I will use exactly the same argument I will have 3 terms which is equal to a constant first term contains only x second term contains only y third term contains only z okay and this must always be equal to a constant irrespective of x y and z and that is possible when all the three terms individually are constants this is what I have written this was the first term which I had written earlier I had divided by capital x capital y capital z so this becomes 1 upon x d 2x dx square 1 upon capital y d 2y dy square 1 upon z capital d 2z dz square this I have divided so it becomes 2m upon h square which I have taken on the right hand side okay so therefore you just find it out that these three terms turn out to be equal to a constant this is purely a function of x this is purely a function of y this is purely a function of z okay if this equation has to be satisfied all three of them have to be individually constant because for example if I just change x okay and if this is not a constant this term will change but this term will not change okay so the result can cannot remain constant if I just change y and do not change x and z okay if this is not a constant this term will change but these two terms will not change so their sum can never remain constant so if their sum has to remain constant only possibility is that all these three terms individually remain constant so this is very very interesting because here in one shot you can separate out the variables generally when we separate out r theta phi coordinates in hydrogen atom case we actually separate them out one by one okay in this case in one shot you are able to separate out the three variables and you get three differential equations out of this particular one differential equation so this is what I have written and for the convenience I have written as minus kx square minus ky square minus kz square the reason is that if you look at this particular equation the sum of this equation has to be equal to minus 2m upon h square because I am looking for a solution which are for which e is positive because potentially is already zero so e is always has to be positive therefore their sum has to be positive therefore I am writing this in the form of minus kx square minus ky square minus kz square such that kx square plus ky is square plus kz square must be equal to 2m e upon h square so this just ensures the writing this x square ensures that this particular quantity is positive this particular quantity is positive this particular quantity is positive and negative sign is taken here taken here and taken here all right so this is the equation which must be satisfied the only thing is now that we have now three different equations and we can solve all these three equations individually this equation happens to be exactly the same way which we have solved for a particle in a one-dimensional infinite box so I will put exactly the same solution which will be of a sin kx plus b cos kx only thing here there are three different kx's here is one kx another is ky another kz so all these three equations have to be written in terms of different kx's so this let us first solve only for x if I solve for x I just write this capital x and this constant just for convenience I have defined as ax this constant I have defined as bx so solution will be a sin kx this I have written because here there was a minus kx square okay these are see in principle these three can vary such that the only condition is that the sum should be equal to this 2 ma upon h square all right now I have to put boundary conditions I put boundary conditions exactly the same way what I want that once along the x direction I go outside the box it means my l value of x becomes less than 0 or value of x becomes greater than l I know the wave function has to be 0 the boundary is at x is equal to 0 and at x x is equal to l so at that particular points I my wave function must be 0 exactly the same way that we have been doing it earlier okay it means wave function at x is equal to 0 irrespective of y and z value must be equal to 0 wave function at x is equal to l irrespective of y and z value must be equal to 0 okay so I just put this condition and I will get exactly the same way as I had got for particle in a one dimensional box capital x is equal to 0 capital x this is only possible because this is the only term which depends on x no other terms depends on x out of the wave function so if this has to be 0 I had to put here at capital x is equal to 0 then only I will be able to solve it I cannot put in y or z because that does not contain x so this condition implies that at x is equal to 0 small x is equal to 0 capital x should be equal to 0 similarly at small x is equal to capital l capital x should be equal to 0 like this before we get I mean they cannot be sin term present so we will have bx is equal to 0 and kx l should be equal to nx pi so up till this point this is exactly identical for particle in a box except for the fact that in a particle in a one dimensional box whatever we are getting as k was directly related to energy but here energy depends not only on kx also depends on ky and also depends on kx. Now using exactly similar arguments I can solve for y component and z components also and we will get exactly same conditions I will get ky l is equal to ny pi and capital y I will get as ay sin ky y similarly we will get kz l is equal to nz pi and therefore capital z will be az sin kz pi all right and now the energy will be given by three terms there will be nx ny and nz cross kx square plus ky square plus kz square must be equal to 2 m e upon h square therefore I find out what will be the value of h h will be has to be equal to h square upon 2 m kx square plus ky square plus kz square and I have said kx must be equal to a an integer multiplied by pi divided by l similarly and ny has to be another integer which is ny pi by l so looking at that particular thing I get this as h square pi square 2 ml square nx square plus ny square plus nz square so only those values of energy corresponding to which nx ny and nz are positive integers okay will be the one which will solve the correct boundary conditions and therefore will represent the energy of the particle see one another interesting thing here is that these nx ny nz are independent okay nx ny nz can take independently whatever value they want except the only thing is that they must be integer they must be positive integers for example this nx could be 1 this ny could be 10 and nz could be 25 that is still an allowed energy there is no restriction on nx ny nz or there is no relationship okay what you will find in the case of hydrogen when you talk about hydrogen atom you get three quantum numbers there also okay which are called n m and ml okay but their values turn out to be interdependent depending upon one particular value of n you can have only certain selected values of ml so then they cannot vary independently okay while in this particular case of particle in a three dimensional box these all three nx ny nz can vary independently my wave function now becomes this now because there was an ax there was another ay here there is another az so I have combined all those things and put this as a so my wave function is a sin of course it depends on what is the value of nx in fact these nx ny nz values are generally called quantum numbers so there are three quantum numbers in this particular problem and depending upon these quantum numbers the wave function would be different and of course by nature of the equation a is always a constant which has to be determined from normalization of course you have to take a three dimensional normalization normalized within the box and put the probability of finding the particle anywhere in the box to be equal to 1 and then only you will be able to determine a which I will not determine because this is rather simple if you are interested you can do it and it is given in almost every textbook okay now my issue was to introduce the concept of degeneracy which I will do it just now so these are the comments they are three interdependent quantum numbers which determine energy so there are three independent quantum numbers nx ny nz okay or then they are independent so not interdependent independent quantum numbers okay they can all vary independently and the corresponding the quantum number which gives you the least energy which is we call as the ground state energy corresponds to nx is equal to 1 ny is equal to 1 nz is equal to 1 we have mentioned earlier also in a particle in one dimensional box that if nx becomes if any one of them becomes 0 then in that case this wave function will become 0 if this is 0 then this is sin 0 and the entire wave function becomes 0 the wave function becomes 0 it means the probability finding particle inside the box becomes 0 okay but we have been told that there is a particle inside a box okay so this is definitely not an acceptable solution okay phi must remain nonzero and therefore the least value of the three quantum numbers allowed is nx is equal to 1 ny is equal to 1 and nz is equal to 1 okay this is the least one which is allowed as a ground state energy now this is what I was calling as a degeneracy there are different set of quantum numbers which give exactly the same energy even though they have different wave function let us look at nx is equal to 2 ny is equal to 1 nz is equal to 1 so I have written as 2 1 1 or nx is equal to 1 ny is equal to 2 and nz is equal to 1 and nx is equal to 1 ny is equal to 1 and nz is equal to 2 let us go back to this particular equation my earlier equation this is my energy so if this becomes 2 this becomes 1 and 1 this term becomes 4 plus 1 plus 1 6 okay if this is 1 this becomes 2 and this becomes 1 then I again get 1 plus 4 plus 1 again I get 6 if this becomes 1 this becomes 1 this becomes 2 again I get 6 all these possibilities nx is equal to 2 ny is equal to 1 1 so 2 1 1 1 2 1 or 1 1 2 all three of them will give me exactly the same value of energy but the wave functions according for us they will be different so here if nx is equal to 2 this particular thing along the x direction will show two loops while along the y direction will show only one loop while this also along the z axis will show only one loop only one wiggle no one half of the wavelength okay while if I take and y is equal to 2 along the x direction it will show only half wavelength along the y axis it will show full wavelength and z axis it will show half wavelength okay so these wave functions are going to be different okay so these are called degenerate states but all these are allowed states of the particle all right now if we come to let us say a electron which always poly exclusion principle to occupy because these happen to be three different states so all these three states individually can be occupied by two electrons one with a spin up another with a spin down all right so if we take spin degeneracy also into consideration of course spin does not come specifically from the solution of Schrodinger equation okay but we know that spin exists and it has its own degeneracy so you will find that all these three states which have the same energy in principle will occupy two let us say fermions of two particles let us say talk of electrons because there is a spin half okay so there are two double degeneracy so these three states can accommodate six electrons even though the energy of all of them will be identical so these are what all which are called degenerate states the states which have sort of different type of wave functions okay but on the other hand they represent the same energy also there is some interest in degeneracy because by application of some external perturbation like electric field magnetic field it may be possible many times to remove the degeneracy I think professor Suresh when he was talking about magnetism might have talked I am not very sure about the removal of degeneracy it is generally possible sometimes that by applying external influence some of the states which are degenerate their degeneracy can be removed it means you can make a slight there is the the energies of these state become may become slightly different degeneracy is a very very important aspect in in quantum mechanics and especially in solid state physics because when we are talking of a very large number of electrons occupying let us say states in metal okay then we do consider these type of things that you know how many states are how many electrons are possible to be occupied in a state with a given energy okay and that becomes very very important to consider degeneracy so let me come back here the first excited state is what we call as a triple degenerate because there are three wave functions which give rise to the same equation okay this triple degenerate excluding spin I mean one has to be very very clear one should not doubt this is a triple degenerate excluding spin they are clear cut three different type of wave functions which give you exactly the same energy these wave functions are different from each other but their energies corresponding to them is same calling them degenerate state now I will essentially end this particular lecture and the quantum mechanics portion other than just giving one particular argument because this is a question which a very large number of students ask and you know I thought I always think of specifying see in earlier equation I have written kx square plus ky square plus kz square should be equal to positive quantity here all right now people many times ask is it not possible that one of these ky square can be negative it means ky square can be negative it means one ky can be imaginary such that overall sum is positive because all I want that the energy should be positive now in principle it should be possible that you have kx square minus ky square plus kz square and have the same energy what I want to show is that a value corresponding to negative energy is not possible to be satisfied by boundary condition so this is the my last two transparencies and then we will just close this particular chapter so can kx be imaginary is there a possibility that one of the ky squares is negative and others are positive such that their sum is positive we do not want to always think mathematically okay let us assume that is kx square which is negative it means kx is imaginary in that case the solution of x would be slightly will be different and the solution will be of the form ax e raise power alpha x plus bax into e raise power minus alpha x because I have said kx is imaginary so therefore alpha will be real so then the solution I will not write as a sine and cosine term but I will write like this okay but all I want to show that this type of solution is not possible because if this solution existed then I must apply boundary condition and the boundary condition at x is equal to 0 means the wave function should be equal to 0 so if I put x is equal to 0 I get ax plus bx is equal to 0 so this is what I am writing ax plus bx is equal to 0 now even at x is equal to l this should be 0 so I put ax e raise power alpha l plus bx e raise power minus alpha l to be equal to 0 so this should be equal to 0 I rather ax plus bx is equal to 0 so ax is equal to minus bx so I put bx is equal to minus ax So, this becomes Ax raised to alpha L is equal to Ax raised to minus alpha L. Now, if this condition has to be satisfied, then either Ax should be 0. If Ax is equal to 0, then in that particular case the wave function has to be 0 because you know this is the product by general solution has to be Ax containing a term which is Ax multiplied by A y multiplied by A z. So, this will give 0. Other possibility is that alpha L is equal to 0 and because L cannot be 0 therefore, alpha has to be equal to 0 and if alpha is to be equal to 0, then Ax now because here Ax plus Bx, Ax should be which is Ax plus Bx because in this particular case if alpha is equal to 0, my solution will be Ax plus Bx. If alpha is 0, then Ax plus Bx and I have already said that boundary condition means Ax plus Bx is equal to 0. It means Ax is equal to 0. So, if a solution of this type existed, a solution of this type existed, it means if k i square, one of the k i square would have been negative. In that case, once we apply boundary condition, I would have automatically got the wave function to be equal to 0 because if capital X is equal to 0, only possibility of solution is k x is equal to 0 and that is only possible when entire wave function becomes 0 and particle does not exist because that is not given. So, it is not possible that a solution like this has to is possible and therefore, all individually k x square k y square k z square should be positive or k x k y k z should be real. So, I think I will stop here though my time is over, but you know there are some questions we can entertain question for the 5, 10 minutes. Sir, in case a potential step. Yes. That you are called free step problem e less than v naught. Yeah. We get r is the 0 percent. Does it not wider the wave nature of light? As we know there is a reflection and transmission both. Yeah. Whatever may be the interface. You cannot first of all compare this particular thing with light because light is a different system, this is a particle system. Sir, I am saying anyway. Yeah. If anyway meet to the interface. Yeah. It will go for molecular reflection and transmission both. That is right. But in this case there is 100 percent molecular reflection and 0 percent transmission. That is right. Does it not violate the nature? No, it does not violate the nature because entire light is reflected. I mean, in what way this particular violates? The only thing is that in case of light, as somebody has pointed out there is also a possibility that light gets absorbed. Well, that is a thing which I am sort of excluding here because these are particles unless there is a process by which these particles can be absorbed and give rise to something else. I am excluding that possibility. So, these particular particles can either get transmitted or can get reflected. Only thing their number should be conserved. So, I do not see any way by which the wave nature is sort of all I am trying to say that you cannot directly compare with the light because light as I said in principle photos could be absorbed. There are so many methods by which photos could be absorbed. While here we are talking of some different type of particle, if at all they could be absorbed then they have to be treated in a different fashion. Yes, yes, Institute of Technology, Indore. My question is. Yes. You talk about the particle in a box. Yes. But the potential is symmetric. What about the asymmetric potential? Sorry, your explanation is for asymmetric potential the x is equals to 0 to L. See, I. It is for symmetric from minus L by 2 to L by 2. See, what you are trying to say that in the same particle in a box problem, if I would have made the problem symmetric. Are you saying that thing? Yes, sir. Okay, that poses absolutely no problem. That is what I said. You try to do the exercise yourself. I put let us say minus L by 2, put 0 and put L by 2. You write exactly the same equations which you are writing, a sin kx plus b cos kx. All right. Now, you apply the boundary condition at minus L by 2 and plus L by 2. You will find neither a nor b becomes 0. In one of the case, it will become minus b, another case becomes plus b. You solve the equations, you will get exactly the same answer of energy. Of course, wave function will become different because this particular term, I mean you know what is the wave function first. The first wave function will be this like this. So, this will contain only cos term. The second wave function would be like this, okay, which using this particular origin will contain a sin term. Nothing of that will change. In fact, this normally to first year students, I give is an exercise in one of the tutorials. Say that solve this particular problem by taking minus L by 2 and plus by L by 2 and convince yourself that you will get exactly the same value of energy. In fact, the plots will look exactly identical. But of course, the functional form will look different because my origin has shifted to 0 or my origin has shifted to this particular point rather than this particular point. See, in earlier case, my solutions were all sin. Here, they will be alternately sin and cosine. It makes no difference. When we mention in symmetric case, there will always be a cosine term. No, it will be alternately cosine and sin term. First term will be cosine, second will be sin, then again there will be cosine, then again there will be sin term. In the previous discussion, you said that the ground state energy of hydrogen atom is corresponding to Nx is equal to 1 for the calculation of energy states of that deep potential well and infinite potential well. Yeah. Yeah, for three-dimensional cases, if it is Nx is equal to 0, 1 and Ny is equal to 0 and Nz is equal to 0, then there may be for the possibility. No, no, that is not an allowed state because no, no, if any one of them becomes 0, wave function becomes 0. So, least value is Nx is equal to 1, Ny is equal to 1, Nz is equal to 1. Okay, Nx should have, should it start from 1? It should start from 1, minimum value of Nx. That side, that side. Yeah, okay, okay. This is Amrita. Can you please throw some light on the wave function? Yeah. For the special case of E equals to V0. Infinite potential well? Yeah, in the finite potential well. Yeah. Where one side is finite, the other side is infinite. Okay, okay, okay. The first problem. Okay. I presume you are talking of this particular problem. Is it right? Yeah, correct. Okay. Say first of all, a particular bound state has to exist. Let us imagine that a one bound state appears, then this particular bound state because the wave function has to be 0 here. So, it will start from here. It will be sinusoidal term, but it cannot terminate 0 here because wave function is not 0 here. Okay. In fact, wave function has to exponentially decay. So, this will have a term, sine term, which goes like this and comes somewhere and for time exaggerating here. And then this particular part, it will exponentially decay. Okay. At this particular point, not only the wave function, but its derivative should also be continuous. So, it has to be a smooth curve like this. So, this will be the first, the ground state wave function. If the second wave function, the first excited state wave function also exists, then in that case, remember it has to start from 0 here. So, it will go like this. It will come back like this. And then eventually, it has to decay like this. All right. So, this will not cut here in the halfway, but it will be somewhere on the right-hand side because one full quarter wavelength, half wavelength will come and rest will leak out here at this particular point. So, this is the way wave function will look like. And if you just make a mirror reflection, you will say that this particular thing, if I just put another one, this will be the alternate solutions for a particle in a finite box, whether there is no barrier. Yeah, correct. My question was for the special case of E equals to V0. What will happen to that? E is equal to V0. You just have a solution which is just at this particular point, which is just equal to E is equal to V0. In fact, you could have solved this particular equation just by putting E is equal to V0. And then you would have got exactly the same equation, which I have obtained earlier. The solution is just a straight line. A just a straight line in the second region as well. Yeah, that is right. Because see, that is the only one which will satisfy all the conditions because you want wave function to be 0 here. You also want it to be continuous. Thank you. You can try it out independently yourself. And then if you have any doubt, you can send the question to me. 101. Jaipur Engineering College. What are the possible origins of 0 point energy? See, origin is quantum mechanics. See, as I say quantum mechanics is my basic law. From that particular law, I am getting this particular question. I have no other control. So this is the way nature is. See, like when I say, when I apply a force, why particle does acceleration? It is because there is a Newton's law of motion. Similarly, why there is a 0 point energy? This comes from Schrodinger equation. It is as simple as that. So there is no origin. I mean, quantum mechanics is the origin. Because my quantum mechanics is my law. From that particular law, I am getting that 0 point energy should exist. This is Motilal Nehru of MNIT. Okay. Sir, there is a question regarding the tunneling, sir. Yes, sir. The process tunneling is explained in the order of nanotechnology, nanoparticles. See, there are so many places where the tunneling happens. Now, I am not very sure, when you say nanoparticles, also there are many things happening. But there is a possibility of tunneling in some, I mean, depends on what process are you talking. Okay. So tunneling is whenever they have a potential barrier, okay, which in principle, classically is forbidden. But you can find the particle, if this is small enough, you can find significant number of particles causing that particular thing. So I am not sure whether specifically for some aspect of nanoparticles, whether you can see tunneling. But I will not be surprised if you see tunneling. There is one more question. You have one more question. Okay, please. One more question regarding wave packet. Yes. There are infinite number of waves forming wave packet. That is correct. Then in every case, there are infinite number of waves, then the size of wave packet will be very large. No, no. See, if you have infinite wave, you require infinite waves to limit the length of the size of the packet. If there is only one wave, then wave packet is infinite. When you are mixing many more waves, then only the size is becoming finite. Okay. So a finite size wave packet consists of an infinite number of infinite waves. All right. So this is not correct that I am mixing lot of wave packets. So my way, I mean, a lot of waves to my wave packet will grow. In fact, I have given you example that if I mix two waves, then this particular maximum keeps on going further and further off. If I mix two state, three waves, then it is further off. If you mix infinite waves, okay, this pushed off forever. So it gets limited. Thank you. Thank you. This is MSB. Hello, good afternoon, sir. Yeah, good afternoon. Latur, yes. Yes, yes, sir. Potential, page number 9 if we refer. So here equal to 0 and x is equal to L we have considered. Yes. But when we consider the potential well and it is the bi-section of the potential well. Yeah. Yes. So if the center point we consider x is equal to 0, instead of that if we consider L by 2, then it would be better and we can proceed further with that consideration. Are you saying that this is the way I should consider it? Yes. Okay, this is what I told you that there is absolutely no difficulty in considering this way. In fact, many of the textbook prefer in this particular wave fashion because this gives you a solution which are more symmetrical, all right. So there is no, this is a personal preference. I take 0 to L and give this as a problem to the students to work it out, okay. I could have done the other way. I could have taken minus L by 2 to L by 2 and given them as a problem other one. See, basically where I choose origin, my physics is not going to change, all right. So it depends on what you find much more convenient. See, the advantage of giving 0 to L is that because that is the first problem I solve on quantum mechanics and you get one single solution of sine, which is rather simple to start with, okay. Otherwise, you have to always keep on varying between cosine and sine and you have to keep on giving argument on that particular thing. So, I mean, for the first level course, I find it convenient to talk of 0 to L and then give this as a problem. But it's a matter of choice. You know, sure you are right. I could have done the other way. Hello, sir. Yeah, yeah, please. One more question, sir. Yeah, please go ahead. Sir, my question is on the earlier session that you said that the wave velocity do not depend on the wavelength of the light. Then what is the relation v is equal to n lambda? No, no, no, no, sorry. I think you misunderstood it. See, relationship v is equal to n lambda when I am trying to say, let me put it like that. If I say relationship is v is equal to n lambda and v is constant, all right. Then in that particular case, this is what we call as a non-dispersive medium, all right. In general, this is n which we call as a frequency. In fact, many times we call it new lambda, all right. This new depends on omega and lambda depends on k, all right. In general, if you talk of an arbitrary wave, including the particle wave, the relationship between omega and k is fairly complex. So, if you try to calculate velocity using this relationship, this will not be a correct way of looking it. So, what you do, you actually use this particular expression. See, let me put it like that. What is omega by k? Omega by k is 2 pi by 2 pi nu divided by 2 pi by lambda. So, this is new lambda. So, which essentially means if omega by k is a function of k wavelength, it means v is also a function of k. So, this relationship will hold good for phase velocity. Remember, phase velocity for that particular value of wavelength, because if I change the wavelength, phase velocity will change. But what is important that if this happens, then in that case, this velocity is of not physical significance. Then what you should calculate is d omega dk and calculate d omega by dk at a particular given value of k, because all realistic waves are actually wave packets. And wave packets travel with this speed and not this speed. But on the other hand, if v does not depend on k, then omega by k is constant. Then d omega dk has the same value. In that case, group velocity and phase velocities are same. So, whenever I am saying that a wave travels or rather wave packet travels with the same speed and lambda, where v is constant, group velocity and phase velocity happen to be same. But in general, this need not be. And in that case, I must calculate the speed of wave packet. And speed of wave packet is calculated by using relationship d omega dk and not omega by k. And of course, d omega dk also in general will be a function of k. I am not sure whether I have explained you or if your question is answered or not. Okay, okay. There is another question that. Yeah, please go ahead. In double slit experiment. Yeah, yeah. Now, the phenomena of interference is due to the waves only. But if we replace the wave by photon or the electron, how is it possible? I mean, that is what I am trying to say that, you know, there are certain phenomenon which can be better understood only on the wave basis. If you want to insist on explaining these phenomenon on particle basis, you have issues and then you have to give rise of the idea of what we call as uncertainty, then there are certain things which I cannot understand. This is precisely what I am trying to say that there are certain phenomenon which are better understood only by the wave nature. If you insist that I want to talk only of also the particle nature, then I have issues. Then I will not be able to understand how these particles reach there. That is precisely what I am trying to say. See wave particle duality brings certain amount of complications in the way I look. Because according to me, the classical particle is a particle which is localized. Okay. On the other hand, so long I consider this as a wave, I have no issue. Similarly, in photolithically-referred experiment, if I have to explain this on the basis of wave nature, I have a lot of issues. But on the basis of particle, I can explain it very well. All right. This is precisely what is the idea. Will the particle nature will also show the same interference pattern or not? If you are seeing interference pattern, that can be better explained on the basis of wave nature. Okay. Even though you may like to call this particle as a classical particle, like an electron or a neutron. All right. Okay. I think we will just close this session.