 Now, let us start what we did in our last lecture that is today morning we discuss some other properties of the wave function rather we postulated how to extract some more information for wave functions. For example, how to extract let us say the mean value of a particular quantity a physical quantity let us say x or x square or some people have asked the question what is the physical significance of mean of x and mean of x square we will see the physical meaning of all these things when we come to the uncertainty calculation. It is basically if I am interested in finding out what is the average value of x or what is the average value of x square that is what we are doing I could be interested because of various reasons for example, one example which I just now get that if I have to calculate the standard deviation in the value of x I will need the mean value of x square. Then we solve the simplest problem of particle in a one dimensional box and then see how this particular boundary condition eventually leads to quantization of energy if you remember we had started with this idea that like the wavelengths get quantized when we are talking of vibration within a string and if the wavelengths turn out to be dependent on energy the energy will also turn out to be quantized we have exactly use the same concept here we have put a impose a boundary condition on the particle waves or by what we call as a wave function and in those by by putting those boundary condition we found that only certain values of k which is actually related to wavelength become allowed and because k is eventually related to the energy therefore the energy automatically gets quantized so here we have a theory from which actually we get the quantization of the energy. Now let us go a little bit more on the technical aspect of the quantum mechanics and try to use these wave functions of particle in a box wave function and try to extract something which is more important information because my idea is that I am eventually leading towards the postulates of quantum mechanics. So, I am first giving examples and then coming to the postulates rather than the other way in fact in special theory of relativity I will do the totally opposite way I will first give the postulates and then I will talk about how these postulates are used to actually come to the proper equations here I am doing the other way because the postulates here are sort of not well categorized as I have mentioned earlier also they tend to be little more abstract. So, if we have given an example earlier probably we understand them little better and then we eventually formalize everything so that is what I am going to do I am going to use these particle in a box wave function which I presume that now we understand somewhat better and then use them to do various postulates. Now first thing that I would like to talk is about the superposition of wave functions. See remember when we started doing some formal quantum mechanics we said we rejected the standard wave equation because that would have not allowed the superposition of wave functions and therefore we landed into an equation where we could actually have superposition principle valid and therefore wave functions should be allowed. Just now we have calculated the time independent part of the wave functions using Schrodinger equation for particle in a box. Now question is that in principle a superposition of these wave functions would also be allowed. Therefore in principle a particular particle could have a state could be in a particular state which is a resultant of the superposition of the wave functions. So, remember when I am talking of the wave functions the actual wave function is whatever I have got as a result of solution of the time independent Schrodinger equation then multiplied by the time dependent part which is e raise power minus i e t by h cross. Of course this equation which I have written is somewhat a generic equation but let us assume as of now that the phi n's that we have talked about are actually the wave function that we have obtained for a particle in a box. Of course as I say this equation that I have written is a general even if the problem is different I can still superimpose these wave functions in the following way. Let us just look at this particular subscript that I have written n. We have just seen that if the particle is in the ground state it means corresponding to the lowest energy state then the wave function will be different. If I go to the first excited state the wave function will be different. In fact as I keep on going to different states different energy states in the problem corresponding to each state there will be different wave function. See in the ground state wave function there was just half in oscillations. In the first excited wave function we had one full wavelength and things like that. Therefore this subscript has been put as n. So when I say this is phi nx because I know that wave function depends on whether it is n is equal to 1 state or n is equal to 2 state or n is equal to 3 state. So whatever is that this represents that particular wave function corresponding to nth state. Now corresponding to nth state the energy will be En therefore when I am writing the complete wave function I must write phi nx multiplied by e raised to the power minus i En t by h cross. So this becomes my actually psi xt which is the complete wave function including time dependent part. Now what I do I take a particular value of a particular wave function let us say phi 1 multiplied by some constant c1. It could be a real it could be a complex then take phi 2 multiplied by c2. In general I could take any values of n and sum it over n. If I am interested in just 2 then it will be just 2 terms. I will give you a specific examples in it later. If I am I want to mix 3 wave functions I can mix only 3 wave functions. If I want to mix 10 wave functions I can mix 10 wave functions. So this is most general equation where I have mixed n wave functions. As we have seen earlier this particular wave function which is a linear combination of all these wave functions should also be an allowed wave function. In principle a particular particle could exist in a state psi xt which is a linear combination of all these wave functions. This is what I have written the wave function must be able to superpose. We can see that following is a solution of time dependent Schrodinger equation and is the general equation is probably a general equation of what could happen to the wave function of a particular particle. So a particle in a box for example could have a wave function like this which is a linear combination of individual wave functions corresponding to a specific state n. Now if you are not satisfied you can substitute this particular thing in time dependent Schrodinger equation. The original Schrodinger equation that we have started time dependent Schrodinger equation and then this particular thing will really turn out to be solution of a time dependent Schrodinger equation. You can satisfy yourself that it is really so but it has to be so because that is the starting ground from that particular point only we had started discussing about our quantum mechanics. Of course here what we say here phi nx are the solution of time independent Schrodinger equation corresponding to energy En. We can show that such a state is not a stationary state in general. We can show that if we take this particular type of wave function and if I calculate the probabilities you will not find the probabilities to be time independent. Remember in the morning session we have mentioned that there are certain things which are called stationary state. In the stationary state the averages especially the probabilities will not be function of time. Here in general you will find that these states they are time dependent state the position will be time dependent and this is not a stationary state. So what I have told in the morning that there are certain states which are non-stationary state. This is an example of a non-stationary state because the probability would depend on time. This again you can verify yourself. Now let us go a little bit more interesting things we talk. Let us suppose you put time t is equal to 0 in this equation. So I put time t is equal to 0 because this is valid for all times so I can always put time t is equal to 0. If I put time t is equal to 0 then this whole thing becomes 1 because e raise power 0 is 1. So this wave function psi x 0 now I will call it will be summation of n c n phi n x. So this is also a proper valid state of the solution. So this is also a valid solution at t is equal to 0. So a particular particle at time t is equal to 0 could be in a state let us say summation of c n phi n x for example it could be just c 1 phi 1 plus c 2 phi 2 or c 1 phi 1 plus c 3 phi 3 or mixture of three terms or mixture of five terms or whatever it is. This is also a valid solution. Now question is that if at all a particle possesses this state if a particular particle has this state will it be a solution of time independent Schrodinger equation? You can show that this will not be a solution of time independent Schrodinger equation because it is very very funny and appears to be somewhat contrary to the state because here corresponding to each phi n the energy is different. So when you would like to sum over it you will not be able to sum over and add all this phi's and this will have and this will not be a solution to this particular equation. See in earlier case because there was exponential of e raise power i n time to think which would we should take care and in general that would be a solution of the more general Schrodinger equation which is time dependent Schrodinger equation. On the other hand a state like this at t is equal to 0 will not be a solution of time dependent Schrodinger equation. But we just now said a particular particle could have a such a state. Now if at all the particle has such a state if I make a measurement of energy what will be the value of energy I will get. Question is that if we had just phi let us say phi 1 if there are just one state then I will get always the energy even. If the particle was in a second state which is the first excited state corresponding to n is equal to 2 then I will always get the energy e2. If it is in n is equal to 3 state then I will get energy e3 where in this particular expression of energy I just substitute n is equal to 1, 2 or 3. But now it is in a sort of a mixed state this is in a state which is a linear combination of phi 1, phi 2, phi 3. Now if a particular particle exists in a state like this and if I make a measurement of energy what is the value of energy that I am going to get this is what I have written. But is this a solution of time independent shooting any question the answer is no. What would be the value of energy that we shall get if the particle was really in about state which is a allowed valid state. Now the answer is that it depends on how much you have mixed it up what is the way what is the coefficients that you have mixed it up. Now thing is that if for example if we have only let us say 1 and 2 state we just write it here. Suppose we have a state which is like c1, phi 1 plus c2, phi 2 this is state was corresponding to energy e1 this state was corresponding to energy e2. Now this is a valid state if I make a energy measurement what is the value of energy that I will get will I get mean of e1 or e2. But if I get mean of e1, e2 that is not an allowed energy at all. So therefore we postulate again in quantum mechanics if a particular particle is in a mixed state like this where I have mixed two things when I make a measurement I will either get e1 or I will get e2 I cannot get any other energy state. And because it is mixture of only two states one corresponding to energy e1 another corresponding to energy e2 therefore I will get either e1 or I will get e2 I will not get any other energies. Similarly if I had another mixed state which was given by let us say c1, phi 1 plus let us say c3, phi 3 plus let us say c7, phi 7 this is also an allowed state that in that case if I make a measurement then I could either get e1 or I could get e3 or I can get e7. I will not get any other state because this is a mixture of three states phi 1 or rather superposition of three states phi 1, phi 3, phi 7. So if I make a measurement then I will either get e1 or I will get e3 or I will get e7. But then what is the probability that I get e1 or what is the probability that I get e3 what is the probability that I get e7. This probability will be proportional to this particular individual coefficient cn square. So probability will be proportional to cn square. So that is what is the next postulate that we are coming that if we have a superposition of wave functions then when you make a measurement you can get any of the values of the energy and the probability that you will get that particular energy is proportional to that coefficient square. And if we normalize these probabilities now remember we have to further normalize it I will just now give you an example because I must have c1 square plus c3 square plus c7 square equal to 1. If it so happens then the probability will be just c1 square plus c3 square plus c7 square. I will be giving more examples of this particular thing as we go ahead. Let us go back to the transparencies now. Starting with a normalize psi x0 it means if I normalize such that summation of cn square is equal to 1 cn square is equal to the probability of finding the particle energy to be en. So in a mixed state I will not get a definite value of energy I can either get any I could get any of the en values and this will be proportional to I mean in this case if it is normalized this will be just equal to cn square and of course if cn is 0 the probability of finding that particular energy would also be 0. Let us before I go little bit ahead and give you an example let us take some more properties of the wave function and then we will talk about some of the examples. Now there is one particular property which is called orthonormality of wave function which is a very very important property. We have not talked about the degenerate states yet so those people who know about quantum mechanics and they were talking about the degeneracy let us postpone that discussion. These particular things are at the moment let us assume that because degeneracy introduces some other complex complexity in those things. So let us not at this moment talk about this thing as we are seeing in this particular case there are no degenerate states so as far as particle in a box is concerned whatever I am saying is probably all right. See we can always show that phi n star x multiplied by phi mx if n is different from m will be equal to 0. This you can verify yourself. If you take let us say phi 1 and here is phi 2, phi 1 star phi 2 will always be 0 if we integrate from 0 to l all right. Of course in this case it is real the the phi's are real so this star has no specific significance but even if it had what I am trying to write is phi n star x multiplied by phi mx dx but if m is equal to m and if the wave function is normalized then this must be equal to 1 because that is the way we have normalized it taking phi star phi and shown it to be equal to 1. This is what is generally called a Kronecker delta Kronecker delta delta mn or delta nm is defined this is the definition of Kronecker delta this value is equal to 1 when n is equal to m it is 0 when n is not equal to m. So this is the definition of Kronecker delta so this is equation is conveniently written is what we call as orthonormality condition of wave functions that wave functions tend to be orthonormal it means if n turns out to be equal to m this value will be equal to 1. If n is different from m then this particular value will be 0 but I have just now shown okay let us we will come to this particular condition later. There is another thing I mean I am sort of introducing all these things which are slightly different and normally these things are not given and the textbook corresponding to the first year thing in IIT Bombay we generally introduce all these concepts right in the first year though some of these appear to be somewhat difficult concepts. Okay now this can also be shown I mean which I am not showing I am just giving you some sort of a logic that any function of fx okay this function fx could be any arbitrary function x okay which is defined only between 0 and l because for particle in a box the wave function exists only between 0 and l. So if there is any function which is between x is equal to 0 and l and is 0 everywhere else it can always be expressed as a linear combination of c and phi and x. This is very very interestingly it is almost like what we used to do in Fourier analysis see any function of fx can be considered as a linear combination of phi and x wave functions okay so we can always write any other function it could be just some sin 2x sin 3x whatever it is only condition it has to be made that it has to exist only between 0 and l and it should be 0 outside for a particle in a box wave function then this function can be represented as a linear combination of c and phi and x all right. Now of course c n can always be found out because what you should do if you take this fx multiplied by sin star x then I multiplied by rather phi n star x I multiply by phi nx and integrate then I realize that only term because of the orthonormality condition phi n star phi n will be equal to 0 while phi m star phi n where n is different from m will be 0 and only phi n star phi n will be equal to 1 let me just write. So I have this condition which was summation of c n phi n what I am doing I am multiplying it by phi m star and integrating it once I integrate see it contains a large number of terms okay phi 1, phi 2, phi 3, phi 4 like that until I hit a phi m value. Now if it is phi m let us suppose m is phi just to give an example let us suppose m is equal to phi and let us say and of course we are going to add n so if I take phi m 5 let us say phi phi star c 1 phi 1 and integrate it this will be 0 because of the orthonormatic condition. Similarly if I put c 2 phi 2 that will also be 0 until I hit c 5 and phi 5 if I put phi 5 then this will become 0 and this whole thing will be equal to just c 5. So this summation will just give me c 5 so I have been able to find out what is the value of the coefficient we are not shown that the series converges but at least it gives you idea how do I determine I have to multiply this by a phi n if I am looking at what I should have written c m here so that you can actually calculate the mth coefficient from this particular equation. So this is I multiply by a given value of n then only that particular term will be 1 or other terms will become 0 because of the orthonormatic condition and I will be able to determine what is the value of c n. This is what is called a completeness condition the wave functions actually turn out to be complete. The above can be easily visualized using orthonormatic property of the wave function which I have just now shown. Now this you can first I would like to show the orthonormality condition this proof is very very simple you take phi n star phi m so I have taken sin n x and pi x and sin m pi by x this using the standard geometrical identity can be put as cos n minus m pi x by l and cos n plus m pi x by l this you can show that if n is different from m then this will be equal to 0 otherwise this will be equal to 1 which we have already shown. If it is not there you can just put the conditions here you integrate these things put the condition and you will see that this turns out to be equal to 0. So it is very easy to show this orthonormality condition. Now let us take some example and at least this particular orthonormality condition let us try to take some examples. Now let us suppose we try to mix up two states n is equal to 1 and n is equal to 4 state. Now remember the wave function that we have written earlier was under root sin under root 2 by l this was the normalization constant sin n pi x by l all right. Now n has been equal to 1 so phi 1 will be given by under root 2 by l sin pi x by l this is normalized in its own self because under root 2 by l is the normalization constant. So had the wave function be only this much or have phi 1 only or the state would have containing only this particular state then this was a normalized wave. This in its own self is normalized wave function because this is under root 2 by l instead of n I have put 4 so it becomes sin 4 pi x by l all right. But when I mix these two this no longer remains normalized if I just put phi 1 x plus phi 4 x it means I have taken c 1 is equal to 1 and c 4 is equal to 1 then it is no longer normalized. Follow wave function is not a normalized wave function because I have just multiplied by 1 and I have multiplied by 1 and 1 square plus 1 square is no longer equal to 1. So this has to be further normalized if you are not very sure then you just evaluate this you take psi star psi you will find that this will not turn out to be equal to 1 all right. So in order that we normalize I must have see this has been written as c 1 phi 1 plus c 4 phi 4 where c 1 was equal to 1 this was equal to 1. Now c 1 square plus c 4 square is no longer equal to 1 rather c 1 square plus c 4 square is equal to 2. So if I want to normalize I have to multiply this by 1 upon root 2 phi 1 x phi 1 plus 1 upon root 2 phi 4 then only it will remain normalized. So this has to be further normalized. In order that we normalize this will be the wave function. So this is a properly normalized wave function not if I have just added these two. Now if I say what is the probability of finding the particle in the energy state E 1 this probability will be equal to this square which is 1 upon 2. What is the probability of finding the particle in the energy state E 4 this will be this square which is 1 upon 2. So there is an equal probability that the particle will be found in n is equal to 1 state as n is equal to 4 state. So if we have very large number of boxes and in all those boxes a particular particle is given by the wave function which has been written here at of course this is valid only at time t is equal to 0 because I have put time dependent part to be equal to 0. Then in that case if I make a measurement of let us say energy in all these boxes then in half of the boxes I will find that the particle is in the energy state E 1 and in half of the boxes I will find the particle to win the energy state. So the physical interpretation implies imagining a large number of boxes where the wave function of the particle is given by the above. If a measurement of energy is done in half of them we shall find the particle to win n is equal to 1 and in another half in n is equal to 4 state. So I hope this idea of probabilities become clear the idea of mixing the wave function superimposing the wave function and from that interpreting the probabilities has become different. See in normal standard wave function we are talking only about the probability of finding out the position but that was only a stationary state. Now we are also allowing the superposition of the states and therefore the interpretation can also be extended about finding the probability of the energies. Now let us take one particular question this is sort of a problem and let us try to do this thing. In fact in this while solving this problem I will introduce a few more interesting concepts or let us call it few more you know sort of abstract concepts and then after that at the end of this particular lecture I will sort of introduce some formalization of the postulates of the quantum mechanics. So now here we have a state like this first question is that is this a normalized state this is under root 2 by 5 L sin pi x by L plus under root 8 by 5 L sin 4 pi x by L. Of course here the normalization constant does not depend on n. So let us try to write this particular equation in a slightly different form. So what we had written is under root 2 by 5 sin by x by L 8 by 5 S sin 4 pi x by L. Yes I think come back to here. So this is what it is. So I should rather first take the individual things I can write this particular thing individually normalized functions. So I write I can write this particular thing as because you know under root 2 by L was the one which would have made this particular individual wave function normalized. So what I will write this as under root 1 upon 5 under root 2 by L sin pi x by L. So you can see that these are the same things. Similar here I can write this as under root 4 by 5 but this root 2 by L factor does not depend on value of n that is how let us now see. So this particular thing I can write as phi 1 x this thing I can write as sin pi 4 x. So my C 1 is equal to under root 1 upon 5 my C 4 is equal to under root 4 by 5 and all other C's are 0. So only there are 2 C's present one is C 1 is equal to under root 1 upon 5 another C 4 is equal to under root 4 by 5 and all other C's have become 0. Now you can clearly see that C 1 square of course these are real is 1 upon 5 C 4 square is equal to 4 by 5 and if I add these 2 then I get 1 by 5 plus 4 by 5 this is 1. So summation of C and square is equal to 1. So this is actually a normalized wave function even after mixing this particular wave function is a normalized wave functions. So what we have said that this particular wave function is actually a normalized wave function. Now as I have just now seen that this particular wave function can be written in this particular fashion of course under root 4 by 5 can be written as 2 upon root 5. Now it means if a particular particle is in this state at time t is equal to 0 which is always possible because superposition has to be allowed. In that case if I make a measurement of energy then I will not get necessarily a E 1 or E 4 there is a 20 percent probability because this is a square of this particular quantity this quantity has to be squared. So square of this quantity is 1 upon 5 which is 0.2. So in 20 percent of the boxes you will find the energy to be E 1 and in 80 percent of the boxes you find the energy to be E 4 corresponding to n is equal to 1 and corresponding to n is equal to 4. So I hope the idea of this mixing and how to extract the probability of finding a particular energy is clear. In this case 20 percent of the boxes will give an energy corresponding to n is equal to 1 and in 80 percent corresponding to n is equal to 4. Now let us go a little bit ahead and try to talk about measurement. First we have said what will be the expected value of the energy, what is the mean value of the energy? We have to multiply by the individual probabilities to get the mean value of the energy. So mean value will be given by C n square to E n because corresponding to to energy E n the probability is this and so long we have normalized C n square I enter to further divide by summation of C n square. So this multiplied by the probability that this has the energy they are only two possible states energy E 1 and E 4 corresponding to E 1 the probability is 0.2 corresponding to E 4 the probability is 0.8. So when I take the average of these two I will get some energy which is between E 1 and E 4. Of course that particular energy in its own self will not be an allowed energy because allowed energies have to be only E 1, E 2, E 3, E 4. But if I am taking an average of so many things I can always get something which is in between. So for example if you had equal probability of E 1 and E 4 I would have always got E 1 plus E 4 divided by 2. Though this may not be an allowed energy at all but I am taking a large number of measurements and calculating the average energy this average energy need not be an allowed energy. So for example this energy will not be an allowed energy but the average energy happens to be this allowed energies are only E 1 and E 4. Now what will happen this was the state corresponding to time t is equal to 0 if let us assume that no measurement was done. So I am talking of the measurement at this particular concept remember today morning we were talking about our Copenhagen interpretation and we said that what measurement is likely to do. So at this moment let us assume that we are not doing any measurement. So this state is itself at time t is equal to 0 the question is that what will happen to this particular energy state at a later time t. This particular individual components will evolve according to their own time dependent part and therefore the state will be something like this. This was my original phi 1 which was multiplied by 0.2 this will evolve as E raise power minus I E 1 t upon h cross because correspond to make this particular phi x into a psi x I have to multiply by time dependent part. In time dependent part there is an energy involved and this energy corresponding to this state is E 1. So this will evolve as E raise power minus I E 1 t upon h cross while the second component because the energy corresponding to this is E 4. So this will evolve as E raise power minus I E 4 t upon h cross. So their time evolution will be different here this particular term will evolve with energy E 1 and while this particular thing will evolve with energy E 4. We can check that this would not be a stationary state and the probability of finding the particle at a location would be a function of time. So if I take psi star psi you will always find this I have mentioned even earlier psi star psi will depend on time because when you are taking complex conjugate and multiplying this term multiplying by plus will cancel out but there will always be a term which is E 1 minus E 4 which will have time dependent term which will not cancel it out. So therefore psi star psi will not give you something which is independent of time. This is an example of a non-stationary state this is not a stationary state. Now let us suppose what we do we make a measurement. Let us suppose we make a measurement on one of the boxes once we have made the measurement we have seen that either I will get E 1 or I will get E 4. Now if I have made the measurement and let us suppose in a box that particular energy turns out to be E 4. Now what will happen? Suppose I have made the measurement and I find the energy to be E 4. Now if I measure make a measurement again is it possible that I make it again back even but just now in the morning we had discussed that if I make two measurements just immediately one after another then whatever I have observed earlier must be valid. Otherwise there is no meaning in experimental data if I take perform an experiment I make a measurement and I find something and immediately after I make second measurement I may find something totally different. What is the sense of that particular data? Experimental data has no validity. If I make a measurement immediately after that this energy has to be equal to E 4. But if this particular particle was in the original wave function which was there then what I will find out that I could have found out even and E 4. So how to solve this particular thing we have another postulated quantum mechanics that postulates says that once you have made the measurement now the particle is no longer in that particular state and what we call the wave function has collapsed wave function has changed the state is no longer there that state is no longer there this particle once it has you have made the measurement you have disturbed the system the particle now the wave function has collapsed only into the state E 4. Now whenever you take if you take a measurement immediately after that you will find the energy only to be 4 probability of getting energy even is 0 now. So this is what is the question if a measurement is done in one of the boxes at time t is equal to 0 and the energy is to be found E 4 what will be the wave function at a later time t. Now this has collapsed the wave function now collapses. Now it is only 5 4 because you have made the measurement so the particle particle has just landed up into this state. And now it will evolve as only e raise power minus i 4 t by h cross that term which have us having phi 1 is no longer existing there you have disturbed the system you have made the measurement wave function is collapsed is changed this particular concept of collapse of wave function is a very very interesting concept and many times you will find that in normal modern physics books it is not sort of explained. So time wave function has collapsed and now this is the state what will be the measurement of energy yield in this box at a later time because now it is in a stationary state there is only one term. So as far as this particular case is concerned even if you make a measurement after 5 minutes you will always find the energy to be 4 because that even term has disappeared now wave function has become different. Once you have made the measurement particle has taken a stand it has collapsed the wave function has collapsed and gone to only one particular state. Now it remains in that particular state only the particle is now in a stationary state hence the measurement would leave to the value of e 4. So you can see how in quantum mechanics the measurement concept is evolved in terms of the collapse of the wave functions. Now let us take this particular particle in a box example of course this involves lot of mathematics so I will not go into the all the steps on this particular thing I will just calculate the average I will just tell you how the averages are calculated these things are will be uploaded into your model. So if you are interested you can probably do mathematics yourself I am just going to give you a step. Now let us suppose for a particle in a box I want to find out what is the mean value of x for that I will do you have seen that if I have to find out the mean value of x it means the expected value of x what I will do psi star x psi xt as we have seen the time dependent part in this particular case is of no consequence. Let us suppose that we are talking of stationary states here you take this particular thing then you will just getting x sin square and pi x by L because remember the wave function was under root 2 by L sin pi x by L. So this star does not mean because it is anything because it is real under root 2 by L squared will give you 2 by L and this will give you sin square and pi x by L and you take this as x which is this particular x you just integrate in by parts you just write this sin square as 1 minus cos 2 x and integrate this particular thing this what I have integrated this thing once you integrate this you will get L by 2 this you can just again you have to you will do little bit of mathematics again you have to integrate by parts and eventually find that irrespective of the value of n remember and I have taken as a general thing you will always find that if a particle is in a given state you will always find the average value to be equal to L by 2. Now let us take the average value of x square which is little more complex which I will not do in details if I take the x square mean value but many of you have asked this question what is the significance of x square. So at this particular moment actually I want to derive what is the uncertainty relationship for this particular particle in a box state so that is why I am trying to write this particular thing. So I am interested in calculating mean value of x square because I would use this particular thing to calculate the standard deviation and the value of x I make lot of measurements and then I take the standard deviation of x for that I need the mean value of x and I also need the mean value of x square. So exactly the same thing other than x I have x square here so this will be 2 by L sin square and pi by x by L and then you have to do x square you have to keep on integrating by parts I am not giving the full mathematics the final result will turn out to be like this. Now let us come to the mean value of momentum if I want to calculate the mean value of the momentum verify last step by integrating by parts then in that case remember I have to use an operator which is minus i h cross del del x so this happens to be the mean value of the momentum minus i h cross psi star del psi del x using the operator in fact this in today's morning I had also mentioned what is the px operator or rather yesterday how mean value of px has to be obtained you just do this thing. So this sin term will now become a cosine term because one side differentiates with respect to x and there will be n pi by L which will come out here one side differentiate then you will get this is sin n and this is cosine cos n so this will become as 2 sin n you can integrate this particular thing and when you integrate you are actually integrating over the entire cycle this will turn out to be 0. This again interesting result that the mean value of the momentum turns out to be 0 one can always argue because there is equal probability of finding the particle with momentum plus x and minus x. So I mean there is no specific reason why the momentum should be either plus x or minus x so the only value it can have is a 0 it will average out with equal probability of moving in x direction and minus x direction. So in principle I do expect eventually that the mean value of momentum should be 0 and that is what will happen but the mean value of px square will not be 0 because when psi take negative sin and positive sin and square them that will only give me positive values. So let us calculate the mean value of px square if I calculate the mean value of px square I have to use this operator twice and as we had discussed earlier this will give you minus h square because minus multiplied by minus will give you plus but i square will give you minus 1 and h square and here will be del 2 psi del x square. So this is what we have done del 2 psi del x square so you have to integrate twice and when you integrate twice you get sin square and pi x by L which is the integral which we have used while when we are normalizing it. So it is exactly the same thing and this equation will come out to be n pi h bar by L whole square which by the way happens to be just equal to 2 m by 2 m times En which is also not surprising because the energy in this particular case is purely kinetic energy so you do expect that En must be equal to px square by 2 m. So now what we have done we have calculated the mean value of x we have calculated the mean value of x square I have calculated mean value of x px I have calculated the mean value of px square. So I can calculate what will be the standard deviation in the value of x I can also calculate what is the standard deviation in the value of the momentum of course this is one dimensional problem so we are talking only of px we are not talking of p y and p z. So these are the uncertainties which is defined by standard deviation as under root of px square minus px square this I have just now calculated px turned out to be equal to 0 so delta px will turn out to be n pi h cross by L. Similarly when I take delta x this is mean of x square minus mean of x square mean of x square I have calculated this mean of x we had calculated was L by 2 you take all this particular thing into consideration this is what you will get as delta x I can multiply delta x into delta px and find out what will be the uncertainty product for this particular particle in a box delta x into delta px turns out to be h cross multiplied by n square under root of n square pi square minus 6 by 12 which of course depends on n and you can see that corresponding to n is equal to 1 delta x delta px product turns out to be 0.57 h cross while n is equal to n is equal to 2 turns out to be 1.67 h cross. So remember earlier we had introduced uncertainty principle we had introduced very very qualitatively purely to get an idea of order of magnitude. Now we have defined quantitatively what is delta x we have defined quantitatively what is delta px these things have a specific meaning now in quantum mechanics when we say delta x is the standard deviation of finding out value value of x delta px is standard deviation in finding the value of momentum then I can write delta x into delta px and that is what this will give and of course you can see that this is always larger than 0.5 h cross 0.5 h cross is the one which we get for Gaussian wave packet that is the least value of uncertainty product that you can get for any wave function this also can be shown. Now let us look at some of the postulates of quantum mechanics we are now in a position to somewhat formalize what are the postulates of quantum mechanics. I will give you some of these things I will introduce in while defining these postulates some additional information which as we keep on going ahead we will be telling out of course these again I remind these things are fairly abstract and sometimes to say least shocking but sometimes student always saying that it appears like a magic show that you are trying to talk something which is just not in the real life but that is the way quantum mechanics is I mean we have no other go if you want to understand the nature okay I mean this is the way we have to we have to go I mean unfortunately many things are not that straightforward and so clean as things were in the classical mechanics okay quantum mechanics relative state mechanics are somewhat different okay let us go to the postulates let me again remind you that these postulates are not standard if I say what is the first postulate of quantum mechanics you know people may sort of disagree okay while in the case of let us say Newton's law of motion everyone's if you say Newton's first law everyone knows what is Newton's first law every textbook will define the first law exactly in the same fashion similarly if you say for relativity special theory relativity what are the postulates first postulate second postulate standard everyone talks of the same way but here postulates are somewhat varying and these are only some of the postulates because quantum mechanics is I mean advanced quantum mechanics is not only highly mathematical but keeps on becoming much and much more abstract I have even told that you know the way of what we call as a wave mechanics representing wave functions by waves okay it is in its own self okay can be replaced by talking into going totally to an operator mechanics which you also call matrix mechanics okay so therefore corresponding to what type of theory we are talking the postulates also become somewhat different okay so that is why I am writing I am sort of being rather careful and writing only some postulates of quantum mechanics just enumerating them in one way or the other the first thing which I call is a system description and time evolution this particular postulate I have also mentioned earlier in fact some of these postulates I have already mentioned earlier while discussing and giving the examples some of the postulates will be probably new which I have not discussed so far so this particular postulate I have discussed earlier a particle under a potential vx is described by a wave function psi x which contains the information about all physical properties of the particle this if you remember I have already described it earlier the time evolution of psi x is governed by the time dependent Schrodinger equation this is what I have said the system description and time evolution see first is that there is something which is called wave function and this wave function how it has to be obtained it has to be obtained by the solution of Schrodinger equation. Now let us look at this particular thing in fact part of these things have been used but part of the other things we have not mentioned we say that wave function has to be well behaved when we say wave function has to be well behaved it means it must possess certain properties and what are the properties that we expected to possess is that the wave function must be single valued it means for a given value of x it cannot give you two values it cannot be a function which corresponded to a particular given x or any particular point you get two values of wave function that is not possible because wave function is eventually related to something which is real so wave function on its own is only a complex but this eventually leads to something which is more physical which is more real and a real quantity like for example probability of finding the particle at a given point cannot have two different values it can have only one unique value. So, there is a condition which has been put on wave function to be realistic is that it should be single value the second thing is that it should be always finite it cannot become infinite because if it becomes infinite eventually the probability also becomes infinite and it must always be continuous this of course we have used in a particle in a box that should always be continuous we must maintain that wave function should always be continuous again because this is related to something physical and physical thing I do not expect them to show discontinuity they should anything which is well behaved must always move smoothly. The second part of this particular being well behaved is that the position derivative that is d psi dx should also be continuous which I have not used in this particular case in fact for a particle in a box which I had just now solved in the morning okay d psi dx will not turn out to be continuous because there is another condition that unless vx shows an infinite jump only when vx shows an infinite jump d psi dx need not be continuous because in this particular case we found that vx was turning out to be infinity at x is equal to 0 and x is equal to l so I did not use this particular condition. But if for example if we have a finite box which probably I will be able to solve at the end of this particular wave mechanics part then we have to apply in that case the continuity of d psi dx also. Now I will not be able to talk about this particular thing and give you I mean these are all postulates I can always raise my hand if somebody asks why this particular thing okay by this particular condition okay other than saying that you know this is what I expect physically okay I am quoting one particular statement which has I have which I have taken from a textbook textbook by Crane on modern physics okay which is only to just compare this these particular statements from classical mechanics and it is only to convince you that these conditions these conditions that I have put on wave function to become well behaved makes sense okay I am not proving that this is a postulate this cannot be proven this is what this is our starting point in its own self okay so this is only to convince that this makes sense. So let us just use that particular sentence which I have taken from the Crane's book here there are two things which I have written one with blue color another with the red color the blue color corresponds to the classical mechanics and the red color corresponds to the quantum mechanical part. So this is from Crane's book on modern physics okay what I will be doing is first I will be reading on the only the blue part I will ignore the white part which will pertain to the classical mechanics and I am sure you would agree with whatever I am saying as far as the classical mechanics aspect is concerned when an object moves across the boundary between two regions where it is subject to different forces classical mechanics we will talk about forces when an object moves across the boundary between two regions in which it is subjected to different forces the basic behavior of the object is found by solving Newton's second law. The position of the object is always continuous okay suddenly potential cannot disappear from a point and reappear from some other point it is always continuous in classical mechanics it has to move from one point to another to another continuously from one point to another the position of the object is always continuous across the boundary and the velocity is also continuous velocity also suddenly cannot cannot jump as long as the force remains finite if the force suddenly becomes infinite then at that time velocity can show you a jump I mean remember in realistic situation we never have infinite force we never have infinite change in potential energy but these are only idealized condition but in classical mechanics the force becomes infinite in principle the velocity can show a sudden jump because rate of change of momentum tpt will also be infinite in that sense because force is infinite. Now let us read this particular thing with red thing just to compare only with the quantum mechanics part when an object moves across boundary between two regions in which it is subject to different potential energies remember we talk of potential energies in quantum mechanics when an object moves across the boundary between two regions in which it is subjected to different potential energies the basic behavior of object is found by solving the Schrodinger equation. The wave function remember position wave function has somewhat similar connotations the wave function of the object is always continuous across the boundary and the derivative d psi dx is also continuous as long as the change in potential energy remains finite remember change in potential energy is related to force. So, in quantum mechanics we talk the wave function of the object is always continuous across the boundary and the derivative is also continuous as long as the change in potential energy remains finite. So, this is only just to convince you that these conditions that I have put on the wave function for it being wave well behaved make some sense it matches with our classical ideas. Now the second thing this also I have mentioned earlier but I will introduce here something which is slightly different in fact some people have asked this particular question also which I have not answered because I knew that I am going to discuss this particular thing later. Each dynamical variable that relates to the motion of the particle can be represented by an operator. I had already talked about this particular operator I have given you the well the operator expression for example, for momentum and energy. So, each dynamical variable if we talk about angular momentum we talk of anything this is given by an operator which must satisfy certain criteria. Which one of the criterion is that whenever we are using I am only getting any observable things must always be real. I am not going to discuss about all the complexities of this particular operator all I want to say that this is must satisfy certain criteria. Now the second part is something new which we have not discussed the only possible result of measurement of the dynamical variable represented by an operator is one of the eigenvalues of an operator. I have not even defined what is eigenvalues probably people who are from mathematics they already know I mean eigenvalues have been defined in the case of matrix here we also defined in terms of an operator. So, for example, if g is an operator this is what I have written here and it operates on a function and after operating I get this equal to g n times phi n it is the same function phi n. Remember this is an operator while this is a constant which just gets multiplied by phi n. So, if an operator operates on a function and after operation my result turns out to be a constant multiplied by phi n then these are called eigenvalues of this operator and these phi n's are called eigenfunctions these are just the definitions. So, corresponding to an operator you can have its eigenvalues and you can have its eigenfunctions. So, only for certain specific functions this particular condition will be satisfied it may not be satisfied for every function and corresponding there could be multiple functions for which this condition could be satisfied all of them will be eigenfunctions of this operator okay and corresponding to each of this eigenfunction I must satisfy this condition that when g n operates on phi n I must get a constant times phi n whatever are those constants these will be the eigenvalues. So, the condition is that if we have an operator g n which represents a physical quantity then eventually when it operates and if we make a measurement of this particular quantity I must get only an eigenvalue of this particular thing I cannot get anything other than the eigenvalue of this operator. So, if this eigenman for example this particular expression represents let us say position or a momentum then when it operates on this particular eigenfunctions whatever are the eigenvalues only those eigenvalues are the permitted values that this particular momentum can have. So, instead of momentum I could have any other physical quantity but then observables have to be only eigenvalues only possible result of a measurement of a dynamical variable represented by an operator is one or the other eigenvalues of the operator of course if they have to be observed they better be real. So, eigenvalues are real numbers for the operators representing the dynamical variables. So, it says that eigen I mean the operators must be such that their eigenvalues must be real because if they happen to be complex complex quantities cannot be observed whatever we are going to observe have to be real things therefore in fact there is a condition which I have said earlier this in fact what we call as a Hermitian operators the operator should be Hermitian their eigenvalues must be real because eigenvalues are the one that I am going to observe. Now let us introduce what we call as a Hermitian operator those of you who are familiar with the normal standard classical mechanics course. Hermitian is just defined as sum of kinetic energy and plus potential energy which is defined as p square by 2m plus v. We have already talked about p square operator we have also talked about v operator that is just simple multiplication operator. So, similarly we can define in Hamiltonian operator which is sum of kinetic energy and potential energy this will be defined as p square by 2m plus v. So, if I want to write this eigenvalue equation corresponding to this this will be this particular Hamiltonian this particular operator multiplied by phi n must be equal to a constant times phi n this is the eigenvalue equation. If this operates on phi n whatever phi n that I am getting will be called eigenfunctions and whatever E n that I am getting they will be called eigenvalues. And according to the postulate we have just now mentioned if I want to measure sum of potential energy and kinetic energy what I will be getting is only one of these eigenvalues. Now replace this p square by its own operator and let us see what is this equation. We replace this by their operators you get minus h square by 2m del 2 phi del del x square plus v n psi n is equal to v n psi n. You can see that this equation is nothing but time independent Schrodinger equation. So, what we are doing when we are actually calculating time I mean when we are solving the time independent Schrodinger equation actually I was solving a eigenvalue equation corresponding to Hamiltonian operator. So, whatever I am getting as phi n are actually the eigenfunctions of the Hamiltonian operator and whatever I was getting as the values of allowed energies they were the eigenvalues of the Hamiltonian operator. So, as you can see that this particular Schrodinger equation has now been painted in a slightly different fashion. This has been more generalized in fact what we realize now that in quantum mechanics this eigenvalue equation turns out to be more important equation more general equations and this time independent Schrodinger equation happens to be just one of the eigenvalue equations. Then we talk about completeness this also I have mentioned just now before in the lecture the eigenstates of an operator representing a dynamical variables are complete. It means any admissible wave function can always be expressed in the following way in terms of eigenfunction of any operator. Now the same function for example I could choose this to express in terms of eigenfunction of let us say energy operator or Hamiltonian operator or a momentum operator or a position operator whatever are those operators any function can always be represented as a linear combination of eigenfunctions of any operator. And whatever the set of eigenfunctions or whatever that eigenfunction of an operator that we use we call it a basis. So, I am expressing this let us say in momentum basis it means I am trying to express a given function in terms of eigenfunction of momentum eigenfunctions or Hamiltonian eigenfunctions. So, I am using in that case a Hamiltonian basis. So, any function whatever it is which I have just now mentioned can be represented in fact this phi n could be eigenfunction of an operator or could be of a different operator. Of course the C n's will become turned out to be different. But nevertheless any function can always be expressed in terms of eigenfunction of any other operator of any operator that we decide to choose. Let me read again. Any admissible wave function can always be expressed in the following way in terms of eigenfunction of any operator of our choice. These eigenfunctions are called to be forming the basis. We had discussed this aspect in detail with eigenfunctions of the Hamiltonian operator. The fourth thing is about the probability. Probability that an eigenvalue g n would be observed as a result of measurement is proportional to the square of the magnitude of the coefficient C n square in the expansion of psi. This also I have just now discussed. Basically it means if I have to find out what is the probability of having a given value of momentum, express your wave function, what express your state in terms of the eigenfunction of the momentum operator. Then whatever are the coefficients, take coefficient square that will give you the probability of finding that particular eigenvalue. The proportionality becomes equality if we have normalized wave function. Then we talk about final thing which is the collapse of wave function. If a measurement gives a particular value of eigenvalue psi n, the wave function discontinuously collapses to phi n. This is something which is important which we have said. The effect of measurement is that whatever was your original value of your wave function, at that times once we have made a measurement and we have resulted into an eigenvalue, it will collapse to a value of phi n. Now this state would still be linear combination of some other operator. That is a different question. But specifically to that particular value if I have calculated, if I have expressed and I got an eigenvalue g n corresponding to a given dynamical quantity, then after that the wave function has collapsed just to that phi n. I think I will stop here and another 5 minutes. So let us just take 2 or 3 typical questions. How do we define potential well? Can you give an example in a physical situation? As I said the infinite potential well, this is what I have written here, is actually not realistic because see strictly speaking a discontinuous potential well is also not realistic because whatever happens in the real situation, the potential energy should also vary continuously. There is no nothing like discontinuity. But as I said in physics we always want to start with simplest thing and this happens to be the simplest problem on quantum mechanics and especially at the beginner's course on the quantum mechanics, especially at the first year student. We want to take very, very simple examples. So we can always think, I mean as I have given a comparison with classical mechanics that suddenly there is a change in the type of force that we are talking or something like that you can talk. But they are not really realistic in that sense. We are just assuming that there is a possibility where potential energy can change discretely from one region of the space to another region of space. At all you want to imagine you can assume that there is a very, very hard box in which a particular particle is concerned is kept and the walls are so rigid that there is no possibility of this particular particle to penetrate out. So this I mean the most approximate thing which you can talk about is that a particular particle kept in a very highly rigid box. So my question is for a particle in a well, why N cannot be negative? See because in this particular case the solution turns out to be actually a stationary wave solution and in that case a negative value of N, negative wavelength does not mean anything. See only times when we talk about a negative sort of value of K is when you are associating a travelling wave. In that case there is no travelling wave so only K values which are positive which are allowed. A negative value of K does not mean a negative value of K in a real wave theory would always mean when a wave propagates in minus x direction. So in that particular case because these are not propagating waves we do not allow a negative values of x. In fact this lands into a problem when we are talking in comparison to solid state physics which I think Professor Suresh will take in the next week where we would like an electron to propagate inside solid in both plus x direction and minus x direction. In that case the boundary conditions have to be changed so that we allow both positive values of K and negative values of K. In this particular specific case the type of boundary condition that I have put these are sort of what we call as a stationary state boundary condition. In that case the solution are stationary wave which does not represent a net motion of the wave therefore in that case only thing which you can talk about is a wavelength and negative wavelength does not mean anything. When negative K would mean as I said something which is propagating backwards but that something when actually wave propagates. In that case the boundary condition that has to be used will be different which actually we often do in the case of solid state physics. So difference between Kronecker delta function and Dirac delta function. Can you please give any kind of example to explain them both? So Dirac delta function is a slightly different function which is used only for the continuous variables while this is used for discrete variables because if Kronecker delta are talking of m and n values both of which are integers. So when m is equal to n then it is 1 otherwise it is 0. Well Kronecker delta function is something which is a very very sharp function which exists only at one particular point of x and it does not exist anywhere. So it is slightly different from its connotation. I mean Kronecker delta and both Dirac delta function are often used in quantum mechanics but in this case because I am talking of two discrete values of m and n. m is also integer n is also integer when m and n are equal then this Kronecker delta is equal to 1 otherwise it is 0. Can you tell any specific example like when to use with Dirac delta function? So Dirac delta function is used and I mean I am pretty sure Professor Ghosh would be talking about Dirac delta function in many of the cases because in electromagnetic theory it is often used. In quantum mechanics it is also used sometimes we talk of a wave function which is of the form of Dirac delta function. In fact in solid state physics we often use when we say that a particle actually moves in a box which is something like a Dirac delta function. So we often use this type of condition when you want the potential energy to be very, very sharp which exists only at a particular point and then we talk about Dirac delta function but in this case this situation is slightly different is more mathematical in that sense. If quantum mechanical results have to match classical mechanics at macroscopic dimensions can we get Newton's laws of motion from Schrodinger's equation? Yeah in principle yes the only thing you know these quantities which are we talk is eventually replaced by their average quantities. In fact that is the way operators are defined when we talk of the Newton's law of motion we will talk the average of momentum rather than talk of individual momentum because the particles in principle now would be could have different values of momentum. So we will talk of the average momentum behaving Newton's law of motion or when we talk of conservation of momentum we will be talking of the average momentum which is conserved. So I mean this is very naturally done and normally done replacing these things and eventually come to the in fact normally a quantum mechanics course we do in a different fashion we take Newton's law of motion and replacing that by average quantities and we come back to the operators and say this is the way momentum operator should be. So that is definitely possible. I think we will stop here.