 Good morning participants. Before I start, I will try to answer some of the questions that we have been receiving at the chat mode. I would just like to mention that the number of questions that we are getting unfortunately are too large and it is not possible for me to answer all of them, though I would have liked to do it. So what I am doing, I pick up essentially some questions which are sort of important questions and especially there are one or two points which I think are good questions and they require little bit more discussion. So I will explain them little bit more in detail before we go ahead. So let me start with first question. It says that as we know the double slit interference is the superposition of two single slit diffraction phenomenon. Can we get double slit interference curve by some mathematical operations of two single slit diffraction curves? Well, the answer is yes and no. The thing is that first of all it is not really superposition of two single slit diffraction phenomenon. Had it been, then we would not have this particular issue. You could have always written n12 is equal to n1 plus n2. But there is always a mathematical way in which we can sort of discuss or describe the interference of two waves. Same way it can actually lead to the interference pattern. So there is a standard mathematics available for the Young's double slit experiment which is generally taught at the high school level and that is the similar type of mathematics we will apply also when we are talking of diffraction of electrons. So as I say, as far as we are concerned probably there is no difference between electrons and photons because both are particles except that electron is massive and therefore depending upon the energy, its wavelength would be different. For the same energy, the energy of the photon and electron will be different otherwise we can use mathematically exactly the same technique. Now second question is that can we measure the phase speed that is the velocity of wave packet by experiment? See again I repeat that velocity of wave packet is not a phase speed but it is a group speed and phase speed is purely of mathematical consequence. It has no realistic bearing. So we cannot talk of phase speed or any measurement of phase speed. Whatever we measure is always the group speed. Now the next question is that different color corresponds to different wavelengths which according to de Broglie principle corresponds to different momentum of photon. Now when we are talking of photon, normally we do not apply de Broglie principle to calculate the wavelength because that is given more from relativity. For the case of photon, we have discussed that momentum of the photon is given by h nu by c. This particular thing I will describe more in detail when we are discussing the relativity portion. Basically relativity says that if a particle has a zero rest mass and its energy is e then its momentum must be e by c. So this is from that particular equation that we get the standard equation which if you already know it is alright but if you do not know we will be discussing it when we are discussing relativity. Is this equation? This is the relationship between energy and momentum. And when we say energy, energy is the total relativistic energy which does not have a classical analog. So when we put m not equal to zero, we get e is equal to p c. So therefore the momentum is equal to h nu by c. So this can be written by lambda. So this strictly speaking not de Broglie relationship that we are applying to evaluate the wavelength of the photon which can actually directly be obtained from frequency that is not an issue. The only thing which we say that this expression p is equal to h upon lambda was taken by de Broglie also for massive particles. So when we talk of de Broglie relationship we generally talk it for massive particle not necessarily for photons. Now the question is that different color corresponds to different wavelength. I mean that is as we have mentioned that is different frequencies which according to de Broglie principle corresponds to different moment of the photon. That is perfectly right because the moment of the photon would actually depend on the frequency and the color of a light also depends on the frequency. So there is no contradiction about that. Now next question is about different type of photons which I do not understand the question. I do not in a photon is a photon. Of course photon can have a different energy. Photon can correspond to a different wavelength or different frequency but you know a photon is a photon. So I do not really know what you mean by different type of photons. Now the next question is that who the wave packet is considered for one particle? Is then how come multiple characteristic can exist for single particle with k k plus dk k minus dk? That is what I was always trying to tell that when we associate a wave packet with particular particle the wave packet actually consists of a large number of values of k which essentially means a large value of lambdas. So therefore we cannot associate a fixed value of k corresponding to a particular wave packet and that is how the actual uncertainty principle gives rise to. So I am not sure whether I have understood the question but this is what I mean what I could make it out. The next question is that can you give the physical significance of complex nature of wave function? As I have been mentioning again and again that this is complex but whatever is the observable is the real thing and we have only discussed so far how to obtain the probability of finding the particle but how the wave function is to be used for obtaining certain other information will be discussing subsequent lectures. Now next question is that how size of wave packet changes with momentum of particle giving Heisenberg principle? This is what I have sort of briefly mentioned though not really mathematically probably I will ask our TES to work out one or two simple problems on these things. Basically I have told in a sort of a very very qualitative or hand waving way that you know when you have a larger range of sort of wave vectors combining you essentially get a shorter wave packet. So that is what I would like to say. If it is a wave what is the difference between a sin kx minus omega t and a sin kx plus omega t this I think also I have mentioned earlier that if you are taking a sin kx minus omega t and a sin kx plus omega t both represent sinusoidal wave except that this particular wave travels in plus x direction while this wave travels in minus x direction. So that is the only difference because as I said it depends on the relative sign of x term and t term. So this is t term there is a negative sign here this is x term there is a positive sign here. So if the signs are opposite this will represent a wave which is travelling in plus x direction because if t goes up x also has to go up to keep kx minus omega t constant which essentially means keep keeping phase as a constant. On the other hand if I put a sin kx plus omega t if t goes up then x has to go down in order to keep the same phase it means if you are looking at a point at the which has the same phase okay as t progresses that will have to move in minus x direction. So this will represent a wave which is travelling in minus x direction while this represents a wave which travels in a plus x direction. Now we go to the next question in a double slit experiment with electron will there be any effect of repulsive force on the interference pattern this also had discussed earlier saying that you know this particular interference pattern would be valid even if we take if we put the electrons just one by one probably in Moodylai will give you some websites from where you can download some of the videos of the recent experiments which have been done where you can really see that this interference pattern would be valid even if you are taking electrons just one by one. So effectively the I mean that that would have been the first doubt that you know whether the electrostatic interaction between these particles would have changed things but that doesn't happen also I would like to add that you know even if we do not have electrons we have neutrons which are not charged particles still I can perform the same interference experiments and I would have no worries and I will exactly get similar type of results. Then next question is that in a single slit electron diffraction what happens to the electrons after hitting the screen you know we do not bother what happens when it has hit the screen this could probably just get absorbed and if I have a detector probably this will create some pulses which can be used for counting the total number of electrons. So that is of not no consequence of what happens this electron when it hits the screen I mean normally there will be a detector if I want to check it okay. So this particular electron will detect will create some sort of pulses in that detector which can eventually be used by a counter to count them okay and if there is no detector probably it will get absorbed I mean electrons are fairly reactive particle they can very easily get absorbed in a sort of system. Next is in the electron double slit experiment the electrons interfere like waves yet they interact with the screen like particles. So we can see both and wave and particle nature are manifesting this experiment know that I am not very sure whether this is a correct statement the way you have written. First of all I am not sure when how you are saying that when it hits the screen acts like particle okay it only acts as like it you know also think of a wave packets which are hitting there more over when we are talking of complementarity you know we are talking of a at a single time. So when we are talking you know when we are doing an experiment at a given time whether I can look both at the wave property and the particle property. So I am not fully convinced with this type of question. Now the next question is that in time in the independent Schrodinger wave independent Schrodinger's wave equation how partial derivatives change to perfect derivative. So let me just sort of explain this particular thing. For example if you have term like this del I am just giving an example of let us say f t into phi of x if I take partial derivative by definition of partial derivative it will mean that when I am taking derivative with respect to x time has to be treated as constant. So therefore this f t which is only a function of time and does not contain any x the entire factor has to be treated as a constant. So this I can write almost as c multiplied by phi x where c is a constant. So this f t will just come out of this particular thing so I can write this as f t of del x of phi of x. Now there is a need to write partial derivatives only in the case when you have multiple variables here okay and then you want to insist that I am taking derivative with respect to this variable keeping other variables constant. Now because f t which was the other variable has already come out of the differential sign so and the thing which is in front of this differential sign is only a function of x. So it makes no sense to write the partial derivative because the function on which this particular derivative is operating is purely a function of x and I am also taking derivative with respect to x. So therefore I can always write this as f t of d phi dx that is what was the idea of course you take the second derivative the same thing will apply in this particular case also or when we take derivative with respect to time also it will apply exactly the same thing. Now next question is that uncertainty if the wave packet is large or small then what about the position momentum of the particle this also I thought I had described about uncertainty principle that if the wave packet is large then we say uncertainty in the position is large if wave packet is small then we say the uncertainty in the position is small okay but in order to generate a smaller wave packet I have to have a larger values or larger range of k values okay mixed up so therefore a smaller length of the wave packet implies a larger range of k larger dispersion the larger variation in the values of k it which means a larger variation in the value of momentum. So that is what I meant I am not sure whether I have understood the question very clearly. Now next question is that is a sin kx plus omega t is a wave I just now explained yes it is a wave but it is a wave which is moving in minus x direction is delta x delta p equal to h or h cross or h cross by 2 this also I think I had explained last time that you know and today in fact probably I will work it out taking a actual wave function how delta x and delta p turn out to be because see so long whatever we have talked about the uncertainty principle was fairly weight okay fairly qualitative okay we have not really quantified anything we are just talking again in a hand waving fashion okay but now we are in a position to actually I clearly sort of give an expression for uncertainty and derive the uncertainty principle so which I will be doing today I mean in a given problem in fact normally first year students always ask the question that you know what uncertainty principle should I use when you have been given a particular problem unfortunately many of the problem for uncertainty principles are also fairly qualitative and I also always answer that it does not matter which one you take because all I am talking is an order of magnitude and just for the ease of correction whenever we set the question paper we always give them that use this uncertainty principle so that becomes a little easier for us to evaluate the answer script otherwise everyone uses a different uncertainty principle and then you know for a big class it's very difficult to evaluate the copies now the next question is that is it possible to see electromagnetic waves light waves no this is again a very I am not very sure what the question means I mean accordingly you never see a light or you never see a wave a light wave okay you only feel it okay or you see the you see objects in the light okay but you do not actually see the light but I am not very sure what you mean by this particular question next question is it possible to arrive at Schrodinger wave equation with form other than the third equation in representation of wave I said you cannot obtain using sine and cosine because if you use sine and cosine that will never solve that particular equation because I am taking a relationship between the second derivative of x with the first derivative in time and sine and cosine functions will not be a solution but instead of plus i k x minus omega t I could have taken minus i k s minus omega t which I have also mentioned the Schrodinger equation will change but then when I am interpreting the things I have to always keep this thing in mind that time factor will be plus i omega t there and how does this time factor will you know sort of be useful you know we will see when we are you know solving some of the problems here then you go justification of a photon or a light source to see photons in a third experiment not so clear sorry my the question is also not very clear justification of photon or light source to see electrons see all I am trying to say let me just try to re-explain this thing maybe that will help strictly speaking I do not see the electron but as I said these are sort of a thought experiment when these experiments cannot be performed at this particular level but you know on the other hand you know we just describe them in a more dramatic fashion so as to know convey the idea that we want to actually convey so I mean let us agree that we are not really watching the electrons or we do not have a method of by which we will really watch the electrons the basic thing that I am trying to say is that if at all we are able to perform an experiment where I put a light source here let us say somewhere here and a electron is passing what is something like that that if I have to find out if there are two doors in a room and I have to find out whether a person has gone from this door or from this particular door there has to be light source if it is completely dark I can never find from which particular door the person has entered the room now if there is a light then what will happen as the person starts entering this particular room there will be light reflection from that particular person and then we will see oh this particular person has actually entered the room from this particular door if I see a reflection coming from here of somebody's face from this particular door close to this particular door then we know this particular person has entered the room from this particular door so I am just sort of extrapolating this particular thing to look at the electrons so I am imagining that if an electron goes from this particular slit okay and if I have a light source then okay this particular light reflected from electron will tell me whether this electron is really going from this particular slit okay similarly if there is a reflection of light coming from this particular close to this particular slit I say oh this electron is actually going from this particular slit so what I was trying to say that every electron which reaches a screen okay in principle you can find out from which particular slit that that has come by actually watching them see as I said that if it is completely dark everywhere and there students entering the classroom okay I will never be able to tell from which door they have entered but on that hand if there is a light source then by looking at their face and finding out from which particular door they have come they are coming I can always find out which person has come from which particular door so in that sense in a more dramatic sense I have used this particular idea that I am trying to watch the electrons but I am strictly speaking not watching the electron I cannot watch the electron but this is just to sort of explain a little bit of idea the only thing is that when a student enters the room the effect of photon is very very small okay but in the case of electron electron being a very lighter particle when photon interacts with electron it will always cause a scattering of the electron because their momentum are of similar order in fact when we do Compton effect we will discuss that this is precisely the Compton effect experiment that when a photon actually interacts with an electron it can change the path of electron okay photon interacting with a student probably will not change the path of the the person but on the other hand you know a photon interacting with electron may change its path so what we are saying is that this photon which are coming and interacting with these electrons if at all I want to watch the experiment then these photons have to interact with these electrons and these electrons probably would have changed the way they were actually going had there be no photon and therefore that could be thought as one of the reason why the interference pattern has got destroyed so basically what we are trying to say is that if I can perform an experiment by which I am reasonably sure that electron is going from slit one or slit two we will always find that any electron reaching on this side definitely has come from slit one or from slit two but in that case I will never be able to see interference pattern so long I keep this vagueness I do not ask the question okay I keep perfect dark here so to say dark under coats so that these electrons do whatever they want and I am not asking them question how they are coming okay that only in that particular condition I will be able to see the interference pattern so that is what actually is basically uncertainty principle okay I have to accept this uncertainty that I will not be able to find out from which particular slit the electron goes if I have to really see the interference pattern. The next question is about the group velocity phase velocity this I thought I will explain little bit more in detail because you know there quite a few questions on this particular thing I have purposely avoided the use of term phase velocity for the matter waves though many textbooks talk about it and I will tell you the reason why I am I sort of avoided it but because they have been questions so let me try to sort of answer it now the thing is that see as we know that the phase velocity is defined as omega by k and group velocity is defined as d omega dk. We also mentioned that the two velocities would be same if we have a simple relationship of the type omega is equal to vk all right now let us try to find out the phase velocity of course this particular relationship is nothing but nu times lambda because omega is 2 pi nu and k is 2 pi by lambda so this is just nu by lambda so let us try to find out for the particle waves. So for particle waves if I write the phase velocity lambda will be equal to h by p this is you know standard de-broad relationship the problem is that how do we define the frequency of a matter wave in fact this besides the reason that I did not want to raise this particular issue because this is a little more complex issue see generally in the experiment we always measure the lambda so we do not have a really a unique method of I mean we do not I mean as far as I know cannot talk about the frequency but in order to maintain consistency with whatever we have been talking in physics we can define frequency or omega by the relationship E is equal to h nu where h is the Planck's constant. Now when we are talking of E again I have not done relativity so that was another reason I did not want to talk about it but just I will give you the result that E in principle I have just now written the expression that E is equal to E square P square C square plus m naught square C to the power 4. Now we can also define kinetic energy kinetic energy is different from E and we talk of E is equal to k plus m naught C square which is the rest mass energy assuming that there is no other type of sort of energies are involved. Now because for photon rest mass energy turns out to be 0 so E turns out to be equal to k so in that case whether I write k is equal to h nu or whether I write E is equal to h nu it makes no difference but in the case of massive particle there is a slight ambiguity that I can either write k is equal to h nu or E is equal to h nu both will be consistent with the definition of the photon with the expression that use will be consistent with the expression that we use for the case of photons. So as far as this particular aspect is concerned there is slight ambiguity of how do we define the frequency but let us use any one of them. If I use E is equal to h nu then what you can write this nu this as far as nu we will write E upon h if I write E upon h this h will cancel then we will get E upon p and if I am talking of total energy that in relativity is given by m c square and p given is given by m times v this expression we will derive later I am not really derived but you know we will come across later when we do relativity at the end of this particular workshop okay but if I do that this m will cancel remember this m is not m naught but m which depends on a velocity. So this expression turns out to be equal to c square by v or c square by u let us say where u is the speed of the particle. So what we see because generally u has to be always smaller than c in fact by relativity u has to be always smaller than c so the phase velocity turns out to be larger than the speed of light. So always people ask the question that you know how a deep Broglie wave can travel with speed larger than velocity but remember this is only the phase speed and phase speed has of no physical consequence. If we take the group speed or group velocity that will always turn out to be equal to the particle velocity and not c square by u okay. In fact I have asked our TES to work out this particular problem they will probably work it out today I am not today but tomorrow when they will really show that you know if you try to calculate the group speed it will always turn out to be the equal to the speed of the particle and that is what is of more importance because phase speed as I have said is purely of mathematical consequence okay no particle moves with the phase speed okay. Of course there is also an ambiguity in fact as I said that instead of e I could have used k if I use k then I can write h is equal to p and I can write this as k by h and assuming that the particle is a non relativistic which we do in the case of Schrodinger equation I can write this as h and p I can write as m v and k I can write as half m v square by h then h would cancel then m will cancel this v will go and this turns out to be equal to v by 2. So if I use this particular expression of k is equal to h nu instead of e is equal to h nu I get the phase speed to be equal to v by 2. Now this is also not right in the sense that the phase speed of the v-bongly wave turns out to be half the speed of the particle let us say u by 2 I mean I have used u here so I can use u here also u square so it is u by 2 alright. But as I said though these are only for mathematical delight physics wise it does not make a difference because physics wise only the group velocity which turns out to be of importance and we can show that whether we take e is equal to h nu or whether we take k is equal to h nu the group speed will always turn out to be equal to u and this is what is of importance that v-bongly wave will travel with the same speed as the speed of the particle and this is what we must expect that the wave the speed of the wave which is associated with the particle must travel with the same speed as the particle otherwise it makes no sense that waves move the different speed wave packet goes with a different speed than the particle okay. So this is what we expect and that you can show that this should be true whether you take k is equal to h nu or whether we take e is equal to h nu unfortunately it so happens that when we actually sort of come to Schrodinger equation we always use k is equal to h nu so there as I say there is a small amount of ambiguity say small amount of doubt because if you could perform an experiment and could really measure what is the frequency of the particle we would not have this particular doubt but it so happens that whether we use k is equal to h nu or whether we take e is equal to h nu it does not make a difference so this ambiguity always remains. Our next question is that what is the difference between classical theory and quantum theory means limitation of classical theory and how quantum theory is emerged out please explain in brief this particular thing I think we will discuss little later as we are going ahead in the course regarding wave and particle duality does it mean that only multiple electrons show a property and not a single electron see as I say that this particular wave is a probability wave this also have discussed that if you are talking of one particular electron okay this of course is governed by the wave nature but on the other hand this will I mean this will always arrive at some place okay only when we perform this experiment multiple times in fact I have given that particular example of wave function which I will be discussing today also okay what we actually mean by the probability it is almost like a head and tail that you know if I just toss a coin even though the probability of getting ahead is let us say half but when you toss the coin you will get either head or tail it is also like it is possible that first time you get head second time also you may get head third time also you may get head it does not mean that the probability becomes one okay you have to perform this experiment really over large number of time to get probability to be equal to half so because the wave function is actually a probability wave therefore if you want to really detect the overall nature of the the wave you have to perform the experiment over a very very large number of particles but once you perform only in one of these things okay the particle will always appear somewhere the next question is uncertain relation exists for two cases one for single state diffraction other for the wave can you explain me if there is a generalized uncertain relation see uncertain relation is always a general relationship what I have tried to do is to get an idea of uncertain relationship by two different methods but as I have said they are much better mathematical ways of looking into it okay which I am doing partly here I will be doing partly today and you know some of these things will form actually a little later part of the advanced quantum minus course which is beyond the scope of this particular course next question is that you know we have been talking only diffraction of electron and not of interference and polarization this is a sort of interesting question the thing is that you know diffraction is also so to say a sort of interference so of course you know depending upon what type of experiments we are trying to do is we call it diffraction or interference I mean I am we can definitely perform the interference experiment double slit experiment is generally called as interference experiment single state is generally called a diffraction experiment so both these things are possible with electrons as far as the polarization is concerned this is interesting thing is that you know I mean there is always a spin of the electron and in principle there is always a possible of polarizing the current if at these days in spin tronics we always get spin polarized currents there is I mean in neutron diffraction there is also when we we call I mean there are certain experimental techniques which talk about the polarized electron beam in which you know you can really look at the spin and this particular spin can be used consider as a polarization you know the see like we can consider in which plane in which in which particular direction E and B vectors oscillate that gives you the idea of polarization similarly we can talk of the spin as breaking that degeneracy and calling it a polarization so if you are talking of a spin polarized I mean a polarized beam on neutrons it means basically their spins will be pointing out in a specific direction the next question is that in single state why the electron deflected is only if the slit width is very low actually when slit width is very low that is the time you will see a significant effect of diffraction otherwise diffraction is always present because if you look at the equation this is d sin theta is equal to lambda so therefore only when d is for a fixed lambda only when d is smaller the theta will be the component will be larger if you are talking about wave group moving with same velocity equal to the velocity of the particle now while we are forming the wave equation we have reintroduced the phase velocity see strictly speaking I have not introduced the phase velocity or I talked about the phase velocity when I was talking only in the case of wave equation but you know as we are talking of electromagnetic waves electromagnetic waves in vacuum they are non-dispersive and therefore phase velocity and group velocity are same so as far as electromagnetic waves in vacuum are concerned I can talk about phase velocity because that happens to be equal to the group velocity the uncertainty pressure or position momentum is h that we derive using two approaches how we get the value of minimum uncertainty product that is what I say I will refer you to advance books on quantum mechanics where they have said that the minimum product actually turns out to be for a Gaussian wave packet which turns out to be equal to h cross by 2 kindly elaborate again its significance of considering the solution of size equal to a raise power i k x minus omega t I said that this is the exponential type of the solution is the only possible solution which will solve that equation where we are relating the second derivative of position with respect to the first derivative of x again the question is the wave packet is real but why we take the wave function as complex quantity when we talk of wave packet we we are not specifically talking about only the de Broglie wave but we are talking of a wave in general because whatever you have said about wave packet is a part of wave theory which is valid even for electromagnetic waves or for that matter any other type of wave so wave packet when we are talking we are not really talking about de Broglie but it was a general thing but it so happens that when we are talking about the wave associated with the particle that turns out to be a complex quantity so that is the I would like to say we said about eigen and eigen value that we will come to that later there is a question that what happens when the two waves of equal and opposite amplitude or whatever it is move in opposite direction then we we know that they form standing waves you can always apply superposition principle in a proper way and get the result of what will happen to the disturbance then the other question is that why potential energy does not is not exclusively time dependent I say if it is not exclusively time dependent it could be time dependent then I cannot use time independent shooting in equation that is all I will say why time dependent term is negative in the wave equation that is only a convention I could have always worked out the entire quantum mechanics as with plus i omega t okay the equation would have been different nothing of physics would have changed okay we are just using a convention so that you know we are consistent over all over the entire world okay entire textbook so whenever we are talking of quantum mechanics we generally always assume conventionally that the time dependent term is of the form e raised to minus i omega t I think these are probably most of the questions which I have tried to answer now let us come to the actual lecture okay so let us sort of recapitulate what we did last time using certain logic if you obtain the time dependent shooting equation this is the time dependent shooting equation in three dimension this whole operator can be written in terms of del square psi and the advantage here is that this del square depending on the coordinate system that we use can be expressed either in Cartesian coordinate system or spherical polar coordinates or central coordinates whatever coordinates you want to use then we said for the case when we does not have an explicit time dependence only in that case we succeeded in separating the variables to obtain time independent shooting equation this is the time independent shooting equation then we realize that the actual wave function psi x t will be whatever we obtain from the solution of this particular equation that is phi x then I have to multiply e raised to the power minus i e t by h cross which is actually the solution which has come from the time dependent part of the shooting equation so my wave function will always be given by this particular quantity then we started the about the physical interpretation of the wave function in terms of probability of finding the particle we explain what I mean this d should not be there we explain what we mean by the probability we said that when we trying to calculate the probability of observing a particular particle we do not mean that this particular particle will keep on changing the position if we keep on measuring the position one after another immediately one after another once we have identified the particle and found it to be present somewhere immediately after that if we make the measurement the particle will be only there we cannot find it out that you know particle has gone somewhere else so when we are talking about probability we talk of probability measurement over a large number of particles this is what what I mean emphasizing that we assume that we have very very large number of particles okay which are all given by exactly the same wave function when you try to measure the position of the particle for all of them you not find an identical result for some particle you will find its position somewhere for another you will find somewhere else okay and that is why there is a probability associated with that while unlike quantum mechanics where if you know the conditions precisely you would know the position precisely in quantum mechanics you get well-defined probabilities but you do not get well-defined positions okay now let us ask some of the interesting some of the what we call as somewhat philosophical questions and this is something on which there have been a lot of discussion earlier when the quantum mechanics was being developed now let us ask the question suppose we have made a measurement and we find a particle somewhere okay let us suppose we find a particle at x is equal to x0 okay or very close to x is equal to x0 when we made the measurement now question is that what is the particle there itself earlier or what happened see I have made a measurement at a given time I find the position of particle to be there but what is the particle there before just now okay let us just try to ask this sort of a sort of a I said under puts a philosophical question all right now there were three different approaches which were given by earlier people there are three type of interpretations or you know thoughts which were given to this position so question is that if a measurement was done and the particle was found at x is equal to x0 where was it just before the first view of looking into it is what we call as a condition of hidden variable okay this people who took this particular realistic approach they said that okay let me just try to read and then I will explain indeterminacy could be different from ignorance the particle was there only but we are not knowing some additional information was needed to provide the complete description this is called the condition of hidden variables let me try to explain this particular aspect see when we toss a coin and get head and tail we think that this is sort of random process okay whenever there is a cricket match you know you just toss a coin the empire tosses a coin okay then we will say head or tail and then everyone assumes that this particular getting head or tail is a purely random function but strictly speaking is it really random okay this particular coin which has been tossed okay it has been tossed under a force you know somebody has put a force it goes up and comes down okay if I know forces air sort of pressures or whatever air buoyancy etc everything okay this is a purely a mechanical problem okay in principle I will be able to precisely if I know all the conditions well I can always predetermine whether this will be head or tail the only thing the variables that we are talking are too many variables and they are not really very easily controllable so when I am tossing a coin okay where I am hitting it okay what happens to this particular what type of angular momentum I am providing what type of you know momentum I am providing and things like that and when I am catching the coin how I am catching this particular thing okay there are so many variables involved with that that I can sort of find it very very difficult to control it but suppose I had a control on these things suppose I would have known my all those variables then this is purely a mechanics problem okay the coin goes up and coins comes down it has to have either a head or tail and I can predict whether it will be head or tail if I knew all the conditions or the problem unfortunately I do not know them and therefore I treat this particular phenomenon as random so a set of people thought that actually the particle was there when we take a measurement just before the measurement also okay only thing we are not knowing because we do not know the parameter how the particle came there at that particular point okay I had no idea okay I was no not aware of this particular thing well I am not aware of all the variables only when I performed an experiment then only I knew that my particle was there similar to tossing a coin when I performed the experiment and I put the coin in my hand then only I knew whether it will be head and tail okay otherwise when I am talking of head it was actually the head there itself only thing I was not knowing even just before I catch the coin I knew I mean in principle the head was there but I did not know okay when I performed the experiment then only I knew that the head was there so this is called the condition of hidden variable okay so some people thought that the way this particular way of mechanics works the particle when I have measured particle and found the particle to x is equal to x naught at x is equal to x naught the particle was actually there but we were not knowing it okay because we are not knowing okay all the conditions all the variables which brought the particle there just like tossing a coin unfortunately this approach is not the one which is accepted today the one the approach which is accepted today is the next approach which we call an orthodox approach we say this particular thing is important let us read it the particle was not really there our method of measurement when I made the measurement then particle took a stand to be present there it means when I actually made the measurement I further up the system and therefore the particle decided to come there I mean decided is under quotes the particle does not decide okay particle took a stand okay so by my method of measurement okay change the condition altered the conditions so that the particle could be formed at that particular thing so it is my method of measurement which sort of make the particular particle stand come there so this is what is called orthodox approach this is also called Copenhagen interpretation because as probably you are aware that Nees vor had a lot of things to contribute in this particular area and these and he was living in Copenhagen and this particular the ideas they are brought with that particular set of people which were actually working with Nees vor and this is generally called Copenhagen interpretation and third approach is agnostic which is refusing to answer okay a question which I cannot answer why should I bother to answer so if you want to look about the details of this particular approaches this is given in the Griffiths book of introduction to quantum mechanics as I said Copenhagen interpretation is the one which is accepted these days which says that when you perform the experiment method of performing the experiment has altered the situation and therefore the particle took a stand there okay this is something which I will discuss little later in the later part of the course also okay about what we call about the collapse of the wave function. Now let us come to a something which is a little more mathematical see if you look at this particular equation the nature of the equation is that if psi happens to be some solution of this equation a into psi will also be a solution because if you put a times psi okay a would cancel out from all the things okay the nature of the equation is that if psi is a solution a psi will always turn out to be solution so you will always have a variable a which you will not be able to find it out just purely from the solution of this particular this particular a is generally found out from what we call as a normalization because we know that psi star psi represents the probability of finding the particle and if I find if I look at the probability of finding the particle across the entire universe that has to be one particle has to be present somewhere particle must be somewhere so if I extend my limit you know my condition of finding the particle to write from minus infinity to plus infinity particle must be somewhere there probability of finding some particles somewhere should always be equal to one okay it means if I integrate psi star psi dx from minus infinity to plus infinity okay assuming a Cartesian coordinates system then this should be equal to one remember psi star psi dx will give me the probability of finding the particle between x and x plus dx and I integrate this probability for minus infinity to plus infinity so this gives you the probability of finding the particle anywhere from x is equal to minus infinity to plus infinity that must be equal to one and that something which determines a okay so actually a is found out by what we call as a normalization this process is called a normalization of the wave function so wave function just the solution of this will only give you psi and this a has to be obtained from a different condition which we call as a normalization condition in fact I will give you example of all these things taking particle in a box as an example and probably depending upon time one or two more cases. Now in general for a three-dimensional thing we can write this as psi star psi d tau is equal to one and this is entire space we generally write this as a d tau taking as volume element if I am talking of Cartesian coordinate system this d tau is just dx dy dz then this integral will be triple integral from x is equal to 0 to 0 or rather minus infinity to plus infinity y is equal to minus infinity to plus infinity etc etc if you are using let us say spherical polar coordinate then the d tau can be written as r square dr sin theta d phi and corresponding the limits have to be changed from r is r 0 to infinity and things like that okay. Also let us remember there is also an ambiguity because what you will be able to determine from normalization is only the value of a square because you know when you are taking this is a so you will take only the modulus of a square so you can determine only with an unknown constant phase factor unknown phase factor this I will give an example I think will become little more clear a little later. Now the fact that you can integrate this which is called square integrability it means you take a square which means take a complex conjugate and multiply and then you can integrate it from minus infinity to plus infinity this actually puts a necessary condition of the wave function. So wave function must be of a particular type okay only then this square integrability is possible that puts under certain condition the wave function what we call as has to be well behaved okay it means the it has to possess some conditions and one of these conditions is that it should be square integrable okay. Let me just mention I am not very sure whether I have mentioned that we are using the word wave function which we generally use as the solution of the Schrodinger equation okay wave function is nothing but the wave associated with the particle. So in quantum mechanics we call this as a wave function we have been talking earlier about wave packet so what we are talking of wave function is just the wave packet that is associated with the particle that is the way one has to look at it. I would also like to mention this particular thing this is also interesting point that normalization is strictly not possible for a free wave as the probability particle has equal probability of being found out in the entire space okay in that case one has to look at the problem in a slightly different fashion. So these are what we call as a free state problem about which we will discuss later there normally we do not normalize the thing because our interest in is fine finding out certain different things all together. So I am just enumerating here but all these points I will explain later when I take an example. So for example if you take a wave like that e raise power i k x minus omega t you take complex conjugate of that that will be a star assuming that a in general could be complex then complex conjugate of this term change plus i to minus i so this becomes minus i this will cancel okay then it will come just a star a and if you integrate from minus infinity to plus infinity you just get a equal to 0 okay which does not really make sense okay. So this in principle you cannot find out a by integration for a free particle about that we will discuss later okay what is our interest we do not really intend many times to normalize the problem I mean if at all you have to normalize it there are different ways of doing it I mean strictly speaking there is nothing like in finite way in fact we have said okay so we have to look at these particular things in a slightly different fashion. Now I am just giving a particular example a particular problem so let us assume that the wave function of a particular particle at t is equal to 0 is given by this particular equation I mean just a sort of a problem okay let us not bother about whether it is a solution of Schrodinger equation or not I mean I am just talking of purely from the point of view of giving you the idea of the probabilities okay so let us assume that there is a particle which is represented by a wave function which is given by like this okay I am let me repeat that I am not really talking about the Schrodinger equation just give you idea of the probabilities now a person performs a number of experiments on identical particles which are described by the same wave function at the same time which is at t is equal to 0 and he finds it to be 100 particles of them are found very very close to x is equal to 2a so I say infinitesimally small vicinity of x is equal to 2a so let us say between 2a and 2a plus dx I find 100 particles there see remember this is continuous variable so we cannot talk of exactly precise position so we say 2a and 2a plus dx I find 100 of these particles in how many of these measurements the particle would have been found in the same infinitesimally vicinity of x is equal to a means how many number of particles I would have found between x is equal to a and a plus dx where this dx is the same as in the earlier case so I know the wave function I have just found out that if I make a measurement between 2a and 2a plus dx I find 100 particles okay question is that if I know the wave function how many number of particles would have been found between a and a plus dx it is a simple problem let us just discuss it earlier so as we know the probability of finding the particle is given between psi x square dx okay so if I have to find out what is the probability of finding the particle between 2a and 2a plus dx I just put x is equal to 2a so if I put x is equal to 2a so p 2a da will tell me the find probability of finding the particle between 2a and 2a plus dx so here I put x is equal to psi x I have already been given which is e raise power minus x square by a square and when I take psi star psi psi square so this e raise power minus x becomes twice real function so I take just the square of this thing say if I take square it will become e raise power minus 2x square okay so this is minus 2 for x I have put equal to 2a okay so this just becomes e square into e raise power this is 2 square into 2 so this becomes e raise power minus 8 a square cancels out p of a dx would be given by exactly the same equation except that for x is equal to 2a I would have put x is equal to a so if I put x is equal to a I will get minus 2a square and now a square will cancel so I will get a square into e raise power minus 2dx so this would have been the probability of finding the particle at between a and a plus dx this would have been the probability of finding the particle between 2a and 2a plus x and I know that this is proportional to 100 so I have to find out what will be this number this is what I say p 2a upon dx upon p a is equal to dx so all whatever is the you know sort of coefficient that gets cancelled out so I will get e raise power minus 8 divided by e raise power minus 2 okay I do not know what is the probability of finding at I mean how many particles would have been found at p a so let us assume that this is a m so this e to power minus 8 has to be proportional to 100 okay proportionality constant as I said cancels out so I can just calculate p calculate m m will be equal to I take 100 into 10 power 6 which is approximately 40,000 so 40,000 particle would have been approximately found between a and a plus dx okay remember this is an exponential function so this number is extremely exceedingly large so this is just to give an idea how I am using wave function to find out the probabilities okay there will be some more questions which I think some of the TS will take care of now next thing that we define is what we call expected value expected value sometimes this name expected is somewhat misleading see but why I mean in statistics also we use expectation value this generally means the average value so at least as far as this course is concerned we just take this as a mean value so as we have just now seen in quantum mechanically in quantum mechanics there is always a probability associated with the result obtained after measurement big measurement there is certain probability of obtaining a particular result now this expected value is the average value that one would get after a very large number of measurements were made on identical system with same wave function so like example which I have given you okay you can if I have to find out the mean value expected value of the position it essentially is mean value position if I have let us say 1 million particles I make the measurement of all on all of them all of them are given being given by the same wave function I make the measurement on all of them of the position I will not find them to be exactly identical okay I calculate the average of all of them if I calculate the average of all of them then this is what I will say the expected value of position so expected value is defined as the average and let me repeat this thing which I have been telling number of times is an average value that one would get after a very large number of measurements are made on identical systems okay with the same wave function we must have exactly identical particles with exactly the same wave function and would have made should have made multiple measurements there and then only I can calculate what will be the expected or mean value of any particular dynamical variable at the moment I have talked to only about position but now I will give you if I have to find out let us say mean value of momentum or mean value of any other dynamical quantities how do I go about it so as I said the mean value of x is just this because psi star psi gives you the probability there is a reason why this x is put in between psi star and psi of course here this is a product so I have put this x here also or I could have put x here also do not have any difference but there is a reason because many times when I give a not many times but when I give a general definition of the expected value okay we have to put everything between psi star and psi if I take means of x square expected value of x square instead of x I will put x square now I said let us try to find out any other dynamical quantity I may not be just interested in the mean value of position I may be interested in the mean value of momentum for example so we have talked about the operator okay we have said I mean we will give a general definition about the operators when we talk about the postulates of quantum mechanics today okay we will say that for every dynamical variable there is a operator associated okay now this operator if I know then the mean value of that particular dynamical quantity the operator of which is let us say g is given by this particular thing okay and this is given by of course integral has to be done over the entire space psi star g psi d tau remember d tau is the volume element which I have already talked about it earlier and here is very very important to put g here because g has to operate only on psi not on psi star okay so therefore the general definition is psi star g psi if I have to find out the mean value of any dynamical quantity first I have to look at the operator this operator must operate on psi xt multiply by psi star then you will get the mean value of g okay I must mention that these operators are always chosen in such a fashion or have to be in such a fashion that this mean value turns out to be real though your wave functions are in general complex even the operators in general can be complex okay but on the other hand we said that mean value has to be real because this is something which is measurable okay anything that we measure must turn out to be real okay so therefore when we measure these things must turn out to be real and we can show in fact when we go to the dynamics we have to talk about the postulates okay we will talk about this particular aspect that an operator has to have this particular property that its mean value must always turn out to be real all right the other thing is that one people always ask the question why we define it in this particular fashion okay why is a very difficult question here because you know these are I can always raise my hand and say that these are the postulates of quantum mechanics okay there is nothing that I can do about it but again I would like to sort of give some justification these operators are defined so as that they match with the classical mechanics this particular thing statement I have made even earlier that we know that for macroscopic particle quantum mechanics is not necessary classical mechanics works so in the limit of large energies or particles being large this must end with the classical mechanics okay if you look at take an advanced course on quantum mechanics this is generally shown that this is the way operators have to be defined in order to have consistency with the classical mechanics. Now for example now we have already taken the defined the operator corresponding to x component of the momentum which is minus ih cross del x so if I have to find out the momentum mean value of the momentum particle and if I know the wave function then I have to apply minus ih cross of course ih cross is just a pure multiplication and it is also a constant so minus ih cross I have taken out of this particular integral sign and del del x has to operate on psi xt so this particular expression will give me the mean value of momentum if I have to find out the mean value of px square then I have to operate px twice in fact I have also given you the operator corresponding to px square when we are talking about the Schrodinger equation yesterday it will be minus ih square del 2 del x square again I have taken minus ih square out of the integral sign this being a constant okay this becomes just del 2 psi del x square remember again that this operator operates only on psi psi star is off outside the gambit of this particular operator so you must operate this thing and then multiply by psi star then integrate of course here we are talking about one dimension so I have put dx otherwise in general it has to be d tau and the limits have to be appropriately chosen depending on what type of coordinate system that you are talking about okay you can always show that the expected value of the momentum will turn out to be real this I have mentioned even earlier just now that operators are always chosen such that now the any observable quantity has to be real because mean value of momentum has to be real okay therefore this must give you a real value the operators of dynamical variables are always such that their expected value turns out to be real this is what I have mentioned now there are certain other aspects which I will try to discuss about the difference between you know time dependent time independent Schrodinger equation before I start going into little more interesting aspects about the wave function so let us assume that we have solved we know a particular problem where vx was actually purely a function of x and there was no time dependence involved and I solved the Schrodinger equation and obtained phi x let us try to find out the mean position of any quantity which is just a function of x let us say x or just x square or x cube or x square plus x or whatever it is something which is only a function of x so let us say there is a function of fx okay it could be just simply x okay and I want to find out the mean value of this particular thing using this particular equation and this particular solution of the Schrodinger equation now remember whenever I am writing this solution okay this is not the entire wave function this wave function has to be multiplied by erasova minus iet by h cross so let us try to find out the expected value of f of x this is what I will write remember expected values are always written in the angular bracket this means the mean value expected value I apply the definition psi star f of x psi xt because this is a purely a function of f x this is a purely a multiplication operator I could have written anywhere here it does not matter now I have already known my phi this I have formed by solution of time independent Schrodinger equation I realize that psi must be given by phi of x erasova minus iet by h cross this is the entire wave function now here it has to be psi star so I must take complex conjugate of this particular quantity let us take in general I mean for all that psi could be complex so I put psi star x here which is the complex conjugate of psi then complex conjugate of erasova minus iet by h cross minus has to be changed to plus so this becomes erasova plus iet by h cross okay now everything is multiplication so you can see that erasova plus iet by h cross will cancel with erasova minus iet by h cross so this term time dependent term will cancel out and what you will get is phi star x f of x phi x quite interesting because you find first of all that there is no time dependence here this fx is not time dependent the mean value of f is not really time dependent okay second thing is that for calculating this mean averages I need not have bother about this time dependent term at all I could have just use phi star I could just use the solution of time independent Schrodinger equation and just retain phi star fx phi x totally ignoring the time dependent part now whenever we have a situation like this where this particular time dependent part will cancel it out okay which will happen with a particular solution like this okay then we call such type of states as stationary state as I say this is time independent these are called stationary state whenever they have expected value for example of fx turns out to be time independent we call it stationary state also time dependent part of the wave function is of no consequence in such a case it should not be now is of no consequence in such a case all right now obviously when I say these are stationary states you will always think that probably there are certain states which are non-stationary state okay yes they are and this is what I am going to discuss just now okay let us first try to solve the problem about a particle in rigid box which probably all of you are knowing so let me try to do a little bit hurriedly and then after that I will start discussing about these non-stationary states because that forms a very very interesting part of the I think probably that I will do in the afternoon lecture but let me just quickly go through particle in a rigid box problem which as I say is a very very simple problem the simplest problem in quantum mechanics and let us just try to understand this particular problem okay then I will use the wave functions of this particular problem and then try to define what I mean I will come back to stationary states and non-stationary states at that particular time okay so let us assume that there is a particle okay remember in Schrodinger equation we always talk in terms of potential energy okay I think some people have asked this question so let me again tell that V is potential energy but somehow when we talk in quantum mechanics we are used to calling it potential I mean it is true that when we are talking of potential potential is different from potential energy in the entire electrostatics we talk of potential as different as potential energy okay but somehow it so happens in quantum mechanics we have used to call it we are used to call it as a potential so whenever I am saying potential in the sense of Schrodinger equation we always mean potential energy okay so potential energy of particle is 0 between x is equal to 0 and x is equal to l and is infinite everywhere else we had to find out what is the solution of Schrodinger equation or basically if a particle is contained in this what we call as a rigid box in a rigid one-dimensional box what will be the energies that this particular particle could possess that is what we are let us try to discuss this particular aspect particle is between x is equal to 0 and x is equal to l of course we can show which I will not show at this particular time probably we can do it a little later if we have time that whenever potential energy is infinite the only solution of Schrodinger equation that is possible is that wave function is 0 it means the probability of finding the particle beyond x is equal to l and before x is equal to 0 would be 0 so wave function will be 0 here wave function will be equal to 0 here the wave function will be existing only between x is equal to 0 and x is equal to l so particle can be found only within this particular box it cannot be found anywhere outside this particular box of course remember these cases are somewhat idealistic but at this moment the idea is to use the Schrodinger equation and try to understand the quantum mechanics rather than going into more realistic problems okay I mean as we always say I re-quote my my own self that in physics we always start with the simplest problem okay even if it is not non realistic okay enough then we try to make it as realistic as possible later on okay so as I said that this wave function phi I am calling it wave function but essentially it means phi okay it is 0 for x less than 0 for x greater than l as I have just now mentioned that this is the only solution possible which I am not showing which can be shown little later so the equation time independent Schrodinger equation has to be written only between x is equal to 0 and l and that is what I have written and in this case v is equal to 0 so e minus v the v has become 0 so the equation becomes just like this and this becomes an equation of this particular type now e is supposed to be the energy of the particle which is not supposed to be dependent on x so therefore this is a constant it is very easy to solve this type of equation and the general solution as you know is a linear combination of a sin term and a cosine term you could have also used an exponential term the solution would have been exactly identical but just let us use a simple sin and cosine term so the general solution of this particular equation is a sin kx plus b cos kx is a second order differential equation so you will get two unknown constants one is a another is b okay now what you can do you can substitute this particular equation back here you will really find that there is a solution now this thing is a solution here this particular k will turn out to be equal to k square is equal to 2 me upon h square okay this equation here you will be obtained when you substitute this equation in the Schrodinger equation here in this particular I am sorry in this particular equation when you substitute it here then the relationship k square is equal to 2 me upon h square will be obtained this is rather easy to remember because this tells that e is equal to h square k square by 2m okay using normal standard classical mechanics h square k square upon 2m means kinetic energy because potential energy is actually 0 here so e has to be only kinetic energy so it is easy to remember it in this particular way okay here k has to be real okay in this particular case because we are talking of only positive energies because potential energy is 0 here so I am talking of the energy states which are only about this particular energy now we have to apply boundary conditions and the first in bond I have not talked yet about the boundary condition the first boundary condition which actually pertains to the fact that well wave function must be well behaved is that wave function must always be continuous across the space wave function cannot show discontinuities okay because eventually wave function is related to something which we can observe okay and all realistic things are expected to be continuous okay so if they are sort of continuous the probability of the finding particle has to move constantly as a function of x that they cannot change as a function of x okay now there are two boundaries here at x is equal to 0 and x is equal to l okay as far as function here is concerned a combination of sine and cosine functions will always be continuous between x is equal to 0 and x is equal to l but I must insist that it should be continuous with the wave function on this end and must be continuous with the wave function on this particular end so this is these are the two points x is equal to 0 and x is equal to l where wave function sort of changes therefore I must insist that the wave function is continuous at this edge and at this particular edge this is what I am doing I remember wave function is 0 here wave function is 0 here so I must have phi of 0 should be equal to 0 similarly phi of l should also be 0 because I know that phi for values of x less than 0 is 0 and phi for value of x greater than l is also 0 so at this particular end also they must be 0 you will appreciate this almost like applying boundary condition for vibration of a string that the displacement of this string turns out to be 0 okay if I put phi 0 to be 0 then I substitute in this particular equation which I have just now obtained a sine kx plus b cos kx okay so I put a sine 0 plus b cos 0 sine 0 is anyway 0 so this goes away so this becomes b is equal to 0 because cos 0 is 1 so when I say b is equal to 0 okay if this equation has to be satisfied then the only way is that b should be equal to 0 and this is also obvious because cosine function is one function which at x is equal to 0 is not going to be 0 and I want the wave function to be 0 at x is equal to 0 this is only possible when there is no cos term present there is only a sine term because sine term is the only term for which at x is equal to 0 this particular term would give you 0 so it means there cannot be a cosine term in the wave function so I put b is equal to 0 but now I also want phi l to be 0 okay now out of this my wave function this term has disappeared all that has happened remained is a sine kx so I put a sine kl and put this equal to 0 sorry I am being a little faster here assuming that this particular particle in the box thing has been sort of well studied and well taught by probably all of you now I want a sine kl to be equal to 0 if this is equal to 0 then there are two possibilities that either a is also equal to 0 if a happens to be 0 b has to be anyway 0 it means the wave function is entirely 0 wave function is entirely 0 it means the probability of finding the particle in the box will be 0 because wave function is 0 it means the particle does not exist but I have been told that there is a particle inside a box look at the arguments okay we have already been told that there is a particle in the box and if this particular particle has to exist within the box its wave function cannot be 0 it means I cannot accept the fact that a is equal to 0 okay if a cannot be 0 then sine kl has to be equal to 0 and that is only possible if kl turns out to be equal to n pi now we have already seen the value of k which is given k is actually related to the energy of the particle so I just write this as k as under root 2 m e upon h cross from that equation l must be equal to n pi which means that in case this condition has to be satisfied it means in case the wave function has to well behave in case it has to disappear at x is equal to l then only possibility is that energy takes one of these values where n is an integer which has to be greater than or equal to 1 about this I will discuss little later okay which has to be greater than 1 or greater than or equal to 1 okay only when energy takes one of these values the wave function will be well behaved and the second boundary condition will be met okay essentially it means that energy must possess one of these values where n is an integer then only it is possible that you know the boundary conditions are met and see we have got such a beautiful way the quantization of energy which at the time in the earlier cases you know we are sort of putting only phenomenologically you know there is no logic from which this quantization is coming this boundary condition has imposed this particular they have sort of quantized the energy levels is coming in such a natural fashion by application of the boundary condition. See I must insist that n has to be greater than equal to 1 okay because if n is turns out to be 0 then energy turns out to be 0 if energy turns out to be 0 because basically it means k is equal to 0 and k is equal to 0 again means the wave function is 0 and I said I am not going to accept the fact that wave function is 0 because I have been told that the particle exists so n is equal to 0 state is not allowed it is the least value of energy is corresponding to n is equal to 1 okay this is what we call as a ground state energy the ground state energy is the lowest possible energy available to the system if this comes from a more general principle that any particular particle which is bound in a limited region of space okay its energy is must be quantized and also the lowest energy cannot be 0 this is generally called as 0 point energy the lowest possible energy. Now there are some certain comments we had two unknowns a and b but strictly speaking we had three conditions including the normalization conditions here I have told that this a cannot be determined by another method other than normalization alright so we had three conditions with only two variables now in case of all the three conditions has to be met this will limit the values of k to certain discrete values this is the additional equation which is the one which is responsible for quantizing the energy state the second thing is that the value of n can only be positive and non-zero this we have discussed okay because these are actually solution are stationary waves so actually k can only be always be positive okay there is no meaning of negative k then I say the lowest energy is non-zero which we call as a 0 point energy okay it means if you put a particle in a box its lowest energy is always going to be 0 some people ask a question suppose I put a small ball in let us say a container does it mean that it has a lower I mean its lowest energy is 0 okay according to quantum mechanics yes okay and in this case because potential energy is 0 so that energy must be kinetic so does the particle move let us just try to calculate the lowest energy corresponding to n is equal to 1 okay and let us assume that there is a electron which is kept in a box of one angstrom then you can calculate these numbers all our fundamental constants this will turn out to be 37.6 electron volt so if I put an electron in a box of one angstrom the lowest energy that this electron will have is 37.6 electron volt as you can say this is quite huge one angstrom is typically the distance between the atoms okay so I can always imagine a situation where I mean this is sort of realistic situation where the electron could be in a box of one angstrom and in that case the lowest energy state is 37.6 electron volt which is a fairly large energy but on the other hand if I put a marble you know marble which we the children play which has a weight of let us say 0.01 kilogram and I put in a 0.1 meter box these are typical macroscopic values then its kinetic energy lowest energy will turn out to be 5.488 into 10 power minus 64 joules means essentially it will take 10 power 20 years to move one millimeter so it means if I really put a marble in this particular box and let us say okay I am putting it and therefore it must have a kinetic energy according to quantum mechanics yes it will have a kinetic energy okay but it is so small that will take one millimeter movement in 10 power 20 years obviously such motions are never possible to observe okay so this particular quantum mechanical effect will never be visible in this macroscopic things but will be definitely visible when we are talking at microscopic level when we are talking of particles as tiny as electrons okay and when we are talking of boxes which are typically of the order of inter atomic spacing so therefore quantum mechanics is necessary when I want to describe the behavior of electrons at the solid okay it is not necessary when I want to describe the motion of a marble or a motion of a cricket ball because there the quantum mechanical effects will essentially be negligible if I want to normalize I will just hurriedly go through these transmissions because purely mathematics and I will anyway upload these things on Moodle okay so in order to normalize I must normalize this and I say normalization for calculating the probability again the time dependent part will be of no consequence because that will be canceling out so I have to just normalize only the phi star so if I normalize it this must be equal to 1 if I evaluate this particular integral eventually you will get a square L by 2 equal to 1 so this is what I will be able to determine so you can have multiple possibilities of a so I can always write as a is equal to e to the power i theta under 2 by L this will be the most general solution of this particular equation which I can determine it but normally again conventionally we always take theta equal to 0 and we write a is equal to under 2 by L so this is what we call as the normalization constant which is we write we take the phase angle to be 0 this gives me a is equal to under root 2 by L which gives me the wave function to be this which is lying only between x is equal to 0 and x is equal to L and this will be 0 elsewhere all right now if I plot this particular thing this is what I will be getting this is the wave function and the probability is psi star size and this is a real so it has to be just square I have roughly plotted of course this is not exactly like this because this has to be squared but they look identical so please excuse me for my bad no drawing so essentially it says that if the particle is in ground state which is the lowest energy state then if you take a very large number of boxes and try to make measurements you will find maximum number of particles to be found at the center of the box but if the particle is what we call as the first excited state energy state just after the ground state you will hardly find any particle to be in the center of the box most of the particles will be found either here or here okay on the other hand had it been really a classical problem okay I mean classically do not expect these type of thing these are very very surprising and shocking okay as far as wave mechanics is concerned so I think I will stop here our time is over I today unfortunately I do not have time to take any questions from this thing okay I will suggest that please post your questions okay in the afternoon class if I have some time then I will try to answer questions from this session as well as from the next session