 So, good morning. So, we are going to start today's session. First hour, this first slot is actually to do some problems on quantum mechanics which was actually done by Prasad Shri Prasad. The theory was done by Prasad Shri Prasad. I will do some problems based on the topics that he covered in the last week. After I have a lecture where we will continue with the magnetism. So, right now there is no magnetism, we will be just doing the problems of quantum mechanics. I think there is some problem, the tutorial sheet, the problem sheet is ready. Now, we can upload it. I hope you can see that. Yeah, it is you can download it now. Please download it because I am going to do those problems. All the problems I will be doing it here right now. Please have a look at the problems, all problems on quantum mechanics. I will anyway I will read it out. The questions I will first read it out, then I will try to solve it. So, the problems have been selected in such a manner that it covers most of the things that are that were done by Prasad Shri Prasad on this topic. So, let us start. See the first problem is actually you can hope you can read it. It is a particle in a one dimensional box is in a stationary state whose wave function is given by 0. The wave function is 0 for x less than minus a. It is having a value 1 plus cos pi x by a for the region between minus a and plus a and again it is 0 for the region greater than a. So, this is a question. The particle whose stationary state is represented by this wave function. Now, the questions are of course, it is given that a and small a and capital A are constants. The questions are when is it a well behaved wave function that is number 1. Number 2 calculate the value of a the constant a has to be evaluated of course, in terms of small a. Next is again very important part verify that the uncertainty principle is true in this case that means, you have to show that the uncertainty in x times the uncertainty in p is greater than h bar by 2. So, that is third part. And the fourth part is find out the classically allowed region. So, that region where the particle is allowed to move even classically quantum mechanically there is a difference as you have seen. So, these are the 4 parts of the question in this case. So, you would have seen what is meant by a well behaved wave function. A well behaved wave function the first thing is that it must be square integrable because the integral of the square actually give you the probability density which is very important. And it has to be single valued it has to be continuous the first derivative also also must be continuous. So, that it is fitting into the Schrodinger equation. So, you can see all these conditions are satisfied as we are going to see again all these conditions are satisfied. And hence the first part of the question whether it is well behaved or not it is a well behaved wave function. Now, to find out the constant of course, in the region 1 and region 3 anyway it is 0. So, nothing to be done. So, our region of interest is simply minus a to plus a. So, to find out the constant which actually acts as the normalization constant here to find it out the only way is simply to calculate the probability density and equate it to 1 that will give you the value of a. So, how does we how do we do that? So, as we know. So, the probability density in general is the psi x t square. So, this in our of course, the x this in our case everything is one dimensional. So, it is the integration is only with respect to dx. So, in our case this can be written since the region outside of this that is minus a to plus a is the wave function is 0 integration from minus infinity to plus infinity it is simply minus a to plus a and I can substitute this. So, that becomes this a square times 1 plus cos pi x divided by a square dx this can be written as of course, a square is a constant it will come out this I can expand. So, I have expanded this 1 plus cos pi x by a the whole square and now I can integrate all these three terms between minus a and plus a. So, that I am sure you will be able to do please work it out the answer turns out to be. So, this has to be equal to this integration integral will give you 1. So, this has to be 1 because of the normalization. So, that once I substitute I am integrate and substitute I get the value of a to be please verify you have to integrate these three terms between minus a and plus a and equate that to 1 because of the normalization that gives me a square and from that I will find out my a capital A in terms of small a which works out to be 1 by square root of 3 I am just avoiding these steps I am sure you will be able to integrate and verify that this answer is correct. If you want you can take couple of minutes and try to work it out these integrations are not very difficult cos I hope you know how to integrate it I am sure you will be able to do that cos of course, there is no problem. So, that gives you the normalization constant that is appearing in this wave function. Next part is to verify verifying that is as I mentioned verifying the uncertainty principle in this case of course, this is true in any case, but so here also we are taking this example to show that the uncertainty principle is actually true. So, one has to show this I am sure you have seen in the lectures how to get delta x that is uncertainty associated with the x and then associated with the p that is a in this case it is a x component of the momentum that is a full momentum here because this problem is one dimensional. So, we can find out first thing is to find out delta x we know that this can be written as the expectation value of x square minus expectation value of x the whole square square root of. So, this is the definition of delta x and similarly for delta p is p or in this case is p x is nothing, but p. So, this is true and hence I can to find out delta x I have to find out the expectation value of x square I have to find out the value of expectation value of x and then substitute. So, that I can get delta x similarly I can calculate my delta p by knowing the expectation values of p square and p and remember p is in our case it becomes. So, let us try to find out one can first find out expectation value of x. So, this can be defined as in general it is minus infinity to plus infinity, but in our case as I mentioned everything is over within the range between minus a and plus a. So, the x is operator remember in this case. So, this is a complex conjugate, but in our case everything is real or the wave function is given as 1 plus cos by x by a it is a completely real function. So, this can be written as x times simply the probability density times x will give you this integral. So, this one can find out by actually substituting as we have done earlier I can substitute the square of the wave function that is given I can multiply with x and so on and I can do, but remember you do not have to really do that looking at the form of the function one can actually see whether this is going to survive or not. The integral is from minus a to plus a and the wave function actually is even and now your x is odd. So, you the total thing the integrant actually is going to be odd in this case and when you are integrating from minus a to plus a this will actually give you 0. You can work it out you can convince yourself that this is actually 0. So, the expectation value of x is 0 in this case. You do not have to really work it out if you want you can work it out no problem, but it is not needed many of these cases you should see whether the integral is going to survive if it is by arguments if you can show that it is not going to survive you do not have to waste your time, but still at least in one case if you want to really try please try it and you see that it is actually 0. Then I will go ahead and find out in the similar manner expectation value of x square. So, this is again in the similar manner I can write integral minus a to plus a x square in place of x now it has become x square because I am trying to find out the expectation value of x square here you have to work it out this is an even function. So, you have to work it is not going to 0. So, you have to work it out you have to substitute. So, that gives me. So, expectation value of x square is minus a to a. So, x square now I have this a square 1 plus cos phi x by a dx. So, as we did earlier in the probability density or the normalization condition calculation you have to expand this and then multiply with x square and doing this integration you will get a little complex value this is this works out to be a square divided by 6 pi square to a square divided by 6 pi square 2 pi square minus 15. So, this is the expectation value of x square. So, I have the expectation value of x which of course is 0 in this case and the expectation value of x square which means that I am able to write down what is expression for the uncertainty delta x just by taking the difference between the two. So, delta x therefore, delta x as I have written the formula works out to be. So, one part of this uncertainty principle is calculated. So, delta x is given by this. Now I have to calculate the delta p for which I need to know what is my expectation value of p and expectation value of p square. So, let us find out expectation value of p. So, this again psi star x t when we have minus i h bar d by d x psi x t d x you can actually here also you can use an argument to find tell that this actually will be 0. The reason is you see as I mentioned the wave function is completely real you have an imaginary part here in the operator and hence the whole thing is going to be imaginary if it works out and the expectation value has to be a real quantity. So, which means that this has to be 0. So, the expectation value this is the problem which is going to happen in other cases also for example, in the discussion of magnetism we will have this problem when your wave function is going to be real the expectation values of the various components of angular I mean in this case some momentum here later the angular momentum will be 0 because there also this p is coming into picture. So, you can work it out and this goes to 0. So, the expectation value of p is going to be 0 then calculating the expectation value of p square in a similar manner. So, this becomes so minus i h will come to i. So, it becomes h bar square minus h bar square then this gives you minus a to a psi x of course, t I do not have to worry about it and then this becomes d square psi x divided by d x square d x this of course, you have to differentiate this once and once again. So, it becomes cos becomes sin and again it becomes cos. So, the integration will give you pi square h bar square divided by 3 a square. So, I have found out the expectation value of p and using the same method I calculate the expectation value of p square. Now, with these two as we have seen earlier we can find out the uncertainty in p that is delta p which is given by pi h bar 2 3 a. Therefore, the product of the uncertainties delta x delta p is when I substitute these two values I get h bar divided by 3 square root of 1 minus 15 divided by 2 pi square which is definitely more than h bar by 2. So, you can see that you are multiplying with p pi and this is going to be more than h bar by 2. So, this is proved in various situations like harmonic oscillator all other examples also it is done. So, in this case also this works out and you prove the uncertainty principle to be valid in this case also fine. Now, coming to the next part that is fourth part the question is find the classically allowed region. So, this anyway I have put a one more problem to discuss this classically allowed region you know that in quantum mechanics even in the region where it is not allowed classically you will have certain probability density non-zero probability density or the wave function will be non-zero in those classically forbidden regions also. So, what is then by classically forbidden region? Classically forbidden region I will talk about forbidden region is where your E remember E is the total energy actually is less than the potential energy. E is the total energy E is the total energy V is the potential energy and K is a kinetic energy. So, to just know the boundary between classically allowed and forbidden regions what we have to put is we have to put the condition that E is just equal to V that means your K plus V is equal to V. So, the boundary between the two regions one is classically allowed and other thing is classically forbidden can be found out by equating the total energy to be equal to the potential energy. So, that gives you K plus V equal to V which means that this is the region which is defined by K equal to 0. So, K is implies our second derivative because K is related to P square by 2 m. So, the second derivative so this in this particular case. So, if you find out the second derivative is satisfied at x equal to plus or minus a by 2 since the condition that is going to happen is second derivative is minus pi square by a square a cos pi x by a equal to 0 is a condition to be satisfied. From this I can find out that this is going to be 0 or this equation is going to be satisfied at x equal to plus a by 2 and minus a by 2 that means classically allowed region is between minus a by 2 and plus a by 2. So, that is a probability density there is a wave function in the region outside this also that is a difference between the classical motion and the quantum mechanical motion as you have seen in the lecture I mean different lectures. So, this is again a proof where you see that you have classically allowed region and the classically forbidden region and how the difference happens how one can find out what is the classically allowed region in a given problem how to find out what is the region where it is classically allowed where it is not allowed. Towards the end I will take another example where again I will show the similar idea of classically allowed and classically forbidden. So, I think that finishes the first question. So, there are there were some comments or some questions about use of time dependent Schrodinger equation also. So, usually we do most of the problems on time independent part time independent Schrodinger equation is very much used in the I mean lectures that were given to you in the last week. But of course, it is true that many cases one has to use the time dependent Schrodinger equation in the light of that I have taken one problem where we have some kind of a time dependence very simple form it is there. So, that is what is coming in the question number 2. Question number 2 is consider a one dimensional motion of a particle confined to a region between 0 a. I hope you are able to download the question paper the problem sheet and the wave function is given by sin pi x by a this is a special part, but it has also got a temporal part which is given by e to the power of minus i omega t. The question is find the potential v and the probability of finding the particle in the region between a by 4 and 3 a by 4. So, here it is a time dependent wave function is time in dependent unlike the earlier case there is a time dependent factor here. So, one has to use the time dependent Schrodinger equation. So, which of course is time dependent Schrodinger equation equal to plus v of course, take it as v x and a function of x and t. So, operator then you have psi x t. So, this is the time dependent Schrodinger equation for the wave function given by this. Now, what I can do is I can find out this part this is basically that time derivative of the wave function which can be easily found out because this will be a constant in that case. For finding out this term that is the first term on the right hand side I have to do the differential twice with respect to x. So, that this part becomes a constant. So, I can find it out also. So, that gives me. So, I can find out if I find out of course this is x and t I am not writing. So, it works out to be minus i omega e to the power of minus i omega t is there. So, differential will give me minus i omega. So, the whole thing will become it will just get multiplied. So, it becomes x t. So, minus i omega times the wave function that is the time derivative. Similarly, I can do the spatial derivative that derivative with respect to x. So, once I can do and I do once again. So, that gives me dou square psi by dou x square is worked out to be minus pi square divided by A. You remember this is pi by A will come twice and there is a minus sign because sign becomes cos and then again becomes cos cos becomes sin again. So, that is minus sign that is coming because of that two differentials with respect to x will give you minus pi square divided by A square and the wave function simply repeats. Of course, this is psi x t everywhere. Now, these two things I can substitute in the time dependent Schrodinger equation which will give me i h. So, let me write on substituting these values in the time dependent Schrodinger equation. I will get i h bar psi omega equal to h bar square pi square divided by 2 m A square plus v x t. Now, so the wave function has got cancelled everywhere. So, I have this. Now, you can see there is no time dependence here there is no time dependence here. So, that means, v must be a function of v x only there is no time dependence for the potential. So, the v x therefore, v x is given by h bar omega. So, that is h bar omega is coming because of this h bar omega minus h bar square pi square divided by 2 m A square. So, this is the way one can find out what is the potential that is subjected to this problem. So, what we do is we find out the time derivative, we find out the spatial derivative, second derivative, substitute in the time dependent Schrodinger equation and you can separate out the potential in this form. So, that is the answer to the first part. So, the second part of this thing is what is the probability density in the region between A by 4 and 3 A by 4. So, what we should be done? So, the probability density. So, the probability of finding the particle in the region between A by 4 and 3 A by 4 can be written that is a probability is A by 4 3 A by 4. So, you have to write psi x t square dx since nothing is mentioned about a normalization. So, the region full region is 0 to A psi x t whole square dx. So, here it is between A by 4 to 3 A by 4, this is the full region of the problem that is 0 to A the probability density integrated. I can substitute I know the value of psi x t, I can substitute that gives me 2 plus pi divided by please check the values this is around 0.82. So, the considerable probability is existing in the region between A by 4 and 3 A by 4 as you can see it is 0.82. So, please substitute and make sure that this is right. So, these are two parts one is of course, the probability density as was told to you in the lectures the time dependence does not come into picture because it gets cancelled. So, you do not really realize it, but in the other case you see that one can use the time dependent Schrodinger equation by finding out the time derivative and the spatial derivative and then one can find out what is the potential in the problem fine. Next question is very important in fact, it is I understand that lot of discussion was there during the lecture part on that this is question 3 this is basically regarding the collapse of the wave function. So, you have a super post state initially which is a super position of various states and as told to you during the lecture when you make a measurement the wave function the super post wave function collapses to a particular value and that continues later on as well as the system is concerned. So, the collapse of the wave function is very nicely demonstrated in this problem with a certain few other ideas. So, it has again many parts. So, let me read it out again a particle and remember this is the problem which we are doing in the context of the particle in an infinite square well or the rigid box problem. So, where we know the wave function we know the energies both are needed to do this problem. So, let us recollect it once first I will let me read it out the question and then we will talk about it. A particle of mass m is confined to a 1 d square well infinite square well of course a side a its wave function at a given instant is given like this it has got 3 terms and then various questions are asked. So, let us remember the wave function for a particle in a box infinite square well the wave function is given by square root of 2 by L right. So, this n becomes a quantum number L in this case is A. So, A is the box side of the box. So, this is a normalization constant sin n pi x by A is the wave function normalized wave function. The energy is n square pi square h bar square divided by 2 m A square again n is a quantum number this is the only quantum number that is there in the system because it is one dimensional box we will see what happens if it is 2 dimensions later on. So, E n is given by this these 2 are needed to proceed further. So, the given wave function is given let me call superpose state is psi x t is given by I by 2 2 by A sin pi x divided by A that is a first term and there is a constant times sin 3 pi x divided by A minus 1 by 2 square root of 2 by A sin 4 pi x by. So, this is the superpose state that is given to you as I have written down you have the wave function you have the energy. So, looking at this we can easily find out that this first term corresponds to n equal to 1, second corresponds to n equal to 3, third corresponds to n equal to 4 this is kind of an artificially prepared system superpose state. And as you know any superpose state can be written in terms of the again functions of energy in this case and the corresponding coefficients the square of that will give you the probability of getting that particular energy this corresponds to E 1, this corresponds to E 3, this corresponds to E 4. So, this corresponds to E 1, this corresponds to E 3, this corresponds to E 4. So, the constant that is coming here the square will give you the probability of getting E 1 and similarly for this I remember there is a small problem here because there is an I here which of course, is allowed. So, one has to be careful in dealing with the taking the probability of getting the first energy again value first result. So, before we proceed what we can do is first question is what is A? The first part is value of A, how do you find out the value of A? Remember these probabilities when it is added one should get 1 because that is essentially the normalization condition stated in a different sense. So, these coefficients should give you 1. So, that condition can be used to find out what is unknown A here. So, this can be written. So, first what I will do is I will write to find it out I write psi xt in terms of this standard of form, standard of form is with a 2 by A normalization coming. So, let me write. So, this I can write it as I by 2 this is exactly in the standard of form. So, I will write it as psi 1, this I will write because this is the total constant. So, I will take A prime and then I will write it as psi 3 which is corresponding to n equal to 3 minus 1 by 2 this is 1 by 2 then 2 by A sin 4 pi x by A will give me corresponding psi 4. So, now, this is written this super post state is written in terms of I by 2 psi 1 A prime I am not taking A I am taking A prime. So, that I am taking separating this square root of 2 by A factor it becomes psi 3 minus half that is here and this whole part becomes the psi 4 as I have shown here. So, this corresponds to n equal to 1, n equal to 3, n equal to 4 or this corresponds to energy E 1, energy E 2, energy E 3 and energy E 4. Now, the condition is demands. So, thing is if I take this as this is a condition that has to be satisfied where these are the constants remember when this is a complex number one has to take the complex conjugate of this. So, that will give me. So, the second term and third term there is no problem. So, these are all real in this for example, it becomes 1 by 4 whereas, here it becomes A prime square, but here one has to take the complex conjugate. So, this gives me the complex conjugate is I by minus I by 2 times I by 2 plus A prime square plus 1 by 4 must be equal to 1. So, this is the coefficient of the first term C stars C n star C n or C 1 star C 1 this is C 2 square modulus this is C 3 square modulus. So, this gives me from this. So, of course, this I can find out this gives me A prime equal to square root of 1 by 2 because this is minus I square and I substitute I get A prime is this and we know A is A prime and there is a factor of square root of 2 by A because that is a normalization factor. Therefore, this will be equal to because I have taken this 2 by A in the wave function normalized wave function path. So, to get A this has to be multiplied here. So, that gives me square root of 1 by. So, I found out the first part that is A is given by square root of 1 by A. Now, since I know all the three coefficients I can find out the probabilities. If I make a measurement the probabilities of getting energy E 1 is given by the square of the first one that is actually 1 by 4 because that is minus I by 2 into I by 2 is 1 by 4 because this is C 1 star C 1 this because this is a imaginary number. So, it is giving rise to 1 by 4. So, the probability of getting the value as the energy as E 1 if I make an energy measurement is 1 by 4. Similarly, probability of getting energy as E 3 that is a corresponding to the second term that is E 3 is the number which is corresponding to the second term which of course is A prime we have got it as A prime we got it as 1 by 2. So, this is 1 by 2 square. So, it is square root of 1 by 2 square that is 1 by 2. So, the probability of getting E 3 is 1 by 2. Similarly, for probability of getting E 4 that is last one is our coefficient was minus 1 by 2. So, it is minus 1 by 2 minus 1 by 2. So, that is 1 by 4. It is again as real number just a square. So, the total probability as you can see is 1 by 4 plus 1 by 2 plus 1 by 4 which is 1. So, it is right because anyway this is following this earlier step. So, there is no surprise. So, it is working out alright. And now we know what is the probability of getting a value of E 1 as a result of energy measurement. What is the probability of getting E 3 as a result? Similarly, what is the probability of getting E 4? You can see that the probability of getting E 3 is double compared to that of E 1 or E 4. And we know the values we can find out what is E 1 the value of E 1 as I mentioned n square is 1 square pi square h bar square divided by 2 m a square E 3 of course is 3 square pi square h bar square by 2 m a square E 4 is 4 square pi square h bar square divided by 2 m a square. So, the most probable energy is this one given by this that is E 3. Now, if I want to ask what is the average energy or the mean energy that I am going to get what I have to do is I have to take these individual energies that are possible outcomes of measurement. They have to be multiplied with the corresponding probabilities that means I have to take my energy E 1. It has to be multiplied with the corresponding probability is P 1. I take the second energy that is possible multiply with the corresponding probability that is P 2 and other energy that is result that is possible is E 4 with the probability of P 4. So, this is P 1, this is P 3 and this is P 4. So, I know all these things now I know what is my E 1, E 2, E 3. I know these values I also know the corresponding probabilities namely 1 by 4, 1 by 2 and 1 by 4. So, when I substitute I see that this works out to be 13 pi square h bar square divided by 8 m a square please verify. So, you started with a super post state by finding out the expansion coefficients and making the square of those expansion coefficients to be 1. We found out the one of the missing constants that is a we have found out and then using that we found out the probabilities of getting different energies in the measurement and since we know the corresponding energies also we can find out what is the most probable result that is in this case is E 3 and also one can define an average energy or one can call it as an expectation value of energy which is given by the energy the summation of various possible energies multiplied with the corresponding probabilities I shown in the last line here which in this case works out to be 35 pi square h bar square divided by 8 m a square. So, this is a average energy that is going to happen as a result of this measurement of energy we are only talking about energy in this case fine. Now the next part is if the measurement of energy suppose that measurement of energy yielded the last term that is E 4 suppose your result was all these possibilities are as there as we have seen. Suppose the result was corresponding to the third term that is corresponding to E 4 what is the wave function what is the form of the wave function at a later time they are bringing in the time dependence into picture. So, what is here only the real collapse is going to show up. So, what happens is the result is given that the result of energy measurement is E 4. So, the fact that you got E 4 as energy means the super post state collapsed to psi 4. So, that you got the energy as E 4. So, the super post state has collapsed to this particular state psi 4 which is a stationary state. Therefore, at any later time the wave function psi xt will be purely this psi 4 corresponding to psi 4 that is sin 4 pi x by a this is the time independent part with the normalization constant there is no other term the normalization has to be the full normalization that is 2 by a times this is only the time independent part. But see the difference the normalization has to be completely 2 by a here as this is a pure state because it is a collapsed state it is a pure state it is a stationary state nothing is going to happen to this. And this is going to evolve in time with this usual minus i omega of course is in this case is h bar t. So, this is going to stay for long for time later on this state the original state which had got 3 terms is collapsed. And this is going to continue with this kind of a time evolution given by this time dependent part of the wave function that is coming here this is a usual solution corresponding to the time independent part. So, the time independent part is multiplied with the time with the corresponding energy here you can construct the time dependent Schrodinger equation as you have seen in the lectures. So, this represents the full wave function time dependent wave function at a later time. And now the main point to be noted is that the normalization is again to square root of 2 by a as we talk in terms of a pure state not a super post state because this has to take care of the full normalization. So, this is a very very important problem because this right now this problem talks only about energy measurement this is true for other kinds of measurements. So, because any state can be represented as this combination of energy eigenvalues of various operators here it is energy, but there are other operators as you must be knowing. So, in all these cases angular momentum all these things one has a similar situation that is coming. So, that way this problem is very important. So, there is another problem which again is on particle in a box just to give you an idea about certain numbers this problem is compare the 0 point energies of the following. Three different systems are given one is really a macroscopic system a ball weighing 100 grams and it is confined to a box of something like 5 meters. There is an oxygen atom which is kept in a box of something like 2 angstrom side treat everything as one dimensional problems. And then there is an electron in a box of something like 1 angstrom. So, question is to compare the 0 point energies. The calculation is very simple. So, 0 point energy as you might have heard about it is basically the energy of the system at t equal to 0 the absolute 0. The energy of the system at absolute 0 which you would have seen in the case of harmonic oscillator which is half h bar omega or half h nu is a 0 point energy for a harmonic oscillator. The energy is not 0 even when the temperature is 0 absolute 0. So, this is that particular energy of course, one can relate this one can show that such an existence with the help of uncertainty principle in a very simple manner not going to those details. So, here so this 0 point energy is very important in many situations. So, here we are trying to compare the 0 point energy of a problem which is essentially a particle in a box all the 3 cases which are described here all corresponding to particle in a box. Only thing is that sometimes the boxes of macroscopic dimensions for example, 5 meter is a macroscopic dimension whereas, an angstrom width is of course, in the quantum mechanical regime. So, the 0 point energy is essentially the ground state of the particle in a box which of course, is in our terminology is e 1 corresponding to n equal to 1 which is given by pi square h bar square n square is 1. So, it becomes pi square h bar square by 2 m a square. So, I am using the energy expression corresponding to the particle in a box which we have seen in the earlier problem also. Only difference is now my mass and the dimension a are varying over a large or a wide range and let us see what happens. So, if I do that and of course, I know that one electron volt because better to write everything in terms of electron volt is 1.6 into 10 to the power of minus 19 joule keep this in mind I am sure you are using whenever you teach these things. So, I can find out what is the energy for the first case for the ball is works out to be I mean typically it is coming to something like 10 to the power of 49 electron volts. The same thing for the oxygen atom. So, when you want to do for the oxygen atom you should know the mass of the oxygen atom which can be essentially taken as I mentioned in the magnetism lecture here you can ignore the electron mass you simply take the nucleons the mass of the there are 16 nucleons in this case and the mass of course, is 1.6 into 10 to the power of 27 kilogram. So, we can multiply this you can find out what is essentially the mass of the nucleus which is going to represent as the atomic mass here you do not have to really worry about the mass of the 16 or electrons there. So, you do not have to worry about it if you want to really because we are not going to calculate an exact number we want to get order of magnitude comparison. So, if I do that I just take the oxygen nucleus this works out to be something like 10 to the power of minus 4 electron volt when the box dimension is 2 angstrom 10 to the power of minus 49 now it is 10 to the power of minus 4 electron volt. If I do the same thing for the electron typically this comes to 30 electron volts you see the range. So, this is for a macroscopic object it is very very negligible whereas, for the electron it is found to be about 30 electron volt. What is the significance of this 30 electron volt? If you remember I mentioned in the magnetism lecture also hydrogen atom binding energy is minus 13.6 electron volt. So, if you give positive amount of that the electron is liberated. So, 13.6 is essentially the ionization energy for the hydrogen atom. So, that way this is comparable to that energy. So, this tells you the kind of importance when you are actually going to really quantum mechanical systems where the 0 point energies are really an issue to be bothered. Similar thing is going to happen in the case of helium because then the when the 0 point energy is very large something which is related to this and the 0 point energy is very large the system even the when the temperature is lower the system does not really allow the system to solidify. For example, helium see it has other reasons also. One of the reasons is the 0 point energy for helium is again very large which means that it is it will not be easy to solidify helium just by reducing the temperature. In fact, helium cannot be solidified even at absolute 0 unless you apply a pressure you cannot solidify helium it will remain a liquid. Of course, various forms of helium liquid is very interesting and not going to the details, but getting a solid helium is impossible just by reducing the temperature. One of the reasons is that this kind of a large 0 point energy associated with it because even at such a low temperature the 0 point motion corresponding to this particular energy will not allow to freeze it into a solid form and hence it remains as a liquid. So, this comparison of these numbers is very interesting how you really go from a classical regime to a quantum mechanical regime by the study of this particle in a box example and seeing what happens to the ground state energy which is called the 0 point energy across these three different systems. Even in the oxygen atom case it is really a quantum mechanical situation in some sense. Next problem is slightly different. This problem is not on particle in a box this is on scattering. The beam of particles of mass m is incident on a potential step essentially step potential defined as v equal to 0 this is question number 5. So, we have a potential of this kind I will draw the picture this is x equal to 0 and here v equal to 0 here v equal to v naught. So, you have essentially two regions in this region v equal to v naught it goes all the way up to plus infinity and here 0 extends down to minus infinity. And what is given is a beam of particles of mass m is incident on the step potential and the incident wave function incident beam wave function is given by 2 a e to the power of minus i k 1 x here is a catch. Usually we always take the positive axis to be in this direction this is your positive x axis and our e to the power of i k x terms those beams go in the right direction the post along the positive x direction. In this case the beam that is incident is represented by e to the power of minus i k 1 x which means that the beam is actually in this direction this is the incident beam. The incident beam is coming from right to left how do we know that we know from this form of this time dependent part. So, this incident beam is coming from right to left and so let me call this region as pi 2 this region as pi 1. So, I can write my pi 2 is essentially a sorry this is a it is what is given yeah this is incident is correct. So, this region I call it as pi 2 and this I region because this is my from left I am coming. So, let us take this as pi 1 this is pi 2. So, incident region I take it as pi 1 this I take it as pi 2. So, if I want to write I can write pi 2 it is given that the transmitted beam that is the region 2 whatever beam is coming that is a transmitted beam that is given that is region 2. So, this is region 2 this is region 1. So, region 1 is the incident region region 2 is a region to which it is actually transmitted. So, pi 2 I can write that is given that is a e to the power of minus i k 2 x again it has to be minus because it has to transmitted means it is going in the same direction as the incident beam. Now, what are the values of k 1 and k 2? k 2 of course, is the region where is no potential it is square root of 2 m e as was defined during the lectures. k 1 of course, there is a potential here and it is coming to the right which means that your e is more than v. So, it is e minus v naught. So, that is able to come in this direction in the form of a beam that is why this has to be positive and since it is coming e has to be more than v naught. So, it is k 1 is given by 2 times m times e minus v naught remember here it is there is no potential it is a potential free region the region 2 where it is k 2. So, it becomes 2 m e. So, I have the wave function in the region 2 given by this one. Similarly, region 1 phi 1 can be written as the original beam that has come to a e to the power of minus i k 1 x and I should expect some beam which gets reflected because the whole thing has not been transmitted then definitely I can expect a beam to get reflected and go back in the towards the right direction. So, that I can take it as b which I do not know e to the power of now the reflected beam from here from the step will go in the right direction and hence it has to be e to the power of minus i k 1 x naught minus because e to the power of i k 1 x represents a beam that moves towards the right. So, you can see in the first region that is an incident region the wave function is given by the incident beam which is 2 a e to the power of minus i k 1 x which is given and I know from the problem that there has to be some reflected beam which of course we do not know the constant here. So, I put it as b. So, this b times e to the power of something which moves to the right which is represented as i k 1 x because k 1 has to be the same because that is the same region. So, it becomes minus i k 1 x becomes i k 1 x in the region 2 where it is a transmitted region where the potential is not there it is a e to the it is given already it is a e to the power of i k 2 x where k 2 is given by square root of 2 m e and k 1 is 2 m times e minus v naught. Remember e minus v naught has to be positive. So, that this is a wave that is moving in the towards the left. Now, what we should do is to find out using this now we have two regions and the corresponding wave functions. As you know there are certain conditions regarding the these wave functions to satisfy at the boundary. So, they have to satisfy the wave functions must be continuous at x equal to 0 here. The first derivatives also must be continuous at x equal to 0 here. So, if I do that continuity of wave function at x equal to 0 gives I am not writing all the steps it is 2 a plus b which means that a is minus b. Similarly, continuity of first derivative at x equal to 0 gives again similar manner differentiate you will get minus i k 2 a will be given by minus i k 1 times 2 a plus i k 1 b. From these we get k 2 is 3 k 1, but k 2 of course, we know that is square root of 2 m e is given by this is 3 times 2 m e minus v naught. Therefore, that is e is given by 9 e minus v naught or e equal to 9 by 8 v naught. You can see e is more than v because this is 9 by 8. So, as I told you that is how the beam was able to come from the right side to the left side because energy is more than the potential otherwise it would not have come in this direction. So, that is what is seen here as value of energy more than the potential that is the region v naught. So, this is the energy with which the beam is coming that is the incident beam energy is given by this. Now, what is asked is the reflection coefficient as I mentioned sorry is a part of the wave it is getting reflected at the boundary. So, the reflection coefficient is given by the reflected part divided by the incident part its square. We substituting this we know the value of b and a relative values of b and a we know. So, this is 0.25. Therefore, the transmittance is given by 1 minus r which is 0.75. So, most of the wave is actually in this case the values have been adjusted in such a manner that your transmitter is something like 3 times that of the reflected. Most of the things are actually transmitted because that is determined by the energy and in this case it has been adjusted. So, that the transmittance is something like 0.75 whereas, the reflectance only is 0.25. So, this is something which is related one can have various problems of this kind where you have two regions there is a potential step. Basically in all these cases you have to match in as we have done here you have to match the wave functions and first derivatives and try to find out various things by defining your k values as we have done in this case. Then once you know the reflectance and transmittance you can actually check. So, that the total r plus t must be 1. So, let us take question number 6. Question number 6 is again on classically forbidden region but actually it is in a different problem. This is on a harmonic oscillator problem, simple harmonic oscillator. Simple harmonic oscillator as you must be knowing the potential is of this kind, parabolic potential is half m omega square x square potential. So, what is asked is what is the probability density for the ground state what is the probability density that this the total probability of getting seeing the system in the classically forbidden region. So, first of all we have to find out in this problem as we have done in the earlier case the first problem we have to find out what is meant by classically forbidden region here. Remember the ground state energy as we have seen is half h bar omega. So, that is here not 0 it is here and you know the wave function has this particular shape it goes something like this right it has a tail it goes out it is not 0 just outside. So, it goes out. So, we have to find out what is the probability of getting this in this region of course, here means it is here also you have to find out that is what is asked. But for that first thing you to find out is what is the classically forbidden region as I mentioned classically forbidden region means the energy E must be less than the potential potential energy. In our case the energy we are talking about the ground state. So, that is half h bar that is half h bar omega is less than your potential which corresponds to half m omega square x square or the critical distance x value that is x c is such that my half h bar omega is equal to half m omega square x c square. So, when my x c equal to x c I see that these two are equal for any value of x c more than this or the negative side also you will see that the potential is going to be more than this and hence that region becomes classically forbidden. So, from this expression that is half h bar omega equal to half m omega square x c square I can find out my x c therefore, my x c is given by plus or minus h bar divided by m omega. So, this tells me. So, this is a region which is the starting of the classically forbidden region it goes all the way up to infinity. So, classically forbidden region is plus h bar to infinity on one side and similarly from minus infinity to minus h bar divided by therefore, the probability of finding the system in these regions is I know the range. So, I can integrate the probability density which of course, is given by I because these are equal. So, I can take twice I will take only from one part. So, it becomes h bar divided by m omega to infinity the wave function is given of course, I can find out you can work it out it is not very simple integration, but I am sure you will be able to manage. So, basically the point is to get into this idea. So, of course, the ground state wave functions I x is given as a e to the power of minus alpha x square is given. So, using this wave function if I substitute and integrate between this I am multiplying this with 2 because I am taking only this whereas, I have an equal probability density for this side also that is from minus infinity to h bar by m omega. So, that is why this 2 is coming. So, this integration will tell you what is the total probability of finding the system in the region which is supposed to be classically forbidden. So, the classically forbidden region we do not expect anything, but quantum mechanically you see that it is a nonzero this result is nonzero which is expected in quantum mechanics which is definitely not expected in the classical ideas. So, that way this is another problem where you see what is meant by classically forbidden and classically allowed. The first problem we have seen where there was no specific mention about what kind of a system, but here we are taking a very well known system of harmonic oscillator where we know the ground state, we know the form of the potential and so on. So, that way this is also a very important thing to differentiate between what is meant by classically allowed and classically forbidden region. Question number 7, this is again a particle in a box problem, but only thing is that now it is a 2D box. It is a problem of a two dimensional box which is little more realistic than a one dimensional case. Of course, 3D is the most ideally used many examples. So, the tells you that it is x y and t not only x, but now y is also coming and your box is something like this is given by a sin k 1 x sin k 2 y e to the power of minus i omega t. So, we always know what we have from the one dimensional box, we know that psi x is square root of 2 by L sin k 1 x where of course, my k 1 will be n pi by L. Now, in our case the dimensions are different one is L and other thing is 2 L. So, this is L along the x is this is y, this is x, this is L and this is 2 L. Therefore, my psi x y t should be of this form which means that it should be square root of 2 by L sin k 1 x. I should have for this one 2 by 2 L because the length is 2 L that is only difference along the y direction sin k 2 y e to the power of minus i omega t. Therefore, A is square root of 2 by L times square root of 2 by 2 L will be square root of 2 by L. Just by comparing the one dimensional problem we are trying to identify what should be the value of A. Now, the energy, energy is of course, n x square pi square h bar square divided by 2 m L square plus n y square pi square h bar square divided by 2 m 2 L square whole square. N x is the quantum number corresponding to the x axis, n y corresponding to y axis we are taking two of them separately. So, this will give you the ground state quantum numbers n x n y minimum that is possible for particle in a box is 1 and 1. So, this will give you the energy ground state is equal to e 1 1 which is given by if I substitute I get a value of pi pi square divided by 8 m L square. So, substitute in this one you see that you will get pi pi square divided by 8 m L square. If the lowest quantum number is not 0, but it is 1. 0 will make if I make one of them 0 then it becomes essentially one dimensional problem our problem actually is two dimensional. Now, what is a psi of the ground state is you know the normalization sin pi x by L sin pi y divided by 2 L e to the power of minus i omega t. So, I can find out the probability is 2 by L square I have to do x is from 0 to L by 2 I can find out sin square pi x by L dx times then I will take y equal to 0 to L by 2 sin square pi y divided by 2 L d y. So, I have two integrations one along about x and other thing is over y. So, one can find it out. In fact, this works out to be last part of this question is what is the what are the values of n x n y you have to give a set of two numbers right because you need n x and you have to completely specify the problem. What are the two different sets of n x n y values? So, that you see double degeneracy that means, you have the same energy corresponding to two sets of quantum numbers for that what we are looking for is a lowest such energy. Higher energies are possible, but you want to find out what is a lowest energy where you have this kind of a double degeneracy that is you should have two sets of n x n y values where you get a same energy. So, how do we find out? So, you can actually work it out if your L was the same in both the cases let us look at that way. Suppose you are L the sides were the same it was a kind of a square box now it is something like a rectangle suppose it was a square along x axis and along y axis if it was the same length L what would I mean the situation your ground state is as we have seen it is 1 1 always the quantum number is n x n y is 1 1, but when you go to the next one if whether you take it as n x is 2 and n y is 1 or the other way the energy is going to be the same it is determined by your n x square plus n y square does not matter. So, there is a double degeneracy associated with 2 1 and 1 2, but that is not going to happen here because your lengths are different here. So, one can actually see the you can work it out the two possibilities try these two 2 2 and 1 4 combination. So, this is n x n y is 2 take 2 and 2 n x n y you take it as 1 and 4. So, this first one corresponds to n x this is n y this is n x this is n y if I do you see that in all both the cases the energy happens to be 5 pi square divided by 2 m l square you take this combination of 2 2 or take the combination 1 4 you will see that you will get the same energy. In fact, you cannot get any such two sets of numbers which will be degenerate at means we will give you same energy which is lower than this. So, this is the lowest energy where you can get the double degeneracy. So, compare the case with the square case which I was mentioning the square case you do not have to really think 1 1 is non degenerate 2 1 1 2 then 2 2 of course will not be degenerate non it will be non degenerate then it will go on. So, for example, 2 3 3 2 all these possibilities will give you a double degeneracy, but here you cannot just do that you have to really work it out and find which combinations of n x and n y pair will give you the same energy and in this case the lowest happens to be when I take 2 2 and 1 4 because if I substitute in the expression I can easily verify that in both the cases I will get 5 pi square divided by 2 m l square. So, I thought at least one problem on 2 dimensions if not for 3 dimensions most of the problems where one dimension that is one where you have the concept of degeneracy comes when you have 2 dimensions onwards. So, that is why this problem will give you some idea about the degeneracy aspect also. This is a I put it just as a kind of a qualitative problem which is again on a harmonic oscillator what I am asking is we have just now drawn in other case we have drawn the potential for a harmonic oscillator and I also shown this is our energy this is the potential this is the energy that is shown half m omega square x square this corresponds to E 1 which is half h bar omega. You know that the wave function has this form as I mentioned it has this form it does not die out I mean now it is just goes out it goes out that is it is going into the classically forbidden region as we have shown there is a finite probability density here as well as here. So, the next one will be something like this. So, it has so this is maximum here it has something like this. So, the wave function is something like this here. So, it actually is 0 for the next state n equal to 2 this is n equal to 1 ground state first excited state and so on the question is what happens when I have a half harmonic oscillator that means this part is not there. Half harmonic oscillator means by potential is half m omega square x square on the right hand side this part the potential is infinity. That means this as in the case of an infinite square well this will go all the way up. So, that means this side there is no since this is infinite discontinuity here as in the case of infinite potential well the wave function or the probability density will not be there on this side with this idea in mind this is a very important requirement of for various functions to be the actual wave functions to represent a physical system using this idea is a very important idea that when you have an infinite discontinuity the wave function cannot be there in the other region it cannot penetrate into the region. However, if it is finite definitely you will see that there is some kind of a probability density here. In this case when you tell that it is semi harmonic oscillator half harmonic oscillator that means this is infinity that means this potential here this side or left side of this line is completely having an infinite potential. If you see textbooks you can see the various wave functions they are the form of various functions. Looking at that I want you to think and find out what will be so in the harmonic oscillator case for the full harmonic oscillator it is n plus half h bar omega this is n equal to 0 and this is n equal to 1. So, this n equal to 0 1 2 all this is right n equal to 0 is the ground state here 0 is the ground state. What is asked is half harmonic oscillator what will be I am giving you the result partially what is the condition. Look at the case here can you have an n equal to 0 state in this case n equal to 0 it is look at the thing whereas, you take n equal to 1 you are already having 0 I mean wave function or the probability density here. If you go to the next one again you see that you are getting into a situation where the probability is actually not wave function is not really 0 at this x equal to 0 position. So, this alternates. So, using this just thing I am not giving the full answer just think about what kind of an expression you will get for the energy eigenvalues when your harmonic oscillator is only half. So, full you have the expression that is n plus half h bar omega where n equal to 0 1 2 3 all integers are allowed. What modification is going to happen when I make this side an infinite barrier just think about it. So, this finishes tutorial sheet the problem sheet that I prepared for this part. So, I have not written down all the mathematical steps, but they are not at all difficult. I want you to please go I mean work it out and let me know your response. Sir my question is related to spinning motion of the electron. I want to know whether it really exists or it is a mathematical concept. Spin you are talking about a spin of the electron spinning motion of the electron. Okay. Whether it really exists or it is a mathematical concept. Yes spin I mean spin definitely exists, but not the way we think. Spies first of all you see you have seen the quantum mechanics lectures you have seen that there is the particle concept that we generally think about the spin I mean the electron is completely wrong. So, it is essential I mean at the most you can think of it like a wave packet. So, the spin is not in the usual sense that it is rotating about its axis or something like that that is not possible. Spin is another angular momentum it is called a intrinsic angular momentum anyway I am going to discuss in the magnetism part that is magnetism is essentially contributed by the spin as I mentioned in the last first lecture. So, this is purely an intrinsic property, but no mention I mean one should not really associate with any kind of rotation about axis and things like that about the electron. Because electron is not a classical particle like what we think about a ball. In fact one of the reasons I compared the different quantities the zero point energy in this case for a ball for oxygen atom and electron that was the reason. So, you cannot really compare that way what happens in the case of the macroscopic classical world and what happens in an atomic system where the electron is there. So, it is just an additional intrinsic angular momentum that is all I can tell. And remember in all our discussion magnetism we use spin and the projection of this total spin which of course, is a spin quantum number is a positive quantity. The projection along the so called axis of quantization which is called the which is represented by the corresponding quantum number m s it can be positive or negative, but a spin is positive. Electron spin is half what you tell is plus half or minus half or what you show with an up down arrow or down arrow that is nothing but the projection which is represented by m s. Usually my upper arrow is representing my m s is plus half the other thing is minus half. This plus half and minus half are the m s values corresponding to the allowed, they are the allowed values corresponding to my s equal to half. So, just take it as an additional or intrinsic angular momentum. It has got angular momentum property that is all one should associate. Sir, I tell you sir, 7000 like experiment they proved that the spinning motion actually exists. Yeah, no, no it does not talk about spin motion. I mean do not call the spin motion. It only talks about it has got a spin magnetic moment. And as I told you in the last lecture, any magnetic moment can be related to an angular momentum. What happened is in the famous Stern-Gerlach experiment, the silver was used, silver atoms were used and electrons were in an orbital situation where the orbital angular momentum was 0. If the orbital angular momentum is 0, the orbital part of the magnetic moment must be 0. But still what was found when you applied a Stern-Gerlach magnetic field was applied, of course the field gradient was there for a different reason. The magnetic field was there, it was actually sensing some magnetic moment. So, but we know it was known very clearly that it cannot be the orbital contribution because orbital contribution of the angular momentum was 0. So, we have to really look for some other contribution, some other origin where you can still get a magnetic moment for the silver atom. They will be found that it is nothing but the spin angular momentum associated with the electron which is responsible for the magnetic moment. So, in general you have both the contributions. In this particular case, only the spin contribution was there. But, again again the connection is between the angular momentum and the magnetic moment, spin angular momentum and spin magnetic moment. No other motion one should associate. What about the value x cross by 2? Yeah. So, that you see remember at that time this was theoretically predicted by Dirac, the spin and the spin angular momentum was actually I mean theoretically found out to be h bar by 2 which was experimentally shown, experimentally shown with this help of this Stern-Gerlach experiment because h bar by 2 means you are corresponding and that is the first time angular momentum quantum numbers were getting a non integer value. These are all half integers. So, this half integer half is the first simplest case. So, half if you remember last lecture when it is any quantum number correspond angular momentum the corresponding number of magnetic quantum numbers will be 2 s plus 1, 2 j plus 1, 2 l plus 1. In this case s equal to half you should get 2 half plus 1 is 2. So, Stern-Gerlach experiment when it was done unexpectedly to some sense they could get 2 different splitings. One corresponding to the plus half, the other corresponding to minus half, m s plus half, m s minus half which means that s must be half that is what which was proving the theoretical prediction. While solving question number 7 sir you have taken the limit of probability from x equal to 0 to l by 2. Yeah. So, that was asked no. So, the question is asking what is the probability of finding the system between x equal to 0 and l by 2 and y equal to 0 and l by 2. I will just check whether it is missing in the question. So, the question was giving the limit I will check again whether it is whether it was not typed or not I will just take. So, basically say two dimensional box you have range for x and range for y both are treated as I mean taken as 0 to l by 2 and 0 to l by 2 both x and y otherwise it has to be 1 ok. We will meet in the afternoon.