 In the last module we had said that we are going to discuss what happens when there are finite barriers but then I had a second thought and so in this module let us see what happens or how uncertainty principle is manifested by these wave functions that we are now familiar with in particle in a box. Just to recapitulate these are the things that we have learned so far the Schrodinger equation is exactly solvable boundary conditions lead to quantization this is the most important lesson that we have got in the last 3 modules and the other useful rule of thumb that has a reason is that more nodes you have in the wave function higher is the associated energy. This is going to come handy time and again especially when we talk later about atomic and molecular wave functions. Then we have talked a little bit about eigen function of the linear momentum operator we are going to revise that once again today. And the reason why particle in a box is interesting and the reason why it is taught in chemistry is that even though it is so simple it starts as it acts as or serves as a very good first approximation for some rather complicated chemical systems as we have seen already. One more lesson that we have got is that when we have more than one dimensions then we have to perform separation of variable like what we have done between spatial and temporal variables earlier this is again going to come very handy when we talk about say hydrogen atom. Symmetry and degeneracy go hand in hand this lesson goes a long way in understanding systems like d orbitals that are degenerate in a bare metal ion but get classified into two groups intertahedral or octahedral field and split further upon things like yantela distortion. One way of thinking of it is that we are going from a more symmetric to a less symmetric system progressively and that leads to decrease in degeneracy. As we have seen in particle in two dimensional box the moment we go from a square box to a rectangular box levels that were degenerate earlier no longer remains so. Also when we talked about three dimensional box we understood that we need a four dimensional function xyz for the dimensions of the box and the dimension of the wave function itself. So how is one supposed to draw it? We have exactly the same problem later on when we talk about things like orbitals but let us wait until then. So to conclude this part of the discussion particle in a box model has provided ground for provided a testing ground for more sophisticated treatment that we are going to see once again later on when we talk about approximation methods. And the question we stopped with in the last module is that what happens if the potential barrier is finite instead of being infinite. But as I said in this module we are taking a rain check on that and rather we are going to focus on something that we all know very well but perhaps do not understand all that well the uncertainty principle. The product of uncertainties in position and momentum for example has to be greater than or equal to h cross by 2. Let us see how this is manifested nicely using the wave functions of particle in a box and then we will digress a little bit and we will talk about how we will talk a little more about some general aspects of uncertainty principle. So here goes. Now we have already learned what the expectation values of position and linear momentum are for particle in a box for nth level. For position no matter what n is the expectation value is L by 2 that means if you perform a large number of experiments we are going to find that the average of these experiments is going to lead to L by 2. Let us not forget that in each experiment we find we expect to find a different value of position. It is just that the average value is L by 2 and very soon we are going to define what average value means in a little more formal way and average value of linear momentum is 0. Does it mean that the particle is not moving? Not really. What it means is that the probability of the particle moving towards positive x direction is exactly equal to the to that of the particle moving in the negative x direction. So average value of momentum is 0. But before we go any further let us make sure that all of us are on the same page as far as the definition of average and standard deviation uncertainty these things are concerned then we are going to use them. So let us go back and let us talk about little bit of statistics. What is the meaning of mean value? Well we have learnt how to calculate mean value when we are in class 6 or class 7 maybe or maybe earlier but a more formal definition using probabilities is this. That mean value is summation of j qj multiplied by pj which means we perform a large number of measurements of the quantity q for each measurement we get some value. So j is the identifier of the number of measurements right for first measurement j equal to 1, second measurement j equal to 2, 12th measurement j equal to 12 and so on and so forth. So every time we get some value and then let us say we perform a large number of measurements set it to the power 10 number of measurements. Then what will happen is it is not as if for every measurement we will get a different value some values will occur more times some values will occur less times and from the number of occurrences we can figure out the probability of a particular value qj to occur. Probability if we might remember is nj divided by sum over j nj where nj is the occurrence of the jth event probability is nj number of occurrence of the jth event divided by sum over j nj. That means the total number of measurements total number of occurrences. So the average value q in angular brackets is equal to sum over j qj multiplied by pj this is exactly the same thing that we have learned in school. If you have if you make some measurement 3 times I get 5 twice I get 7 once I get 2 what is the average multiply each value by the number of occurrences sum them up and divide by the total number of occurrences it is basically the same thing that we have written here. pj then is the probability of occurrence of the jth event and it needs to say perhaps that sum over j pj has to be equal to 1 total probability must be equal to 1 and it follows quite easily you keep on adding all the probabilities you are going to get sum over j nj divided by sum over j nj which is equal to 1. Now let us define something else let us define what we have called the expectation value expectation value in quantum mechanics of some quantity a is given by integral psi star a psi a hat psi d tau so does this expression and this expression do they mean the same actually they do why because we can think that for every measurement of using an operator means using a particular measurement right doing a particular measurement for every experiment you make this operator operate on psi and you get some value so that will come as an eigenvalue equation for that measurement. So what will you be left with a j multiplied by psi star psi so we add a large number of that measurements you get an integral of this sort okay so essentially the operator operating on psi star psi that is what the average value or expectation value is this particular aspect is discussed very nicely in Atkins physical chemistry book I recommend that we go through it and get our idea clear about what expectation value is and of course if the wave function is not normalized we have to normalize it also by dividing by integral psi star psi this expression that we have written is valid only for your normalized wave function. In our assignments we do have some problems where we are required to normalize wave functions and find expectation values once we work them out I think we will be very clear about this let us move on right now we are concerned not only with the mean value but also with the uncertainty so what is the meaning of uncertainty uncertainty is represented by a statistical quantity called standard deviation but to arrive at standard deviation it is better to talk about what is called variance first variance is defined as sum over j q j minus average value of q whole square multiplied by pj and it is written as sigma q square why because ultimately we are going to work with its square root which is standard deviation so let us make sure we understand what we have written here let us forget the summation for the time being let us only focus on what is there inside the bracket let us only focus on this part q j minus q so q j minus average value of q what is that for every measurement the jth measurement how much by how much does the value differ from the average value that is q j minus mean value of q suppose I do a summation of this q j minus mean value of q what do I expect to get let us say I do something like this sum over j q j minus mod well average value of q excuse my bad handwriting especially as I am using a stylus well we can also write pj here because we are performing a large number of measurement we should write probability so what we get is the mean value of deviation what do we expect this to be for a good distribution let us say I draw something like this this is the mean value that is going to remain invariant for all measurements on x axis I have j on y axis I have q j what do I expect for one measurement I should get something like this for the next measurement I might get something like this another measurement it might be like this this one might be like this and so on and so forth for a large number of measurement q j minus average value of q for any value of j is this so it can be positive it can be negative also so this summation we can expect that it is going to be 0 for a good distribution of data that summation is going to be 0 but then that is going to be 0 for all measurements provided we have performed a large number of measurements so it is useless that is why we work with the square of it how does it help if I make a square of it then this quantity let us say square of it is here and now use bigger circles this negative quantity its square will also be positive this will be positive this one will also be positive this will also be positive I am assuming that all deviations are less than 1 this is positive this is also everything is positive so when I do this summation now what do I get I get a sum of all these positive quantities essentially the area under the curve right so that is why it is better to work with sum of squares rather than the deviations themselves let me erase this so that it looks a little neater okay now we are all set so I hope you have understood what the significance of variance is variance would give me a positive number it is a sum of positive numbers and smaller the variance better is better it is I mean thinner is the spread but then it is still the square of the deviations we want to come down to its square root to do that let us expand this here q j minus average value of q whole square is of the form of a minus b whole square that anybody can see very clearly and we know that a minus b whole square is equal to a square minus 2 a v plus b square so let us write that what will this sum be the summation over j will be summation over j q j square minus 2 into q j multiplied by average value of q plus average value of q square whole thing multiplied by pj right now the summation of a sum well summation of a linear combination let us say is a linear combination of summations that is very easily understood right so instead of writing sum over j q j square minus 2 q average value of q plus average value of q square multiplied by pj I can write it as sum over j q j square pj minus 2 sum over j q j into average value of q and when I do that average value of q is a constant quantity so that average value can come outside the summation and the third term will be again since square of average of q is a constant quantity it can come out and the third term will be square of average value of q multiplied by sum over j pj let us write it this is what it is sum over j q j square pj comes from the first term here minus 2 average value of q average value of q as we said is a constant quantity will come outside the summation sign average value of q multiplied by q j multiplied by pj plus again square of average value is also constant quantity will come outside the summation square of average values or square of means multiplied by sum over j pj now see we can simplify this without much of a hassle because we have defined already that sum over j q j pj is equal to the mean value of q so the second term immediately becomes minus 2 mean value of q multiplied by mean value of q or in other words minus 2 into square of the mean value of q what about the third term sum over j pj equal to 1 so the third term reduces to square of mean value of q what about the first term the first term is sum over j q j square pj now look at this definition of mean value of any quantity that it is sum over j q j multiplied by pj now instead of q if I wrote q square then the summation would be sum over j q j square pj and there is exactly what we have here so this turns out to be the mean of squares of q so the difference between this and this needs to be understood what we have here is square of means what we have here is mean of squares they are not one and the same it is trivial for us to work out take any set of numbers 2 3 5 6 7 so what did I eliminate 4 2 3 4 5 6 7 work out the mean and take its square on the other hand take squares of each numbers and work out the mean of these squares of numbers they are not going to be the same and that forms the basis of the definition we are heading to so now let us write the expression for sigma q square what does it reduce to the first term reduces to mean of squares of q second term reduces to minus 2 square of mean value of q third term reduces to square of mean value of q so we are going to get mean of q square minus 2 multiplied by square of mean value of q plus square of mean value of q of course minus 2 plus 1 is equal to minus 1 so this reduces to a very simple expression difference between the mean of squares and square of means that is variance standard deviation is the square root of variance so that then turns out to be square root of mean of q square minus square of mean value of q so that is what we have arrived at something that is pretty well known and one can study these from well any textbook Macquarie and Simon is a physical chemistry textbook that has discussed it in some detail so one can actually look at that so now the stage is set for us to go into the study of to understand what is the uncertainty uncertainty now means standard deviation in position what is the uncertainty in momentum linear momentum for a particle in a box I think what we will do is we will talk about uncertainty in position and then we will take a break in the next module perhaps we are going to talk about uncertainty in momentum and continue from there let us see depends entirely on how much time we spend on the next part of the discussion all right so what is uncertainty in position by the definition that we have just constructed uncertainty in position is square root of the difference between mean square position mine and square of mean position and out of these two quantities we know one already we know that the mean position which is this mean value of x for particle in a box is l by 2 that is established previously already now let us work out what is this average value of squares that will take a little bit of time and we are going to use integration in parts since we do not know who is there on the other side of the camera what is the level of mathematical preparation I am going to go slow in this part and we are going to show you every step of this actually used integration by parts already but then that time perhaps I went a little fast so let us make up for that by showing every step in this one derivation later on when we use integration by parts I am going to jump steps all right so average value of squares x square is equal to 0 to l x square multiplied by the square of psi n x dx this is from the same expectation value that we have got the operator here is x square which simply means multiply by x square is a simple as that to go further we are going to need to use the expression for the wave function and this is what it is psi n of x is equal to root over 2 by l multiplied by sine n pi x divided by l this is something that we know very well already right let us plug in that expression then this is what we get 2 by l because sine square psi square right so since it is psi square 2 by l root 2 by l multiplied by root 2 by l is 2 by l comes outside the integration integral 0 to l x square sine square n pi x divided by l dx now when we have an integration like this then usually we like to simplify it by going from the square of trigonometric term to something that does not have a square and the formula that one can use very easily in this case once again going back to class 11 mathematics is the formula that we want to use is cos of 2 theta is equal to 1 minus 2 sine square theta I think all of us would actually know this formula cos 2 theta equal to 1 minus 2 sine square theta so what is sine square theta then sine square theta is not very difficult to see is a half multiplied by 1 minus cos 2 theta so here what is theta theta is n pi x by l so by substituting we get something like this the integral becomes integral x square 1 minus cos 2 n pi x by l this was n pi x by l so 2 n pi x by l and there is a half outside that half has cancelled this 2 in the numerator so this is what we have got now let us simplify this a little bit we will write it as a linear combination of 2 integrals first one is very simple 1 by l integral x square between limit 0 and l second one is minus 1 by l integral 0 to l x square multiplied by cos 2 n pi x by l dx so what we can do is we work out the first one quickly that is very simple I do not even have to say perhaps x square integrates to give you l cube by 3 between limit 0 and l and then of course in the next step l cube by l will be written as l square so this first integral essentially becomes l square by 3 what about the second one we write the second one as minus 1 by l 2 n pi multiplied by integral 0 to l x square d dx of sin 2 n pi x by l dx now why did we do that first of all let us convince ourselves that this is correct what is d dx of sin 2 n pi x by l we expect a 2 n pi by l outside we do not have it that is why we have written l by 2 n pi that part is fine but why do we have to write it in this manner in the first place because the formula for this integration by parts is this integral well since we are doing a standard integration we have written the formula for standard integral also well not standard definite integration we have used the formula for definite integral as well integral between limits l 1 and l 2 u dv dx into dx right dv dx that is why we have to write it as dv dx u into dv dx dx is equal to product of u and v substituting the limit l 2 minus product of u and v for the limit l 1 okay minus integral l 1 to l 2 v du dx dx so as you can see we will have to do it in one more step that we will come to later on but right now what do we have the first term has become l square by 3 and using this formula we can expand the second term by using integration by parts what do we have here integral x square d dx of sin 2 n pi x by l dx so u is equal to x square v equal to sin 2 n pi x by l there is a definite recommended order of priority when one performs integration by parts the algebraic terms polynomials have highest priority for being used as u the trigonometric functions comes second that is why we have used u equal to x square and not v equal to x square all right but all that is well 11 12 11 mathematics so let us go ahead let us work out what is u v at limit l 2 minus u v at limit l 1 what is the limit l 1 0 so what is the value of u u equal to x square so for the limit l 0 this u v is going to be 0 so the second term here is going to be 0 first term is u multiplied by v for l 2 for l 2 x equal to l so x square is l square no problem with that but v is sin 2 n pi x by l that is sin 2 n pi what is that 0 so this u v at l 2 minus u v at l 1 essentially becomes 0 minus 0 what about the second integral v equal to x square and well sorry u equal to x square so du dx becomes 2 x and v is simply the sin term here so this is what we get minus 2 x sin 2 n pi x by l dx so clearly once again we have to do integration by parts okay so we try to do that before that we clean it up a little bit and we get l square by 3 plus 1 by n pi integral 0 to l x sin 2 n pi x by l dx and now it is simple because I have done it once already so I leave it for you to do it yourself and convince yourself that this is the result we get okay we will go to the next one so when we do integration by parts again this is what we get integral 0 to l x d dx cos of 2 n pi x by l dx go to the next step again this time what will happen is v is equal to a cos term right v is equal to a cos term and cos at x equal to l cos 2 n pi x by l is actually equal to 1 and x equal to l there so the first term is not 0 first term is l the second term is still 0 because you are multiplying by x which is 0 and then we get this integral cos 2 n pi x by l dx which is now everybody knows how to work this out and we will do it and we use the conditions that we have learnt already that at the boundaries the wave function is equal to 0 remembering that the wave function is essentially the sin term we get l minus 0 minus 0 inside the bracket and we get l cube by 3 minus l square by 2 n square pi square this is the what have we worked out then let us not forget that we have worked out the average value of x square what are we looking for we are looking for the uncertainty or standard deviation square root of average value of x square minus square of the average value of x this is what we are looking for so let us stop here in the next module we are going to start from here and then we will get the final expression after that we will work out the expression for the uncertainty in momentum and we will convince ourselves that their product is indeed greater than h cross pi 2