 This is where we stopped in the last module. We have worked out an expression for uncertainty in position and that turns out to be L square by 3 minus L square by 2N square pi square. Well to be very honest we have not worked out the uncertainty in position yet. What we have worked out is the average value of X square which is the first in the linear combination under square root sign in the expression for uncertainty in position. The saving grace is that we already know the average value of X from a previous module that is L by 2. Now let us work out the expression under the square root sign average value of X square minus square of the average value of X. This is what we need. We work it out and then put it under square root sign. The final answer that we get is L multiplied by square root of 1 by 12 minus 1 by 2N square pi square. I strongly recommend that all those who are taking this course should work this out by themselves because it is impossible to understand it completely or be convinced just by listening to me and seeing me work this out especially as we are not writing on a board. We are using electronic media. Please work out this expression by yourself. Then only I think you are going to be convinced. So crux of the matter is we have this expression for uncertainty in position that we had looked for. Now let us seek the expression for uncertainty in linear momentum and that is going to be much easier than uncertainty in position. This is the expression that we have got and now uncertainty in momentum what is it? Let us see. Well we already know that out of these two terms under square root that we need to find for uncertainty in momentum we already know that average value of Px is actually equal to 0. So then this expression becomes a little simple. It turns out to be square root of average value of Px square. Please do not think that the square root of average value of Px square the square root and square will take care of each other and it will turn out to average value of Px it will not. Remember the task I gave you in the last module take a set of numbers work out the average of squares and then work out the average of the numbers themselves take square you will see they are not the same. So it has to be written as square root of average value of Px square it is a root mean square quantity analogous to what we have studied when we talked about say speeds of gas molecules in kinetic theory there we talked about root mean square velocity is not it? So this also is a root mean square expression do not think that the square and the square root will take care of each other they will not. But the good thing is all we need to do is we need to find this average value of Px square how do we do it like this we know already how to find the expectation value. So we write the wave function and in this case we are working with a real wave function so it is so psi star equal to psi the wave function multiplied by Px square operating on root over 2 by L sin n pi x by L take this integrate between 0 to L and your home well almost because you still have to take a square root. So what is Px that we know h cross by i d dx what is the operator for Px square that would be making the same operator operate twice that is the meaning of operator Px square. So make it operate twice what do you get h cross by i multiplied by h cross by i that gives us minus h cross square and then d dx operated twice gives us d 2 dx 2. So this might ring a bell actually because when we talked about Schrodinger equation we got something very similar to this remember the Hamiltonian operator the first term the kinetic energy operator was minus h cross square by 2m why because kinetic energy is Px square by 2m okay so all our discussion is in sync. So Px square operator is minus h cross square d 2 dx 2 now so now what we have to do is we have to take this expression for the operator plug it in the expression for the plug into the integral let us do that this is what we have minus h cross square comes out of the integral sign d 2 dx 2 of root over 2 by L sin n pi x by L I am intentionally not taking this root 2 by L and root 2 by L out because if I can keep the wave functions then there I know what the integral is when the limits are 0 to L the integral for psi x square is equal to 1 now and also the good thing is here we have a sin function. So what happens when I take second derivative of sin I get back the same sin function the only thing here is that you have to take care of n pi by L for every differentiation this n pi by L will come and we can take it outside the bracket. So this is what we get n square pi square h cross square divided by L square integral 0 to L root over 2 by L sin n pi x by L multiplied by root over 2 by L sin n pi x by L dx what just happened where did I get n square and pi square and L square well because remember we differentiate it twice. So first derivative of sin kx is equal to k multiplied by minus k multiplied by cos kx that is why minus sign also came out in the first differentiation and that took care of the minus sign that was already there and then when you differentiate the cos kx function again 1 k comes out that is why you get k square outside and you get back your sin kx so far so good. What about this integral as we said a few seconds earlier this integral is equal to 1 because we are working with a normalized particle in a box wave function this is equal to 1. So the expression for the mean of square of linear momentum turns out to be n square pi square h cross square divided by L square. What is sigma px? Sigma px is just the square root of this let us find it out it comes out to we have very simple expression n pi h cross by L. So earlier we have determined the uncertainty in position now we have determined uncertainty in momentum the next task is to multiply them together multiply them with each other and see whether uncertainty principle is satisfied by this system of particle in a box that we are talking about sigma x remember is L by 2n pi square root of n square pi square by 3 minus 2 sigma px is equal to n pi h cross by L let us multiply them this is what we get what happens here L and L cancel each other n pi and n pi cancel each other so you are left with h cross by 2 outside the square root sign inside the square root sign you have n square pi square by 3 minus 2 this is what you get. Let us try to see whether we can get a simpler expression here what is the value of this n square pi square by 3 for the smallest value of n what is the smallest value of n smallest value of n is 1 we are talking about particle in a box. So if the smallest value of n is 1 then what is the value that we expect here pi square pi is 3.142 in fact so square root of that is going to be well square of that is going to be more than 9 you do not need to remember the exact value so more than 9 divided by 3 what is that that is more than 3 so more than 3 minus 2 is more than 1. So what you have under the square root is a real number which is more than 1 does not that prove that for any value of n this product is always going to be greater than h cross by 2 you see in this expression h cross by 2 is multiplied by this quantity under square root which we have shown is always more than 1 that is why the product of uncertainties of position and momentum for particle in a box turns out to be always greater than h cross by 2 which is nicely in agreement with uncertainty principle. So we have demonstrated that uncertainty principle is obeyed for a particle in a box remember non-conformity with uncertainty principle sounded the death knell of Bohr theory. So it is comforting to know that the first system that we have studied using the wave mechanical treatment of Schrodinger conforms with uncertainty principle. So a very major problem that was there in Bohr theory is not there at least so far. So next question we want to ask is that is this a specific case that only for particle in a box uncertainty principle is valid or do we expect for this wave mechanical treatment that it is not going to be it is going to be valid for everything is there a general way in which we can look at uncertainty principle itself. So this discussion might sound to be a little bit of digression because after all the topic under which we are performing the discussion is particle in a box but this digression if I can call it digression is definitely worth it so bear with me. So let us see let us understand uncertainty principle a little bit better to do that let us revise what kind of wave functions we have for your Px. So Px again we know is h cross i by i d dx and a good wave function for this is psi equal to e to the power plus minus ikx the wave function is an imaginary wave function but it is important to note that the eigenvalue is not imaginary eigenvalue is definitely real because when you differentiate e to the power plus minus ikx with respect to x you are going to get plus minus ik multiplied by the same function so that i and this i are going to cancel each other. So Px operating on psi gives you plus minus k h cross e to the power plus minus ikx we have touched upon this in one of the previous modules but let us make sure that once again everybody is on the same page with this. So this as you see is an eigenvalue equation with e to the power plus minus ikx as the eigenfunction is it normalized we are not trying to normalize yet maybe normalized may not be normalized let us not worry at the moment what is more important is is that it is an eigenfunction of the linear momentum operator and the eigenvalue is a real eigenvalue k h cross well the way we have written it plus minus k h cross what does it mean it means that the linear momentum associated with the system associated with this wave function is plus minus k h cross that plus minus k h cross is the value of the linear momentum no surprise root over 2 Me divided by h cross right that would better be the linear momentum now see what does this plus or minus sign signify k h cross is a magnitude we understand fine plus means that direction of motion is along positive x and minus means the direction of motion is along negative x so sign of the wave function specifies the direction of the motion of the particle remember we are right now talking about one dimensional space right so if it is just plus minus ikx then direction of motion if it is plus ikx then your momentum is defined very very well if it is minus ikx then you know you have zero uncertainty in momentum okay let us say I am not saying plus minus let me say it is plus ikx then what is the value of linear momentum plus k h cross one value no deviation this is not average value every time you do the experiment you are going to get this eigenvalue k h cross right so for this wave function e to the power ikx no matter whether the sign is plus or minus momentum is defined very nicely is defined precisely what about position position x can take up any value right because you work out the expectation value of position also here what will it be position is x multiplied by the wave function x hat operator is just x multiplied by wave function what do you get you get x x is a variable so the uncertainty in position is infinite it is defined by the boundaries so what we see here is that the moment the momentum is specified precisely in the wave function position becomes infinitely uncertain this is the philosophy of uncertainty principle but let us go ahead and before we make that statement let us see what happens if you can get a situation where position is better defined what happens to your momentum then to do that let us take this wave function we are very familiar with this wave function cos kx sin kx sort of particle in a box like wave function that as we know can be written as a linear combination of exponential imaginary terms okay let us say we have chosen a and b in such a way that we get either k cos kx or sin kx actually in the diagrams that we are going to show you we are going to use cos kx it is a little more convenient but let us not forget cos kx and sin kx are basically same the only difference is the phase shift right where cos kx is equal to 0 sin kx equal to 1 but the shape is the same so you just have to give an offset to x then sin can become cos okay so let us say we work with this wave function if this is a wave function then what does it mean it means that your linear momentum can be either ik or minus ik so there is a little bit of uncertainty in linear momentum two values are there px can be plus k h cross or minus k sorry i said ik that was a mistake plus k h cross or minus k h cross what about position position we can understand position by looking at the probability probability is mod of psi x square so if you plot this function mod of psi x square here it is just psi x square what do we get we get some regions of more a greater probability density some regions of zero probability density some points at zero probability density so if you just take a small slice dx then in this position and this position and this this this and wherever there is a maximum in this plot of psi square you get maximum probability of occurrence so that is the position so what we see is that position gets a little better defined at least we can say that the particle is more probable to be here here here and here and less probable to be here earlier position could have been anything all values of x had the same probability when you took one of the exponential terms so we have demonstrated that if we introduce a little bit of uncertainty in momentum position becomes more precisely defined let us now go way further let us take a large number of e to the power i k x and e to the power minus i k x for different values of k and let us add them up like this let us now work with sum over j a j e to the power i k j x plus b j e to the power minus i k j x coefficients a j and b j are chosen in such a way that you get say c j cos k j x that kind of a wave function so right now we are just making a wave function so everything is in our hands what happens then what happens to momentum now there are many values of j right I have not said how many values of j are there maybe there are 1 lakh values of j in the expression that I am going to show you I have taken a combination of about 10 values of j so momentum has become more uncertain depending on how many values of j we are working with you can have k 1 k 2 so what is momentum k h cross so you can have plus minus k 1 h cross k 2 plus minus k 2 h cross 1 and so forth depending on how many terms we have taken so in when we take this linear combination we get this kind of a wave function where your position your momentum has become significantly more uncertain what about position to get position once again we are going to take a square of this sum it is easier if you take a square of this sum over j c j cos k j square when we do that we get a plot like this now see what has happened now the other positions here there is a local mean maximum of probability density but it is so small same here the clear global maximum of mod psi square appears at this position of x where all the cosine functions are instead for x equal to 0 for all cosine functions the value is 1 the maximum value that is why here all these cosine waves are in step as you go further what happens is there is a phase difference between all right so the waves start getting out of phase this is a phenomenon that is made use of in things like Fourier transform time domain spectroscopy as well you have waves of different frequency at one point they are all in step then they start getting out of step you get destructive interference so what you get finally is that your linear momentum in this case is highly uncertain as we have said position gets reasonably well defined and for the records this kind of a situation where we have taken a large number of waves that are in step at a particular point this is called a wave packet and one can even experimentally generate wave packets and wave packet dynamics is something that has been of interest to physical chemists, spectroscopists, quantum chemists for several decades now and then they provide a very important information about how things proceed in ultra short time scale in systems we have discussed this in more detail in the NPTEL course on ultrafast spectroscopy in chemistry that was offered last semester whoever is more interested can have a look at those lectures but for now what we have got is that the moment we made linear momentum uncertain we generated a wave packet which had significantly better significantly more precise information about location of the particle so what we have shown here is that in uncertainty principle if linear momentum is well defined position becomes ill defined automatically in situations where position is better defined linear momentum becomes ill defined automatically. So the wave mechanical treatment here of Schrodinger is absolutely in line with uncertainty principle and using it we also get to realize that uncertainty principle is really the threshold of nature beyond which we cannot go here let me acknowledge my teacher Professor Rupindranath Banerjee of Jadapuri University who actually taught me this expression threshold of nature when I was in class 11 it is because of great teachers like them that students like me learn certain things that we perhaps would have not understood so well otherwise. So it is not possible to make a better instrument if you think that we will make a better instrument and we will determine linear momentum and position correctly it is just not going to happen it is not about instruments it is an inbuilt law of nature that in microscopic world if position is well defined momentum cannot be and vice versa and it is not restricted to just position and momentum also it is a feature that you see for all conjugate properties energy and time for example. Now since we are talking about uncertainty let us talk a little more about the workings of quantum mechanics and let us make this statement when operators do not commute properties cannot be determined simultaneously what is that supposed to mean? Commutation means you have two operators let us say x and px you make px operate first on a wave function you get whatever you get another function make x operate on it you get some value then invert the sequence of operation if you get the same quantity then x hat dot px hat minus px hat dot x hat would be equal to 0 which means the commutator is equal to 0 we will see it right now. So these are the wave functions we are working with let us do x hat px hat psi which means when you write like this px hat operates on psi first then x hat operates on it so x hat px hat on psi what is it? It is x multiplied by h cross by i d psi dx now we are not using particle in a box wave function it is a general statement x multiplied by h cross by i d psi dx let us do the commutation let us invert the sequence of operation px hat x hat psi what would that be x hat operating on psi is just x multiplied by psi so now you have to make this h cross by i h cross by i d dx operate on the product of x and psi we all know what d dx of u v is it is this px hat x hat psi gives us h cross by i x d psi dx plus psi very simple I am not going through this step. So now what is px hat x hat psi minus x hat px hat psi it is not 0 it is h cross by i psi and we can write it as an eigenvalue equation we can write it like px hat x hat minus x hat px hat psi is equal to h cross by i psi and in fact this operator is often written like this that x hat comma px hat in third bracket that means the commutation commutator of x hat and px hat that is equal to i h cross so what we see is that it is not equal to 0 and then postulate of quantum mechanics is that when operators do not commute properties cannot be determined simultaneously we will have some examples of operators that do commute in the assignment right. So this is what we have discussed so far we have considered a very simple model of particle in a box we started with particle in a one-dimensional box and then we extended it to two dimensions and three dimensions and then from this very simple model we have learned so many things over the last four modules I think we have even learned how we can see that wave mechanical treatment in general is in keeping with uncertainty principle. So sometimes students complain that whenever we try to study quantum mechanics or quantum chemistry this particle in a box model keeps coming back and is taught many times there is good reason for it in fact even in this course we are going to refer to the particle in a box model once again when we talk about say variation principle. This provides a good testing ground for sophisticated treatments as we had said but for now we are done with that discussion of particle in a box where the particle is confined within a box where the walls are infinitely high potential energy outside is v equal to infinity. Let us now move on to a variant and let us now finally tackle the question what happens if the potential barrier is not in finite what happens if the potential barrier is finite. When I put it like this perhaps it sounds like and a question that has arisen out of just curiosity from doing so much of theory that is not completely true. In fact, this question arose out of an experiment in the very early days of quantum mechanics early 20th century it was found that when people worked out the and not taking the name of the people you work out itself it involves people like Gamow and Rutherford and other stalwarts the kinetic energy of alpha particle emanated from a radioactive nucleus was worked out and the potential barrier immediately outside the nucleus was also worked out. And it turned out that well first of all this potential is finite and it turned out that the energy kinetic energy of the alpha particle is actually lower than this potential barrier. So, now the question came how does this come out? It appears that when the potential barrier is finite the particle can somehow come out of that overcome the barrier even though its energy is less than the energy associated with the barrier. How does that happen? The question of finite potential energy barrier arose from here and that is what we will discuss in the next module.