 at different times. However, we can do this as an infinitesimal transformation. So, we can attempt to calculate Dirac picture state x t plus delta t x t this little kernel of a small time evolution. And this is equal to because this is t plus delta t is here and Dirac picture is e raised to plus i h t, but it is this way. So, it becomes equal to x t still Dirac, but e raised to minus i h delta t. So, this however is also not very useful is momentum in this because ultimately x will not the trying to build this up using only x will not work. So, what we do is we compute p t plus delta t x t equal to and this way approximate by writing by expanding this exponential. So, what we are doing is assembling the pieces of the whole kernel and we will put it together in the end. So, this is also the time dependent momentum basis done in the same way. Then 1 minus i h delta t over h cross dot acting on x t Dirac. Now, we come to the point of why we put p and q p and x. The point is this is an operator h, but now I can replace it with its Eigen value between p and x and to do it inside the h I take all the p to the left all the x to the right. Then the p's will act to the left x will act to the right they will become numbers ok. So, and this is the logic behind doing p x. This is the normal ordering prescription remember postulate number 5 quantum kinematics. Quantum kinematics said that we have a great convenience of expressing any operator in terms of the basic canonical set q p and except for the for the little irritation that when we try to do it for operators that are higher powers of x and p there is a problem of ordering the x's and p's. So, that same problem occurs here as well. You will find in literature some people saying because the path integral is just some some over some paths and it is all classical expression there is no operator ordering problem that is not true. If you just follow everything properly this is the precise point at which operator ordering also enters in path integral. This makes the path integral formula exactly equivalent to the Heisenberg formulation no more and no less ok. So, we have to do some operator ordering here. So, after we do this normal ordering we get it equal to it becomes c number and we can assemble this to be equal to. Now, we know the q p overlap this identity operator become 1 now and then inaccuracies in the next order in delta t times my favorite q p overlap x t and this I know this we are defined it is equal to e raise to minus i x p over h cross. So, expect that we can write p x. So, we reduced this transition amplitude the infinite decimal one between x and p to a c number. So, this is going to be 1 key component of the final part of the derivation which now becomes very simple except for the complication of writing it out. So, then I will just tell you the answer I mean what is done next. So, what was instead of just saying finally, consider let me see let me say that thus introduce a large number of intermediate time points with values. So, t initial. So, we will introduce n of them less than t f. So, I am going to write the formula here, but already to give you a picture what we are doing is time slicing. This of course, was done later by Feynman, but the time slicing was done by Dirac, but he did not draw any pictures. So, we well in Feynman's world we draw x in this direction and t in that direction and we have some x i x f and t i and t f and what we are doing is introducing for convenience let us put 3 of them and. So, between this point and that point there are these 3 slices time slices t 1, t 2, t 3 what turns out is that you have to introduce one more q or one more p than the q's because we want to write. So, there is an integral sign which is going to become a gigantic integral sign in the end x f, t f now I insert here p n plus 1 t f p n plus 1 t f right this is the complete set of states. So, and then I write over here p n plus 1 t f no sorry you are right. So, we need yes and then next step is the q n right the next. So, I can reduce p from a later time to x at an earlier time. So, there is x n t n right. So, here I introduced p n plus 1 t f complete states at t f I will next introduce x n at t n complete set of states, but then now a step down p n minus 1 sorry between x and p I cannot take a step down. So, and I have p n plus 1 here. So, this is p n t n now I take a step down, but the p I have to retain at the same point doing this I finally, reach x i t i where I have. So, I will reach finally x 1 t 1 p 1 t 1 p 1 t 1 x i t i right. So, there is of course, this as well, but and everything is Dirac picture a somewhat crucial detail that most people forget to write because these are basis states they are not wave functions. So, that normally they cannot have time dependence they are just basis, but this is the time dependent basis it is actually a sort of a motivation for Dirac's later introduction of the interaction picture you know the interaction picture where you split h 0 and h prime and then the basis are all evolved using h 0 the time dependent basis in which you calculate this matrix. So, right. So, this is the main trick and now you can see what is happening because I have two kinds of contributions I probably even have a colored chalk here. So, I have this which is x p overlap at the same time or I have this which is a p x overlap with a time step, but which I have already computed. So, I insert all these factors as c numbers and I assemble a gigantic formula which is all c numbers. So, the transition amplitude is then given this kernel is then given which anyway is a c number because it is an overlap of two states reduces to a c number. So, and yes and I have put an integral sign here and to have all these things to be meaningfully you know resolvent of identity I have to have product over. So, I will have the p n plus 1 will stand out, but I have p product over 1 to n of d x d 3 x i if you like I have d 3 x i d 3 p i is what I have that is what the integration is over. We get the answer. So, we will have to write. So, we assume that these are at the same time. So, I get e raise to the x p overlap. So, let me write the final answer you can assemble all this it becomes equal to the x f t f x i t i equal to limit n going to infinity of integral I like to display this because it looks. So, you will get this from this x p overlap right. So, you get a 1 over square root 2 pi h cross and then product over j equal to 1 to n of d well this is 3 if you like then it is raised to 3 by 2 and then d 3 x j d 3 p j divided by 2 pi h cross I will fix it and times exponent of i over h cross sum over j and then delta t. So, if you take this orange and this 1 then you will find that the this x p overlap just gives you e raise to i p n dot x n over h cross divided by square root 2 pi h cross and then this overlap which gives e raise to minus i p n, but x n minus 1 over h cross and minus i h p n comma x n minus 1 times delta t. So, times t n minus t n minus 1 over h cross and the whole thing divided by another square root of 2 pi h cross n not square root sorry because I have 3 dimensions now raise to 3 by 2. So, 1 unit of this step generates for me this thing gets squared 2 pi h cross becomes cubed and there are p at the same point, but x n minus x n minus 1 right and minus h and because you are normal ordered the p is of a later time and x is of an earlier time. So, p j plus 1 x j and then t n minus t n. So, this is the unit of the whole long formula which gets repeated and so, we can reassemble it in this form which is very nice and suggestive because all I have to do is put a delta t below this and take delta t out and then it looks like p x dot minus h ok. So, so, then we stop quarrelling about all this one extra p and all this limit and all this and say well it is exponent because now we have all these intermediate integrations over d x as well as d p's and if we just focus on x and ignore p for the time being then what we are saying is I go from here to here this is my infinitesimal propagation, but then I have to integrate over all possible t 1's ok. Then I could go to some other value of x at time t 2 then some third value of x at time t 3 and then land up there, but I have to integrate over this entire thing because that is what this complete set of states means right where we introduced it this p sorry this p n t n. So, we have to integrate over the entire spectrum of eigenvalues of x at every instant of time which is what Feynman called sum over parts right. If you introduce enough things then it is amounting to and the values of x range over the entire spectrum of x eigenvalues although you may just be going x i to x f. So, if I send the ball from here to there it is going to also explore America and Romida galaxy and go there. So, we will stop here today and this is essentially the time slice path integral formula we just write it symbolically like this and it is very elegant because it just says e raise to i o. So, lot of people think that this is the well quite correctly this is the action the usual action the functional of p and x, but remember that it is origins are in the Hamilton principle function which is an integral over time and this thing is symbolic and if you ever run into legal problems of some leftover things you can always fall back on its technical definition which is this. So, if there are any faults then go back to this and try to repair it. So, this is the fine print behind the symbolic integration over parts. Now Feynman did not struggle with all this Feynman was much cleverer than us we did not I do not think he probably ever read about this. So, he just latched on to the answer and said yeah and of course, this is the Lagrangian in when you this is same as q q dot x x dot if there is time. So, you can think of this as the Lagrangian and so he said yeah it is e raise to i h cross integral of the Lagrangian and just got to calculating and the way he argued to himself and I mean the way he explains it simply is that this detailed sum is simply the statement that the amplitude to go from here to here is made up of a product of amplitudes to go from here to here, but at every point it is a complete set of states. So, you have to integrate over the intermediate set of Eigen values which just make sense and then you do not have to struggle over all this. But at the same time I want to show you that there is a very precise derivation which makes this formula exactly as good or as bad as the Hamiltonian as the Heisenberg formulation which is based on canonical variables. So, if you read Feynman then you will not see any of this he will simply say oh it is this and then you will only say look, but if you are going from here to here you could be going from here and then going there, but then you may introduce more and then it will look like this. So, but why it would be the action he does not take the responsibility of answering what he does is that he differentiates this expression and shows that it satisfies Schrodinger equation. So, it is the kernel of the Schrodinger operator and so, you have the answer. So, that is Feynman's way of doing it which we will effectively do next time. So, he then cannot get the proportionality constant right. So, then he fixes it in some way. So, if you read Feynman and Hibbs then all those clever tricks are given which are cleverer, but mysterious because you do not know where why they work, the reason why they work under the bonnet it is this all this is the wiring ok. So, we will do probably one more turn of this.