 Suppose, we want to we start with the simplest thing possible which is try to calculate the kernel for a free particle. So, we return to the sorry this nodes are written in terms of classical mechanics terminology q rather than x. So, I will not try to change back to. So, we had this p q dot minus h, but what I am going to do now is to write the Lagrangian version of the thing directly. However, the p slice p integrals are there and that is what this unknown n is for it is actually not an unknown you can calculate it exactly. So, exercise calculate n. So, ignore all this technicality right now. What I want to demonstrate is that this thing can be calculated from an expression like this and the key trick is to use or rather abuse the Gaussian integral many many times. So, and note that this is equal to i over h cross one half m velocity square which would have d t square, but there is integral d t. So, that cancels one delta t. So, this is what we have now what we are going to do is to this looks somewhat easy to do because instead of q I should use differences of q then it is just of a Gaussian form right. It is exponent d say suppose it is q j plus 1 minus q j as the integration variable then it is integral d y e raised to y. So, y j etcetera we should have this kind of integral y j square times half m and that is bit like the Gaussian integral if between friends we imagine that i is actually a minus sign. So, we say that oh that is minus sign times a minus i. So, I will treat the minus i as some constant and just do a Gaussian integral. So, if we do this Gaussian integral what answer do we get? So, let us do this right you understand we agree that that is what will come out of this we just have product of these lots of these. So, let us do this elementary Gaussian. So, recall that we have e raised to yeah. So, is square root of pi over sigma. So, in this case we should get and let me write this quite clearly minus infinity to infinity d what do you can write y j d y j e raised to and now I have i m and there is a delta t as well y square m over delta t. So, i m over 2 h cross delta t times y j squared should become equal to square root of pi and then divided by. So, times 2 h cross delta t over minus i m. So, that is what we use as the basic tool and there is yeah. So, if there is t j then you put t j. So, we go here and do this, but there is one minor technical problem which turns out later to be interesting. It is that the number of differences is one larger than the number of integration variables right because you have from 0 to n. So, if I have 3 differences, but then there are 4 points. In other words this integration goes over only 1 this and this, but then there are 3 differences. So, there and also I am sorry to say yes. So, there is also q j plus 1. So, this is actually right. So, this is actually n plus 1. So, if this is i and this is f then it is 1 to n in between that is right. So, the number of slices and the i and f are fixed. So, the number of intervals is one more than the number of integration variables. So, 0 becomes q i and you go up to n then you get q n plus 1. So, you will get the right. So, 0 is equal to same as q i which is x. So, I should probably be writing yeah because of that yes ok. So, that is actually not an integration variable, but in summation you have to put it because otherwise you will not get the thing. So, what we do is that we introduce one more auxiliary y. So, the first one will be. So, we define like this y j is equal to q j plus 1 minus q j. So, there will be a y 0 and we put a delta function to show that that extra one is actually not genuine. So, let y 0 be equal to. So, firstly let y j be equal to q j plus 1 minus q j i e y 0 is equal to q 1 minus q i right. And now we got n plus 1 y j is to integrate over. So, j goes from 1 to n no sorry 0 0 to n that is how I get right I have to have j go from 0 to n. So, we also insert and how are the constraint well if you add all of them then you will get q f minus q i right sum over all the y's is just q f minus q i. So, minus q f plus q i. So, we insert an auxiliary p and this we write as equal to the delta function we write as an exponential representation. Yes. So, for the delta function we simply wrote the well known formula well known representation what is this? This is sum over sum over y good. And we inserted a gratuitous 1 over h cross and divided by 2 pi h cross. So, that we match up with all the other h crosses that are occurring here. So, this has to be inserted and once we do this we get this. So, ignore the limit n going to infinity part, but we write integral product over instead of i to n I now write 0 to n and write d y j right I traded the q's for this by and I need to put the Jacobian of this transformation. So, that we can calculate here Jacobian of transformation from d q i's to d y j's is 1 actually if you do it because if you take. So, which is inverse of d y by d q this is what I am putting because I am going to insert it here. So, that Jacobian is 1 because well. So, it is inverse of d y j by d q i's and that is easy to compute because d y by d q d y j by d q j is equal to minus 1, but d y j by d q j plus 1 or any other j is equal to 0. So, you just get a product of minus 1's actually which does not really matter because that is an overall because Jacobian is after all modulus of this. So, it is equal to 1. So, the Jacobian of transformation is 1. So, that is not needed here oh I am. So, sorry not infinity very good thank you thanks right. So, all you have to do is start writing y n equal to q f minus q n y n minus 1 equal to q n minus q n minus 1 and going up to y 0 equal to q 1 minus q 0 q i, but if you add then these things will keep cancelling and. So, you accumulate that sum of 0 to n of q of y j is equal to simply q f minus q i. So, that is what that is and here again you can see directly how the Jacobian will work out because the derivatives are all either plus or they are all minus signs, but you do not have to worry about accumulating minus sign because Jacobian is always modulus of that thing Jacobian as the interpretation of a volume. So, it has no sign. So, thanks for pointing out the errors ok. So, let us just see. So, we have this integral and then we have integral. So, that gets all of this including this then we can put a 1 over 2 pi h cross and then we can put an integral minus infinity to infinity d p then we can put in that expression. So, exponent of i over h cross sum over 0 to n. So, in this sum over 0 to n I will try to include this piece as well. So, we put 0 to n m over 2 and 1 over delta t j's which is just t j plus 1 minus you know it is difference of the successive time values times y j squared and in this summation we also have p y j and finally, outside the summation we have i over h cross p times q f minus q f plus q i. So, plus i over h cross p into bracket minus q f plus q i. So, this i p q f minus q i can all come out. So, we can write it as 1 over 2 pi h cross and we have forgot the big n outside and unknown n is lurking there, but not to worry about it or you can write it as minus i p q f minus q i right. Now, we have a product of various integrals this d y j and then this Gaussian integral. So, let us just call them i i j's product over j i j and i j is equal to integral minus infinity d y j x i over h cross and we need to write only 1 j term. So, into m over 2 1 over delta t j y j squared plus p y j. Now, this we do by completing the squares and converting it into a Gaussian right there is a y j squared and there is a y j. So, yes now to do this we need to add and subtract no sorry m over 2 will have put a 2 there anyway. So, I do not need anything here good right and minus p j square multiplied by t j over 2 m. So, now, what we have to do is to shift in this in this integral we have to shift the y j by this amount which is some constant, but the integration range is minus infinity to infinity. So, it does not matter and we get a Gaussian with this as the 1 over sigma squared we had written somewhere yeah sorry equal to sigma we put sigma x squared yes. So, if there is sigma here 1 over square root sigma occurs. So, here this is what is sigma. So, 1 over square root of that will occur and this will just come out because it is not in the integration. So, i j is therefore, equal to e raise to just the minus p squared t j of delta t j over 2 m and there is an i of course, i over h cross in front of everything p j squared over 2 m sorry p squared over 2 m we like that expression because it looks familiar minus i over 2 p squared over 2 m times delta t j and times square root of a pi factor and this is the sigma squared. So, 1 over that. So, 2 delta t over and then there is the i h cross multiplying it. So, h cross over and we have to put a minus i. So, that is what i j is. Now, we do not have to be frightened by all this because all this masala will cancel that n in front. So, if you do the p integral in the exercise correctly then you will exactly cancel this strange accumulating factors. Remember that this is going to get a product an infinite product of delta t j's in the limit that n is going to infinity, but it is all because it is just an overall factor and actually you will find that if you have done the exercise correctly that the first p integration generates exactly these things and they just all cancel ok. But we are left now with a nice piece e raise to minus i over h cross p squared by 2 m times a delta t j in the numerator and then there is a product of these. So, there is a summation in then. So, it will just become t f minus t i and. So, the answer is so that you do not form the impression that like a genius we are writing the answer here. It is equal to n times this product over 0 to n 2 pi h cross delta t j over i m there is a minus i was it minus required yes yes we should put minus i m, but that does not matter because it is ok. So, this strange factor, but we are going to read these two together and as I told you it will go away most of it will go away if you do this n times a very nice integral the integral is minus infinity to infinity d p over 2 pi h cross e raise to minus i over h cross into p f into p times q f minus q i that is that one right e raise to minus i q f minus q i and then from the summation of the n delta n's we get plus p squared by 2 m and this is the answer. This is the answer in the form I like it you can because now you have become experts at doing this do this one last integral one it is just one more Gaussian there is d p there is p squared and there is a linear term. So, complete this square and then you can beat it down to a final answer. So, actually this answer amusingly is given in so, can be so, exercise number 2 this part as I am telling you will become 1 if you do exercise number 1. So, if you do this exercise number 2 you will find an answer which is given in Schiff's book as an exercise ok, but not through path integral. So, Schiff is a very clever very very clever book it has a lot of hidden gems in it. Schiff and Merzbacher these two books are really very nice books for theorists to read.