 Now, we are on to a developing a formalism which brings out what field theory is about and how to deduce results from field theory. So, last time we start. So, what we are doing now is forced harmonic oscillator or rather Green's function. So, we use the Lagrangian formalism and the Lagrangian is of course, integral d t of and we had set m equal to 1. So, half q dot square plus half omega squared q squared q times f this is what it was. And then what we do here is that we propose a Fourier transformed variable for q and similarly for f right. Our definition will be g equal to is the convention for the Fourier transform. So, if we do this then we can see that half q dot square. So, this is our definition of Fourier transform convention. So, do not worry what the g is any function g this is the convention. Now, we try to calculate what happens to the half q dot square or to q dot square. So, q dot square will require you to take something like this and differentiate it under this integral sign. So, if you carry out this exercise all you have to do is take this expression twice I am very sorry d e d e right. So, we will have d e d e prime and because you differentiate you get a I e down q tilde of e and here e raised to I e prime t q tilde of e prime right we are just squaring q dot. But from this we get integral d e d e prime then minus e times e prime and then the q tilde e q tilde e prime and then e raised to I e plus e prime t. So, if we look at this under the integral d t sign. So, then under integral d t this becomes delta function of e plus. So, this becomes integral d e d e prime over I need not well let us write it down anywhere times the minus e e prime and then q tilde e q tilde e prime, but delta e plus e prime. So, using up one of the e integrations we will get d e over 2 pi and 2 pi delta thank you. So, 2 pi is gone here and minus e squared e squared times q tilde e q tilde minus e and that is it. Now, we have to put it alongside this omega squared q squared it has to be added on to omega squared q squared, but q squared is just going to have without any I e appearing this thing twice. So, it does not seem to have e and minus e. However, you can always reverse the integration direction in one of them because it does not change the meaning of what this expression is and then you can write the q squared also as if it is integral of. So, integral d t q squared and there is an omega squared q squared this also becomes equal to omega squared times integral d e and times simply q tilde e q tilde minus e. So, you can recast it in that same form and then we had added damping factor which we make proportional to this. So, we make it proportional to q squared. So, we had a term minus epsilon integral d t of q squared where these integrals are all from minus infinity to infinity because this can be between arbitrary times, but eventually we want in the Fourier transform we certainly need e going from minus infinity to infinity because otherwise we do not have Fourier transform and yes therefore, these integrals are all otherwise we do not get delta function. So, these are all integrals from minus infinity to infinity. So, this is also integral minus infinity to infinity. So, the point being that this thing added to the exponent in the path integral will add a positive definite quantity with a negative constant in front and it will cause damping, damp at large values of t. So, it is damp at large values of q if you go to large values of this q squared then this begin becomes large and will damp everything out. So, we do not have to worry about very large q's and you can focus on a finite region. So, thus the transition amplitude q f infinity q i minus infinity this transition amplitude becomes integral d q of e raise to i we have dropped h cross now times integral minus infinity to infinity of d t of sorry and we are going to rewrite it in terms of e. So, becomes d e of one half this e squared and then we have sorry this is Lagrangian not Hamiltonian. So, this is minus sign. So, there is that minus sign, but that minus sign comes multiplied with this i. So, if I have to recast this in that same line I have to write. So, I write this minus sign as i times i take one i out. So, I will get minus omega squared, but minus i epsilon times the q tilde e q tilde minus e and now we also have to introduce these terms this term q times f and that q times f we write as equal to we put an overall half and we write for symmetry q tilde e forcing function tilde of minus e that will come only once, but to keep an overall factor half outside we also write q tilde minus e times f tilde e and I can now put a big square bracket outside ok. So, the half multiplies all of this. Ultimately it can also be again made a Gaussian. So, if you can just look at the structure of this aside from a time derivative there which in the energy domain will just become multiplication by energy. This is quadratic in q or q tilde variable and this is linear in q. So, all you have to do is a term that is of the form f squared and you can make it into a complete square. So, that is what we do by choosing q tilde e to be equal to small q tilde divided by right. So, we have this in front you pull this out it will go below these. So, you need let us fix what this has to be let us see what all is required to get it right. So, if I just take q tilde e and q tilde minus e big q tilde minus e then does the product work out to be that. So, I pulled out this. So, it will give this term correctly and the cross term has to be q tilde e f tilde minus e divided by this which is correct and this divided by this times this. So, I will get both these terms right. So, this is the correct definition. So, we will have to just subtract the square part of this term. So, we find integral of this q f infinity q i minus infinity propagator becomes a kernel becomes integral d q of e raise to i times half into q tilde e. So, this we have to change where this we have to change to q tilde eventually, but let us write it like this right now q tilde e q tilde minus e q tilde e times. So, this way which we had taken out of course, will multiply the whole thing. So, into e squared minus omega squared plus i epsilon times q tilde minus e. So, that is one term d e and so, we will try to make it q tilde e, but with some Jacobian. So, let me just say times a Jacobian, but here we get. So, that got cleaned up and all we get is a this squared term has to be subtracted and so, in the exponent minus. So, we will we have to draw some or we can write times because it does not matter times i times and no and the half factor because that half factor is existing here as well. So, i times a half times integral d e of and this f tilde e f tilde minus e and actually the half factor will go divided by e squared minus omega squared plus i epsilon. So, put minus sign here and now I have to argue to you that this Jacobian. So, actually that this Jacobian is one that is what we want to check right. So, what we observe is that this shift amounts to I can now Fourier transform it back. So, then it will become q of t equal to all q of t. So, now, note what we have done is just take Fourier transform of this back. So, we get capital Q of big q of t equal to the small q of t and plus this Fourier transform d e over square root 2 pi e raise to i e t f tilde e over the e squared. So, this Jacobian is one because this has no dependence on q. So, one in whichever dimensional space you want to think of some kind of infinite dimensional space, but it is still identity and therefore, likewise well I need not actually be too ambitious right now. All I have to do is make sure that I have changed variables from I can still write without any loss of generality q capital Q of t that is all I write. We know that this is function of e and this is t, but we know there is a relation. So, we can always work out main thing is that we want to move over to doing an integration over big q rather than small q. If we try to go from time domain to the energy or frequency domain, then again the Jacobian will be variation of this with respect to variation of g tilde which will leave behind a delta function of 0 or something. You know some I will define quantity, but which does not depend on q itself. So, whatever that Jacobian is is a constant ok. So, this constant overall constant I am not going to worry about we will leave it outside. I do have some note here that says d q d t that is done in time domain. So, I will not check what happens if you do the better idea. The point is you leave this measure in the time domain this you can always convert back and it will give you the usual t expression, but in q good. So, we do this only this the big Jacobian we do only in the known variables and known time domain and now observe that the front first factor the integral f tilde f tilde over e squared minus omega squared does not depend on of q and the q part. So, that or we can write the whole formula. So, that this expression q f infinity q i minus infinity is just going to be equal to this expression right. Because here I can certainly convert back to time domain without worrying and it will give me back my old Lagrangian except in the variable big q instead of small q and times this additional exponent e raise to i times the this term. We will next try to convert that term also to time domain. This front factor is simply seen to be the answer when forcing function is 0. So, this is with forcing function f you can put it here and we have dropped our memory of Dirac some point, but there are Dirac bases. And so, that becomes equal to i over 2 and now we claim that this is going to be in time domain equal to f right f of t f of t prime. So, both d t and d t prime where the d is the time domain propagator. In other words just thinking in terms of Fourier transform this thing which is 1 over e square minus omega square plus i epsilon we think of that as d tilde of E ok. This is easy to check right because there is both d t and d t prime. So, if you if you take this d t minus t prime and insert it here then the denominator will not depend on the t or t prime, but you can take e raise to i e t to this with this f of t and e raise to minus i e t prime with f t prime and do the d t and d t prime integrals you will just get f tilde of t and f tilde of minus f tilde of E and f tilde of minus E and be left with the denominator as is and 1 d E integration. So, this is same as this and so, that is an elegant way of recasting and this is our answer sort of and this for the force harmonic oscillator. So, sometimes we will write a short form for it which is also equal to this arrow goes there. So, notation is here f of argument 1 d of 1 and 2 and second f of argument 2 and then integral over t and t prime. So, this we will use as a notation for this and we can we are still we still seem to be stuck with this q f and q i which look arbitrary. So, what is actually done is also to extend the space range. So, we then define yes. So, what we do is introduce a vacuum to vacuum amplitude just for the possibility that after you have done some drama the two vacuum may differ by some overall phase. So, they may not be exactly identical, but they will be same they will both we normalize to 1, but this then is integral d q f d q i omega infinity q f infinity then the q f infinity q i minus infinity n times q i minus infinity q i minus infinity 2 omega minus infinity. This vacuum to vacuum amplitude is a very important concept. So, that is so, we can show that this also becomes equal to similar kind of thing and overall thing without any j. So, this in the presence of any forcing function becomes. So, suppose we have we do this with the presence of f then we would have the presence of f over here, but f is supposed to switch off at plus and minus infinity. In fact, f is in time domain. So, we should have said it there at t equal to and therefore, it does not appear here does not afflict the overlap over there. So, then that factor that we have there the f d f just comes out of the whole thing and we get that this equals e raise to i over 2 minus i over 2 what I have minus i over 2 f d f, but this is what we wanted because we had made that proposal that any averages can be calculated by varying this now with respect to this external current. So, f is sometimes well right now it is not a current it is just a forcing function in field theory it becomes a current. Many of these ideas were actually developed by Schwinger and they have eventually come back into the although in early developments for sometime it was ignored. So, this idea of constructing vacuum to vacuum amplitude writing a generating functional and by introducing external forcing function with respect to which you can vary now d f varying with respect to d f will bring down a q. So, you can calculate a q q correlator by varying with respect to f twice. So, this can be used. So, this is the this is now the average this is not quantum mechanics. So, and i is because there are i minus i square and if you have any operators all you have to do is write whatever power series you have in q you just construct that same power series in variations with respect to f. So, you will get higher powers of variation and now that we have committed so many sins we even if you have a transcendental function all you do is replace wherever you see a q by i d by d f is the same answer. So, you can calculate the average value of any operator or products of such operators by doing this.