Answer by Qmechanic for Getting Feynman propagator using path integral
Consider a free/quadratic action$$S_2[\phi] ~=~\frac{1}{2} \phi^k (S_2)_{k\ell} \phi^{\ell}~=~\frac{1}{2}\int\!d^4x\int\!d^4y~\phi(x) S_2(x,y)\phi(y). \tag{1}$$Here we are using DeWitt condensed...
View ArticleAnswer by Prof. Legolasov for Getting Feynman propagator using path integral
Yes, it can be easily derived.First some notation.We denote by $\left< \Omega \right>$ (the quantum expectation value) the path integral$$ \left< \Omega \right> = \int \mathcal{D} \phi \,...
View ArticleGetting Feynman propagator using path integral
In QM using Feynman path integral(FPI) we derive the propagator of free particle which comes out to $$(f(t))e^{iS_{cl}/\hbar}.$$But in QFT the Feynman propagator is derived using the differential...
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