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  Problems while Wick rotating the path integral

+ 3 like - 0 dislike

I have engaged in the derivation of the Euclidean path integral performing the Wick rotation in full detail. Unfortunately I am facing some trouble and I come here seeking advice.

For simplicity I work on 1 dimension and in god-given units. The amplitude for a spinless particle of unit mass to go from the point $x_i$ to the point $x_f$ in a time interval $T$  is given by
\int D[x]\,e^{i\int_0^Tdt\,\mathcal{L}(t)}
=\int D[x]\,e^{i\int_0^Tdt\,\big\{\frac{1}{2}\left(\frac{dx}{dt}\right)^2-V(x)\big\}}
let's focus on the integral on the exponent
to get the Euclidean path integral i gotta Wick-rotate this. In order to do this i write the Lagrangian for a general complex variable $z=t+i\beta$
and I consider the contour

I also assume (maybe naively) that there is no pole in bothering us for $\mathcal{L}(z)$. Cauchy's theorem allows us to write
Let's go one by one. For $L_R$ I parametrize $z(t)=t$
For $L_I$ I have $z(\beta)=i\beta$
for $C$ i have $z(\phi)=Te^{i\phi}$
by Cauchy's theorem then
if I plug this in the path integral I get
\int D[x]\,e^{i\int_0^Tdt\,\big\{\frac{1}{2}\left(\frac{dx}{dt}\right)^2-V(x)\big\}}
=\int D[x]\,e^{\int_0^Td\beta\,\big\{\frac{1}{2}\left(\frac{dx}{d\beta}\right)^2+V(x)\big\}}e^{-i\int_{C}dz\,\mathcal{L}(z)}
and you see the problem here. I lack a minus sign in the first exponential, and the second one shouldn't be there. Maybe I can get the correct expression by manipulating the second exponential but I right now I don't see how. Can anyone right my wrongs?

asked Aug 5, 2016 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ revision history ]
retagged Aug 5, 2016 by Dilaton

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