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How can I show that $$S_E[x]=\int_{t_i}^{t_f} dt \left(\frac{m}{2}\dot{x}^2+V(x)\right),$$ starting from the definition of transition amplitude $$A=\langle x_f\,|\,e^{-\frac{i}{\hbar}(t_f-t_i)\hat{H}}\,|\, x_i\rangle?$$

Here you are my try. I am completely following the Feynman derivation for the phase space path integral. Starting with Wick rotation $t \rightarrow -i\tau$ I got $$A=\langle x_f\,|\,e^{-\frac{1}{\hbar}(\tau_f-\tau_i)\hat{H}}\,|\, x_i\rangle=\langle x_f\,|\,e^{\left(-\frac{1}{\hbar}\frac{(\tau_f-\tau_i)}{N}\hat{H}\right)^N}\,|\, x_i\rangle$$

and using $\epsilon=\frac{\tau_f-\tau_i}{N}$ I got $$\int\left(\prod^{N-1}_{k=1}dx_k\right)\langle x_k|e^{-\frac{\epsilon}{\hbar}\hat{H}}|x_{k-1}\rangle=\int\left(\prod^{N-1}_{k=1}dx_k\right)\left(\prod^{N}_{j=1}\frac{dp_j}{2\pi\hbar}\right)\langle x_k|p_j\rangle \langle p_j|e^{-\frac{\epsilon}{\hbar}\hat{H}}|x_{k-1}\rangle$$ then I use the approximation $e^x=1+x+O(x^2)$ neglecting $O(x^2)$ so it becomes $$\int\left(\prod^{N-1}_{k=1}dx_k\right)\left(\prod^{N}_{j=1}\frac{dp_j}{2\pi\hbar}\right)\langle x_k|p_j\rangle \langle p_j|x_{k-1}\rangle\left(1-\frac{\epsilon}{\hbar}H(x_{k-1},p_j)\right)\approx\int\left(\prod^{N-1}_{k=1}dx_k\right)\left(\prod^{N}_{j=1}\frac{dp_j}{2\pi\hbar}\right)e^{\frac{i}{\hbar}p_jx_{k}}e^{-\frac{i}{\hbar}p_jx_{k-1}}e^{-\frac{\epsilon}{\hbar}H(x_{k-1},p_j)}=\int\left(\prod^{N-1}_{k=1}dx_k\right)\left(\prod^{N}_{j=1}\frac{dp_j}{2\pi\hbar}\right)e^{\frac{i}{\hbar}p_j(x_{k}-x_{k-1})-\frac{\epsilon}{\hbar}H(x_{k-1},p_j)}$$ and so I have the exponential in the form: $$e^{\frac{i}{\hbar}p_j(x_{k}-x_{k-1})-\frac{\epsilon}{\hbar}H(x_{k-1},p_j)}=e^{-\frac{1}{\hbar}\epsilon\left(p_j\frac{(x_{k}-x_{k-1})}{-i\epsilon}+H(x_{k-1},p_j)\right)}$$

EDIT: I just noticed that using the Wick rotation the lagrangian becomes exactly what I was looking for since $\partial_t\rightarrow i\partial_{\tau}$ and this means $$\frac{m}{2}{\partial_t x}^2-V(x)\rightarrow -\left(\frac{m}{2}{\partial_{\tau} x}^2+V(x)\right)$$ but how could I handle the minus sign?

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    $\begingroup$ Well done for trying. Where's your attempt? $\endgroup$
    – Rumplestillskin
    Dec 31, 2016 at 22:53
  • $\begingroup$ I added my attempt. Hope you're satisfied now :) $\endgroup$
    – ft1993
    Jan 1, 2017 at 0:37

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A Wick rotation of the action is fairly straightforward; there are some subtleties and a deeper meaning, but it is just a coordinate transformation. In particular, starting with,

$$S = \int dt \left[ \frac12 m \left( \frac{dx}{dt}\right)^2 - V(x)\right]$$

we perform the transformation, $t' = ct$ for $c \in \mathbb C$. Then $dt = \frac{dt'}{c}$ and notice $\frac{dx}{dt} = c \frac{dx}{dt'}$. Thus,

$$S = \int \frac{dt'}{c}\left[ \frac12 m \left( \frac{dx}{dt'}\right)^2 c^2- V(x)\right] = \int dt' \left[ \frac{c}{2} m \left( \frac{dx}{dt'}\right)^2 - \frac{1}{c}V(x)\right].$$

If we now choose, $c = i$, and relabel $t'$ as $\tau$, we have,

$$S = i\int d\tau \left[ \frac12 m \dot x^2 + V(x)\right] := iS_E.$$

This is why in the path integral, $e^{iS} \to e^{-S_E}$.

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  • $\begingroup$ Thank you for your answer, but what I don't understand is why the euclidean action is $S_E=\int d\tau \left[\frac{1}{2}m\dot{x}^2+V(x)\right]$... is this just a definition? $\endgroup$
    – ft1993
    Jan 1, 2017 at 0:48
  • $\begingroup$ @ft1993 Well, in the path integral, you end up with an overall minus sign times something that is basically an action, so it makes sense to take the minus out and define that quantity as the Euclidean action. This is not totally arbitrary as the Euclidean action is actually used to compute things, like in the study of false vacuum decay. In a theory where $S$ is based on a metric tensor, going from Minkowski space to Euclidean space means changing the metric to $\delta_{\mu\nu}$ and this certainly motivates calling $S_E$ the Euclidean action. $\endgroup$
    – JamalS
    Jan 1, 2017 at 0:56
  • $\begingroup$ Ok, so in this case you define $S_E$ as Euclidean action since it looks like an action, but when you properly do your calculation you actually see that it is the Euclidean action? Thank you! $\endgroup$
    – ft1993
    Jan 1, 2017 at 1:01
  • $\begingroup$ @ft1993 Well in quantum field theory you pass to the Euclidean metric by making a Wick rotation. In this case, you can interpret it as the same thing in a $1+0$-dimensional theory. It is the Euclideanised action. $\endgroup$
    – JamalS
    Jan 1, 2017 at 1:08
  • $\begingroup$ @JamalS You left out a crucial step — changing the contour of integration (i.e. $\tau\in (-\infty,\infty )$, or equivalently $t\in (+i \infty,-i \infty)$). This is arguably the most important part of the Wick rotation, otherwise you've just done a superficial change of variables (i.e. nothing). $\endgroup$
    – Arturo don Juan
    Apr 24, 2020 at 20:45

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