I'm trying to prove the following equality: $$ <x_{f},\, it_{f}|x_{i},\, it_{i}>=\mathcal{N}\int_{\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(t_{f}\right)=x_{f}\wedge x\left(t_{i}\right)=x_{i}\right\} }\mathcal{D}x\exp\left\{ -\frac{1}{\hbar}\int_{t_{i}}^{t_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}-\left(-V\left[x\right]\right)\right\} \right\} $$ where the definition of $ <x_f, t_f|x_i, t_i>$ is given by:

$$ <x_{f},\, t_{f}|x_{i},\, t_{i}>\equiv\mathcal{N}\int_{\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(t_{f}\right)=x_{f}\wedge x\left(t_{i}\right)=x_{i}\right\} }\mathcal{D}x\exp\left\{ \frac{i}{\hbar}\int_{t_{i}}^{t_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}-\left(V\left[x\right]\right)\right\}\right\} $$

What I have done so far:

  • Assume that the domain of integration, that is, $\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(t_{f}\right)=x_{f}\wedge x\left(t_{i}\right)=x_{i}\right\}$ is such that all the functions in this set can be analytically continued to a new domain of integration $ \left\{ x\in\mathbb{C}^{\mathbb{C}}:\, x\left(it_{f}\right)=x_{f}\wedge x\left(it_{i}\right)=x_{i}\right\} $. Is this valid?

  • Plug in the definitions: $<x_{f},\, it_{f}|x_{i},\, it_{i}>=\mathcal{N}\int_{\left\{ x\in\mathbb{C}^{\mathbb{C}}:\, x\left(it_{f}\right)=x_{f}\wedge x\left(it_{i}\right)=x_{i}\right\} }\mathcal{D}x\exp\left\{ \frac{i}{\hbar}\int_{it_{i}}^{it_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}-\left(V\left[x\right]\right)\right\}\right\}$

  • Now to compute $ \int_{it_{i}}^{it_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}-\left(V\left[x\right]\right)\right\} $ make a change of variable (is this valid?? Don't you need Cauchy's theorem and also to assume that the time boundaries go to infinity?) $t \mapsto -it$ to get: $i \int_{t_{i}}^{t_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}+V\left[x\right]\right\} $ so you get the correct exponent.

  • But how do you prove that $\mathcal{N}\int_{\left\{ x\in\mathbb{C}^{\mathbb{C}}:\, x\left(it_{f}\right)=x_{f}\wedge x\left(it_{i}\right)=x_{i}\right\} }\mathcal{D}x = \mathcal{N}\int_{\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(t_{f}\right)=x_{f}\wedge x\left(t_{i}\right)=x_{i}\right\} }\mathcal{D}x$? Is it even the same $\mathcal{N}$?
  • $\begingroup$ I think you might've forgotten to change the measure: $dt=id \tau$ $\endgroup$
    – Danu
    Mar 30 '14 at 16:31
  • $\begingroup$ Sorry, I might have confused you. I edited the change of variables to reflect what I originally meant. $\endgroup$
    – PPR
    Mar 30 '14 at 20:29
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    $\begingroup$ Just a fair remark, be cautious with Wick-rotations, when you have vectorpotentials $q\vec{x}\cdot\vec{A}$ this won't give a neat, simple real action! $\endgroup$
    – Nick
    May 4 '14 at 14:40
  • $\begingroup$ It seems like this was answered here for QFT: link.springer.com/article/10.1007/BF01645738 and a series of related articles. $\endgroup$
    – PPR
    Jul 26 '14 at 8:38

Carefully following Feynman's procedure one actually finds: $$\langle x''|e^{-i\frac{(z''-z')}{\hbar}H}| x' \rangle $$ $$=\langle x'', z''|x', z'\rangle =\lim_{N\to \infty\: \epsilon \to 0} \left[\frac{m}{2\pi i \hbar \epsilon} \right]^{N/2}\int_{-\infty}^{+\infty}\cdots \int_{-\infty}^{+\infty} \left(\prod_{i=1}^{N-1} dx_i \right) \exp\left\{\frac{i\epsilon}{\hbar} \sum_{j=0}^{N-1} \left[ \frac{m}{2}\left(\frac{x_{j+1}-x_j}{\epsilon} \right)^2-V(x_j)\right]\right\}\quad (1)$$ where $z',z'' \in \mathbb C$, and $\epsilon = \frac{z''-z'}{N}$.


  • the procedure encompasses the general case of complex time lapse $z''-z'$;

  • $\epsilon$ is a complex number of the same nature as that of $z''-z'$ and this fact is responsible for both the change of sign in front of the kinetic energy, passing from Lorentzian to Euclidean formalism, and the disappearance of the overall factor $i$ in front of the action in the same situation.

As the formula (1) holds for generally complex time $z$, the "Wick rotation" is automatic: It is nothing but the specification of the nature of $z$, real or imaginary.

In (1), there are no true paths parametrized by the parameter $z$, however you are free to interpret $x_{j}$ as a possible position at complex time $z_j = z'+ j\epsilon$. Actually an effective (powerful I might say!) interpretation is that the sum is computed along the class of all such "broken" paths, in view of the integrations in $dx_i$ connecting in all possible ways $x_j$ and $x_{j+1}$.

In the limit as $N\to +\infty$ one expects that these paths become smooth (actually the story is different, since the set of smooth paths has zero measure...) and, formally, one writes down the said limit as $$ <x'',\, z''|x',\, z'>=\mathcal{N}\int_{\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(z''\right)=x''\wedge x\left(z'\right)=x'\right\} }\mathcal{D}x\exp\left\{ \frac{i}{\hbar}\int_{z'}^{z''}dz\left[ \frac{1}{2}m\left(\frac{dx}{dz}\right)^{2}-V\left(x\right)\right] \right\}\:.$$

You see, in particular, that the Euclidean factor $\mathcal{N}$ has to be interpreted as the analytic continuation of the Lorentzian one since $$\mathcal{N} = \lim_{\epsilon \to 0} \left[\frac{m}{2\pi i \hbar \epsilon} \right]^{N/2}$$ and $\epsilon$ depends on the nature of the considered notion of time.

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    $\begingroup$ Thanks for your answer. It has certainly helped me understand a few things, I guess the most important of which is that the functional-integral notation is merely symbolic and acquires concrete meaning only when discretizing time and taking the limit. However, from your answer it would appear as if there is no non-trivial step happening when going to Euclidean spacetime, which is something I was under the impression is completely non-trivial. For instance, when would going to Euclidean spacetime breakdown, and how do you see it in this way of developing the path-integral? $\endgroup$
    – PPR
    May 11 '14 at 13:24
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    $\begingroup$ Actually there are many nontrivial steps passing from Lorentzian to Euclidean formalism. For instance, you see that each integral in $dx_i$ must be interpreted, in general, in the distributional sense, since the integrand function is oscillating but not, absolutely integrable, in the Lorentian case. Conversely, it is a true integral in the Euclidean sense as the function rapidly vanishes for large arguments in the exponent. This case can be handled with a true infinite dimrnsional measure. Generally speaking, the functional integral is a good mathematical object if time is complex but not real $\endgroup$ May 11 '14 at 17:42
  • $\begingroup$ An effective procedure is computing the functional integral for complex time, using such regularized propagator in computations, and removing the imaginary part of time just as last step. $\endgroup$ May 11 '14 at 17:46

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