# From Minkowski to Euclidean Time in Path Integrals

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 $\mathcal{N}$?
• I think you might've forgotten to change the measure: $dt=id \tau$
– Danu
Mar 30 '14 at 16:31
• Sorry, I might have confused you. I edited the change of variables to reflect what I originally meant.
– PPR
Mar 30 '14 at 20:29
• 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!
– Nick
May 4 '14 at 14:40
• It seems like this was answered here for QFT: link.springer.com/article/10.1007/BF01645738 and a series of related articles.
– 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 $\epsilon = \frac{z''-z'}{N}$.

Therefore,

• 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.

• 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?
– PPR
May 11 '14 at 13:24
• 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 May 11 '14 at 17:42
• 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. May 11 '14 at 17:46