Edit
It seems that I haven't written my question clearly enough, so I will try to develop more using the example of quantum tunnelling. As a disclaimer, I want to state that my question is not about how to perform a Wick rotation in the path integral formulation!
Let's look a the probability of quantum tunnelling in the path integral formulation. The potential is given by $V[x(t)]=(x(t)^2-1)^2$, which has two minima at $x=\pm x_m=\pm1$. Given that the particle starts at $t=-\infty$ at $x=-x_m$, what is the probability that it is at $x=x_m$ at $t=\infty$. The probability amplitude is given by
$$K(x_m,-x_m,t)=\langle x_m|e^{-i \hat H t}|-x_m \rangle$$
The usual trick is to Wick rotate $t\to-i\tau$, compute everything in imaginary time using a saddle point approximation and at the end of the calculation rotate back to real time. I understand how it works. No problem with that.
What I want to understand is
- how can I do the calculation without using the Wick rotation?
- how does this solution connect to the Euclidean formulation?
In principle, we should be able to do the calculation with the path integral formulation in real time
$$\int Dx(t) e^{i S[x(t)]/\hbar}$$
In the stationary phase approximation we look for a complex path $x(t)$ which minimizes the action, and expand about this point.
Choose $m =1$ for simplicity. The equation of motion is
$$\ddot x-2 x+2x^3=0$$
which has no real solution, i.e. no Newtonian (classical) solution. But there is a complex function that solves it: $x_s(t)=i \,\tan(t)$. One problem is that it behaves pretty badly. If anyway I accept this a correct solution, I should be able to compute the gaussian fluctations, add up all the kinks/antikinks, etc. and recover the correct result (usually obtained with the euclidean action and $\tau\to -it$). Am I right?
So my question is: is it possible to do the calculation that way, and if so, how is it related to the trick of going back and forth in imaginary time?
Original
I have a question on the mathematical meaning of the Wick rotation in path integrals, as it is use to compute, for instance, the probability of tunneling through a barrier (using instantons).
I am aware that when computing an ordinary integral using the Stationary Phase Approximation
$\int dx e^{i S(x)/\hbar}$
with $x$ and $S$ real, one should look at the minimum of $S(z)$ in the whole complex plane, which can be for instance on the Imaginary axis.
In the case of a path integral, one wants to compute
$\int Dx(t) e^{i S[x(t)]/\hbar}$
and there is a priori no reason that the "classical path" from $x_a(t_a)$ to $x_b(t_b)$ (i.e. that minimizes $S[x(t)]$) should lie on the real axis. I have no problem with that. What I don't really get is the meaning of the Wick rotation $t\to -i\tau$ from a (layman) mathematical point of view, because it is not as if the function $x(t)$ is taken to be imaginary (say, $x(t)\to i x(t)$), but it is its variable that we change !
In particular, if I discretize the path-integral (which is what one should do to make sense of it), I obtain
$\int \prod_n d x_n e^{i S(\{x_n\})/\hbar}$.
where $S(\{x_n\})=\Delta t\sum_n\Big\{ (\frac{x_{n+1}-x_n}{\Delta t})^2-V(x_n)\Big\}$
At this level, the Wick rotation applies on the time slice $\Delta t\to -i\Delta \tau$ and does not seem to be a meaningful change of variable in the integral
I understand that if I start with an evolution operator $e^{-\tau \hat H/\hbar}$ I will get the path integral after Wick rotation, but it seems to be a convoluted argument.
The question is : Is it mathematically meaningful to do the Wick rotation directly at the level of the path-integral, and especially when it is discretized?
Best Answer
I think the path-integral is a complete red herring here! I'll try to convince you that Wick rotation yields completely equivalent way of writing the Lagrangian in classical field theory.
Consider a classical action
$$S[x] = \int L[x(t)] dt$$
where $x:\mathbb{R} \to \mathcal{M}$ for some target manifold $\mathcal{M}$. The Lagrangian is schematically given by
$$L[x(t)] = \left(\left.\frac{dx(s)}{ds}\right|_{s=t}\right)^2-V(x(t))$$
where $V(x(t))$ is some polynomial in $x(t)$ which (critically) involves no derivatives.
Now analytically continue $x$ to a function $\tilde{x}: \mathbb{C}\to\mathcal{M}$ by defining $\tilde{x}(c) = x(|c|)$ which is obviously analytic. Relabel $\tilde{x}$ as $x$ for simplicity. Define a new variable
$$\tau = it$$
inside the integral, and substitute. (Warning: there are mathematical subtleties about complex substitutions, which should be dealt with using Jordan's lemma). Ignoring the subtleties, the resulting integral is
$$S[x] = \int L[x(-i\tau)](-i)d\tau$$
Now let's examine what happens to the Lagrangian more carefully. Looking at the first term we have
$$\left(\left.\frac{dx(s)}{ds}\right|_{s=-i\tau}\right)^2$$
Change the differentiation variable to $u = is$ and this term becomes
$$\left(i\left.\frac{dx(u)}{du}\right|_{u=\tau}\right)^2$$
Relabelling $u\to s$ we see that the first term has the same form as originally, but with $t$ replaced by $\tau$ and an extra minus sign, viz.
$$-\left(\left.\frac{dx(s)}{ds}\right|_{s=\tau}\right)^2$$
Now on to the potential term. This is much simpler because $x(-i\tau)=x(|-i\tau|)=x(\tau)$ by definition so the potential term is just
$$V(x(\tau))$$
which is of exactly the same form as originally. Now we define a Euclidean Lagrangian
$$L_E(x(\tau)) = \left(\left.\frac{dx(s)}{ds}\right|_{s=\tau}\right)^2+V(x(\tau))$$
Putting it all together we find
$$S[x] = i\int L_E[x(\tau)]d\tau$$
Finally defining
$$S_E[x] = \int L[x(t)]dt$$
we see that it's mathematically equivalent to calculate the path integral as
$$\int \mathcal{D}x\exp(iS[x]) \textrm{ or } \int \mathcal{D}x\exp(-S_E[x])$$