Path integral and imaginary time in Quantum and Statistical Mechanics I have come across the path integral formulation of quantum mechanics, and have found plenty of websites, papers and book chapters explaining the relation to statistical mechanics. The general reasoning (to my understanding) seems to be:
Quantum Mechanics states that the probability of a particle transitioning from $A$ to $B$ is, by slight abuse of notation, proportional to 
$$
\left(\sum_{\text{all paths $A\rightarrow B$}} \exp(\frac{\mathrm i}{\hbar} \mathcal S[A\rightarrow B])\right)^2
$$
where $A\rightarrow B$ is the path, and $\mathcal S$ is its action. The argument being that paths of non-stationary $\mathcal S$ cancel out.
Statistical Mechanics state that the probability of a system being in state $B$ is proportional to
$$
  \exp(-\frac{1}{k_{\mathrm B} T}E(B))
$$
where $k_{\mathrm B}$ is Boltzmann's constant, $T$ is temperature and $E(B)$ is the energy or Hamiltonian of B.
The connection between the two is usually established in one of two ways:


*

*The Wikipedia article states, that in evaluating the QM equation, it is convenient to forget about interfering waves and just replace the $\mathrm i$ by $(-1)$, letting the high $\mathcal S$ values vanish by small numbers, instead of by interference. I can see how that is convenient, but I don't see why it is correct or a valid approximation. The best I could come up with as a metric of how the effect of interference scales with $\mathcal S$ is $p(\mathcal S) \propto 1/\mathcal S^2$, in various ways, but most basically in the $\operatorname{sinc}(x)$ pattern found in diffraction.

*The most common approach seems to be, that to make the two equations match, we have to switch to "imaginary time" and replace time $t$ by inverse complex temperature $\hbar/\mathrm i T k_{\mathrm b}$, labeled as Wick rotation. Now, again, I can see why this is very convenient indeed, but asides from the mathematical effect that it is some operation on a formula that gives the desired result, I fail to see any physical justification.
My basic problems are these:


*

*I understand that using time as a complex variable is justified in relativity. The above models however, to my understanding, are non-relativistic.

*Even if it were complex, why would be be proportional to temperature? In statistical mechanics, $T$ is most often considered constant in a given scenario. $t$, in contrast, is never constant. Is replacing time by temperature (complex or not) not the same as requiring temperature to always vary?

*The point of Wick's rotation seems to be not to change the meaning of a formula, but to change the format. So the Minkowski metric "$x^2-t^2$ becomes the Euclidean metric $x^2 + (\mathrm i t)^2$. We changed $-$ to $+$ and $t$ to $\mathrm i t$, and the formula remains the same. In the above example, it looks like we are only changing $t$ to $\mathrm i t$, but leaving the rest of the formula the same - hence the "rotation" does not cancel out, it is a different formula. E.g. the Wick rotation is not $[x^2-t^2] \rightarrow [x^2 - (\mathrm i t)^2]$, why should be it $\exp(\mathrm i t E) \rightarrow \exp(-t E)$? 

*Even if all this was possible, I don't see why the Hamiltonian can just be replaced by the Lagrangian in the general case (nonzero potential).


It seems that whatever text I found about this issue is simply content with making the expressions match by whatever transformation is necessary. There is never (as far as I found) any physical justification or interpretation, only that the result matches (or maybe the connection is obvious and I fail to see it). To me it seems arbitrary.
In essence, what puzzles me is that I would be happy to use the expression
$$
p(A\rightarrow B) \propto \exp(-\text{const} \cdot \mathcal S[A \rightarrow B]),
$$
and it seems that everybody (= lots of people? some?) does use that, but I can't seem to find out what the justification is, or under what limiting (but valid) assumptions this could be true and why.
Any help or suggestions would be most appreciated. Thanks!
 A: 
Blockquote
  In essence, what puzzles me is that I would be happy to use the expression
  $$p(A\rightarrow B) ∝ \exp(−const⋅\mathcal{S}[A\rightarrow B])$$

This is not very accurate, you should try to link the partition function of a classical system with the path integral of a quantum system. So the correct expression would be:
$$S ∝_{up \ to \ wick \ rotation} \sum_{paths} exp{ \frac{i}{\hbar}S[path]}$$
How do we obtain such a result ?
Let us start from the partition function of some system.
$$Z=\sum_{states}exp{ -\beta E(state) }$$
If we instroduce some complete basis for the states $|\psi>$ than this can be rewritten as:
$$Z = \sum_{|\psi>}<\psi|exp({-\beta H})\ |\psi> = \int d{|\psi>} <\psi|exp({-\beta H})\ |\psi>$$
now we will perform the wick rotation, if we close our eyes to the details this simple means substituting $\beta = \frac{1}{kT} \rightarrow it$ which is just a simple change of variables. We find:
$$Z = \int d|\psi><\psi|exp(-iHt)|\psi>$$ 
But Feynmans sum over histories path integral tells us that: $<x|e^{-iHt}|x'> = \int_x^{x'}[dx]e^{i\int_0^t dt' L(t')}$ such that we can rewrite this as:
$$Z = \int d|initial conditions \ \psi>\ \int_{|\psi>}^{|\psi>}d[full configurations\ |\psi>]e^{i\int_0^{\beta} dt' L(t')}$$ 
The last step is some interpretation. 
We see that the configuration at $t=0$ and $t = \beta$ must be identical as we evolved from $|\psi>$ at t=0 to $|\psi>$ at $t=\beta$. This is implemented by requiring time to be a periodic (=compactified) dimension with radius $\beta$.
We also see that we are integrating over all posible intial conditions, this is typically implemented into the path integral with the notation $\int d|\psi>\int_{|\psi>}^{|\psi>}[d|\psi>] = \int_[d|\psi>]$.
We also replace $it \rightarrow t_E$ which indeed corresponds to changing to euclidian signature henche the subscript E. 
$$Z = \int[d|\psi>]e^{-\int_0^\beta L_E dt_E}$$
Where $L_E$ is the Euclidian lagrangian obtained via the subsitution $t=it_E$ for example:
$$T = \frac{dx}{dt}\frac{dx}{dt} \rightarrow T_E = \frac{dx}{idt_E}\frac{dx}{idt_E} = -\frac{dx}{dt_E}\frac{dx}{dt_E}$$
$V \rightarrow V_E = V$
Such that, as you meantioned: $L_E = -H_{Minkowski}$
Disclaimer: I might have missed some signs here and there, but the above explanation should definitely help you to understand what is going on !
