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In quantum mechanics, the amplitude of wave function propagation can be found using the Feynman's path integral $$ \langle z'|e^{-itH/\hbar}|z\rangle=\int\limits_{x(0)=z\\x(t)=z'} Dx(t')\: \exp\left\{\frac{i}\hbar\int_0^t dt'\:\left[\frac{m\dot{x}^2(t')}2-V(x(t'))\right]\right\}. $$

In the (quasi)classical limit $\hbar\rightarrow0$, the leading contribution to the integral comes from the classical trajectory $$ \frac{d}{dt'}\left[\frac{m\dot{x}^2(t')}2\right]-\frac{d}{dx}\left[V(x(t'))\right]=0, $$ where the action is minimal, and fluctuations around this trajectory provide quantum corrections to the result.

In quantum statistical physics, the path integral can be used to calculate matrix elements of a thermal density matrix by switching to the imaginary time $\tau=it/\hbar$: $$ \langle z'|e^{-\beta H}|z\rangle=\int\limits_{x(0)=z\\x(\beta)=z'} Dx(\tau)\: \exp\left\{-\int_0^\beta d\tau\:\left[\frac{m\dot{x}^2(\tau)}2+V(x(\tau))\right]\right\}. $$

What is the physical meaning of a least-action trajectory in the imaginary time? What do fluctuations around this trajectory mean and how do they qualitatively affect the resulting matrix elements?

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2 Answers 2

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Here we will comment on one aspect of OP's question, which can be phrased as follows:

What is the connection between the WKB approximations for the path integral in Minkowski vs. Euclidean time?

That's a great question. Let us consider the semiclassical limit $\hbar\to 0^{+}$.

  1. One one hand, in Minkowski time, the path integral $$\begin{align} Z_M&:=~\int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar} S_M[\phi]\right\} \cr &\sim~\sum_{\phi_{\rm cl}}\frac{1}{\sqrt{\det(\ldots)}}\exp\left\{\frac{i}{\hbar}S_M[\phi_{\rm cl}]\right\} \quad\text{for}\quad\hbar~\to~0^+ \end{align}\tag{1}$$ is dominated by stationary configurations $\phi_{\rm cl}$, i.e. instantons, cf. the stationary phase approximation. See also this related Phys.SE post. The square root determinant in the denominator indicates a Gaussian integral of quantum fluctuations around each instanton.

  2. On the other hand, in Euclidean time, the path integral $$\begin{align} Z_E&:=~\int\!{\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{- \frac{1}{\hbar} S_E[\phi]\right\} \cr &\sim~\sum_{\phi_{\min}}\frac{1}{\sqrt{\det(\ldots)}}\exp\left\{-\frac{1}{\hbar}S_E[\phi_{\min}]\right\} \quad\text{for}\quad\hbar~\to~0^+ \end{align}\tag{2}$$ is apparently dominated by the global minima $\phi_{\min}$ for the Euclidean action, cf. the method of steepest descent. Everything else is exponentially suppressed.

The two methods 1 & 2 seem quite different: Stationary points are not the same as global minima points.

However, according to physics lore, granted pertinent analytic properties of the integrand, they can be connected via Wick rotation. Famously, this is easier said than done, see e.g. this Phys.SE post.

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  • $\begingroup$ Thank you for the answer! However it does not address directly my question: what is the physical meaning of trajectories where the Euclidean action is minimal? It seems that, acting similarly to derivation of Lagrange equations (only the sign of $V(x)$ changes), we can get the equation for this trajectory like $m\ddot{x}=V'(x)$, which looks weird. $\endgroup$ Commented Oct 16, 2017 at 19:20
  • $\begingroup$ The Euclidean Lagrangian looks like a standard Lagrangian (i.e. kinetic term minus potential term), with an apparent potential equal to minus $V$, see e.g. this and this Phys.SE posts. $\endgroup$
    – Qmechanic
    Commented Oct 16, 2017 at 20:10
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I think you should have a better understanding of the imaginary time formalism first. Below is my understanding of this problem.

When we deal with finite temperature field theory problem, we usually go to the imaginary time domain and use Matsubara frequencies further. How to understand it physically? Roughly speaking, we can say that we convert thermal fluctuation to quantum fluctuation. First, we need to understand what is thermal fluctuation. Without thermal fluctuation, the configuration is unique. Because of temperature, the system can have other configuration with probability: $$p\propto e^{-\beta E} $$ This is what we mean by thermal fluctuation. Note there is no time-dependent because we care about the equilibrium properties. But for quantum fluctuation, we have dynamics and it says that $\varphi(x,t)$ has probability: $$ \varphi(x,t)\propto e^{iS(\varphi(x,t))/\hbar} $$ So, roughly speaking, when we go to imaginary time, we convert thermal fluctuation($T$) into quantum fluctuation($\text{ dynamics: } \tau$): $$ \text{Tr} e^{-\beta H }=\int \mathcal{D}(\bar{\psi},\psi) e^{-\frac{1}{\hbar} \int^{\beta \hbar} d\tau L(\tau)} $$

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  • $\begingroup$ Thank you! However I'm interested here in more specific things: 1) what trajectory provides a major contribution to $\langle z'|e^{-\beta H}|z\rangle$ in the classical ($\hbar\rightarrow0$) and zero-temperature ($\beta\rightarrow\infty$) limits? 2) What are the physical meaning of fluctuations around this trajectory? 3) Quantum fluctuations above which state? How are they entangled/disentangled in the imaginary-time path integral? $\endgroup$ Commented Aug 27, 2018 at 15:28

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