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# Dominant paths in the propagator and density matrix

\underlineLet us first consider the real time quantum propagator. The quantity appearing in the exponential is an integral of

${1 \over 2}m\dot{x}^2 - U(x) \equiv L(x,\dot{x})$

which is known as the Lagrangian in classical mechanics. We can ask, which paths will contribute most to the integral

$\int_0^t ds \left[{m \over 2}\dot{x}^2(s) - U(x(s))\right] =\int_0^t ds L(x(s),\dot{x}(s)) = S[x]$

known as the action integral. Since we are integrating over a complex exponential $$\exp({iS \over \hbar})$$, which is oscillatory, those paths away from which small deviations cause no change in $$S$$ (at least to first order) will give rise to the dominant contribution. Other paths that cause $$\exp({iS \over \hbar})$$  to oscillate rapidly as we change from one path to another will give rise to phase decoherence and will ultimately cancel when integrated over. Thus, we consider two paths $$x (s)$$ and a nearby one constructed from it $$x(s) + \delta x (s)$$ and demand that the change in $$S$$ between these paths be 0

$S[x+\delta x]-S[x]=0$

Note that, since $$x (0) = x$$ and $$x (t) = x'$$, $$\delta x (0) = \delta x (t) = 0$$, since all paths must begin at $$x$$ and end at $$x'$$. The change in $$S$$ is

$\delta S = S[x+\delta x]-S[x] =\int_0^t ds L(x+\delta x,\dot{x}+\delta\dot{x}) - \int_0^t ds L(x,\dot{x})$

Expanding the first term to first order in $$\delta x$$, we obtain

$\delta S = \int_0^t ds \left[L(x,\dot{x}) + {\partial L \over \partial \dot {x}} \delta \dot {x} + {\partial L \over \partial x } \delta x \right ] - \int _0^t L (x, \dot {x} ) = \int _0^t ds \left [{\partial L \over \partial \dot {x}}\delta\dot{x} +{\partial L \over \partial x}\delta x\right]$

The term proportional to $$\delta \dot {x}$$ can be handled by an integration by parts:

$\int_0^t ds {\partial L \over \partial \dot{x} }\delta\dot {x} = \int _0^t {\partial L \over \partial \dot {x}} {d \over dt} \delta x = {\partial L \over \partial \dot {x}} \partial x \vert_0^t - \int_0^t ds {d \over dt}{\partial L \over \partial \dot{x}}\delta x$

because $$\delta x$$ vanishes at $$0$$ and $$t$$, the surface term is 0, leaving us with

$\delta S = \int_0^t ds \left[-{d \over dt}{\partial L \over \partial \dot {x} } + {\partial L \over \partial x}\right]\delta x = 0$

Since the variation itself is arbitrary, the only way the integral can vanish, in general, is if the term in brackets vanishes:

${d \over dt}{\partial L \over \partial \dot{x}} - {\partial L \over \partial x}= 0$

This is known as the Euler-Lagrange equation in classical mechanics. For the case that $$L=m\dot{x}/2 - U(x)$$, they give

 ${d \over dt}(m\dot{x}) + {\partial U \over \partial x}$ $$\underline {0}$$ $\underline {m \ddot {x} }$ $$- {\partial U \over \partial x }$$

which is just Newton's equation of motion, subject to the conditions that $$x (0)$$, $$x (t) = x'$$. Thus, the classical path and those near it contribute the most to the path integral.

The classical path condition was derived by requiring that $$\delta S = 0$$ to first order. This is known as an action stationarity principle. However, it turns out that there is also a principle of least action, which states that the classical path minimizes the action as well. This is an important consideration when deriving the dominant paths for the density matrix, which takes the form

$\rho(x,x';\beta) = \int_{x(0)=x}^{x(\beta\hbar=x'}{\cal D}x ( \cdot ) exp \left [ - {1 \over \hbar } \int _0^{\beta \hbar} d\tau \left({m \over 2}\dot{x}(\tau) + U(x(\tau))\right)\right]$

The action appearing in this expression is

$S_E[x] = \int_0^{\beta\hbar} d\tau \left[{m \over 2}\dot{x}^2 + U(x(\tau))\right]=\int_0^{\beta\hbar}d\tau H(x,\dot{x})$

which is known as the Euclidean action and is just the integral over a path of the total energy or Euclidean Lagrangian $$H ( x, \dot {x} )$$. Here, we see that a minimum action principle is needed, since the smallest values of $$S_E$$ will contribute most to the integral. Again, we require that to first order $$S_E [ x + \delta x ] - S_E [ x ] = 0$$. Applying the same logic as before, we obtain the condition

 ${d \over d\tau}{\partial H \over \partial \dot{x}} - {\partial H \over \partial x}$ $$\underline {0}$$ $\underline {m\ddot{x} }$ $${\partial \over \partial x}U(x)$$

which is just Newton's equation of motion on the inverted potential surface $$-U(x)$$, subject to the conditions $$x (0) = x$$, $$x (\beta \hbar ) = x'$$. For the partition function $$Q (\beta)$$, the same equation of motion must be solved, but subject to the conditions that $$x (0) = x (\beta \hbar )$$, i.e., periodic paths.