Why the deepest potential well need not be the most populated

Assign a weight to each configuration

Picture a thermally agitated bead wandering across two valleys. A sequence of photographs shows which locations occur most often. At equilibrium, a small cell of low energy usually appears more frequently. Finding the population of an entire valley requires adding the probabilities of all its cells.

The configuration coordinates $x,y$ increase to the right and upward. They may represent scaled molecular deformations. The potential energy is $U(x,y)$, the temperature is $T$, and $\beta=1/(k_{\mathrm B}T)$, with Boltzmann constant $k_{\mathrm B}$. A probability density becomes a probability after multiplication by a small area. The partition function $Z$ normalizes the total probability.

Source: PLUMED Masterclass 21.2 · Michael R. Shirts and John D. Chodera

Two boxes are enough to see entropy

Worked example: box A contains two distinguishable configurations of zero energy. Box B contains eight, each higher by $k_{\mathrm B}T\ln 2$. Box A has total weight two. Each configuration in B has weight one half, giving total weight four. The higher-energy box therefore has probability two thirds.

For $g$ equal-energy configurations the entropy is $S=k_{\mathrm B}\ln g$. Free energy $F=E-TS$ compares the total weights. Here B has free energy lower than A by $k_{\mathrm B}T\ln 2$. The negative difference reflects its greater population.

The width of a continuous valley plays the same role as the number of discrete configurations. Width means volume in configuration space. Entropy depends on actual configuration-space volume; line thickness is a drawing choice.

Add a whole slice when projecting

A camera reporting only $x$ combines every possible $y$ at that coordinate. This sum gives the marginal density $p_x(x)$. Its negative logarithm, multiplied by thermal energy, defines the projected free energy $F_x(x)$. An arbitrary constant $C$ sets the vertical zero.

Suppose locally $U(x,y)=V(x)+\kappa(x)y^2/2$ and the transverse domain is wide enough to extend the integral over the real line. A Gaussian integral gives $F_x(x)=V(x)+(k_{\mathrm B}T/2)\ln\kappa(x)+C$. A stiffer, narrower cross-section raises the free energy. The minimum-energy curve $V(x)$ omits this contribution. A finite box instead requires its actual integration limits.

A different coordinate asks a different question

Rotate the viewing direction through angle $\theta$ and define $s=x\cos\theta+y\sin\theta$. A collective variable summarizes configurations with a few numbers. Each value here labels a slanted slice. One direction separates two valleys; another can stack them on top of each other. A flatter projected curve may then conceal a slow crossing.

Even a monotone change of scale matters. If $z=h(s)$, conservation of probability gives $p_z(z)=p_s(s)/|h\prime(s)|$. State the coordinate measure before comparing pointwise barriers. Integrated probabilities of corresponding regions survive this relabeling.

The two-coordinate landscape used along the course

Our numerical examples use the reflecting square $\Omega=[-1.8,1.8]^2$. Coordinates, energy and time are dimensionless, with default thermal energy one. Parameters $a,b$ set two double-well barriers, $c$ couples the coordinates and $d$ tilts the landscape. Defaults are $a=b=3$, $c=1.5$ and $d=0$. This teaching model can be integrated independently.

A reference curve sums Boltzmann weights over a two-dimensional grid. A trajectory estimate uses the positions actually visited by the bead. Their agreement tests the selected projected occupancy. Hidden conditional distributions require another check.