Probability at potential minimum Let's imagine a particle in a potential with a single minimum, like the Lennard-Jones type. The most probable position of the particle would be the potential minimum, so I have attempted to argue it as follows:
In equilibrium statistical mechanics, the probability density to find a certain state is a function of the energy of the state, $P = P(E)$, according to Boltzmann. Let's assume that the kinetic part of the energy is small enough (because of small mass, for example) to see the effect of the potential clearly. Then, $E=E(x)$ and I have to rewrite $P$ as a function of $x$ in the following way:
$$P(x) = P(E(x)) \left| \frac{dE(x)}{dx} \right|.$$
The problem is that $dE/dx$ will vanish at the potential minimum, which means that $P(x)$ is simply zero at this point, contrary to my expectation.
Is it a misconception that the particle will probably be found at the potential minimum? Or, is there anything wrong in the above argument?
 A: Classical Boltzmann distribution gives the probability density of finding a particle in a particular point of the phase space. E.g., If we have Hamiltonian
$$
H= \frac{p^2}{2m} + V(x)
$$
the distribution is given by
$$
W(p,x) = Ce^{-\frac{H(p,x)}{k_BT}}, C=\int dp dx e^{-H(p,x)}
$$
The momentum can be integrated out, so the distribution that you need is
$$
w(x) = C_1e^{ -\frac{V(x)}{k_BT}}
$$
One can also arrive to this distribution by solving a Fokker-Planck equation for a particle (where instead of temperature we have a combination of friction and diffucion coefficients).
Update
We can obtain the distribution of energies from the canonical distribution as
$$
W(E) = \int dpdx \delta\left(E-H(p,x)\right)Ce^{-\frac{H(p,x)}{k_BT}}=
\int dpdx \delta\left(E-\frac{p^2}{2m} - V(x)\right)Ce^{-\frac{E}{k_BT}}=
Cg(E)e^{-\frac{E}{k_BT}},
$$
where $\delta(.)$ is the delta function and
$$
g(E) = \int dpdx \delta\left(E-\frac{p^2}{2m} - V(x)\right)
$$
could be called the density-of-states.
If we again integrate out the momentum, the contributions to the integral come from the points where
$$
p =\pm\sqrt{2m\left(E-V(x)\right)}
$$
note that the real roots exist only when $E\geq V(x)$. We thus have
$$
g(E) = \sqrt{2m}\int \frac{dx}{\sqrt{E-V(x)}}\theta\left(E-V(x)\right),
$$
where $\theta(.)$ is the step function.
One could also look at the distribution of the potential energy, $\epsilon=V(x)$:
$$
P(\epsilon)=\int dx \delta\left(\epsilon-V(x)\right)C_1e^{ -\frac{V(x)}{k_BT}}=
C_1 e^{ -\frac{\epsilon}{k_BT}}\int dx \delta\left(\epsilon-V(x)\right)
$$
Update 2
Calculations for a Harmonic oscillator potential give:
$$
W(E) = \frac{1}{k_B T}e^{-\frac{E}{k_BT}},\\
P(\epsilon) = \frac{1}{\sqrt{\pi k_B T \epsilon}}e^{-\frac{\epsilon}{k_BT}}.
$$
Since the OP dispenses with the kinetic energy, it is the latter distribution that appears there, and it has a singularity at the potential minimum, which resolves the paradox.
