I couldn't understand the concept of free energy landscape. Usually free energy is defined in the following way. $$F=-k_{B}T\ln Z=-k_{B}T\ln\sum e^{-\beta \epsilon}$$ where $\epsilon=\sum\frac{p_{i}^{2}}{2m}+V(\vec{x}_{1},\vec{x}_{2},\vec{x}_{3},...,\vec{x}_{n})$. Here the summation over all the $\{\vec{p}_{i}\}$ and $\{\vec{x}_{i}\}$. So $F$ seems like a function of $\beta$ {so, $F=F(\beta)\}$. So where does the concept of landscape come here? The derivation of probability distribution comes in the following way$$P(\epsilon) \propto \Omega_{\text{system}}(\epsilon)\Omega_{\text{reservoir}}(E-\epsilon)$$ where $E=E_{\text{reservoir}}+\epsilon$, with $E$ total energy of system+reservoir. Expressing in terms of entropy we get $$P(\epsilon)\propto\exp\bigg(\frac{S_{\text{system}}(\epsilon)}{k_{B}}\bigg)\exp\bigg(\frac{S_{\text{reservoir}}(E-\epsilon)}{k_{B}}\bigg)$$ Taylor expansion of $S_{\text{reservoir}}(E-\epsilon)=S_{\text{reservoir}}(E) - \epsilon\bigg(\frac{\partial S_{\text{reservoir}}}{\partial E}\bigg) +O(\epsilon^{2})\sim S_{\text{reservoir}}(E)-\frac{\epsilon}{T}$ \begin{align} P(\epsilon) &\propto \exp\bigg(\frac{S_{\text{system}}(\epsilon)}{k_{B}}\bigg)\exp\bigg(\frac{S_{\text{reservoir}}(E)-\frac{\epsilon}{T}}{k_{B}}\bigg) \\ &\propto\exp\bigg(\frac{S_{\text{system}}(\epsilon)-\frac{\epsilon}{T}}{k_{B}}\bigg)\exp\bigg(\frac{S_{\text{reservoir}}(E)}{k_{B}}\bigg)\\ &\propto\exp\bigg(-\frac{F}{k_{B}T}\bigg) \end{align} So, \begin{align}F&\propto-k_{B}T\ln P(\epsilon)\\ &\propto-k_{B}T\ln P(\epsilon(\vec{p}_{1},\cdots,\vec{p}_{n};\vec{x}_{1},\cdots,\vec{x}_{n} ))\\ &=-k_{B}T\ln P(\vec{p}_{1},\cdots,\vec{p}_{n};\vec{x}_{1},\cdots,\vec{x}_{n} ). \end{align} I ended up this result because of degeneracy of the system in the given energy $\epsilon$ If I assume, there is no degeneracy of the system then, $P(\epsilon)\propto\Omega_{\text{reservoir}}(E-\epsilon)\propto \exp(-\beta\epsilon)$ . Free energy becomes functions of all the momenta and position coordinates only when there is a degeneracy in the given energy of the system. I have two questions,

  1. Is free energy landscape a function of momenta and position coordinates? If so, is my derivation correct?
  2. Again the dependence of position and momenta coordinates comes into play in the free energy expression only when there is a degeneracy in the energy level of the system. Can I say free energy surface makes sense only when there is a degeneracy in system's energy levels?

I am asking in the context polymer/protein folding problem where one encounter free energy landscape.


1 Answer 1


The idea of a free energy landscape is directly related to the idea of a coordinate variable (CV) (also known as reaction coordinate). CVs are simply arbitrary projections $\vec{s}(\vec{x})$ of coordinate space $\vec{x}$ to a lower dimensional space. This is commonly done to simplify otherwise intractable problems, where one cannot handle the high-dimensional nature of $\vec{x}$ in a meaningful way.

In the absence of such variables, the free energy $F$ of a system is directly related to the integral of the Boltzmann weight of the energy $E(\vec{x},\vec{p})$ of each combination of positions $\vec{x}$ and momenta $\vec{p}$:

$$e^{-\beta F(\beta)}=\int_{\{\vec{x},\vec{p}\}} e^{-\beta E(\vec{x},\vec{p})} d\vec{x}d\vec{p}$$

where $\int_{\{\vec{x},\vec{p}\}}d\vec{x}d\vec{p}$ denotes discrete or continuous summation over all available states and $\beta=\frac{1}{k_B T}$. This equation is very nice, because you can directly see what the free energy actually is: the free energy of a macrostate has the same function as the energy of a microstate. As such, you can think of it as an effective energy of a particular collection of samples in phase space.

While the regular free energy $F(\beta)$ is the effective energy of all possible phase space samples at a particular temperature, you are free to define a macrostate in any way that is convenient for your purpose. One way in which you can do that is to express the free energy as a function of a low-dimensional (usually between 1 and 3 dimensions) CV, i.e. $F(\beta,\vec{s})$. This allows you to obtain a free energy landscape, i.e. a way to assign each combination of CV values an effective (free) energy $F(\beta,\vec{s})$. Formally, this is expressed as:

$$e^{-\beta F(\beta,\vec{s})} = \int_{\{\vec{x},\vec{p}\}} e^{-\beta E(\vec{x},\vec{p})} \delta(\vec{s}-\vec{s}(\vec{x})) d\vec{x}d\vec{p}$$

with $\delta(\cdot)$ being the Dirac delta. In the case of protein folding, if you have a single CV which goes continuously from an unfolded to a folded protein, such a free energy profile (i.e. 1-dimensional free energy landscape) allows you to determine the relative effective energy of each of these intermediate states.

I am not sure I understand the degeneracy question, but I think it's much easier to think about CVs in a continuous space, rather than discrete space (which is the case for the strictly classical protein folding models). In such a setting, every energy level is degenerate to some degree, and the mutiplicity of this degeneracy is determined by the Dirac delta function, which "picks out" the relevant combinations of $\vec{x}$ that satisfy the constraint $\vec{s}(\vec{x})=\vec{s}$ (with some slight abuse of notation for brevity). In all cases, both the positions and the momenta are important, since they are an inevitable part of the summation.


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