# Free energy of a one-dimensional harmonic oscillator

The potential energy of a one-dimensional harmonic potential can be expressed as $$U(x)=\frac{1}{2}K(x-x_0)^2$$, where $$K$$ is the force constant and $$x_0$$ is the equilibrium position. I'm wondering how one can calculate its (1) free energy and (2) free energy profile along $$x$$.

I first calculated the partition function as $$Z=\int^{+\infty}_{-\infty}\exp\left (-\frac{1}{2}\beta K (x-x_0)^2\right)dx=\sqrt{\frac{2 \pi}{\beta K}}$$, where $$\beta=1/k_B T$$ is the inverse temperature. Then, I calculated the free energy as $$F=-\frac{1}{\beta}\ln Z=-\frac{1}{\beta}\ln\left ( \sqrt{\frac{2\pi}{\beta K}} \right)$$ While I'm relatively confident that the problem-solving logic is correct, I don't really understand the what this free energy represents. Compared to this, a free energy profile makes much more sense to me as it tells us which region in the space of $$x$$ has lower energy and is therefore more probable.

At the same time, I'm not entirely sure what the free energy profile should look like. By definition, the free energy profile can be obtained by taking the negative logarithm (with a multiplying factor of $$\beta^{-1}$$ of course) of a partially integrated partition function obtained by integrating $$\exp(U({\bf x}))$$ over all variables except for the variable along which we want to plot the free energy profile. However, in this case, $$x$$ is the only degree of freedom, so I assume that no integration needs to be carried out, namely, the free energy profile along $$x$$ should be simply $$F(x)=-\frac{1}{\beta}\ln\left(\exp(U(x))\right)=-\frac{1}{\beta}U(x)=-\frac{K}{2\beta}(x-x_0)^2$$ However, this looks weird to me.

Also, I sometimes saw that people calculate the free energy profile as $$F(x) = -\frac{1}{\beta}\ln P(x)$$, but I'm not sure if this is correct or if I should use this formula in this situation. I guess I'm confused when to use $$F(x) = -\frac{1}{\beta}\ln Z(x)$$ and when to use $$F(x) = -\frac{1}{\beta}\ln P(x)$$.

• $U$ is part of the Hamiltonian so it depends on space. $F$ is a statistical quantity related to the average value of the energy (i.e. it is related to the expected behaviour of the particle in the harmonic potential in equilibrium with a heat bath at temperature $T$). Mar 12 at 9:47
The basic idea of the canonical ensemble is that the temperature $$T$$ (or its scaled inverse $$\beta$$) is fixed and the energy follows a Boltzmann (or exponential) distribution: $$P[E\in[E',E'+dE]]= Z e^{-\beta E'}dE.$$ Where $$Z$$, the partition function, is the normalization constant. If the system is classical, you can do a change of variables and compute the probability of finding the system in a particular microscopic state defined by positions and momenta: $$P[x\in[x',x'+dx],p\in[p',p'+dp]]\propto e^{-\beta H(x',p')}\frac{dxdp}{h}.$$ If you are interested only in the position probabilities, then you have to integrate the momentum, $$P[x]=\int dp P[x,p].$$ Proceeding like this you obtain the probability of finding the particle in a particular position for the one-dimensional harmonic oscillator: $$P[x\in[x',x'+dx]]=Z' e^{-\beta\frac{K}{2}(x'-x_0)^2}dx.$$ Which I think you refer to as the energy profile. Note that $$Z\ne Z'$$.
A different thing is the Helmholtz free energy ($$F$$), which is the Legendre transform of the internal energy ($$U$$), and can be shown to follow: $$F=-\frac{1}{\beta}\log(Z).$$ The Helmholtz free energy is the work you can extract through isothermal processes. Moreover, all macroscopic quantities (e.g. the pressure) can be computed from the knowledge of $$F$$ at fixed $$T,N,V$$.
Lastly, note that the internal energy is another macroscopical quantity, which can be computed both from $$F$$ or from the Boltzmann distribution: $$U=\langle E\rangle=\int Z e^{-\beta E}dE.$$