Thermal expansion of Al

We will now calculate thermodynamic quantities, and in particular the Helmholtz free energy, defined in Eq.  and  for the classical and the quantum case, respectively, which we rewrite as free energies per atom:

$$F_c(V,T) = U_0(V) + k_{\rm B}T \frac{1}{N_{\bf q}} \sum_{{\bf q}; s=1}^3 \ln \frac{\hbar \omega_{{\bf q},s}(V)}{k_{\rm B}T},$$ (9.2)

and

$$F(V,T) = U_0(V) + \frac{1}{N_{\bf q}}\sum_{{\bf q}; s=1}^3 \frac{... ...1}^3 \ln \left(1- e^{ -\frac{\hbar \omega_{{\bf q},s}(V)}{k_{\rm B}T}}\right ),$$ (9.3)

where $U_0(V)$ is the energy per atom of the perfect crystal, computed in Sec. , and the sum over q runs over $N_{\bf q}$ points in the BZ, chosen for example by uniformly dividing the BZ. To this with phon you need to add the following setting to the INPHON file:

LFREE =.TRUE.
TEMPERATURE = 300
QA = 16; QB = 16; QC = 16

where QA, QB and QC define the grid of points in the BZ to calculate the sum Eq. . Again, we need to be sure that the results are converged w.r.t. the density of these points, which you can easily establish by playing with these parameters and see how the free energy changes.

Note that the energy $U_0(V)$ is not included in the free energy outputted by phon, as this information is not coded into the forces in FORCES. This is obvious, as any constant added to the energy does not have any effect on the forces induced by displacing the atoms from their equilibrium positions as these are due to energy differences, and so this constant needs to be added to the vibrational free energies explicitly.

We can now find the equilibrium volume at every temperature, looking for the point where the free energy  is minimum. We see that in the quantum case there is a contribution to the free energy even at zero temperature, known as zero point energy, so we can already obtain a quantum corrected equilibrium volume by minimising $F(V,0)$ w.r.t. $V$. At $T > 0$ both $F$ and $F_c$ display a temperature dependence and minimisation w.r.t. $V$ provides corresponding equilibrium volumes. From the temperature dependence of the equilibrium volumes we can calculate the zero pressure coefficient of thermal expansion:

$$\alpha_0 = \frac{1}{V}\left (\frac{\partial V}{\partial T}\right)_{P=0}.$$ (9.4)

We see that, as the difference between $F$ and $F_c$ is reduced by increasing temperature, also the equilibrium volumes become more similar at high temperatures, but at low temperature there is an important failure of the classical description. ^9.5