How to evaluate pressure

Following the seminal paper of O.H. Nielsen an R.M. Martin Ref. (81), the pressure can be expressed as the negative trace of the stress tensor $T_{\alpha \beta}$ :

$$T_{\alpha \beta} = \frac{\partial }{\partial \epsilon_{\alpha \be... ...vert \Psi_\epsilon \rangle }{\langle \Psi_\epsilon \vert \Psi_\epsilon \rangle}$$ (4.20)

where the $\Psi_\epsilon$ is the wave-function "stretched" by the transformation on each particle $r_{i\alpha} \rightarrow r_{i\alpha} + \sum_\beta \epsilon_{\alpha \beta} r_{i \beta}$ where $\epsilon_{\alpha \beta}$ is a symmetric strain tensor. The pressure is then defined through the negative trace of the stress tensor:

$$3PV=-Tr [ T_{\alpha \beta} ]$$ (4.21)

For an isotropic system, as the one we studied, the pressure can be easily written as:

$$3 P V = \frac{\partial \langle E_l \rangle }{\partial V} = \frac{\partial \langle E_l \rangle }{\partial L} 3 L^2 .$$ (4.22)

where $V = L^3$ and $L$ is the edge of our cubic box. To evaluate this derivative it is convenient to write the energy using rescaled distances that are invariant for stretching of the simulation box, namely $r' = r/L$ . After this transformation the expectation value of the energy can be written as:

$$\langle E_l \rangle = \frac{\langle E'_{kin} \rangle}{L^2} + \frac{\langle E'_{pot}\rangle}{L},$$ (4.23)

where $E'_{kin}$ and $E'_{pot}$ are the kinetic and the potential energy in the new coordinates. Thus the pressure will be:

$$P = \frac{1}{3 V} \left [ 2 \frac{\langle E'_{kin} \rangle}{L^2} ... ...\left [ \langle O_L H \rangle - \langle O_L \rangle \langle H \rangle \right ],$$ (4.24)

where $O_L$ is the logarithmic derivative of the wave-function respect to the simulation box size $L$ :

$$O_L = \frac{\partial_L \Psi_T}{ \Psi_T}$$ (4.25)

Notice that the box size $L$ appears in the one-body terms and in the two-body terms due to the cusp condistions, see Appendix C. The first part of the expression 4.24 is the usual Virial Theorem and the second one is given by the contribution due to the Pulay stress tensor. In fact although we used rescaled coordinates, it is not possible to cancel the dependencies of the wave-function from the simulation box size. This is due to the nuclear and electronic cusp conditions that depend explicitly on $L$ (see Appendix C).
In the expression 4.24 only the electronic kinetic energy is considered. When we perform a dynamic on the ionic part, we have to add the pressure due to the momentum flux carried by the ions. This part can be easily calculate with the standard kinetic theory (see Appendix B of Ref. (82)) of gas, and it reads:

$$\langle P_{ionic} \rangle = \frac{2 N}{ 3 V} \langle E_k^{ionic}\rangle,$$ (4.26)

the total pressure will be the sum of the ionic 4.26 and electronic part 4.24.