Confusion in PIEZOCON

R.Zosiamliana

Hello,

I’m performing a 2D slab calculation using CRYSTAL17 for a hexagonal monolayer system (a = b = 6.81998 Å, γ = 120°, c = 500 Å to simulate vacuum). Geometry is read from an external file. I enabled piezoelectric property calculation.

CRYSTAL printed the piezoelectric tensor (in |e|·bohr units) as:

| 0.000 0.000 -3.931 |
| -3.931 3.931 0.000 |
| 0.083 0.083 0.000 |

My questions:

1) For slab models, should the full 3D volume be used in the unit conversion to C/m2, or just the in-plane area?

2) Are these piezoelectric constants physically meaningful for 2D systems, or is there a standard post-processing correction needed?

Thank you in advance for your help.

Best regards,
R.Zosiamliana
PSRC
3yrs+, CRYSTAL17 user.

aerba

Hi,

Direct piezoelectric constants of a 3D lattice in CRYSTAL are defined and computed as:
$$
e_{ci}^{3D} = \left( \frac{\partial P_c}{\partial \eta_i}\right) = \frac{1}{V}\left( \frac{\partial^2 E}{\partial E_c\partial \eta_i}\right)
$$
that is as first derivatives of Cartesian components of the polarization (c=x,y,z) with respect to strain components, or, equivalently as second derivatives of the energy density (V is the volume of the 3D lattice cell) with respect to Cartesian components of an electric field $E_c$ and strain components, where the strain $ \eta $ is dimensionless and thus the direct piezoelectric constants have units of $ \textup{charge/length}^2 $.

For 1D and 2D periodic lattices, as the volume (V) is not uniquely defined (or not defined at all in some cases), one may divide by the length $l $ and area $ A$ of the lattice cell instead:
$$
e_{ci}^{1D} = \frac{1}{l}\left( \frac{\partial^2 E}{\partial E_c\partial \eta_i}\right) \quad \textup{and} \quad e_{ci}^{2D} = \frac{1}{A}\left( \frac{\partial^2 E}{\partial E_c\partial \eta_i}\right)
$$
that would thus be expressed in units of $ \textup{charge} $ or $ \textup{charge/length} $ for 1D and 2D lattices, respectively.

However, in CRYSTAL for 1D and 2D lattices we do not divide by $l $ or $ A$ , and just define and compute the piezoelectric constants as:
$$
e_{ci}^\textup{1D and 2D} = \left( \frac{\partial^2 E}{\partial E_c\partial \eta_i}\right)
$$
with units of $ \textup{charge}\cdot\textup{length} $.

Yes, these constants are physically meaningful for 1D and 2D systems. For a 2D monolayer system, for instance, depending on what you need to compare with, you can do one of two things:

keep them as they are printed in the CRYSTAL output (units of $ \textup{charge}\cdot\textup{length} $)
divide the values you get in the CRYSTAL output by the area of the 2D cell (and thus express them in units of $ \textup{charge/length} $)

I would not divide by a volume because I would not know the physical meaning of the volume of a 2D monolayer system.

Hope this helps,