[Pw_forum] unit of the polarization (Paolo Umari)

[umari at democritos.it]

Dear Chen,

The electric field is treated in Rydberg atomic units in pw.x
in this way:

1)The polarization P(\Psi) is calculated as sqrt(2.)*<\Psi|r|\Psi>/Omega
where \Psi is the slater determinant of the KS wavefunctions, Omega is 
the valume of the simulation cell,
and the expectation value of the position operator r  is calculated
through the Berry's phase formulation. This assures that for the same
wavefunctions the "numerical value" of the polarization is sqrt(2.) times
larger then that calculated for the case of Hartree atomic units.

The KS Hamiltonian in the presence of the electric field E  contains a term
proportional to Omega*E*\partial P(\Psi) /\partial \Psi
= E* sqrt(2) *\partial *<\Psi|r|\Psi> /\partial \Psi
This is the factor sqrt(2.) appearing in the routine h_epsi_her_set.f90

If you want to calculate the ratio P/E you must use
both values in Rydberg atomic units: the electric dipole as reported
by the code (in Rydberg atomic units) and the input numerical value of 
the electric field which is also given in Rydberg atomic units
(see Example 31)

Note that to obtain the corresponding  numerical value of the electric
field in Hartree atomic units you must divide by a factor of sqrt(2.)

Best regards,

Paolo Umari, DEMOCRITOS


> Dear PWSCF users,
>     I have a question about the polarization unit which has confused me for
> a long time.
>     The atomic unit of electric field is: e/a0^2 where 'e' is the electron
> charge and a0 is the Bohr radius. The unit of polarization (i.e. dipole per
> unit volume) should have the same unit as electric field. However from the
> source code c_phase_field.f90(version4.0.2), the output of dipole per cell
> has a factor sqrt(2) which is the electron in the Ry units. So 'e' becomes
> sqrt(2).
>    Then if I want to calculate P/E which is dimensionless, shall I convert
> the output of P into P/sqrt(2). Note when we input the electric field E, E
> is in the atomic unit, i.e. 'e' is not converted into sqrt(2).
>    I am just confused that since you use atomic unit, why you explicitly
> convert 'e' into sqrt(2) rather than implicitly retain it in the unit?
>
> Hanghui Chen
> Department of Physics,
> Yale University


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