[Wien] A question about unit of electric field in in0

Martin Gmitra martin.gmitra at gmail.com
Fri Jul 3 08:23:25 CEST 2009


Dear Prof. Blaha,

According to WIEN2k v09.1 and v08.3, if the EFIELD variable in
case.in0 is set to 1.0,
then difference of total potential (I generated it using lapw5 ATU VAL
options) with
EFIELD=1.0 and without EFIELD line, shows up a zig-zag variation in a
unit cell from
0.5 to -0.5 (for analytic triangular ramp). My case.in0 is asking 40
Fourier coeficients
for analytical triangular ramp:
-----------
TOT   13    (5...CA-LDA, 13...PBE-GGA, 11...WC-GGA)
R2V      IFFT      (R2V)
  30  30 243    1.00    min IFFT-parameters, enhancement factor
40  1.0
-----------

Am I right that the electric field, let say at z=0.25c, is then equal to
2 EFIELD/c (where c is the lattice constant along z-direction) ?

Does it not mean that in your previous discussion from May 23, 2008, please
see below, the value of the electric field is two times the 173 kV/nm and
not four times? What have you meant by the external potential of 100mRy
in your PRB 63, 165205 (2001)?


With the best regards,
Martin Gmitra
 <wien%40zeus.theochem.tuwien.ac.at?Subject=%5BWien%5D%20A%20question%20about%20unit%20of%20electric%20field%20in%20in0&In-Reply-To=1211472633.9101.48.camel%40zhao>

*Fri May 23 08:29:30 CEST 2008*

You must multiply by 4 ! (because field varies from +100 to -100 and
since it is a zig-zag, it varies in half the period)

zhao schrieb:

>* Dear Prof. P. Blaha,
*>*
*>* In your paper, PRB-63-165205(2001), you say that external potential
*>* 100mRy corresponds to 700 kV/mm. When I tried to get this unit
*>* transformation, I was confused by something. The following is my
*>* problem.
*>* --------------------------------------------------
*>* c=8*sqrt(3)*a0
*>* a0=5.65 Angs
*>* Efield= 0.1Ry/c=0.1*13.6V/(8*sqrt(3)*a0)= 173 kV/mm
*>* --------------------------------------------------
*>*
*>* I cannot get 700 kV/mm. Can you tell me what's wrong with myself?
*>* Thank you very much.
*>*
*>*
*>*
*>* Best wishes,
*>*
*>* Sincerely,
*>* Yong-Hong Zhao*
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