[QE-users] Unit for the output of average.x
Dingfu Shao
dingfu.shao at gmail.com
Fri Oct 26 21:53:52 CEST 2018
Hi, Giovanni,
Yes, I confirm the fermi energy should be right. I actually checked this
twice. In the scf and nscf calculations, I used occupation = 'fixed', and
got the " highest occupied level (ev): -1.0818" from the output of scf.
Then for the calculation of pp.x, I did two tests:
1. set emin = -65 and emax = -0.5 for plot_num =10
2. set emin = -65 and not set emax, since according to the input
description, for plot_num =10, if emax is not specified, emax=E_fermi.
Both tests leads to same result.
Moreover, since I used occupation = 'fixed', no states above EF is
calculated. Therefore, even if I set a wrong EF, it should only leads to a
smaller integrated value compared to the total number of electrons. But now
what I got a integrated value of 29, which is much larger than the 26
electrons in MoS2 monolayer.
For your convenience, I put my input files here. I used QE-6.3 for those
calculations.
MoS2.nscf.in:
&control
calculation='nscf'
restart_mode='from_scratch',
prefix='MoS2',
pseudo_dir = '../pp',
outdir='./tmp/'
disk_io='low'
/
&system
ibrav=4 ,
a = 3.13079 ,
c = 20,
nat = 3 , ntyp= 2,
ecutwfc= 60.0,
ecutrho= 600.0,
occupations='fixed',
/
&electrons
electron_maxstep=300
conv_thr = 1.0d-7
mixing_beta = 0.7
/
&ions
/
&cell
/
ATOMIC_SPECIES
Mo 95.94 Mo.pz-spn-rrkjus_psl.0.2.UPF
S 32.07 S.pz-n-rrkjus_psl.0.1.UPF
ATOMIC_POSITIONS (crystal)
Mo 0.000000000 0.000000000 0.000000000
S 0.333333333 0.666666667 0.077787748
S 0.333333333 0.666666667 -0.077787748
K_POINTS {automatic}
100 100 1 0 0 0
MoS2.pp.in:
&inputpp
prefix='MoS2',
outdir='../tmp',
filplot = 'MoS2.charge'
plot_num= 10
emin = -65
/
MoS2.avg.in:
1
MoS2.charge
1.D0
2000
3
37.7945
Note the awin=37.7945 in MoS2.avg.in might not be appropriate, but it will
not influence the planar average.
Thank you very much.
Ding-Fu
> *Giovanni Cantele* giovanni.cantele at spin.cnr.it
>
> <users%40lists.quantum-espresso.org?Subject=Re:%20Re%3A%20%5BQE-users%5D%20Unit%20for%20the%20output%20of%20average.x&In-Reply-To=%3CAE0AD77F-2745-4B15-9DF1-E8491D64AAF9%40spin.cnr.it%3E>*Fri
> Oct 26 21:08:23 CEST 2018*
>
> Are you sure about you Fermi level? How did you get it?
>
> GIOVANNI
> Inviato da iPhone
>* Il giorno 26 ott 2018, alle ore 19:33, Dingfu Shao <dingfu.shao at gmail.com <https://lists.quantum-espresso.org/mailman/listinfo/users>> ha scritto:
*
>
>* Dear Giovanni and Paolo,
*
>
>* Thanks very much for your suggestions.
*
>
>* I reconsidered the definition of the local density of states (say, LD(x,y,z,E)). Since it is "local", the unit of it should be states/eV/bohr^3 or electrons/eV/bohr^3. Therefore, the integration of it within an energy window should lead to the charge density in this energy window: electrons/bohr^3 . Therefore, if we choose the energy window from the lowest energy to the Fermi energy, we should get exactly the total charge density.
*
>
>* So I did some tests following Giovanni's suggestion, using a simple case of momolayer MoS2, which has "number of electrons = 26.00". The area of its xy plane is S. For those tests, the previous scf and nscf calculations are the same, with LDA USPP, the occupation= 'fixed', and k-points of 40*40*1 for scf and 100*100*1 for nscf.
*
>
>* 1. I calculated the total charge density (rho(x,y,z)) using pp.x with plot_num = 0, and then calculate the planar average of it (rho_avg(z)). Then I integrated it by \int (S* rho_avg(z)) dz and I got 25.89. It is close to 26.00, maybe a more accurate value can be obtained by a calculation with denser k points.
*
>
>* 2. I calculated the integrated local density of states ( ILD(x,y,z)) from -65 eV (this energy is below the lower band) to Fermi energy using pp.x with plot_num =10, and and then calculate the planar average of it ( ILD_avg(z) ). When I integrated it by \int (S* ILD_avg(z)) dz, I got 29.25, which is much larger than the total number of electrons of 26.
*
>
>* So it seems the the test 2 doesn't correctly reflect the reality. I am not sure it is due to something happened with plot_num = 10 in pp.x, or just I understand this incorrectly.
*
>
>* Any suggestions? Thank you very much!
*
>
>* Best,
*
>
>* Ding-Fu
*
>
>
>
>
>
>
>>* From: Giovanni Cantele <giovanni.cantele at spin.cnr.it <https://lists.quantum-espresso.org/mailman/listinfo/users>>
*
>>* To: Quantum Espresso users Forum <users at lists.quantum-espresso.org <https://lists.quantum-espresso.org/mailman/listinfo/users>>
*
>>* Cc:
*
>>* Bcc:
*
>>* Date: Fri, 26 Oct 2018 09:51:40 +0200
*
>>* Subject: Re: [QE-users] Unit for the output of average.x
*
>>* Dear Ding-Fu,
*
>>* as far as I remember there is a surface factor that you need to adjust units.
*
>>* For sure on the ascissa axis the coordinate is in bohr.
*
>>* The planar average give you back a quantity with the same units as the averaged quantity
*
>>* (e.g. if you star from charge density in electrons/bohr^3 you get an averaged electron density in electrons/bohr^3),
*
>>* being defined as (let us suppose that you average in the plane defined by a1 and a2 vectors):
*
>>* rho_avg(z) = ( 1 / S ) * integral( dx dy rho(x,y,z) )
*
>>* That means that if you perform
*
>>* integral( dz rho_avg(z) )
*
>>* you get
*
>>* number of electrons / S
*
>>* If you need number of electrons than just multiply by S with
*
>>* S = cross_product( a1, a2 )
*
>>* (in bohr^2)
*
>>* Just try, better if you do it with the total charge density, to check if the integral returns you
*
>>* the number of electrons.
*
>>* I’m sorry but I cannot check directly if I remember correctly at the moment, but
*
>>* this should work.
*
>>* Giovanni
*
>>* --
*
>>* Giovanni Cantele, PhD
*
>>* CNR-SPIN
*
>>* c/o Dipartimento di Fisica
*
>>* Universita' di Napoli "Federico II"
*
>>* Complesso Universitario M. S. Angelo - Ed. 6
*
>>* Via Cintia, I-80126, Napoli, Italy
*
>>* e-mail: giovanni.cantele at spin.cnr.it <https://lists.quantum-espresso.org/mailman/listinfo/users>
*
>>* gcantele at gmail.com <https://lists.quantum-espresso.org/mailman/listinfo/users>
*
>>* Phone: +39 081 676910
*
>>* Skype contact: giocan74
*
>>* Web page: https://sites.google.com/view/giovanni-cantele <https://sites.google.com/view/giovanni-cantele>
*
>>* On 26 Oct 2018, at 03:31, Dingfu Shao <dingfu.shao at gmail.com <https://lists.quantum-espresso.org/mailman/listinfo/users>> wrote:
*
>>* Dear QE developers and users:
*
>>* I am wondering what should be the unit of the planar average data got from the average.x
*
>>* I am calculating the planar average of charge density within a energy window. What I did is firstly using pp to get the integrated local density of states (ILDOS) of that energy window with plot_num=10, then using average.x to get the planar average.
*
>>* In this case, what is the unit of the second column (say, rho(z)) of the output file? I thought since the DOS has a unit of states/eV, the integration of DOS within a energy window should get some states or electrons. Then the unit of rho(z) should be electron/bohr. But seems it is not. In my case the energy window I concerned contains one electron, However, if I directly integrate rho(z), I can only get a very small value. If I assume the unit is electron/(bohr^3), the integretion of rho(z)*A is also smaller than one (here A is the area of xy plane).
*
>>* Can you help me about it? Thank you very much!
*
>>* Best,
*
>>* Ding-Fu
*
>>
>>
>>* Ding-Fu Shao, Ph. D.
*
>>* Department of Physics and Astronomy, University of Nebraska-Lincoln
*
>>* Lincoln, NE 68588-0299
*
>>* Email: dingfu.shao at gmail.com <https://lists.quantum-espresso.org/mailman/listinfo/users>
*
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