A question about the variable "weight" in selfen_el.f90

[Vahid]

Dear EPW Users,

I have extracted the individual self-energy contributions from a 20x20x20 q grid to a list of k points for the diamond example case with no fsthick set. For a random k point (0,0.1,0.45), I considered the contributions that 3 equivalent q's make to the k point. The q points are

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0.0000000   0.0000000   0.0500000
0.0000000   0.0000000   0.9500000
0.0000000   0.0500000   0.0000000

Their six phonon energies are nearly the same

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  8.478513578340556E-004  8.690353269845230E-004  1.371788110062204E-003
  9.805869334826509E-003  9.807700191526363E-003  9.976014814358882E-003

  8.478507370144750E-004  8.690347206693874E-004  1.371787569949127E-003
  9.805869388560562E-003  9.807700244964069E-003  9.976014888855781E-003

  6.409839467282880E-004  6.409839747712517E-004  1.300855935719668E-003
  9.823412612494933E-003  9.823412614324758E-003  9.985488830814850E-003

But the weight variable calculated from

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weight = pi * wqf(iq) * ( (wgkq+wgq)*w0g1 + (one-wgkq+wgq)*w0g2 )

and their contributions to the k's self energy (2nd column) are quite different as shown below:

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2.840020921411006E-002  0.454447554214972 
7.031015610841913E-004  0.163020527100226     
4.599807289615457E-011  5.734701548147320E-010

Since the three q's have nearly the same energy while interacting with the same k electron, shouldn't the weights be the same?

Thank you,

Vahid

Vahid Askarpour
Department of Physics and Atmospheric Science
Dalhousie University,
Halifax, NS, Canada

[Vahid]

Hi,

I just would like to clarify that the weight I reported for the three q points were the last weights from the ik,imode,ibnd,jbnd loop. Otherwise, there are 96 weights in that loop for each ik and iq point which I did not report.

The self energies were calculated using the following matrix just before sigma_all's definition:

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sigmai_tot(ibnd,ik+lower_bnd-1,iq)=sigmai_tot(ibnd,ik+lower_bnd-1,iq)+g2*weight

So the three self energies contributed by the three q's were summed over the bands and modes.

Cheers,
Vahid

[sponce]

Hello Vahid,

You should check that the sum of the 96 weight is correctly 1 (or eventually 2).

If the point are truly random, the weight is likely to be 1.0/96 indeed.

Best,

Samuel

[Vahid]

Hello Samuel,

Thank you for suggesting to check the sum of the weights. For the first q point and the k point I reported in my first post, I get these 96 weights which come from the 6 modes and 4 bands:

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 7.762864806541902E-004  1.900714716071483E-088  1.900714716071483E-088
  1.900714716071483E-088  1.900714716071483E-088  1.021064846680064E-004
  1.900714716071483E-088  1.900714716071483E-088  1.900714716071483E-088
  1.900714716071483E-088  3.449724804790480E-003  3.624470068423790E-015
  1.900714716071483E-088  1.900714716071483E-088  1.392555733007761E-013
  1.162868088328380E-002  7.658697008806053E-004  1.855903177856707E-088
  1.855903177856707E-088  1.855903177856707E-088  1.855903177856707E-088
  9.988768852914349E-005  1.855903177856707E-088  1.855903177856707E-088
  1.855903177856707E-088  1.855903177856707E-088  3.365972178129029E-003
  3.608760629646159E-015  1.855903177856707E-088  1.855903177856707E-088
  1.402167782315355E-013  1.142397687925616E-002  6.597723509642105E-004
  1.205415093427186E-088  1.205415093427186E-088  1.205415093427186E-088
  1.205415093427186E-088  6.851838511100885E-005  1.205415093427186E-088
  1.205415093427186E-088  1.205415093427186E-088  1.205415093427186E-088
  2.138126564063455E-003  3.884992399302110E-015  1.205415093427186E-088
  1.205415093427186E-088  1.951103389053494E-013  8.807905983448494E-003
  1.137642909376745E-002  4.219810241638788E-089  4.219810241638788E-089
  4.219810241638788E-089  4.219810241638788E-089  2.757438543751118E-005
  4.219810241638788E-089  4.219810241638788E-089  4.219810241638788E-089
  4.219810241638788E-089  1.248181676748954E-004  1.651540243905585E-012
  4.219810241638788E-089  4.219810241638788E-089  3.572659187442366E-009
  2.806613303790095E-002  1.138197790051863E-002  4.219763605397296E-089
  4.219763605397296E-089  4.219763605397296E-089  4.219763605397296E-089
  2.756630180454055E-005  4.219763605397296E-089  4.219763605397296E-089
  4.219763605397296E-089  4.219763605397296E-089  1.247326477863427E-004
  1.653477104954179E-012  4.219763605397296E-089  4.219763605397296E-089
  3.579755515211524E-009  2.806986024830347E-002  1.189778235600534E-002
  4.215664645084622E-089  4.215664645084622E-089  4.215664645084622E-089
  4.215664645084622E-089  2.681959948975107E-005  4.215664645084622E-089
  4.215664645084622E-089  4.215664645084622E-089  4.215664645084622E-089
  1.170590554948513E-004  1.840659337997957E-012  4.215664645084622E-089
  4.215664645084622E-089  4.293959613693773E-009  2.840020921411006E-002
  0.454447554214972     

They do not add up to 1 or 2. 0.4544 is the largest with the sum just over 0.56.

I obtained these weights by defining a new matrix wght as follows right after the definition of weight in my first post:

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wght(ibnd,jbnd,imode,ik+lower_bnd-1,iq)=weight

The q points are not random. The fine q grid is 20x20x20.

So now there are two issues:

1. the weights do not add up to either 1 or 2.
2. I was expecting the three (symmetrically equivalent) phonons to make almost equal contributions to the self-energy of the k point but they are vastly different as I reported in my first post. May be this is due to incorrect weight values.


Best,
Vahid

[carla.verdi]

Dear Vahid

I think there was some slight confusion about the weights - they do not need to add up to 1 or 2 as they are the self-energy factors without the matrix element.
I understand that the q points you considered are equivalent, but I suppose the k+q points are not, so the eigenenergies at k+q will not be the same (hence the 'weight' quantity will be different).

Best
Carla

[Vahid]

Dear Carla,

Thank you for the clarification. I was under the impression that symmetrically equivalent q phonons will interact with a k electron and result in symmetrically equivalent k+q states which naturally have identical energies based on energy conservation. I guess I was wrong.

Best wishes,

Vahid