# [QE-users] How is the wavefunction supposed to be normalized?

John McFarland jm18ec at my.fsu.edu
Mon Dec 24 22:56:35 CET 2018

```Dear Stefano,

I tried to write some code that is like the code you wrote, using both s_1psi() ans s_psi(), and I tested it with a wavefunction I believe is normalized.  The result is a matrix that resembles the identity matrix, except that the diagonal elements are not one but range from about 0.2 to 0.3.  I got a similar result when I tested the subroutine check_ortho() in ortho_wfc.f90.  Is there something I'm missing?

Here is an example of some code I tested:

call calbec( npw, vkb, evc, becp )
call s_psi(npwx, npw, nbnd, evc, tddft_spsi)

do ibnd = 1, nbnd
do jbnd = 1, nbnd

! calculate <psi_i| S |psi_j>
norm = 0

do ipw = 1, npw
norm = norm + conjg( evc(ipw,ibnd) ) * tddft_spsi(ipw,jbnd)
enddo

write(stdout,*) " i " , ibnd , " j " , jbnd , "<psi_i| S |psi_j>" , norm

enddo
enddo

Best regards,

John

________________________________
From: users <users-bounces at lists.quantum-espresso.org> on behalf of Stefano de Gironcoli <degironc at sissa.it>
Sent: Tuesday, December 11, 2018 10:26:15 AM
To: users at lists.quantum-espresso.org
Subject: Re: [QE-users] How is the wavefunction supposed to be normalized?

yes, except the S matrix is never stored anywhere but is applied to the wfc by the s_1psi routine,

so more something like

sum = 0.d0

call s_1psi( npwx, npw, evc(1,b), spsi )

DO  i = 1,npw

sum = sum + conjg(evc(i,b)) * spsi(i)
END DO

stefano

On 11/12/18 15:36, John McFarland wrote:

Dear all,

As I understand it, the wavefunction for each k-point is given by ecv(:,:), with the first index representing a single particle basis and the second index representing each band.  I'm guessing that when a normalized wavefunction is contracted over the S matrix, the result should equal one for each band.  What I think this would look like in code is:

sum = 0

DO  i = 1,npwx

DO j = 1,npwx

sum = sum + evc(i,b) * S(i,j) * evc(j,b)

END DO

END DO

and sum should equal 1 for any band b.  Is my understanding correct?

Best regards,

John McFarland

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