[Pw_forum] Left and right going Bloch's states from PWCOND
Alexander Smogunov
smogunov at sissa.it
Thu Oct 29 16:27:15 CET 2009
Dear Manoj
On Wed, 2009-10-28 at 01:49 -0400, Manoj Srivastava wrote:
> Dear Alexander,
> You are correct, left and right moving states are in general not
> related. I was wrong about trying to create a pair. Thanks for
> clarification.
> I have a few more questions.
> 1. I want to find out the reflection coefficient in the code. In
> transmit.f90, transmission coefficient is |tmat(ig,n)|^2, where tmat is
> vec1(ntran-n2d-npol*nocrosr+ig,n). (line 206, of version 4.0.5).
> Reflection is not explicitly calculated in the code, but it should be
> modulus square of vec1(2*n2d+npol*norbs+ig,n) (line 234). Am I correct?
in the 4.1 version there is output of both t and r.
>
> 2. In the code reflection and transmission coefficients are calculated
> for one direction lets say t++ and t+-, which is transmission and
> reflection coefficient for Bloch' state moving from - to + , or in other
> words, left to right. To calculate transmission and reflection for Bloch's
> state going from right to left, what should I do? Should i just rotate my
> system 180 degree, that way leads get interchanged and also scattering
> area has orientation 180 different from before.
>
again in the 4.1 version you have possibility to calculate t and r for
both directions, in the do_cond.f90 the transmit routine is called with
an extra parameter which is true (false) if you want to calculate
scattering states propagating from the left (from the right).
> 3. In the subroutine sunitary.f90, is raux measure of conservation of
> flux? If I sum all the transmission and reflecion coefficient for
> a particular Bloch's state, I should get 1, and computationally a number
> close to 1. How large could it be for S matrix not to be unitary, i m
> getting raux more than one e.g. 1.0011 but code still considers smatrix
> to be unitary.
in the case when |R + T - 1| > 1.d-4 for some band, it should print out
the value of R+T but continues to run anyway.
All the best,
Alexander
>
> Regards,
> Manoj Srivastava
> University of Florida, Gainesville, FL
>
> <pw_forum at pwscf.org>
>
>
>
>
> On Thu, 22
> Oct 2009, Alexander Smogunov wrote:
>
> > On Thu, 2009-10-22 at 11:34 -0400, Manoj Srivastava wrote:
> > > Dear Alexander,
> > > Thanks for your answer. I just want to make sure. Imagine we have total
> > > number of channels in the left lead 2, so total number of Bloch's state
> > > are 4. 2 of them left going say a and b, and 2 right going say c and d.
> > > So, are you saying that for left going state a, the corresponding right
> > > going state is c? Are they ordered this way?
> >
> > what do you mean by corresponding? Left and right moving Bloch
> > states are in general not related one to another, you can even have
> > different number of them ... Only if you have some symmetry S which
> > brings kz to -kz conserving k_parallel, then the state with
> > \psi_{-kz} will be S \psi{kz}. This is true for example at 2D G point
> > when you have time reversal operation.
> >
> > Now the code simply arranges the propagating states in the order of
> > increasing |k_z|...
> >
> > Regards, Alexander
> >
> >
> > >
> > > Regards,
> > > Manoj
> > >
> > >
> > > On Thu, 22 Oct 2009, Alexander
> > > Smogunov wrote:
> > >
> > > > Dear Manoj.
> > > >
> > > > The output of complex k vectors is performed in
> > > > summary_band.f90 routine. If you want to see all
> > > > the complex k vectors, not only propagating ones,
> > > > you can change at the end of this routine:
> > > >
> > > > -------------------
> > > > do i = 1, nchanl
> > > > WRITE( stdout,'(3f12.7)') DBLE(kvall(i)), AIMAG(kvall(i)), eev
> > > > enddo
> > > > -------------------
> > > >
> > > > to
> > > > -------------------
> > > >
> > > > do i = 1, 2*nstl
> > > > WRITE( stdout,'(3f12.7)') DBLE(kvall(i)), AIMAG(kvall(i)), eev
> > > > enddo
> > > > -------------------
> > > >
> > > > Altogether there are 2*nstl (or 2*nstr) Bloch states in the left
> > > > (or right) lead. First half, [1,nstl], are propagating or decaying to
> > > > the right states, another half, [nstl+1,2*nstl], - propagating or
> > > > decaying to the left. In each group, first nchanl states are propagating
> > > > states.
> > > >
> > > > The propagating states are normalized by the current and are arranged in
> > > > the above order at the end of jbloch.f90 routine, after the following
> > > > lines:
> > > >
> > > > !
> > > > ! Right ordering (+, >, -, <)
> > > > !
> > > >
> > > >
> > > >
> > > > Notice, that in the last versions the code gives in output
> > > > both propagating to the right and to the left states.
> > > >
> > > > Hope this helps,
> > > > Alexander
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > On Wed, 2009-10-21 at 14:13 -0400, Manoj Srivastava wrote:
> > > > > Dear All,
> > > > > I am trying to figure out the left and right going Bloch's states in the
> > > > > lead from PWCOND. For a given (kx,ky)and energy we get kz. The code only
> > > > > prints out Bloch's state moving in one direction. eg. in one of the
> > > > > calculation-
> > > > > k//=(0.375,-0.375)
> > > > > Nchannels of the left tip = 1
> > > > > k1(2pi/a) k2(2pi/a) E-Ef (eV)
> > > > >
> > > > > 0.3157801 0.0000000 0.0000000
> > > > >
> > > > > Now if I want Bloch's state moving in right as well as left direction, I
> > > > > can go to kbloch.f90 subroutine, and print out all the eigen values of
> > > > > AX=exp(ikd)BX, and out of those the ones with real solution would be our
> > > > > Bloch's state, so I get for each channel two solutions-
> > > > > kval (-0.275409421993275,1.823688001395235E-010)
> > > > > kval (0.315780119742506,-3.611201785292708E-012)
> > > > >
> > > > > To figure out the direction, I can calculate current associated with these
> > > > > Bloch's sate and if the current is +ive it is right moving , and if '-'ive
> > > > > its left moving Bloch's state. I can print out current from jbloch.f90
> > > > > subroutine which are -
> > > > > current eigenvalue -1.86502143831863 1.59149029314457
> > > > >
> > > > > So, clearly the first state with kval=-0.2754094 is left moving and the
> > > > > other one right moving. Upto here its clear to me how to identify left and
> > > > > right moving states.
> > > > >
> > > > > I get confused when for a given (kx,ky,E), I have more than one Bloch'
> > > > > state. In another calculation where i get multiple Bloch's state-
> > > > > Nchannels of the left tip = 5
> > > > > k1(2pi/a) k2(2pi/a) E-Ef (eV)
> > > > >
> > > > > -0.0746301 0.0000000 0.0000000
> > > > > 0.1205527 0.0000000 0.0000000
> > > > > 0.3112908 0.0000000 0.0000000
> > > > > 0.4200218 0.0000000 0.0000000
> > > > > -0.4935150 0.0000000 0.0000000
> > > > >
> > > > > so i did the same trick i did above to first print out kz and then
> > > > > current, which gives me -
> > > > > kval (-0.420023481074359,1.979595081419732E-010) (call it a)
> > > > > kval (0.420023367986768,2.500979698670295E-011) (b)
> > > > > kval (-0.306507431678779,-1.236804629184431E-011) (c)
> > > > > kval (-0.125376071175573,-6.134512510438736E-011) (d)
> > > > > kval (-7.945001124706894E-002,6.683546930037856E-011)(e)
> > > > > kval (0.106554601758169,-6.427946951285107E-011) (f)
> > > > > kval (8.866867725358024E-002,8.342250371574646E-011) (g)
> > > > > kval (0.325333314672671,1.260810749228185E-011) (h)
> > > > > kval (-0.488725859521576,1.769197678346003E-010) (i)
> > > > > kval (0.479509832763231,1.765499400037283E-010) (j)
> > > > >
> > > > > current eigenvalue -9.31389492882581 -1.24296522993488
> > > > > -1.21324078359658 -1.11950286753963 -1.08166842367443
> > > > > 1.08187482164864 1.11973146584263 1.21295295042188
> > > > > 1.24280031534940 9.313897787790
> > > > >
> > > > > So, the first 5 are left moving and rest are right moving. But I dont know
> > > > > the pairs. for example for left moving state a, what is the corresponding
> > > > > right moving state whether its f or g ... j ?
> > > > >
> > > > > Any help would be appreciated.
> > > > >
> > > > > Regards,
> > > > > Manoj Srivastava
> > > > > University of Florida, Gainesville.
> > > > >
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> > > > home-page: http://people.sissa.it/~smogunov
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