why?

[alan]

why the ldos of pefect graphene nanoribbon is not periodically?
in my opinion, the pefect graphene nanoribbon is periodically,so its ldos should also periodically.but the ldos i got is not periodically as i have attched.

[Nordland]

It is a wild guess here, but perhaps you have artificial scattering due to a too short electrode. Try to double the depth of your electrode and see how the the ldos looks like .....

[alan]

Is there any problity that it was caused by the optimize?
i want to know if i double of the probe's depth ,the vaule of dos and transmission will also be defferent from the calculate vaule with a shot probe?

[Anders Blom]

At which energy is this taken? What is the transmission at this energy?

If you did an optimization, the system is no longer perfect, I guess

[alan]

i calculate the ldos at fermi energy ,the transmission is 0 while the dos is not 0 but is close to 0.
the ldos should be peridoically all the same , thougt the system is no longer perfect.i mean that the scattering regin is periodically so its ldos should be periodically while i don not add the votage.is it true?
i calculate the transmission and dos,i want to analysis with the ldos at fremi energy.but the ldos looks like something matter,at first i can find its periodicity. i think there is something wrong in my system.but i dot know what is the problem.
can that we set the probe will break the graphene nanoribbon's periodicity or anything else?

[nori]

Was SCF calculation converged at reasonable state?
It seems that electrons in central region disappear...

[Anders Blom]

It's worth noticing that there is no periodicity in a two-probe system in the Z direction, even at zero bias. This is embedded in the scattering boundary conditions, used for the Green's function.

Thus, if the transmission is zero, it means electrons from the left electrode cannot reach the right electrode. Hence, the LDOS will not extend from the left to the right.

[alan]

I find that if the periodicalty is dispear because i have doing the spin calculate.so the ldos is not periodically in the scattering region.(attched fig.2,3  they are for up and down) 2 cells of both the left and right is the probe.the 8 cells on center is scatting region.
while i do not add the spin to the syetem ,the ldos appers well periodicalty.(attched fig. 1.)
so i want to know whether the ldos i got fig 2,3 is reasonable?
 for some one who studies in Green function mathord said that there would be some mistake in my ldos.because he think the two-probe system  with addtion of 0 votage should be equall to the infinite graphene nanoribbon.so he think the ldos of two-probe system at fermi energy should  also show its periodicalty in its scatting region.just like the fig 1 shows.
that confused me.and thanks a lot for your help.

[Anders Blom]

When you add spin to a graphene ribbon like this, a band gap opens up around the Fermi level. Therefore, your LDOS for E=0 shows a non-conducting state, which does not extend across the structure. In the case without spin, the transmission is finite at the Fermi level, and so you get a conducting state, as your first figure shows.

[Nordland]

If the system is perfect and in equilibrium, then any electrons moving in a bulk eigenstate of the electron will not be scattered and therefore if this electron is transmitted throught the central region, it will generate a perfect uniform local density of states.

When calculating the local density of states you can choose the energy of the incomming electron. However it is often the case, that there is no bulk eigenstates for the certain energies. If this is the case, then the electrons at this energy will be in a state with a finite lifetime, and hence they will not obey
bloch theorem ( and hence not be periodical in space), and will always be of a evanescene nature.

Assuming that your calculation is correctly performed, the local density of states reveals the true nature of the system.
In the spin unpolarized calculation there are electrons ( bulk eigenstates) at the fermi energy, and therefore these eigenstates will generate a perfect periodical local density of states as they propergate unpertubated through the system.

However in the spin-polarized calculation, a small band gap arises at the fermi level and there are no electrons/eigenstates at the fermi level (E=0), and therefore electrons moving at this energy will be of a finite lifetime and be evanescene and not periodical, exactly as you find.

To check that this is correct, do the following two things:
1) Calculate the transmission spectrum or a band structure for the unpolarized system, find an energy where there is no transmission or no bands, calculate the local density of states at this energy, and you will see the exact same behavior with a non periodical local density of states.
2) Calculate the transmission spectrum or a band structure for the polarized system, find an energy where the transmission is different from zero, or an energy where there is some bands, calculate the local density of states for this energy, and you will see that it is perfect periodical.

I hope that somebody understands my ramblings.

[alan]

Thank you very much. 
It is very valuable for me.

[alan]

I am sorry ,i came agian.
1.Calculate the transmission spectrum or a band structure for the polarized system, find an energy where the transmission is different from zero, or an energy where there is some bands.the  local density of states for this energy is fig.1(spin up)and fig 2(spin down). I found that is seems like centrosymmetric.it is not perfect periodical.what does it mean( have i  done an wrong calculate)?
2.For the unpolarized system, there does not exit the enery at which the transmission is 0.so it  is still not been worked out now.

[Nordland]

Hey alan.

Can you perhaps send me the file of the geometry you are using then I will try to see if I can see what is the problem?

[alan]

I have sent the files to you by e-mail.

[Nordland]

Thanks I will look into it!