# The lengths of scattering region and spin polarization.

Asked by Jianjun Mao on 2019-04-12

Dear developers,

1. Recently, I have calculated several examples of IV curves using transiesta. I find that different length scattering regions make a large difference to the currents values with the same electrodes. How do I konw my scattering region is long enough or reasonable. For example, if I add some gas molecues on the scattering region to see the variation of currents before and after their adsorption.

2. What is the relaitonship between the total I and Spin ploarizaiton I. I= I( spinup)+ I (spindown). ? My result show that they are not equal or the average of I (spinup) and I (spindown). Is there anything wrong with my calculaitons? No error throughout the whole calculation.

Many thanks~!

## Question information

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Status:
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For:
Siesta Edit question
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Solved by:
Nick Papior
Solved:
2019-04-23
Last query:
2019-04-23
2019-04-23

## This question was reopened

 Nick Papior (nickpapior) said on 2019-04-12: #1

1. This suggests to me that you haven't screened off your electrodes sufficiently. TranSiesta is based on *bulk* electrodes, and thus the electrostatics needs to be bulk-like at the electrode region of the scattering device. I.e. I(V) should not be length dependent if you simply add more electrode layers.

2. The total current is I = I(spin-up) + I(spin-down). In general I(spin-up) need not be equal to I(spin-down).

 Jianjun Mao (jjmao) said on 2019-04-12: #2

Many thanks for you quick answer, Nick Papior~

1. I will make more test calculaions.

2. When I open the spin polarizaions, the I(spinup)+ I(spindown) is not equal to the I (total) without spin polarization.

For example, same structure;

Under bias voltage: 1.5V
no spin polarization: I (total) = 0.8062E-05

Currents (ensure entire Fermi function window):
Left -> Right, V [V] / I [A]: 1.50000 V / 0.806203E-05 A
Left -> Right, V [V] / P [W]: 1.50000 V / -.602797E-05 W

open spin polarization: I (spin up)= 0.68E-05A I (spin down)=0.73E-05

Currents (ensure entire Fermi function window):
Left -> Right, V [V] / I [A]: 1.50000 V / 0.680710E-05 A
Left -> Right, V [V] / P [W]: 1.50000 V / -.519714E-05 W

Currents (ensure entire Fermi function window):
Left -> Right, V [V] / I [A]: 1.50000 V / 0.732193E-05 A
Left -> Right, V [V] / P [W]: 1.50000 V / -.547095E-05 W

I wonder that is there anything wrong with my calculations?

 Nick Papior (nickpapior) said on 2019-04-12: #3

1. Good.

2. Of course, it is a different Hamiltonian, the electronic structure for the non-polarized is not the average up/down. Secondly, TBtrans writes out the current for non-polarized as though it was one channel, i.e. you need to multiply by two for non-polarized calculations.

 Jianjun Mao (jjmao) said on 2019-04-12: #4

Thanks Nick Papior, that solved my question.

 Jianjun Mao (jjmao) said on 2019-04-18: #5

Dear Nick,

Does the output current I of the non-polarized calculation of the magnetic systems meaningful?( In my test, the output current cof a magnetic system with non-polarized calulation is not equal spinup plus spindown or ethier of them) Or a spin-ploarized calculation of magnetic system is a must and the Total current = I (spin up) + I (spin down) ?

 Jianjun Mao (jjmao) said on 2019-04-18: #6

Dear Nick,

Does the output current I of the non-polarized calculation of the magnetic systems meaningful?( In my test, the output current cof a magnetic system with non-polarized calulation is not equal spinup plus spindown or ethier of them) Or a spin-ploarized calculation of magnetic system is a must and the Total current = I (spin up) + I (spin down) ?

 Nick Papior (nickpapior) said on 2019-04-22: #7

They are not equal.

 Jianjun Mao (jjmao) said on 2019-04-23: #8

Dear Nick,

Here, I still have this doubt: If a spin-ploarized calculation of magnetic system is a must? Or, the right way to get the total currents of a magnetic system under bias voltages?

Thanks you again~!

 Nick Papior (nickpapior) said on 2019-04-23: #9

Of course, if it is a polarized system you need to do a polarized calculation.

 Jianjun Mao (jjmao) said on 2019-04-23: #10

Thanks Nick Papior, that solved my question.