Where is the equlibrium fermi level ?

[yangzw1985]

hellow,everyone! can anyone tell me where is the equlibrium fermi level of a two probe system. and how to determine the fermi energy of the two probe system from a calculated transmission spectrum? Thanks!

[Anders Blom]

Fermi level = 0 eV in two-probe when looking at transmission, MPSH eigenstates, etc.

[yangzw1985]

    Thank you, Anders Blom! I have told the answer to my pro, but he didn't agree with you.Can you give me sufficient reasons to help me to make my pro agree with you?
     If the frimi level=0, then what is the real mean of the fermi energy in the exported transmission spectrum is? As can we see from the exported transmission spectrum , for  Electrodes Calculation
#
 ----------------------------------------------------------------
# sc  0 : Fermi Energy =    0.00000 Ry
#-------------------------------------------------------------------------------
# Mulliken Population for sc 0
#-------------------------------------------------------------------------------
 0 Li     q =  1.00000 [ s:  1.000, py:  0.000, pz:  0.000, px:  0.000 ]
 1 Li     q =  1.00000 [ s:  1.000, py:  0.000, pz:  0.000, px:  0.000 ]
 2 Li     q =  1.00000 [ s:  1.000, py:  0.000, pz:  0.000, px:  0.000 ]
 3 Li     q =  1.00000 [ s:  1.000, py:  0.000, pz:  0.000, px:  0.000 ]
#-------------------------------------------------------------------------------
# Total Charge =    4.00000
#-------------------------------------------------------------------------------
# sc  1 : Fermi Energy =   -0.41765 Ry  dRho = 4.6383E+000
#-------------------------------------------------------------------------------
# Mulliken Population for sc 1
#-------------------------------------------------------------------------------
 0 Li     q =  0.99999 [ s:  0.476, py:  0.186, pz:  0.153, px:  0.186 ]
 1 Li     q =  1.00001 [ s:  0.476, py:  0.186, pz:  0.153, px:  0.186 ]
 2 Li     q =  1.00000 [ s:  0.476, py:  0.186, pz:  0.153, px:  0.186 ]
 3 Li     q =  1.00001 [ s:  0.476, py:  0.186, pz:  0.153, px:  0.186 ]
#-------------------------------------------------------------------------------
# Total Charge =    4.00000
#-------------------------------------------------------------------------------
# sc  2 : Fermi Energy =   -0.41572 Ry  dRho = 3.4473E-003
#-------------------------------------------------------------------------------
# Mulliken Population for sc 2
#-------------------------------------------------------------------------------
 0 Li     q =  0.99999 [ s:  0.476, py:  0.186, pz:  0.153, px:  0.186 ]
 1 Li     q =  1.00001 [ s:  0.476, py:  0.186, pz:  0.153, px:  0.186 ]
 2 Li     q =  1.00000 [ s:  0.476, py:  0.186, pz:  0.153, px:  0.186 ]
 3 Li     q =  1.00001 [ s:  0.476, py:  0.186, pz:  0.153, px:  0.186 ]
#-------------------------------------------------------------------------------
# Total Charge =    4.00000
#-------------------------------------------------------------------------------
# sc  3 : Fermi Energy =   -0.39825 Ry  Etot =   -3.70151 Ry  dRho = 3.4600E-002
#-------------------------------------------------------------------------------
# Mulliken Population for sc 3
#-------------------------------------------------------------------------------
 0 Li     q =  0.99999 [ s:  0.480, py:  0.185, pz:  0.150, px:  0.185 ]
 1 Li     q =  1.00001 [ s:  0.480, py:  0.185, pz:  0.150, px:  0.185 ]
 2 Li     q =  1.00000 [ s:  0.480, py:  0.185, pz:  0.150, px:  0.185 ]
 3 Li     q =  1.00001 [ s:  0.480, py:  0.185, pz:  0.150, px:  0.185 ]
#-------------------------------------------------------------------------------
# Total Charge =    4.00000
#-------------------------------------------------------------------------------
# sc  4 : Fermi Energy =   -0.39824 Ry  Etot =   -3.70151 Ry  dRho = 2.3260E-005  dEtot = 1.3067E-008 Ry
#-------------------------------------------------------------------------------
# Mulliken Population for sc 4
#-------------------------------------------------------------------------------
 0 Li     q =  0.99999 [ s:  0.480, py:  0.185, pz:  0.150, px:  0.185 ]
 1 Li     q =  1.00001 [ s:  0.480, py:  0.185, pz:  0.150, px:  0.185 ]
 2 Li     q =  1.00000 [ s:  0.480, py:  0.185, pz:  0.150, px:  0.185 ]
 3 Li     q =  1.00001 [ s:  0.480, py:  0.185, pz:  0.150, px:  0.185 ]
#-------------------------------------------------------------------------------
# Total Charge =    4.00000
#-------------------------------------------------------------------------------


AND AS FOR Equivalent Bulk Calculation (Initial Density for TwoProbe)

sc 23 : Fermi Energy =   -0.35530 Ry  Etot = -7301.92848 Ry  dRho = 3.8679E-004  dEtot = -6.0960E-006 Ry

can you explain the feimi energy for me ?   

thanks!!
     

[Nordland]

The equlibrium energy in our case is -0.35530 Ry, however when calculating the transmission spectrum and other quantities that require you to enter an energy.
Then this energy is relative to the fermi level, therefore in the transmission spectrum the E=0 the correspond to the your fermi level (-0.35530 Ry)
The clever thing about this, is that if your script write the energy range -2 eV to 2 eV, then it is always centered around -2 eV and 2 eV relative to the fermi energy, and hence you can use this generic range.

[Anders Blom]

Energies are always relative. Thus it makes no sense to talk about specific energies unless you also specify the zero point.

The Fermi level in the electrode calculation is reported relative the zero-point of the Hartree potential. This is pretty much an internal thing, and the value as such cannot really be used for anything. ATK will however internally use them to align all energies in the two-probe system to a common energy zero-point.

When the user later specifies an energy for which he wants to compute the transmission spectrum, for instance, and specifies E=1 eV, then ATK will compute the transmission spectrum for an energy 1 eV above the equilibrium Fermi level.

This works the same way also when a bias is applied. However, the bias is "averaged", that is if you apply a bias of 0.5 V on both electrodes (the same bias on both electrodes!), the calculation is still a zero-bias calculation. Only the difference in bias matters.

[yangzw1985]

-0.35530 Ry=-4.7495 ev that means the fermi energy is -4.7495 ev of our two probe systerm. we have calculated the transmission spectrum 500 points from -12ev  to +12 ev. In the transmission spectrum, is the fermi energy is always equals 0 EV? I guess my pro will can not accept the answer and the reasons.

[Nordland]

If you specified the energy range -12 eV to 12 eV to the transmission spectrum, then the transmission is calculated in a interval around the fermi level of -12 eV to 12 eV. Therefore in absolute energy you have calculated the transmission spectrum in a energy window from -16.7495 eV to 7.2505 eV. In terms of energy relative to the fermi energy you have calculated it in a range from -12 eV and 12 eV.

It is quite straight forward, if you write E=1.0 eV to the calculateTransmissionSpectrum, then you will get the transmission spectrum at the absolute energy E = Efermi + 1.0 eV. If you write E = 2.0 eV, you will get the transmission spectrum at the absolute energy E = Efermi + 2.0 eV.

Energies are always relative quantities, and hence the fermi energy is reported relative to the vacuum level, and the transmission energies are reported relative to the fermi energy. Actually, therefore when calculating the transmission spectrum, the fermi level is not important, as all energies are relative to fermi energy.

Another point, it is waste of time calculating the transmission spectrum from -12 eV to 12 eV, as it is only the around the fermi level, the transport occour, and therefore an initial of -2 eV to 2 eV (as it is always relative to the fermi level) is really the optimial choice.

[yangzw1985]

But in some transmission spectrum picture,there is a bias window referred to the energy interval from the chemical potential of the left electrode to that of the right electrode. and it is written as [-v/2,+v/2]where v is the applied voltage. if calculated transinission spectrum from -2 to +2 ,how to determine the bias window?

absolute energy you have calculated the transmission spectrum in a energy window from -16.7495 eV to 7.2505 eV.
 if i want give the transmiaaion spectrum from -6 to +6 at the X axis with E-Ef(ev), where is the fermi level ,0 or -4.7495?

if the -6 to 6ev at the x axis with E-Ef , when the fermi level=0 , Do i have to use the results from -10.7459 to 1.1205.

[Anders Blom]

For all practical purposes, in a two-probe system, the equilibrium Fermi level is zero, and this is the only energy reference point you need. The individual Fermi level values reported from the electrode calculations are useless, since they are only known relative to the vacuum level, and hence they do not mean anything when you are looking at the transmission spectrum. Forget them, they cannot be used for anything anyway

If a bias is applied, the bias window is [-V/2,+V/2] where V is the difference in chemical potential/Fermi level between the two electrodes (this is the definition of the bias), even if you actually apply the bias asymmetrically, like [0,V].

So, whether you run the calculation with bias or not, the reference energy is always zero in the transmission spectrum, and coincides with the average of the left and right electrode Fermi levels.

[bidisa]

I am considering a two-probe system where an open shell molecule is included in the central region. My question is how the equilibrium Fermi level is calculated for sytems where up-spin and down-spin eloectrons are not same in number.

Thankyou in advance,
Bidisa Das

[Anders Blom]

The Fermi level is always computed in the same way, regardless of whether spin is involved or not. The Fermi level is computed such that the total occupation of all level matches the total number of electrons in the system. If the calculation is spin-polarized, some of the levels are spin-up levels and some spin-down, with different energies (they are degenerate if no spin is involved) but there is just one common Fermi level.

[zzhang]

It seems to me that there are three Fermi levels or reference energies we can speak of: Fermi energy of the electrodes only, Fermi level of the equivalent bulk, and Fermi level of the two-probe system. Since physically the equivalent bulk and the two-probe system are different, all the three quantities are different.

For a two probe calculation without bias, there reports a Fermi energy, which is common to both electrodes and the scattering region. This Fermi level is different form that of the Fermi level for the electrode only. I guess one way to see the difference is to calculate the charges in the electrode region, in cases of the isolated electrode and also in the case when the electrode is actually treated as part of the two probe system, up to the Fermi level. The charges thus obtained should be different. 

When calculating the transmission spectra, I guess the bias is relative to the Fermi level obtained in the two probe calculation with ZERO bias, neither the Fermi level in the electrode calculation nor the Fermi level in the equivalent bulk calculation. 

Could somebody correct me if I am missing something here?

[zh]

As we know, the two-probe system consist of  two electrodes and a scattering region. Before these three parts are coupled together, we can imagine that we can separately handle them by the conventional electronic structure calculations to obtain their corresponding Fermi levels (here denoted as [tex]E_F^0(left), E_F^0(right), E_F^0(center)[/tex]). The values of Fermi level for these parts are usually given with respect to the zero-point reference energy of electrostatic potential. Actually, for a given charge density of molecule or bulk system, the electrostatic potential is  obtained by solving Poisson's equation. But there is a constant term in the solution of Poisson's equation for a given charge density. That is to say, an arbitrary constant can be added into the electrostatic potential while it does not change the charge density. This is why the absolute value of electrostatic potential depends on the choice of zero-point reference energy. It also happens to the absolute values of total energy, Fermi level, eigenvalue and so on, for a given system. So when we separately handle the electrode region and the center region, it is meaningless to compare their corresponding Fermi levels with each other. If the two electrodes are same, the Fermi level of left electrode will be equal to the one of right electrode (i.e.,  [tex]E_F^0(left)=E_F^0(right)[/tex]). This can be easily understood.

Now let's think about the situation when center region is coupled with two electrodes. We can also handle this sandwiched structure as an equivalent bulk system by the conventional electronic structure calculations to obtain its Fermi level (here denoted as [tex]E_F^{equivalent}(bulk)[/tex]. This is indeed implemented in ATK.  One may easily realize that [tex]E_F^{equivalent}(bulk)[/tex] will not be equal to [tex]E_F^{0}(center)[/tex], [tex]E_F^0(left)[/tex], or [tex]E_F^0(right)[/tex]. If one insists on exacting the Fermi levels for the left electrode, the center region, and the right electrode in this equivalent bulk system, the values of Fermi levels may be denoted as [tex]E_F^{equivalent}(left), E_F^{equivalent}(right), E_F^{equivalent}(center)[/tex], respectively.  But here he should realize that [tex]E_F^{equivalent}(left)=E_F^{equivalent}(right)=E_F^{equivalent}(center)=E_F^{equivalent}(bulk)[/tex]. That is to say, the Fermi levels for these three parts in the equivalent bulk system should be aligned. The average electrostatic potential in left electrode may be different to the one in center region, but they should be continuous at boundary between the left electrode and center region. This also happens to the right electrode. If the system is homogeneous, e.g, a infinite Li atomic chain, they will be same. If we neglect the difference due to the choice of zero-point reference energy, the difference between [tex]E_F^0(center)[/tex] and [tex]E_F^{equivalent}(center)[/tex] is mainly caused by the coupling between center region and electrodes: the charge transfer between electrode region and center region, the redistribution of charge density in the center region, the changes of energy levels (or electronic structure), and so on. The better way to get the difference between [tex]E_F^0(center)[/tex] and [tex]E_F^{equivalent}(center)[/tex] is to align the layer-projected DOS for the middle atomic layer  of center region obtained by two kinds of self-consistent calculations: the first is the separate self-consistent calculation for center region, as discussed in the first paragraph; the second is the one for the equivalent bulk system, as discussed in the beginning of the paragraph.  Unfortunately, the current version of ATK does not implement the functional of computing layer-projected DOS for bulk system. The problems discussed here are very similar to those when we handle the metal/insulator heterojunction (or other interface system) by the conventional electronic structure calculations.

In the self-consistent calculation of a two-probe system, i.e., in the last kind of self-consistent calculation, as usually observed in the log file, the charge density of whole system is computed by the equilibrium (or non-equilibrium) density matrix formulas (e.g. see PRB63, 245407(2001), PRB65, 165401(2002)) for equilibrium (or non-equilibrium) case. If some one insist on distinguishing the Fermi levels of electrode region, and center region, they  may be denoted as [tex]E_F^{tp,eq}(left), E_F^{tp,eq}(right), E_F^{tp,eq}(center)[/tex], respectively, in equilibrium case ([tex]E_F^{tp,neq}(left), E_F^{tp,neq}(right), E_F^{tp,neq}(center)[/tex] in non-equilibrium case).  Let's discuss the equilibrium case first. In equilibrium case, the definition of [tex]E_F^{tp,eq}(center)[/tex] is given in Eq. (13) in PRB63, 245407(2001), i.e., [tex]E_F^{tp,eq}(center)=E_F^{0}(left)+\Delta \bar{V}_l=E_F^{0}(right)+\Delta \bar{V}_r[/tex]. Here  [tex]\Delta \bar{V}_{l,r}[/tex] is the change of average electrostatic potential in left (right) electrode, i.e., [tex]\Delta \bar{V}_{l,r}=\bar{V}_{l,r}^{tp,eq}-\bar{V}_{l,r}^{0}[/tex].  While the difference between [tex]E_F^{tp,eq}(left)[/tex]  and [tex]E_F^{0}(left)[/tex] also equals  [tex]\Delta \bar{V}_{l}[/tex], similarly for the right electrode. It seems more convenient if we let [tex]E_F^{tp,eq}(left,right)[/tex]  equal [tex]E_F^{0}(left,right)[/tex], i.e.,  [tex]\Delta \bar{V}_{l,r}=0[/tex].  Actually, it may be found that [tex]E_F^{tp,eq}(center)=\frac{1}{2}[E_F^{0}(left)+\Delta \bar{V}_l +E_F^{0}(right)+\Delta \bar{V}_r][/tex] or [tex]E_F^{tp,eq}(center)=\frac{1}{2}[E_F^{tp,eq}(left)+E_F^{tp,eq}(right)][/tex]. And then [tex]E_F^{tp,eq}(center)[/tex] is usually taken as the zero-point reference energy in the energy-axis of transmission spectrum. Finally, let's discuss the two-probe system in non-equilibrium case. In non-equilibrium case, [tex]\Delta \bar{V}_{l,r}[/tex] is defined as   [tex]\Delta \bar{V}_{l,r}=\bar{V}_{l,r}^{tp,neq}-\bar{V}_{l,r}^{0}[/tex] and it will not equal zero because a finite bias voltage is applied on two electrodes. To simplify the discussion, assume that [tex]\bar{V}_{l}^{tp,neq}=\bar{V}_l^{tp,eq}+e*\frac{V_{bias}}{2}[/tex] and [tex]\bar{V}_{r}^{tp,neq}=\bar{V}_r^{tp,eq}-e*\frac{V_{bias}}{2}[/tex] for the finite bias voltage [tex]V_{bias}[/tex]. That is to say, the average electrostatic potential of left electrode is rigidly shifted up by [tex]e*\frac{V_{bias}}{2}[/tex], while the one of right electrode is shifted down by  [tex]e*\frac{V_{bias}}{2}[/tex]. Consequently, the values of Fermi levels of left (right) electrode will also be shifted similarly (i.e., [tex]E_F^{tp,neq}(left)=E_F^{tp,eq}(left)+e*\frac{V_{bias}}{2}[/tex], [tex]E_F^{tp,neq}(right)=E_F^{tp,eq}(right)-e*\frac{V_{bias}}{2}[/tex] ). Similar to the equilibrium case, the value of [tex]E_F^{tp,neq}(center)[/tex] may be also taken as [tex]\frac{1}{2}[E_F^{tp,neq}(left)+E_F^{tp,neq}(right)][/tex] and used as the zero-point reference energy in the energy-axis of transmission spectrum. In the page 245407-5 of PRB63, 245407(2001), the choice of [tex]E_F^{tp,neq}(center)[/tex] has been discussed.  In short,  in the equilibrium or non-equilibrium self-consistent calculation of a two-probe system, the value of Fermi level of center region (i.e., [tex]E_F^{tp,eq}(center)[/tex] or [tex]E_F^{tp,neq}(center)[/tex] ) is just used as the zero-point reference energy in the transmissions spectrum, MPSH spectrum. Usually, [tex]E_F^{tp,eq}(center)[/tex] (or [tex]E_F^{tp,neq}(center)[/tex] ) does not equal [tex]E_F^0(center)[/tex] .