# Transmission function(probability) T(E) in tight binding

1. Jun 22, 2009

### rejinisaac1

I am working on tight binding formulation of CNTs. The transmission function T(E), which is the trace of the product of the lead self energies and the retarded and advanced green's function.
This value is a complex entity. T(E) needs to be calculated at different energy levels and then integrated to find the current flowing. My question is: How does one integrate the complex T(E) over the real axis of energy. Do I take just the real part and integrate or integrate both real and complex part separately and then take the modulus? Please help

2. Jun 22, 2009

### kanato

Do you have a reference on the definitions used here? I am not clear on what exactly you are referring to with T(E). Are you calculating your Green's functions and self energies on the real frequency axis or at Matsubara frequencies?

3. Jun 22, 2009

Transmission function should be real, for both real and complex frequencies. I don't know about the transmission function, but integrating e.g. density over the real axis can be a big problem due to singularities. The solution is to use contour integration. For example, the integration
$$n_i \propto \int_{-\infty}^{E_F} \textrm{Im} G_{ii}^R(E) dE$$
is easiest to do using the half-circle contour
$$n_i \propto \textrm{Im} \int_{C_R} G_{ii}^R (z) dz.$$
Remember to first integrate and then take the imaginary part, not vice versa! The imaginary part of the Green's function is not analytic in the upper half-plane. Here the half-circle $$C_R$$ starts from a point $$E_0$$, below which there should be no states available, and ends at $$E_F$$.

I have always used zero-temperature and linear response such that I have not had to do integration over energy to get the conductance. This is usually enough, since real-life bias voltage cannot be taken into account reliably in the Landauer formalism. I do not know if the reasoning above applies to transmission function, since the advanced Green's function has poles in the upper half-plane.

4. Jun 23, 2009

Oh, and in the trace there should be no self-energies, but hermitian coupling matrices, $$\Gamma _{\alpha} = i (\Sigma_{\alpha}^R - \Sigma_{\alpha}^A)$$. This makes the whole matrix product inside the trace hermitian, and thus the trace is real.

5. Jun 23, 2009

### rejinisaac1

Thanks for the replies. Though I wrote 'lead self energies', it was the coupling matrices that I was referring to. But I still have a doubt

$$\Gamma _{s} = i (\Sigma_{s} - \Sigma_{s}^{\dagger}) , \Gamma _{d} = i (\Sigma_{d} - \Sigma_{d}^{\dagger}) , T(E)=Tr\left[\Gamma _{s}G^{r}\Gamma _{d}G^{a}\right]$$

Though $$\Gamma_{s}$$ and $$\Gamma_{d}$$ are real, the retarded and advanced Green's functions $$G^{r}$$ and $$G^{a}$$ have complex values. So the trace HAS to contain both real and imaginary part.

6. Jun 23, 2009

$$T^* = \textrm{Tr} \left\{ \left[ \Gamma _{s}G^{r}\Gamma _{d}G^{a}\right] ^{\dagger} \right\} = \textrm{Tr} \left[ G^r \Gamma_d^{\dagger} G^a \Gamma_s^{\dagger} \right] = \textrm{Tr} \left[ G^r \Gamma_d G^a \Gamma_s \right] = \textrm{Tr} \left[ \Gamma _{s}G^{r}\Gamma _{d}G^{a}\right] = T.$$