Does tunneling transmission probability depend on the density of states or velocity?

In some quantum text books , the tunneling transmission formula depends only on the density of states of 2 regions (DOS) involved in tunneling. ($T(E) = C \times DOS_1(E) \times DOS_2(E)$, where C is constant). However, in Landauer transmission formula (without tunneling) the transmission depends on both DOS and velocity of carriers ($T(E) = C' \times DOS(E) \times v$). So I am wondering if velocity is important too? If yes, which velocity in which region?!

 For example, see "Introduction to Many-body quantum theory in condensed matter physics", Bruus et al. • Why would the first formula involve a product of DOS instead of a ratio? I don't understand what this product is telling us. (note that I assume here the transmission to be say from 1 to 2). – gatsu Sep 26 '14 at 7:33
• Because if there is no DOS at each side at one energy, there would not be any tunneling current – Hesam Sep 26 '14 at 18:53

I will answer your question why the transmission coefficient depend on velocity in a very naive way. Consider a potential barrier like below, the electron is incident from left with energy $E=\hbar^2k^2/2m$, the barrier width is $a$ and height is $V$. Define $\kappa\equiv\sqrt{2m(V-E)/\hbar}$. The transmission coefficient $D$ can be easily calculated:$$D=\frac{4k^2\kappa^2}{(k^2+\kappa^2)^2\sinh^2\kappa a+4k^2\kappa^2}$$ Making D dimensionless by setting $m=\hbar=a=V=1$, we can plot $D$ vs $E$: Since $E=v^2/2$, we can also plot $D$ vs $v$: Thus you can see that the coefficient is indeed depend on the velocity, because the damping of the wave function in the barrier will be less when it has a larger energy.