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In some quantum text books [1], 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?!

[1] For example, see "Introduction to Many-body quantum theory in condensed matter physics", Bruus et al.

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    $\begingroup$ 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). $\endgroup$ – gatsu Sep 26 '14 at 7:33
  • $\begingroup$ Because if there is no DOS at each side at one energy, there would not be any tunneling current $\endgroup$ – Hesam Sep 26 '14 at 18:53
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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:$$D=\frac{4k^2\kappa^2}{(k^2+\kappa^2)^2\sinh^2\kappa a+4k^2\kappa^2}$$

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Making D dimensionless by setting $m=\hbar=a=V=1$, we can plot $D$ vs $E$:

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Since $E=v^2/2$, we can also plot $D$ vs $v$:

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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.

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