Relation between tunneling current and fermi golden rule (Bardeen model)

Question

I am looking into the paper of Tersoff and Hamann - Theory of the scanning tunneling microscope.

In this one there is the tunnel current written as $$I = \frac{2\pi e}{\hbar} \sum_{\mu ,\nu} f(E_\mu) [1- f(E_\nu +eV)] |M_{\mu\nu}|^2 \delta(E_\mu - E_\nu) $$

So in my lectures the tunneling current is written as $$I_T = \frac{4\pi e}{\hbar} \sum_{\mu ,\nu} |M_{\mu\nu}|^2\delta(E_\mu - E_\nu) [f(E_\mu - E_F^\mu) - f(E_\nu - E_F^\nu)] $$

$\mu$ is the tip and $\nu$ the sample, $E_F$ is the fermi energy.

First of all I am wondering if both equation are equal. As far as I see, the one in my lecture already used the limit of $T \to 0$. But if I try to apply this limit to the formula of Tersoff and Hamann, I still do not get the formula of my lecture.

The other question is about how to apply the tunneling current in this model. In my lectures there was said that one can take Fermi golden rule as a start. How is then the tunnel current related to Fermi golden rule?

Answer 1

The term $ f(E_\mu) [1-f(E_\nu + eV)]$ in a tunneling expression means "we want to tunnel from a state at energy $E_\mu$ to a state at energy $E_\nu + eV$, so we want the first to be occupied and the latter to be empty", which is given by the product of the probabilities $f(E_\mu)$ (occupied at $E_\mu$) and $1-f(E_\nu+eV)$ (vacant at that energy)

At the limit $T\to 0$, these probabilities are either $1$ or $0$. So if we take the difference in the occupancy probability $f(E_\mu) - f(E_\nu + eV)$ it will be $1$ if the first is occupied and the other empty, zero if both are occupied or empty and $(-1)$ if the first is vacant but the second is occupied (in which case the current will flow in the opposite direction). However we can rule this last option out if we assume non-negative $eV$ as $E_\mu \leq E_\nu + eV$ under that assumption, so at zero temperature we will never have $E_\nu + eV$ occupied while $E_\mu$ empty.

Thus we are left with the same interpretation as the first expression, valid at $T\to 0$: a product of probabilities for the tunneling event to be possible, namely the origin state to be occupied while the target state to be vacant.

The last discrepancies are the factor $2$ vs. factor $4$, which I guess in one case took into account spin degeneracy and the other did not, and of course notations of the Fermi energy and voltage bias.

Answer 2

The actual current is proportional to the difference of the currents flowing from left-to-right and from right-to-left (or more complex expression in a setting with more than two terminals), which contains a factor like $$ f(E-E_F^1)\left[1-f(E'-E_F^2)\right] - f(E'-E_F^2)\left[1-f(E-E_F^1)\right]= f(E-E_F^1) - f(E'-E_F^2) $$ As @yyy have correctly pointed out, at zero temperature one of the terms in the left-hand-side is identically zero.