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In quantum mechanics the scattering matrix is often used, to connect the states on the left and right side over some region in the middle. In the trivial case, that the Hamiltonian in the middle is the same as left and right, there is no reflection and the transmission probability for each channel is $1$. Now this probability is independent of energy of course.

In my case, I am interested in an energy independent scattering matrix. I wonder, is there some kind of well defined set of Hamiltonians, that leads to such an energy-independent scattering matrix?

The reason for why I am asking this, is the Landauer formula that relates the conductance of a sample to the transmission probability.

$$G = \frac{e^2}{\pi\hbar} \sum_n T_n $$ Here $n$ is an index describing different channels.

However, this is only valid, if the transmission probability is independent of energy, bringing me back to my question. When are Transmission probabilities independent of energy?

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  • $\begingroup$ relevant: Reflectionless potentials in quantum mechanics $\endgroup$
    – AccidentalFourierTransform
    Commented Apr 14, 2016 at 17:05
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    $\begingroup$ I don't quite get how can $S$ depend on the energy. To me, it never does. Its matrix elements $S_{if}$ depend on the energy of $|i\rangle$ and $|f\rangle$, but $S$ itself does not. How can it even depend on the energy? the energy of what? $\endgroup$
    – AccidentalFourierTransform
    Commented Apr 14, 2016 at 17:11
  • $\begingroup$ @AccidentalFourierTransform There two energies, the relative energy (difference) of the two states $|i>$ and $|f>$ and their average energy. You are right, I did not clearify this. I meant energy conserving scattering, which leaves only one energy, the average. I refer to this energy as the energy of the scattering matrix. $\endgroup$
    – physicsGuy
    Commented Apr 15, 2016 at 0:01

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Transmission probability is independent of energy if the density of states of the scattering region is approximately constant near the Fermi energy, ie. if it is metallic. The transmission probabilities in the formula given in your posting actually correspond to the energy dependent transmission function taken at the Fermi energy $T_n=T_n(E_f)$. Also note that the formula may be generalized to give the current as a function of applied bias $eV=\mu_L-\mu_R$: $$ I=\frac{e}{\pi \hbar} \sum_{n}\int dE\, T_n(E)\left(f(E-\mu_L)-f(E-\mu_R)\right) $$ where $f(E)$ is the Fermi function and $\mu_L$ / $\mu_R$ is the electrochemical potential of the left/right side, respectively. This formula reduces to Landauer's formula for the linear conductance $G$ (with $I=GV$ ) if the $T_n(E)$ are constant in an interval of width $eV$ around $E_f$ and the temperature is low $T\ll E_f$.

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  • $\begingroup$ This helps a lot to clarify what I actually meant. So you are saying, as long as I stay close to the fermi-energy, the Transmission will stay indpenendent of the energy. Can you maybe comment on what to expect, if I have say a metal with random impurities. Would you expect the transmission decrease or increase if go to lower energies [i.e. lower wavelengths]. $\endgroup$
    – physicsGuy
    Commented Apr 15, 2016 at 0:08

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