Physics Asked by highly oscillatory integrand on June 25, 2021
In the context of electron transport, it is stated in many references that the elastic scattering does not destroy phase coherence, but inelastic scattering is the source of the phase loss of electrons.
However, this statement seems to be contradicting with the usual relation that is used. In mesoscopic physics, phase coherence length is written as $L_{phi}=sqrt{Dtau_{phi}}$ where $D$ is diffusion coefficient and $tau_{phi}$ dephasing time. Diffusion coefficient is given by $D=v_F^2tau_{el}$ where $v_F$ is Fermi velocity and $tau_{el}$ is elastic scattering time. (Dimensionality factor is ignored for convenience) From these two, we get a relation of phase coherence length that depends the elastic scattering time, $L_{phi}=v_Fsqrt{tau_{phi}tau_{el}}$, which suggests that phase coherence length depends on elastic scattering.
In your expression, you are forcuing on the dephasing length $L_phi$ in a form of typical diffusion process: $$ L_phi^2 = D tau_phi $$ Where the diffusion constant indicates how fast the electron passing through lattice structure in diifusive motion. The diffusion constant has nothing to do with the dephasing. The information of dephasing is carried in another parameter $tau_phi$, which is also there in your final expression. The elastic scattering relaxation time $tau_{el}$ comes in this expression through the diffusion constant $D$, which is again nothing to do with the dephasing process.
Therefore what you should be asking is the mechanisms which are relavent to the dephasing time $tau_phi$, not focuing on the diffusion constant $D$.
Consider another similar case - the conducting mean free path: $$ l_c = v_F tau_p $$ The Fermi-velocity $v_F$ indicating how fast the electron moves at the Fermi level, but indeed the momentum relaxation $tau_p$ is the mechanism for condutivity. The diffusion constant $D$ plays the same role as $v_F$ here.
Answered by ytlu on June 25, 2021
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