# How does thermal broadening of the Fermi Function cause electron coherence loss?

Generally, there are two ways for electrons to lose their wave-like properties in a solid material. One is by way of collisions that cause changes in the energy and momentum of the electron. The other way is by temperature effects on electron physics, expressed through the "smearing" of the Fermi Function.

From Quantum Heterostructures by Mitin:

...At high temperatures, when electrons with significantly different energies take part in kinetic processes, the coherence in the system is destroyed as a result of a large spreading of the wave function phases.

I have not been able to find a relationship between the spreading of the Fermi Function and phase shifts in the constituent electrons in the system. Could somebody clarify this?

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Thermal broadening by itself does not cause coherence loss, in the sense that if you put a fermion into a free fermi gas at any temperature, it stays coherent forever, because it doesn't interact with the gas.

What people usually mean by "loss of coherence due to thermal broadening" is two things which both act together:

• The distribution of thermal electrons is completely random, in that the density matrix has no off-diagonal matrix elements in the energy representation.
• The elecrons are strongly interacting, and the interactions do not affect the thermal distribution, but instead spread the stochasticity from the thermal electrons to the non-thermal electrons.

But the reference you are citing is saying something else, which is less precise. This is talking about electrons of different energies scattering off the same time-dependent potential. You can't know which electron is which, you get a random phase for the scattering, depending on the thermally random phase of the electron coming in. So if you are trying to ask what the phase of one electron only is going to be, you get a random answer.

This is not really decoherence, in principle, because you haven't entangled information about the electron state in an outgoing quantum, which can be thought of as doing a measurement. But you can still lost effective coherence anyway, because you won't be able to track the coherent electrons through the scattering event, because you won't be able to extract the crazy basis in which the density matrix has good non-random off-diagonal elements.

This is very closely analogous to the classical problem of a thermal gas of X atoms with one X atom that you know for sure is close to a certain point, with close to a certain velocity. In principle, you never lose this information, because the Liouville theorem guarantees that the entropy is constant. But after a few scattering events, even just in an external potential, you will not be able to use this information effectively, and you will lose your extra X particle to the thermal background.

If you could describe what type of decoherence calculation this type of statement is used for, it would help make the answer more precise.

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The coherence calculation this is used for is the thermal diffusion length. $$L_T=\sqrt{D\tau_T}=\sqrt{\frac{Dh}{k_BT}}$$ At distances (or device dimensions being inspected) become longer than the thermal diffusion length, the electrons in the system can no longer be considered coherent. –  Caedar Sep 27 '11 at 15:02