# Fermi "surface" at finite temperature and its measurement in the lab

As we increase the temperature, we know the sharp Fermi surface at zero temperature becomes smeared out at finite temperature $$T>0$$. (Just think of the Fermi-Dirac distribution, there will be no more a sharp kink when $$T>0$$.)

Would this smeared-out Fermi surface affect the lab measurement such as using the de Haas–van Alphen effect or the Shubnikov–de Haas effect? How can the Fermi surface be measured precisely at finite $$T$$?

• Fermi level can be measured using photoelectron spectroscopy Commented Nov 27, 2016 at 22:51
• even if the Fermi level is smeared? at finite T.
– user32229
Commented Nov 27, 2016 at 23:53
• There is a "thermal broadening" of the peaks in a V - 1/H graph. If the sample was at 0K the peaks would be like Dirac deltas without being infinite in height, of course. One can still determine the maximum of a sine, or a cosine, even though it isn't very "peaky". For this reason, it is possible to obtain the Fermi surface even at finite temperature, provided that the corresponding effects take place. Commented Sep 3, 2020 at 14:37

## 2 Answers

There is a "thermal broadening" of the peaks in a resistivity - $$1/H$$ graph, where $$H$$ is the applied magnetic field. If the sample was at $$0K$$ the peaks would be like Dirac deltas without being infinite in height, of course. One can still determine the maximum of a sine, or a cosine, even though it isn't very "peaky". For this reason, it is possible to obtain the Fermi surface even at finite temperature, provided that the corresponding effects take place. I.e. provided that the sample is pure enough, at a low temperature enough, and that a strong enough magnetic field is applied, and that either its magnetization or resistivity is being measured as the magnetic field's strength is swept across a large enough range.

As you say, for $$T>0$$ the occupation of states $$f(E)$$ will not have a sharp cutoff at $$E_F$$ but it will fall to zero over an energy interval of ~$$k_B T$$. The question is then how big is $$k_BT$$?

Well at room temperature, $$k_B T \approx 25$$ meV while electronic energies are on the order of eV. So yes, $$f(E)$$ is blurred at the Fermi energy but it is a small effect at non-extreme temperatures.

• The OP is asking about laboratory conditions regarding the de Haas Van Alphen and the Shubnikov-de Haas effects in particular. The Fermi surface is generally measured by going at very low temperatures compared to room temperature. So your answer isn't a real answer as it is. Commented Sep 3, 2020 at 14:27
• Furthermore the finite temperature has a big impact on the measurements permitting the obtention of the Fermi surface, which goes against your conclusion. This measurement cannot be done at room temperature for any metal, as far as I know. Commented Sep 3, 2020 at 14:42