Fermi level is defined as the energy level where the probability of finding an electron is 50%. Why should Fermi levels of two metals/semiconductors in contact be the same? Why don't the energy levels where the probability of finding and electron is x% line up?


The Fermi level of a solid (metal, semiconductor, etc.) is synonymous to the total chemical potential of a body, which is a thermodynamic quantity but also plays a role in statistical physics as the reference energy of the Fermi-Dirac energy distribution. It is a thermodynamic law that upon contact of two systems the total chemical potentials align, so that the combined system has again one constant total chemical potential. This is similar to the temperatures of two systems equilibrating when brought into contact.

  • $\begingroup$ So it is shear coincidence that chemical potential is the same energy value as a energy level where probability of finding an electron is 50%? $\endgroup$ – Kurburis Sep 20 '16 at 22:22
  • $\begingroup$ No, this is not a coincidence. It follows from the relation between statistical mechanics and thermodynamics in thermodynamic equilibrium. It is known that the Fermi-Dirac energy distribution function of electrons in equilibrium, which are fermions, has a reference energy (Fermi level EF) which is equal to the (total) chemical potential per electron 𝜇. And the Fermi-Dirac distribution function is exactly 0.5 at the Fermi level, meaning that an energy level at that energy has occupation probability of 50%. See en.wikipedia.org/wiki/Fermi_level . $\endgroup$ – freecharly Sep 20 '16 at 22:47
  • $\begingroup$ I should have maybe framed the question better. I know about that relation and that from it we get the 50%. Let's say, if we started from the thermodynaimcal law that you mentioned, could the Fermi-Dirac statistic be derived? If we started with \mi that was in that law, would that \mi manifest it self in the FD statistic? Or maybe you can think about it like this. If in some strange universe chemical potential doesn't align, but some other energy value, would that energy value be the 50% energy value? $\endgroup$ – Kurburis Sep 20 '16 at 23:08
  • $\begingroup$ It is a consequence of thermal equilibrium that chemical potentials of electrons align upon contact and that this energy level (also called Fermi level) is the reference energy for the FD distribution where the occupation probability is 50%. Therefore, if the chemical potentials would not align but some other energy levels this would contradict thermodynamics and statistical thermodynamics. $\endgroup$ – freecharly Sep 20 '16 at 23:58

Because being the Fermi level in one metal higher than the other, it means that electrons in the metal with the highest $E_F$ are able to go to lower energy states in the other metal, but not the other way around. Thus electrons will flow until both metals(or semiconductors, or combination) have the same Fermi energy, breaking this flow. Also you could think that at 0 Kelvin the probability of being above $E_F$ is 0%.

Is very similar to connecting to water containers, one being a little more high than the other, so that the water flows until the height is the same.

  • $\begingroup$ Yeah, but couldn't we say the same if the probability was 30% not 50%? $\endgroup$ – Kurburis Sep 20 '16 at 22:19

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