Hey, I was thinking about the question in class today about how the applied field does not change the Fermi energy or number of electrons in the bulk matter. There was some confusion about the approximation of integral,
(at zero K) as
In particular I am writing about why g(E_f) appears in the second term. This should clearly be there purely from the definite integral being evaluated close to the Fermi energy. At this point we have not assumed the Fermi energy under the applied field is the same as the Fermi energy of the free metal (no applied field). The fact that this second term (i.e. the first order expansion of the integral) is zero regardless of the Fermi energy of the free metal, is the proof the Fermi energy is constant under the applied magnetic field. Recall the integral from E= 0 up to Ef of the density of states times the F-D distribution function at zero Kelvin must be equal to the number of particles (electrons in this case). Note this is exactly the first term in the above equation, where the upper bound is exactly the Fermi energy of the free metal. However the above equation is the expression for the number of particles under the applied field, hence the act of applying the field does not change (to first order) the Fermi energy.
oh, sorry about the equations, just click on them for see the full pic.
ReplyDeleteI was thinking more in terms of our dummy variable in the integral was E, so when we have evaluated our integral, we shouldn't have E in our final expression (unless E was in our bounds for the integration).
ReplyDeleteOne could also think that all of our excitement happens at the Fermi energy, thus the final expression should be around the Fermi energy.
It is good that the number of electrons is conserved because I don't see how applying a magnetic field would change the number of electrons in the system.
ReplyDeleteIts nice that the Fermi energy does not change to first order which keeps the equation for the Pauli magnetisation simple.
Yeah, that's right Ann E should have been integrated out, but the fact that you replace it with E_f does not mean you're assuming the Fermi energy is constant.
ReplyDeleteAnd Joseph, I think the fact that number of electrons remains constant is needed to prove the Fermi energy is constant. If they were not constant, the expression we derived would only say the Fermi energy of the new number of electrons is the same as the Fermi energy of the old number of electrons
Yeah, also, if you have a variable number of electrons, then you can't have a constant Fermi energy...
ReplyDeleteThen we've replaced our dummy variable E with variable E_{f}!!
ReplyDelete