Posted by Steve Simon on April 28, 2014, 5:58 pm, in reply to "I don't believe in the derivation of the magnetic spin susceptibility"
Well, you had me worried, but I think it is correct as it is. Let me see if I can convince you. First, I do note that this factor of 1/2 does show up if you look at E as a function of B. Generally we expect E(B) = E  (1/2) chi B^2 However, I do think that the formulae I have written are correct. For simplicity, let us assume g(epsilon) is a constant independent of epsilon (as it is in 2d). This special case is easy to evaluate and we can check to make sure the factors of 2 are in the right place. In any magnetic field both the up spins and the down spins are filled to the same chemical potential (to avoid confusion call chemical potential C and mu is the bohr magneton so muB is the bohr magneton times B). So we can write Ntotal = integral^C_{muB} dE g + integral^C_{muB} dE g Etotal = integral^C_{muB} dE E g + integral^C_{muB} dE E g Delta N = integral^C_{muB} dE g  integral^C_{muB} dE g The lower limit on the two integrals are both the k=0 states, but these have energies muB and muB for the two spin species. The integral s are easy to evaluate so we have Ntotal = 2 C g Delta N = 2 g muB Etotal = Constant  g (muB)^2 The claim was dE/dB = mu 2 g muB = mu Delta N which comes out as I claim. More generally when g is a function of E, we have to assume it does not vary too much near the fermi surface, and we will get the same result, although it is a bit tricky to show it. To get you started though, I would write Ntotal = integral^C_{muB} dE g(E) + integral^C_{muB} dE g(E) and similar for Delta N and E. Now shift the two integrals so that both of their lower limits are at zero. Then when you differentiate with respect to B you are only messing around with the upper limits. If you are careful you will get the same result. 
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