 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2011/paper-1/section-c/c11/a-internal-energy/solution

For independent conduction-electron states labelled by band, wavevector and spin, combine those labels into

j

. Then

U_{e l} = j \sum ϵ_{j} f_{FD} (ϵ_{j}; μ, T) = \int g (ϵ) ϵ \frac{d ϵ}{e ^{(ϵ - μ) / k_{B} T} + 1} .

Here

g

counts the same spin and band states as the sum, and

μ (T)

is fixed by the number constraint

N = \int g f_{FD} d ϵ

. The energy zero is a fixed band reference. Use the total DOS for total internal energy or the molar DOS for molar internal energy; these conventions must not be mixed.

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