Problem 7.8 (20 pts.), 7.15 (20 pts.), and 7.28 (40 pts.)

Meeting with visiting candidates (10 pts. each; more details to come)

• Lunch Monday November 16, 2026

• Lunch Tuesday November 17, 2026

• Lunch Wednesday November 18, 2026

Hints:

• Problem 7.28:

  • You may find it easier to do part (b) first and find the density of states g(ϵ) for the two-dimensional electron gas. You should find that g(ϵ) is a constant,

    

  • Once you have the density of states, it is easy to calculate N and U. Note, for example, that at T = 0,

    N = ∫ 0ϵF g(ϵ)dϵ
    U = ∫ 0ϵF ϵg(ϵ)dϵ

    The first equation means that you can rewrite g(ϵ) = .

  • You may skip part (c). At T = 0, μ = ϵ_F . At low, but non-zero temperature in 2 dimensions, it turns out that μ is still ϵ_F .

  • For part (d), note that the chemical potential is effectively defined by Eq. 7.53 for a given N and T. Part (d) involves a messy integral. If you define

    

    then you should have an integral which evaluates to

    ∫ x0∞ = ln(1 + e-x0 )

    (If you want to prove that result, multiply the top and bottom by e^-x, and then make the substitution y = 1 + e^-x. Or you could just use the result given above.)

  • For part (e), you should get a result that looks similar to Eq. 6.93, but you need to replace the ratio by an analogous result for 2 dimensions: .