 Problem 1  2D electron gas in a magnetic field
In the third problem of Problem Set 1 you considered a two dimensional gas of free electrons confined to the xy plane, and found the density of states g(ε)=m/π(hbar)^{2}, the Fermi energy as a function of electron density n, ε_{F}=π(hbar)^{2}n/m, and the total energy per unit area, u=π(hbar)^{2}n^{2}/2m.
Now assume that a uniform magnetic field H, pointing along the z axis, is applied to the system. Ignore the interaction between H and the intrinsic electron spin. If n is such that the lowest p Landau levels are completely filled, and the (p+1)^{st} Landau level is a fraction λ filled, find the total energy per unit area u(H). [Give your answer as an expression involving the parameters p and λ.]
Compute the change in energy Δu=u(H)−u(0) and sketch this as a function of electron density n. [i.e. now you have to write your answer in terms of n, rather than p and λ, and then sketch it.]
 Problem 2  resistivity tensor of filled Landau levels in 2D
Consider the geometry we used when discussing Landau levels in lecture: a two dimensional system of length L and width W. We take periodic boundary conditions along the x direction, and restrict y to the interval [0,W]. A uniform static magnetic field H is applied along the z direction. Now assume that there is also a uniform static electric field E in the y direction. In the presence of E, the Landau level wavefunctions that we found in lecture get modified in the following simple manner. The wavefunction with energy (hbar)ω_{c}(n+1/2) remains of the form,
ψ_{nk}(x,y) = C e^{ikx}φ_{n}(y−y_{0})
where C is a normalization constant and φ_{n}(y−y_{0}) is the n^{th} level harmonic oscillator wavefunction centered at y_{0}. However, now y_{0} is related to the quantum number k by,
y_{0} = (c/eH) [ (hbar)k − (mcE/H) ]
[You can read the derivation of this result here.]
Using the above wavefunction, and recalling that this solution was obtained using the magnetic vector potential A_{x}=Hy, A_{y}=0, compute the current in eigenstate ψ_{nk},
J_{nk,y} = − e⟨v_{nk,y}⟩= − 
e m 
⟨ψ_{nk} 
hbar i 
∂ ∂y 
ψ_{nk}⟩ 
  
J_{nk,x} = − e⟨v_{nk,x}⟩= − 
e m 
⟨ψ_{nk} 
hbar i 
∂ ∂x 
− 
e c 
Hy 
ψ_{nk}⟩ 
Now consider a completely filled Landau level of electron states, say the n^{th} level, and compute the total current density from these states,
j_{n} = (1/LW) ∑_{k} J_{nk}
The current is related to the electric field via the resistivity tensor, E = ρ⋅j, with
ρ = [  ρ_{xx} ρ_{xy}  ρ_{xy} ρ_{xx}  ] 
For a single completely filled Landau level, show that
ρ_{xx} = 0 and ρ_{xy} = h/e^{2}
What are ρ_{xx} and ρ_{xy} for s completely filled Landau levels?
When the magnetic field H is such that the s lowest Landau levels are completely filled and all higher levels are completely empty, the above result shows that the magnetoresistivity ρ_{xx} vanishes while the Hall resisitivity ρ_{xy} is finite (it is called the Hall resistivity since the Hall coefficient R_{H} = −ρ_{xy}/H). In this case the current flows exactly perpendicularly to the electric field, so j⋅E = 0, and the flow is dissipationless!
The calculation above hints at what is known as the Integer Quantum Hall effect. This effect is actually much more complicated than the above result because it requires one to consider what happens when random impurities are present in the metal. For a discussion of the Integer Quantum Hall effect, you may see my notes here.
If you did the above calculations correctly, you would have found that J_{nk,y} = 0 for all the Landau level states nk. Hence one might conclude that the total j_{y} = 0, and so ρ_{xx} = 0, even for a partially filled Landau level. This is NOT true! Explain why, in a real material, one can only expect the above calculation to give the right results for a completely filled Landau level.
 Problem 3  screening in a 2D electron gas
Consider a thin metalic film, thin enough that only the lowest k_{z}=0 state is occupied and so the free electron states are specified by their two dimensional wavevector k=(k_{x}, k_{y}) in the xy plane. Imagine that the film is freely suspended with vacuum on both sides.
a) If an external electric potential φ_{ext} is applied to the film, there will be an induced charge density, δρ(r_{⊥})δ(z). Here δρ is the surface charge density in the film, with r_{⊥}=(x, y), and the δ(z) indicates that the charge density is confined to the film at the plane z=0.
Such an induced charge density produces its own induced electric potential δφ, that is related to the induced charge density by,
−∇^{2}δφ = 4πδρ(r_{⊥})δ(z)
By taking appropriate Fourier transforms, solve the above Poisson's equations and show that the induced potential in the film is given by,
δφ(k_{⊥}, z=0) = 2π  δρ(k_{⊥}) k_{⊥} 
where k_{⊥}=(k_{x}, k_{y}). [Hint: You have to solve Poisson's equation in full three dimensions, even though the charge density is confined to the two dimensional plane.]
b) The total electric potential in the film will then be φ_{tot} = φ_{ext} + δφ. If the electric susceptiblity is defined by
χ(k_{⊥}) = − δρ(k_{⊥})/φ_{tot}(k_{⊥})
Show that
φ_{tot}(k_{⊥}, z=0) =  φ_{ext}(k_{⊥}) ε(k_{⊥)
} 
where
ε(k_{⊥}) = 1 + 2π  χ(k_{⊥}) k_{⊥} 
[Hint: just follow the steps in lecture  not much is different in this 2D case.]
c) Following the derivation of the Lindhard dielectric function done in lecture, show that
χ(k_{⊥}) = e^{2} ∑_{q} f_{q} [ 
1 e_{q+k⊥ }− e_{q} 
+ 
1 e_{q−k⊥ }− e_{q} 
]

where the sum is over two dimensional q = (q_{x}, q_{y}) space, and e_{q} are the single electron energies.
d) Following the steps done in lecture, approximate the above sum for χ(k_{⊥}) and derive an expression for &epsilon(k_{⊥}) in the limit of small k_{⊥}. What is the screening length r_{o}? Assuming a free electron model, how does r_{o} vary with electron density? Express your answer for r_{o} in terms of appropriate atomic length scales such as 1/k_{F} and a_{o} (the Bohr radius).
e) For a point charge Q in the plane of the film, what is the screened electric potential in the plane of the film, φ_{tot}(k_{⊥}, z=0). Fourier transform to get the real space potential φ(r_{⊥}, z=0). How does it behave in the limits r_{⊥} << r_{o} and r_{⊥} >> r_{o}?
[Hint: you can't do the integral involved in the Fourier transform to real space in terms of simple analytic functions. Look in some integral tables for the answer in terms of special functions, or try to estimate the integral in the appropriate limits.]
f) How would the above results change if the metallic film were deposited on top of a semiinfinite dielectric insulator with dielectric constant ε_{o}, with the vacuum above the film?