- 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, εF=π(hbar)2n/m, and the total energy per unit area, u=π(hbar)2n2/2m.
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)ωc(n+1/2) remains of the form,
ψnk(x,y) = C eikxφn(y−y0)
where C is a normalization constant and φn(y−y0) is the nth level harmonic oscillator wavefunction centered at y0. However, now y0 is related to the quantum number k by,
y0 = (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 Ax=Hy, Ay=0, compute the current in eigenstate ψnk,
| Jnk,y = − e⟨vnk,y⟩= − |
e
m |
⟨ψnk| |
hbar
i |
∂
∂y |
|ψnk⟩ |
| | |
| Jnk,x = − e⟨vnk,x⟩= − |
e
m |
⟨ψnk| |
hbar
i |
∂
∂x |
− |
e
c |
Hy |
|ψnk⟩ |
Now consider a completely filled Landau level of electron states, say the nth level, and compute the total current density from these states,
jn = (1/LW) ∑k Jnk
The current is related to the electric field via the resistivity tensor, E = ρ⋅j, with
For a single completely filled Landau level, show that
ρxx = 0 and ρxy = h/e2
What are ρxx and ρxy 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 ρxx vanishes while the Hall resisitivity ρxy is finite (it is called the Hall resistivity since the Hall coefficient RH = −ρxy/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 Jnk,y = 0 for all the Landau level states nk. Hence one might conclude that the total jy = 0, and so ρxx = 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 - surface plasmons
Consider a metal with a surface in the xy plane, i.e. z > 0 is metal and z < 0 is the vacuum. The surface plasmon is a solution to Maxwell's equations of the form,
Ex = Aeiqx e−Kze−iωt, Ey =0, Ez = Beiqxe−Kze−iωt for z > 0
Ex = Ceiqx eK'ze−iωt, Ey =0, Ez = DeiqxeK'ze−iωt for z < 0
with q, K, K' real and K, K' positive (so that the amplidue of the wave decays as z moves away from z=0).
Assume that there is no charge density induced in the bulk of the metal, i.e. ∇⋅E=0, (but there may be induced surface charge).
By requiring the above expression to satisfy Maxwell's equation, and to satisfy the usual electromagnetic boundary conditions, tangential component of E continuous and normal component of εE continuous, find equations relating the unknown amplitudes A, B, C, D to each other, and relating q, K, K' to each other and to the frequency ω.
Use for the metal the dielectric function ε(ω) = 1+4πiσ(ω)/ω, with σ(ω) the ac Drude conductivity. You may make the assumption that ωτ >> 1 so that ε(ω) ≈ 1 − (ωP/ω)2, where ωP=4πne2/m is the plasma frequency.
Use your results above to solve for q as a function of ω. Plot q2c2 vs ω2, and indicate for what range of ω there is indeed a solution with the desired properties.
Compute the surface charge density induced on the surface of the metal? (Hint: remember your electrostatics!)
What is the dispersion relation ω(q) in the limit of small q? Show that as q gets very large (i.e. qc >> ω) there is a solution at frequency ω = ωP/√2. What is the polarization of this wave when ω = ωP/√2?
- Problem 4 - screening in a 2D electron gas
Consider a thin metalic film, thin enough that only the lowest kz=0 state is occupied and so the free electron states are specified by their two dimensional wavevector k=(kx, ky) in the xy plane. Imagine that the film is freely suspended with vacuum on both sides.
a) If an external electric potential φext 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,
−∇2δφ = 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⊥=(kx, ky). [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 φtot = φext + δφ. If the electric susceptiblity is defined by
χ(k⊥) = − δρ(k⊥)/φtot(k⊥)
Show that
|
φtot(k⊥, z=0) = | φext(k⊥)
ε(k⊥)
|
where
[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⊥) = e2 ∑q fq [ |
1
eq+k⊥ − eq |
+ |
1
eq−k⊥ − eq |
]
|
where the sum is over two dimensional q = (qx, qy) space, and eq 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 ro? Assuming a free electron model, how does ro vary with electron density? Express your answer for ro in terms of appropriate atomic length scales such as 1/kF and ao (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, φtot(k⊥, z=0). Fourier transform to get the real space potential φ(r⊥, z=0). How does it behave in the limits r⊥ << ro and r⊥ >> ro?
[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 semi-infinite dielectric insulator with dielectric constant εo, with the vacuum above the film?
