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PHY 521/321: Condensed Matter Physics I
Prof. S. Teitel stte@pas.rochester.edu  Spring 2008
Problem Set 1
Due Thursday, February 14, in lecture.
 Problem 1 [25 points]
Assume a metal has a density n of mobile charges of mass m and charge e, and a density n* of mobile carriers of mass m* and charge +e. Assume that there is an immobile background of ionic charges so that the system is electrically neurtral. Using arguments like in the Drude model, compute:
a) The d.c. electrical conductivity σ_{dc}
b) The Hall coefficient R_{H}
c) The magnetoresistance ρ
Try to set up the problem generally, to see that the results of part (b) and (c) will now depend on the strength of the magnetic field H. Then work out the result in the high magnetic field limit, ω_{c}τ>>1 and ω*_{c}τ>>1, where ω_{c} and ω*_{c} are the cyclotron frequencies of the negative and positive charges respectively. You may assume that the collision time τ is the same for both the positive and negative charges.
Hint: You have to assume that only the total current perpendicular to the length of the wire is zero, but not the separate currents from the two types of carriers. This may sound somewhat artificial, as it is not clear where the carriers go when they hit the side edges of the wire. We can get around this by imagining that the two types of carriers can annihilate each other at the edges. This is actually a model for conduction in a semiconductor, where the negative charges are electrons that get excited into the conduction band, and the positive charges are the holes they leave behind in the valence band.
 Problem 2 [25 points]
Consider a metal with a concentration n of free conduction electrons of charge e in a static magnetic field H that points in the z direction.
An oscilating electric field, E(t)=Re[E_{o}e^{iωt}], is applied with E_{o} in the xy plane.
a) By solving Drude's equation of motion, show that the current density oscilates, j(t)=Re[j_{o}e^{iωt}], with an amplitude j_{o} in the xy plane determined from E_{o} by a frequency dependent conductivity tensor σ(ω),
j_{o}=σ(ω)⋅E_{o}
that is
j_{ox}=σ_{xx}E_{ox}+σ_{xy}E_{oy}, j_{oy}=σ_{yx}E_{ox}+σ_{yy}E_{oy}
where
σ = [

σ_{xx} σ_{xy} σ_{yx} σ_{yy}

]

is the conductivity tensor. Solve for all components of the conductivity tensor.
b) Consider now a transversely polarized simple harmonic electromagnetic wave of frequency ω propagating with wavevector k in the z direction. Making the same approximation as we made in lecture, that the wavelength λ of the wave is much larger than the typical electron displacement due to E, show that Maxwell's equations impose the following condition on the amplitude E_{o},
(c^{2}k^{2}ω^{2})E_{o} = 4πiωσ(ω)⋅E_{o}
(Hint: You do not really have to do much for this part, just follow the corresponding steps in the derivation in lecture. The only thing new here is that the conductivity is a tensor instead of a scalar.)
c) The solutions to the equation given in part (b) will be for vectors E_{o} in the directions of the two eigenvectors of the conductivity tensor. The two eigenvalues of the conductivity tensor will then determine the dispersion relations that give k in terms of &omega. These eigenvectors and dispersion relations define the two normal modes of transverse electromagnetic wave propagation parallel to the applied uniform magnetic field H. Show that the eigenvectors are (E_{ox}, E_{oy}) = (1, ±i). These are just the basis vectors for left and right handed circularly polarized waves. Show that left and right handed circularly polarized waves will in general travel with different speeds through the metal.
d) The plasma frequency ω_{p} is defined by ω_{p}^{2} = 4πne^{2}/m, and the cyclotron frequency is defined by ω_{c} = eH/mc. Assume that one is in the large magnetic field limit so that ω_{c}τ >> 1. Assume also that ω_{p} >>ω_{c}.
For the high frequency limit, ω>>&omega_{c}, show that the dispersion relation for left and right handed circularly polarized waves are given by,
c^{2}k^{2}=ω^{2}−ω_{p}^{2}±ω_{c}ω_{p}^{2}/ω
Thus for ω>ω_{p} one has transparent wave propagation, and the left and right handed circularly polarized waves travel at different speeds. Since a planar polarized wave can always be written as a linear superposition of left handed and right handed circularly polarized waves, the different speeds result in a rotation of the polarization of a plane polarized wave as it propagates through the metal. This is known as the Faraday Effect and was one of the early experiments Farady did in order to establish that light was an electromagnetic phenomena.
Now take the opposite low frequency limit, ω<<ω_{c}. Show that the dispersion relations for left and right handed circularly polarized waves are now given by,
c^{2}k^{2}=ωω_{p}^{2}/ω_{c}
i.e. ω~k^{2}, which is rather different from the ω~k that one has in a vacuum or transparent dielectric. Such waves are known as helicons.
 Problem 3 [20 points]
Consider a gas of free noninteracting electrons in two dimensions. This might be a model for electrons in a thin metallic film, or a semiconductor inversion layer. Applying quantum FermiDirac statistics to describe the gas of electrons:
a) What is the relation between the density of electrons per unit area n_{2d} and the Fermi wavevector k_{F}?
b) What is the number of single electron states per unit energy per unit area at energy ε, i.e. what is the density of states g(ε)?
c) What is the total energy of the ground state of N electrons?
d) What is the pressure (in two dimensions, pressure is force per unit length) in the ground state?
e) What is the bulk modulus B?
f) (Kittel problem 6.10)
Consider a square sheet of side L, thickness d, and electrical resistivity ρ. The resistance measured between the opposite edges of the sheet is called the surface resistance,
R_{sq}=ρ×(length)/(crosssectional area)=ρL/Ld=ρ/d
which is thus independent of the area L^{2} of the sheet. (R_{sq} is called the resistance per square and is expressed in ohms per square, because ρ/d has the dimensions of ohms). If we use the Drude result for ρ, then
R_{sq}=m/(nde^{2}τ)
where n is the density per volume, which is equal to n_{2d}/d. Suppose now that the minimum value of the collision time is determined by scattering from the surfaces of the sheet, so that τ≈d/v_{F}, where v_{F} is the Fermi velocity. Show for a monatomic metal sheet one atom in thickness that R_{sq}≈(hbar)/e^{2}=4.1 kΩ.
Hint: use your computation for the 2d electron gas to get v_{F} for the sheet, and you may assume that the spacing between atoms is the same length scale as the size of an atom.

