Statistical physics (2005)

(b)Show that the density of states of this branch is given by

…0

< 0 : D−( ) = 2D0( )

+ 0

II.4. In the rest of the problem, the temperature is taken as zero.

(a)Qualitatively explain how the Fermi energy changes with the electrons number N . Show that if N is small enough, only the branch E− is filled.

(b)What is the value of the Fermi energy F when the E+ branch begins to be filled? Calculate the number of electrons N at this point, versus

Lx, Ly, m, and 0.

(c) Where are the wave vectors k corresponding to the filled states F < 0 located ?

II.5. Assume in what follows that F > 0.

(a) What are the wave vectors k for the occupied states in either branch

E± ?

(b) Calculate the Fermi wave vectors kF+ and kF− versus F , k0, 0, m and .

(c) For N > N , calculate N − N and deduce the relation :

N = LxLy m π 2

II.6. A weak magnetic field is now applied on the sample.

(a)Using the result of I.5, explain why the Fermi energy does not vary to first order in B.

(b)Calculate the magnetization along z of the electrons gas, to first order in B.

II.7. Now one analyzes the possibility to maintain a magnetization in the system for some time in the absence of magnetic field. At time t = 0, each electron is prepared in the spin state |+ z . The magnetization along z is thus M (0) = N /2. For a given electron, the evolution of sz (t) in the absence of collisions was calculated in I.4. Now this evolution is modified due to the e ect of collisions. Because of the elastic collisions of an electron on the material

defects, its k wavevector direction may be randomly modified. These collisions are modeled assuming that :

– they occur at regular times : t1 = τ , t2 = 2τ ,..., tn = nτ ,...

– the angle φn characterizing the k direction between the nth and the (n + 1)th collision is uniformly distributed over the range [0, 2π[ and decorrelated from the previous angles φn−1, φn−2, ..., and thus from the coe cients a±(tn).

– the angles φn corresponding to di erent particles are not correlated.

Under these assumptions, the total magnetization after many collisions can be calculated by making an average over angles for each electron. One writes s¯(t) for the average of sz (t) over the angles φn.

Show that s¯(tn+1) = cos(2ωτ ) s¯(tn).

II.8. One assumes that ωτ 1.

(a)Is there a large modification of the averagespin s¯ between two collisions?

(b)One considers large times t with respect to the interval τ between two collisions. Show that the average spin s¯(t) exponentially decreases and express the characteristic decrease time td versus ω and τ .

(c)Take two materials : the first one is a good conductor (a few collisions take place per unit time), the second one is a poor conductor (many collisions occur per unit time). In which of these materials is the average spin better “protected,” i.e., has a slower decrease? Comment on the result.

(d)For τ = 10−12 sec and ω=5 µeV, estimate the spin relaxation time td.

II.9. To maintain a stationary magnetization in the population of N spins, one injects g± electrons per unit time, of respective spins sz = ± /2 into the system. For each spin value, the electrons leave the system with a probability per unit time equal to 1/tr. Besides, one admits that the relaxation mechanism studied in the previous question is equivalent to the exchange between the + and − spin populations : a + spin is transformed into a − spin with a probability per unit time equal to 1/td, and conversely a − spin is transformed into a + spin with a probability per unit time equal to 1/td. One calls f±(t) the populations of each of the spin states at time t, td is the spin relaxation time.

(a)Write the rate equations on df±/dt. What are the spin populations f± in steady state?

f+ − f−

(b) Deduce the relative spin population, that is, the quantity P = f+ + f− . In what type of material will the magnetization degrade not too fast?

Exercise 2000 : Electrostatic Screening

I.1. The density N/Ω and the chemical potential µ are related by

I.2. In the high temperature and low density limit, the exponential in the denominator of the Fermi-Dirac distribution is large with respect to 1, so that

I.3. The total energy in presence of a potential is written

ε= 2k2 + V (r) 2me

At high temperature and low density

I.5. If βV (r) 1 for any r, n is uniform, i.e.,

N n = n0 = Ω

II.1. The potential energy of an electron inside the electrostatic potential Φtot(r) is −e Φtot(r), the energy of a hole +e Φtot(r). The electron and hole densities are respectively equal to

n(r) = n0 exp[−β(−e Φtot(r)]

p(r) = p0 exp[−β(+e Φtot(r)]

N with n0 = p0 = Ω .

II.2. The total density of charges at r is equal to e[p(r) − n(r)]. The charge localized at the origin also enters into the Poisson equation :

In the high temperature limit and to the lowest order

p(r) − n(r) = n0 e−βeΦtot(r) − eβeΦtot(r)

−2n0 eβΦtot(r)

Whence the equation satisfied by Φtot(r) :

II.3. If NΩ = 1022 m−3 T = 300 K, εr = 12.4, then λ = 30 nm.

II.4. The e ect of the charge located at the origin cannot be felt beyond a distance of the order of λ : at a larger distance Φtot(r) is rapidly constant, the electron and hole densities are constant andopposite. The mobile electrons and holes have screened the charge at the origin.

z= e−βε0 + e−β(ε0+λ+γB) + e−β(ε0+λ−γB)

= e−βε0 [1 + e−β(λ+γB) + e−β(λ−γB)]

For the whole system : Z = (z)N

F = −kB T ln Z

This situation is analogous to a).

Exercise 2002 : Entropies of the HC Molecule

A molecule is spontaneously adsorbed if µ2D < µ3D.

The average 3D distance is of 3.3nm, it is 0.11nm between molecules on the surface (which is less than an interatomic distance, but one has also to account for the large corrugation of the surface).

lecule, such that its physical situation is identical after a 180˚rotation. But HC is not symmetrical

8. Rotation entropy of HC :

Frot = −N kB T ln z2rot

(the factor for indistinguishability is included inside the translation term).

Srot = 33.3 J · K−

Problem 2001 : Quantum Boxes

and Optoelectronics

I.1. Owing to the spin degeneracies one has