Introduction To Modern Solid State Physics(477s).pdf: eKnigu

Introduction To Modern Solid State Physics(477s).pdf

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Date Dec 9, 2005

Figure 1.1: On the determination of rotation symmetry Inversion, I : Transformation r → −r, ﬁxed point is selected as origin (lack of inversion symmetry may lead to piezoelectricity); Reﬂection, σ : Reﬂection across a plane; Improp er Rotation, Sn : Rotation Cn , followed by reﬂection in the plane normal to the rotation axis....

Figure 1.14: The powder method. ∆k = 2k sin(φ/2) = G. So one can determine the ratios φ : 1 sin sin 2...

Figure 2.3: Vibrations of a linear one-atomic chain (spectrum). λmin = 2π /qmax = 2a. The maximal frequency is a typical feature of discrete systems vibrations. Now we should recall that any crystal is ﬁnite and the translation symmetry we have used fails. The usual way to overcome the problem is to take into account that actual...

At very long waves (q → 0) we get (Problem 2.3) Pac = 1 , Popt = − m2 m1 (2.25) next to synchronously, like in the optical mode; the gravity vibration produce alternating The situation is illustrated in...

Consequently we come to the Schr¨dinger equation o − q 2 ∂2 12 ˆ (P , Q) = ˆ H − ωj (q)Q2 (q) j 2 2 ∂ Qj (q) 2 j...

Also remains the motion along z-axis with the velocity vz = ∂ ε/∂ z . Now we discuss one very useful trick to calculate the properties of electrons in a magnetic ﬁeld. Namely, let us introduce the time of the motion along the orbit as d c p t1 = eH v⊥...

Degenerated Electron Gas
Now we discuss the important limiting cases. The ﬁrst one is the case for good metals or highly doped semiconductors in which the density of conduction electrons is large. The large density of conduction electrons means that ζ ∗ 1. That leads to the following approximate presentation for the Fermi function f0 (ε) = Θ(ζ − ε) where Θ(x) = 1 , if x > 0, 0, if x < 0...

Paramagnetism of Free Electrons (Pauli Paramagnetism).
Now we are prepared to discuss the magnetic susceptibility of a gas of near free electrons. In this case the orbital moment l = 0, j = s = 1/2 and gL = 2. Consequently one could...

4.4. MAGNETIC PROPERTIES OF ELECTRON GAS. where we keep only linear in the ﬁeld H1 terms. Solving this equation we get χx = where χ0 = Mx χ0 = Hx 1 − (ω /ω0 )2 M 0z , Hz...

95 semiconductor physics. Substituting the Hamiltonian p2 /2m + eϕ = p2 /2m − e2 / r with ˆ ˆ the eﬀective mass m to the SE we get the eﬀective Bohr radius aB = m0 2 m0 ˚ = a0 = 0.53 , A. B 2 me m m...

For a typical metal p ≈ /a, and we get a. There is another important criterion which is connected with the life-time τϕ with the respect of the phase destruction of the wave function. The energy diﬀerence ∆ε which can be resolved cannot be greater than /τϕ . In the most cases ∆ε ≈ kB T , and we have kB T τ.
ϕ...

Substituting (6.5) into the expression for matrix elements we get v i ) −1 Vpp = V (r − ri ) ei(k−k r u∗ (r)uk (r) d3 r k =V
−1...

The function K cannot be derived from classical considerations because the typical spatial scale of the potential variation appears of the order of the de Broglie wave length /p. We will come back to this problem later in connection with the quantum transport. The function K (r) reads (in the isotropic case) c . p3 os(2kF r) sin(2kF r) F K (r) = −g ( F ) − (π )3 (2kF r)3 (2kF r)4 We see that the response oscillates in space that is a consequence of the Fermi degeneracy (Friedel oscil lations). These oscillations are important for speciﬁc eﬀects − but if we are interested in the distances much greater than kF 1 the oscillations are smeared and we return to the picture of the spheres of atomic scale. So one can use the Thomas-Fermi approximation to get estimates....

After substitution of the cross section in the deﬁnition of the transport relaxation time (6.11) we get √ 2m2 v 3 2m 2ε3/2 τtr = = (6.28) 2π e4 ni Φ(η ) π e4 ni Φ(η ) where Φ(η ) = ln(1 + η ) − η , 1+η η=
2 2 4m2 v 2 rs 8mεrs = . 2 2...

Combining with the expression for the emission and absorption we get (a1 − a2 ) [f2 (1 − f2 ) (N + 1 − f1 ) + f1 (1 − f1 ) (N + f2 )] . kB T Fragments in the square brackets are e = f1 eν 1 −x = N f1 x+ν + eν − eν + 1 N, ν −1 e e +1 f2 e = 1 1 f2 + x+ν = N f2 x+ν + 1 + eν − 1 (N + 1) N + f2 = ν e −1 e +1 f1 where x = (ε − ζ )/kB T , ν = ω/kB T . Finally, we get in the brackets N + 1 − f1 = N f1 (1 − f2 ) + (N + 1)f2 (1 − f1 ) = 2N f1 (1 − f2 ) and the integrand in the collision integral becomes proportional to 2N f1 (1 − f2 ) (a1 − a2 ). kB T We see that only thermal phonons are important since the integrand of the collision operator decreases exponentially at ν 1. As a result, we have proved the estimates made above. Unfortunately, the relaxation time approximation is not exact in this case and one should solve the Boltzmann equation numerically....

The temperature dependence of the resistivity of semiconductors is more tricky because the electron concentration is temperature dependent. We will come back to this problem later. It is also important to know, that at very low temperatures quantum contribution to resistivity becomes important. This contribution cannot be analyzed with the help of the Boltzmann equation and we will also discuss it later....

Optical Phonons
In non-polar materials one can also characterize the interaction as He−ph = Λ0 u where u is the relative displacement of the atoms in the basis. Much more interesting is the interaction in polar crystals where, as we have seen, optical phonons produce electric ﬁelds. In these materials the displacement s = u+ − u− , see Sec. 2.2, page 33 for notations, creates the polarization (Problem 6.8) N 2 0 Mr ωl (u+ − u− ) (6.44) P= 4π ∗ 1 1 1 = −. ∗ ∞ 0 We take into account only longitudinal vibrations which eﬀectively interact with the electrons. Then, 4 π N0 Mr ωl2 2ϕ = 4π div P = div (u+ − u− ). ∗ Then, as usual, we expand the displacements in terms of the normal co-ordinates q 1 uk (r) = √ ej k (q)bj (q,t)eiqr (6.45) j N Mk j where...

Thermomagnetic Eﬀects.
It is clear that the temperature gradient also produces electric currents, and a magnetic ﬁeld leads to oﬀ-diagonal transport. As we have seen these currents are produced by the...