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SOLID-STATE PHYSICS 3, Winter 2009 O. Entin-Wohlman 1. ELECTRONIC TRANSPORT PROPERTIES 2. WEAK LOCALIZATION 3. ANDERSON LOCALIZATION, THE METAL-INSULATOR TRANSITION 4. TRANSPORT BY LOCALIZED ELECTRONS: PHONON-ASSISTED HOP- PING 5. LINEAR RESPONSE AND THE FLUCTUATION-DISSIPATION THEOREM 6. COMMENTS ON SECOND-QUANTIZATION Second-quantization of single-particle operators. Here we consider particles whose number is conserved, e.g., electrons, and ignore the spin index (for the time-being). Let us imagine a system of many particles which do not interact with one another. The Hamiltonian is H= X Hi , (6.1) i where i runs over all particles, and Hi ≡ H(ri ) = − ~2 2 ∇ + U (ri ) . 2m i (6.2) (It is assumed that all particles are identical, e.g., they have the same mass, m.) Since the particles do not interact with each other, we may consider just H(ri ), and omit the particle index i. Using the field operators Ψ and Ψ† , the Hamiltonian is Z H = drΨ† (r)H(r)Ψ(r) . (6.3) Let us denote the eigen functions and the eigen values of the single-particle Hamiltonian by φα and α , respectively (where α includes all quantum numbers), H(r)φα (r) = α φα (r) . 1 (6.4) For example, when the particle has just the kinetic energy, i.e. U = 0, then φα (r) ≡ φk (r) = p 1/Ω) exp(ik · r), and α ≡ k = ~2 k 2 /2m. We now write the field operator in terms of the basis φα , Ψ(r, t) = X φα (r)cα (t) , Ψ† (r, t) = α X φ∗α (r)c†α (t) . (6.5) α The operator cα destroys a particle in the state α, and similarly, the operator c†α creates a particle in the state α. For fermions, they obey anti-commutation relations c†α , cα0 ≡ c†α cα0 + cα0 c†α = δαα0 , cα , cα0 ≡ cα cα0 + cα0 cα = 0 . (6.6) For bosons, they obey commutation relations, aα , a†α0 ≡ aα a†α0 − a†α0 aα = δαα0 , aα , aα0 ≡ aα aα0 − aα0 aα = 0 . (6.7) The Schrödinger equation is i~ X X ∂Ψ X ∂c α φα (r)cα φα (r)i~ α = HΨ = Hφα (r)cα = = ∂t ∂t α α α ⇓ i~ ∂cα =α cα , ∂t (6.8) (since φα form a complete set) and similarly −i~ ∂c†α = α c†α . ∂t (6.9) The Hamiltonian itself, Eq. (6.3), is now Z X X X † H = dr c†α φ∗α (r) H(r) φα0 (r)cα0 = α cα cα . α0 α (6.10) α For example, the free electron gas (without any interactions), in terms of the wave vector basis, is just the kinetic energy Hfree = X ~2 k 2 k 2m c†k ck . (6.11) Example: The density operator. In terms of the field operators, the density operator, ρ(r) is ρ(r) = Ψ† (r)Ψ(r). Using the expansion (6.5), ρ(r) = X φ∗α (r)φα0 (r)c†α cα0 . αα0 2 (6.12) For example, the density operator of the free electron gas (namely, when there no interactions at all) is ρ(r) = 1 X i(k0 −k)·r † e ck ck 0 , Ω kk0 (6.13) and in Fourier space Z Z X † X † 1 X i(k0 −k)·r † iq·r ρq = dre ρ(r) = dreiq·r e ck ck 0 = ck+q ck = ck+q/2 ck−q/2 . Ω kk0 k k (6.14) Example: The current density operator. The current density operator is ~ † Ψ (r)∇Ψ(r) − ∇Ψ† (r) Ψ(r) . 2im (6.15) ~ X † ∗ cα cα0 φα (r)∇φα0 (r) − φα0 (r)∇φ∗α (r) . 2im αα0 (6.16) j(r) = Using the expansion (6.5), j(r) = The Fourier transform of the current density of the free electron gas is then Z Z ~ X q ~ X † 1 i(q+k0 −k)·r 0 iq·r ck ck 0 dre ik + ik = (k − )c†k ck−q jq = dre j(r) = 2im kk0 Ω m k 2 ~ X † kck+q/2 ck−q/2 . (6.17) = m k ♣Exercise. Verify the (operator) continuity equation for fermions, ∂ρ/∂t + ∇ · j = 0, (using the above two examples). Example: The thermal average of the current of free fermions. We have found above that the Hamiltonian of free fermions, in second quantization, reads (ignoring spin components for simplicity) Hfree = X ~2 k 2 k 2m c†k ck , (6.18) and that the (Fourier component of the) current density operator is jq = ~ X † kck+q/2 ck−q/2 . m k 3 (6.19) Let us find the average current of the free fermion system (obviously, we must obtain that there is no current flowing in the system). The expression for the thermal average [written for the Hamiltonian (6.18)] can be put in the form Tr e−βHfree jq ~ X = khc†k+q/2 ck−q/2 i . hjq i = −βHfree m k Tr e (6.20) Since the Hamiltonian (6.18) is diagonal in k−states, then only the q−component of jq may give a non-vanishing contribution to average current. Note that this means that if there is an average current, it will be uniform in space. The q = 0 component of hjq i is hjq=0 i = ~ X khc†k ck i . m k (6.21) The thermal average of the number operator c†k ck with the free Hamiltonian (6.18) is just the Fermi distribution, and hence ~ X 1 ~2 k 2 hjq=0 i = k , k = . m k eβk + 1 2m (6.22) Since k depends solely on |k|, the sum over k here vanishes (each k−contribution is cancelled by the contribution of −k) and consequently there is no average current in the system described by the free Hamiltonian. ♣Exercise. Find the thermal average of the density in a system described by the free Hamiltonian (6.18). 7. COULOMB INTERACTIONS AND THE FORMALISM OF SECOND QUAN- TIZATION The Coulomb interactions in a system of electrons is Hc = X X 2πe2 1 X e2 = eiq·(ri −rj ) , 2 ij,i6=j |ri − rj | i,j,i6=j q q 2 (7.1) where ri is the location of the ith electron. As the electrons are located at ri , rj , etc., we can define the electron density, ρ(r) at a general point r, in the form ρ(r) = X i 4 δ(r − ri ) . (7.2) The Fourier transform of the density is then Z X X ρq = dr δ(r − ri )eiq·r = eiq·ri . i (7.3) i It therefore follows that the Coulomb Hamiltonian (7.1) can be written in the form Hc = X 2πe2 q2 q ρq ρ−q − n , (7.4) where n is the electronic concentration. [This term takes care of the term i = j that we have introduced into Eq. (7.1) in order to use Eq. (7.3). From now on we omit it, since it is just a constant.] We have already derived the second-quantization form of the electronic density–it is given in Eq. (6.14), ρq = X c†k+q ck . (7.5) k We therefore have that the second-quantization form of the Coulomb Hamiltonian is Hc = X 2πe2 X q q2 c†k+q ck c†k0 −q ck0 . (7.6) kk0 We can easily generalize this form to include the spin component, Hc = X 2πe2 X X q q2 kk0 c†k+qσ ckσ c†k0 −qσ0 ck0 σ0 . (7.7) σσ 0 We will describe in class the Hartree and the Hartree-fock approximations of the Coulomb energy. 8. LINEAR RESPONSE–THE KUBO FORMULA We wish to study the way in which a system responds to a certain perturbation applied to it, which breaks the symmetry between k and −k. When the perturbation is sufficiently weak, one may expect the response to be linear in the perturbation (hence the name ‘linear response’). Here we will calculate the current generated in an electronic system which is placed in an electric field, and obtain the Kubo formula for the electrical conductivity. 5 Electrical conductivity. Let us denote the Hamiltonian of the system itself (without the electric field) by H0 . Then, in the presence of the field, the Hamiltonian of the system is H = H0 + Hext . (8.1) It is convenient to describe the electric field by introducing a vector potential, A(r, t), such that the electric field, E(r, t), is E(r, t) = − 1 ∂A(r, t) . c ∂t (8.2) In order to find the explicit form of Hext , we remember that in the presence of a vector potential, the kinetic energy of the electron, in terms of field operators, is changed as follows Z Z ~2 e ~2 e † (8.3) dr∇Ψ · ∇Ψ ⇒ dr ∇ + i A Ψ† · ∇ − i A Ψ , 2m 2m ~c ~c and the current density operator is (here we add the electron charge, −e, to our definition) J= −e~ † e e e2 Ψ (∇ − i A)Ψ − [(∇ + i A)Ψ† ]Ψ = j + An , 2mi ~c ~c mc (8.4) where n ≡ Ψ† Ψ is the density operator, and j is the current density operator without the vector potential. In deriving this expression we have used the gauge choice ∇ · A = 0, namely, the vector potential is a transverse field (since in Fourier space it is normal to the wave vector, i.e., q · A(q) = 0). Returning now to the kinetic energy, we see that it can be written in the form Z ~2 1 dr ∇Ψ† · ∇Ψ + j · A + O(A2 ) . 2m c (8.5) The quadratic term in the vector potential is not needed, since we are interested in linear response only. It therefore follows that the perturbation term in the Hamiltonian is Z 1 Hext = drj · A . c (8.6) We now wish to find the average current density induced in the system. (Note: in the following, in order to shorten the notations, we will use units in which ~ = 1, namely, the dimension of time is 1/energy.) Suppose that the field is switched-on at t → −∞. Formally, this means that we have to multiply Hext by eηt , with η → 0+ , so that at t → −∞ the external perturbation is zero. In the following we will omit this factor, but in actual calculations it must be inserted. Just before the field is switched-on, the system was in one 6 of the eigen states of H0 . Let us denote that state by φα . Since after switching on the field that state is no more an eigen state, it evolves in time according to i ∂φα (t) = (H0 + Hext )φα (t) . ∂t (8.7) As we know that for Hext = 0 the solution for this equation is e−iH0 t φα , we write the solution in the presence of Hext in the form φα (t) = e−iH0 t U (t)φα . (8.8) (Note that for any operator O, eiOt commutes with O.) Inserting the form (8.8) into the Schrödinger equation (8.7), we obtain an equation for the operator U , dU (t) = Hext (t)U (t) , dt (8.9) Hext (t) = eiH0 t Hext e−iH0 t , (8.10) i where gives the time-dependent external Hamiltonian in the interaction picture. The formal solution of Eq. (8.9), with the initial condition U (−∞) = 1, may be found by iterations, Z t 0 U (t) = 1 − i 0 Z t 0 dt Hext (t ) − −∞ t0 Z dt00 Hext (t0 )Hext (t00 ) + . . . . dt −∞ (8.11) −∞ The full formal solution is usually written in the form of a time-ordered product, T , such that Rt U (t) = T e−i −∞ dt0 Hext (t0 ) . (8.12) The time-ordering operator T arranges the operators according to their time arguments, such that those of later times appear to the left. However, for the purpose of the present discussion we need just terms linear in the perturbation Hamiltonian Hext , so we write Z t dt0 Hext (t0 ) , U (t) = 1 − i (8.13) −∞ and consequently φα (t) = e −iH0 t Z t 1−i −∞ 7 dt0 Hext (t0 ) φα . (8.14) The expectation value of the current density, Eq. (8.4), in the state φα is hence Z t Z t 0 0 iH0 t −iH0 t dt Hext (t ))e Je (1 − i dt00 Hext (t00 ))|αi hα(t)|J|α(t)i = hα|(1 + i −∞ −∞ Z t Z t dt00 Hext (t00 ))|αi , J(t) = eiH0 t Je−iH0 t . (8.15) dt0 Hext (t0 ))J(t)(1 − i = hα|(1 + i −∞ −∞ Since our calculation is valid only to first order in Hext ∝ A, then [using Eq. (8.4)] Z t e2 hα(t)|J|α(t)i = hα|j|αi + dt0 hα| Hext (t0 ), j(t) |αi . Ahα|n|αi + i mc −∞ (8.16) (Note the appearance of the commutator of two operators at different times.) The expression (8.16) has been derived for a certain eigen function of the Hamiltonian H0 . Let us now sum over all eigen states, giving each one of them its statistical (Boltzmann) weight. The result will be the thermodynamic average of the current density operator, i.e., Z t e2 hJ(t)i = hji + Ahni − i dt0 h j(t), Hext (t0 ) i , (8.17) mc −∞ P −βE P −βE α hα|O|αi/ α , where E where for any operator O, hOi = α are the eigen αe αe energies of the system Hamiltonian, H0 . Inspecting Eq. (8.17), we see that the first term is the average current in the absence of the field. Obviously, this quantity is zero. The second term includes just the density of the electrons, and this quantity obviously does not depend on the temperature (for metals). The last term gives the real current induced by the electric field. In order to see clearly that it is the third term of Eq. (8.17) that gives the actual current, we need to introduce the perturbation Hamiltonian into Eq. (8.17). We will simplify a bit the derivation by assuming that the vector potential depends on time, but not on the P coordinates. (This assumption can be easily relaxed.) Then Hext = (1/c) ν jν Aν , where the index ν runs over the coordinates, ν = x, y or z. Equation (8.17) now takes the form Z e2 iX t hJν (t)i = (8.18) Aν (t)hni − dt0 Aν 0 (t0 )h jν (t), jν 0 (t0 ) i . mc c ν 0 −∞ (The vector potential can be taken out of the commutator since it is not an operator.) It is now convenient to define the current-current response function, Rνν 0 (t − t0 ) = −iΘ(t − t0 )h jν (t), jν 0 (t0 ) i . 8 (8.19) Here, Θ is the Heavyside function, 1 , x>0 Θ(x) = 0 , x<0 , (8.20) . Note that the response function (8.19) depends only on the time difference. This is because the average cannot depend separately on the time arguments (homogeneity of time). Since the Heavyside function takes care of the fact that in the integration t0 is always smaller than t, we can extend the integration upper bound to ∞ and write Z 1X ∞ 0 e2 A (t)hni + dt Aν 0 (t0 )Rνν 0 (t − t0 ) . hJν (t)i = mc ν c ν 0 −∞ (8.21) Going over to Fourier space, this result reads hJν (ω)i = e2 1X A 0 (ω)Rνν 0 (ω) . Aν (ω)hni + mc c ν0 ν (8.22) Remembering now that the Maxwell equation gives [see Eq. (8.2)] E(ω) = − iω A(ω) , c (8.23) we may write the result (8.22) in the form of Ohm’s law, hJν (ω)i = X σνν 0 (ω)Eν 0 (ω) , (8.24) ν0 in which the frequency-dependent (i.e., ac) conductivity tensor is σνν 0 (ω) = i i e2 hni δνν 0 + Rνν 0 (ω) . mω ω (8.25) This is the Kubo formula for the electrical conductivity. The dc conductivity is obtained upon taking the limit ω → 0. A remarkable property of this formula is that the response function (8.19) is calculated from the eigen spectrum of the Hamiltonian of the system in the absence of the field, H0 . The Kubo formula for the susceptibility tensor. The spin operator of the electron is customarily described using the three Pauli matrices, 1 S= σ , 2 9 (8.26) where σ = x̂σx + ŷσy + ẑσz , (8.27) with σx = 0 1 1 0 , σy = 0 −i i 0 , σz = 1 0 0 −1 . (8.28) The Pauli matrices satisfy σ`2 = 1 , σ` σm = iε`mn σn , (8.29) where here 1 is the unit matrix and ε`mn is the full antisymmetric (Levy-Chivita) tensor. Thus, S 2 = 3/4 as it should, since for the electron S = 1/2. The spin-density operator of the electrons, using field operators, reads S(r) = 1X † Ψσ (r) σ σσ0 Ψσ0 (r) . 2 σσ0 (8.30) Therefore, in second quantization (in terms of creation and destruction operators in k−states), its Fourier component is S(q) = 1 XX † σ 0c c 0 . 2 σσ0 k k+q/2,σ σσ k−q/2,σ (8.31) [Note the similarity with the (Fourier component of) the density operator, Eq. (6.14).] In particular, the z− component of the spin-density operator is 1 XX † 1 X † † z S z (q) = ck+q/2,σ σσσ c c − c c c = 0 k−q/2,σ 0 k+q/2,↑ k−q/2,↑ k+q/2,↓ k−q/2,↓ . (8.32) 2 σσ0 k 2 k The transverse components are conveniently written as S + (q) = S x (q) + iS y (q) X † 1 XX † y x ck+q/2,↑ ck−q/2,↓ , = ck+q/2,σ σσσ 0 + iσσσ 0 ck−q/2,σ 0 = 2 σσ0 k k (8.33) and S − (q) = S x (q) − iS y (q) X † 1 XX † y x = ck+q/2,σ σσσ ck+q/2,↓ ck−q/2,↑ . 0 − iσσσ 0 ck−q/2,σ 0 = 2 σσ0 k k 10 (8.34) It is seen that the operators in S + (q) are such that a down-spin is destroyed and an up-spin is created, while in S − (q) it is the other way around. We also see that S z counts the number of up-spins minus the number of down-spins. Let us now consider a system of electrons (described by an Hamiltonian H0 which is not specified at the moment) that is paramagnetic, namely, there is no net magnetization at equilibrium. In other words, hSi vanishes. However, when one applies a magnetic field H on this system, the situation is changed. We will consider here only the coupling of the magnetic field to the spins of the electrons, and will neglect the coupling of the field to the orbital motion (i.e., we will disregard the Lorentz force). The interaction of an electric field with the electron spins is the Zeeman interaction. It adds to the system Hamiltonian the term Z HZ = − drH(r, t) · S(r) . (8.35) [We absorb into the definition of the magnetic field the Bohr magneton µB .] One can now repeat the linear response formalism to calculate the thermal average of the spin variables. This gives us the linear relation between the thermal average of the α− component of the pins, hSα (r, t)i (where α = x, y or z) and the γ− component of the magnetic field, Hγ , (where γ = x, y, or z) such that Z XZ 0 hSα (r, t)i = dt dr0 χαγ (r − r0 , t − t0 )Hγ (r0 , t0 ) , (8.36) γ where the susceptibility tensor is χαγ (r − r0 , t − t0 ) = iΘ(t − t0 )h[Sα (r, t), Sγ (r0 , t0 )]i . (8.37) [It depends only on the difference r − r0 and on each of them separately, since we assume the system to be spatially homogeneous.] ♣Exercise. Derive the expression for the electronic spin-susceptibility, Eq. (8.37). In an isotropic system, or when the system has a cubic symmetry, one expects that the susceptibility tensor has only diagonal matrix elements, and that they are equal, namely, χxx = χyy = χzz and χxy = χyz = χzx = 0. It turns out that it most convenient to calculate χ−+ , χ−+ (r − r0 , t − t0 ) = iΘ(t − t0 )h[S− (r, t), S+ (r0 , t0 )]i . 11 (8.38) Since the susceptibility depends only on t − t0 , we may also put t0 = 0. The susceptibility is then [see Eqs. (8.33) and (8.34)] X 0 χ−+ (r − r0 , t) = e−iq·(r−r ) iΘ(t)h c†k+q↓ (t)ck↑ (t), c†k0 −q↑ ck0 ↓ i . (8.39) kk0 q The q−Fourier component of the susceptibility is X † χ−+ (q, t) = iΘ(t) h ck+q↓ (t)ck↑ (t), c†k0 −q↑ ck0 ↓ i . (8.40) kk0 Adding the Fourier transform with respect to the time t we have XZ ∞ χ−+ (q, ω) = i dtei(ω+iη)t h c†k+q↓ (t)ck↑ (t), c†k0 −q↑ ck0 ↓ i , (8.41) 0 kk0 where at the end of the calculation we have to take the limit η → 0+ . Pauli susceptibility–the spin-susceptibility of free electrons. Let us find the spin P susceptibility of a system of free electrons, whose Hamiltonian is just Hfree = kσ k c†kσ ckσ , with k = k 2 /2m. In that case c†kσ (t) = eik t c†kσ and Eq. (8.41) becomes XZ ∞ Pauli dtei(ω+iη)t ei(k+q −k )t h c†k+q↓ ck↑ , c†k0 −q↑ ck0 ↓ i . χ−+ (q, ω) = i kk0 (8.42) 0 Since now the operators in the commutator do not depend on time, we can calculate the commutator explicitly, to find XZ Pauli χ−+ (q, ω) = i k ∞ dtei(ω+iη)t ei(k+q −k )t hc†k+q↓ ck+q↓ − c†k↑ ck↑ i . (8.43) 0 The thermal average of the number operator for free electrons does not depend on the spin index, and is just the Fermi function, hc†k ck i = f (k ), where f (k ) = 1/(eβk + 1) (we take the chemical potential as zero). Performing also the time-integration in Eq. (8.43), we find χPauli −+ (q, ω) = X f (k+q ) − f (k ) . − − ω − iη k k+q k (8.44) We first consider the static limit, ω = 0. In that case, we can send η to zero, and then X f (k+q ) − f (k ) χPauli . (8.45) −+ (q, ω = 0) = k − k+q k In the particular case q → 0 as well, an expansion [f (k+q ) = f (k ) + (k+q − k )∂f (k )/∂k ] gives the simple result χPauli −+ (q X ∂f ( ) Z ∂f () k = 0, ω = 0) = − = dN ()(− ' N (0) , ∂ ∂ k k 12 (8.46) where N () is the density of states at energy , and N (0) is the density of states at the Fermi level. The final step in Eq. (8.46) holds at temperatures lower than the Fermi energy (which practically always holds for metals). The density of states (per spin) at the Fermi level for a three-dimensional free electron gas, is kF m/(2π 2 ), where kF is the Fermi wave vector. Example: The zero-temperature static spin susceptibility at finite wave vectors. We first write Eq. (8.45) in the form χPauli −+ (q, ω = 0) = X f (k ) k X 1 1 1 − = 4m . f (k ) 2 k−q − k k − k+q q + 2k · q k (8.47) At zero temperature, f (k ) = 1 for k < kF , and is zero otherwise. Hence χPauli −+ (q, ω 2π = 0) = 4m (2π)3 Z kF dk k 2 0 Z π dθ sin θ 0 q2 1 . + 2kq cos θ (8.48) Performing the angular integration, χPauli −+ (q, ω 2π = 0) = 4m (2π)3 Z 0 kF dk k 2 q + 2k 1 ln . 2kq q − 2k (8.49) It is convenient now to write q = 2kF x and k = kF y to obtain [denoting the density of states at the Fermi level by N (0) = mkF /(2π 2 )] χPauli −+ (q, ω Z 1 x + y 1 = 0) = N (0) dyy ln 2x 0 x−y 1 1 − x2 1 + x = N (0) 1 + ln . 2 2x 1−x (8.50) Note that as x ≡ q/(2kF ) tends to zero, χPauli −+ tends to N (0), while when x becomes very large, it tends to N (0)/x2 . Thus, for a parabolic energy band, the Pauli spin susceptibility is largest as q → 0. This is very important when interaction effects are included. 13