Download SOLID-STATE PHYSICS III 2009 O. Entin-Wohlman Thermal equilibrium

SOLID-STATE PHYSICS III 2009
O. Entin-Wohlman
1.
ELECTRONIC TRANSPORT PROPERTIES
Thermal equilibrium
At thermal equilibrium, the number of electrons having energy E is given by the Fermi
distribution,
f (E) =
1
eβ(E−µ)
+1
,
(1.1)
in which β ≡ 1/(kB T ) is the inverse temperature (kB is the Boltzmann constant) and µ
is the chemical potential. At zero temperature (β → ∞) the chemical potential is equal
to the Fermi energy, EF , and the Fermi function becomes a step-function, such that all
states with energy E ≤ EF are full, and all states with energy above EF are empty. In the
grand-canonical ensemble, where the chemical potential is fixed, the number of electrons
is temperature-dependent. The electronic density (number of electrons per unit volume) is
given by
∞
Z
dEN (E)f (E) ,
N=
(1.2)
−∞
where N (E) is the density of states (number of states having energy E per unit volume).
Calculation of integrals containing the Fermi function
In metals, the relevant energies are around the chemical potential (' the Fermi energy),
which is much larger than the thermal energy kB T at room temperature (and below). Therefore, one may calculate explicitly the integral in Eq. (1.2). We present now the general
procedure for calculating such integrals.
Suppose that we need to calculate
Z
∞
I=
dEg(E)f (E) ,
(1.3)
−∞
where the function g contains the density of states, and is a well-behaved function of the
RE
energy E. Integrating by parts, such that G(E) = −∞ dE 0 g(E 0 ), we have
Z
∞
I = G(E)f (E)
−
−∞
∞
−∞
1
dEG(E)
∂f (E)
.
∂E
(1.4)
At very low negative energies there are no states and therefore the density of states (which
is included in the function g) tends to zero, while at very high positive energies the Fermi
function, Eq. (1.1), vanishes. Therefore, the first term in Eq. (1.4) vanishes. To treat the
second term there, we expand the function G around the chemical potential,
1
G(E) = G(µ) + G0 (µ)(E − µ) + G00 (µ)(E − µ)2 + . . . .
2
(1.5)
Then,
I=
X
n
Z ∞
1 (n)
d
1
In , In = − G (µ)
dE(E − µ)n
n!
dE eβ(E−µ) + 1
−∞
Z
i
h1 ∞
xn
1
dx x/2
.
= n G(n) (µ)
β
n! −∞ (e + e−x/2 )2
(1.6)
The term in the square brackets is a number, which is zero for odd n’s. For even values of
n it can be calculated, to give
I = G(µ) +
π2
(k T )2 G00 (µ) + . . . .
6 B
(1.7)
For a free (three-dimensional) electron gas, the density of states is given by
N (E) = N (0)(E/EF )1/2 , E ≥ 0 ,
(1.8)
where N (0) is the density of states at the Fermi energy. Returning to our calculation of the
electronic density, we find
G(E) = N (0)
2 E 3/2
,
3 EF1/2
(1.9)
and consequently, assuming that µ ' EF ,
2 π 2 h kB T i2
N = N (0) EF 1 +
+ ... .
3
8 EF
(1.10)
∗ ∗ ∗Exercise. Find the total energy of the free electron gas.
The Boltzmann equation
The Boltzmann equation (also called the transport equation) is an equation for the distribution function of the electrons, which replaces the Fermi function when the electrons are
out of equilibrium. The electrons are out of equilibrium since they experience external fields
(i.e., perturbations). The Boltzmann-equation approach assumes that those perturbations
2
change slowly enough in space and in time. Under such conditions, we may use semiclassical
considerations.
Let us first ignore the collisions of the electrons (with lattice imperfections, other electrons,
etc.) and assume that the motion is due only to external fields. When the perturbations
are smooth enough, the r (position) and the k (momentum, using units in which ~ = 1)
coordinates of every electron evolve in time according to the semiclassical equations of motion
1
ṙ = v(k) , k̇ = −e E + v(k) × H ≡ F(r, k) .
c
(1.11)
Here, v is the velocity, e is the electron charge, E is the electric field, H is the magnetic field,
and F is the Lorentz force. Considering an infinitesimal time interval dt, then an electron
which at time t is at r with momentum k must have been at time t − dt at r − v(k)dt, with
momentum k − Fdt. In the absence of collisions, the electron distribution, g(r, k, t) at time
t is identical to the distribution function at time t − dt,
g(r, k, t) = g(r − v(k)dt, k − Fdt, t − dt) ,
(1.12)
since no electron has been scattered off (or scattered into) the classical trajectory. Expanding
in the infinitesimal time interval dt, we find
∂g
∂g
∂g
+v·
+F·
= 0 , no collisions .
∂t
∂r
∂k
(1.13)
Collisions give an additional source for the change of the distribution function with time.
Now g is reduced due to ‘scattering-out’ of electrons away from the classical trajectory, and
it is enhanced due to ‘scattering in’ of electrons,
∂g ∂g ∂g =
−
,
∂t coll
∂t in
∂t out
(1.14)
∂g
∂g
∂g ∂g +v·
+F·
+
=0.
∂t
∂r
∂k
∂t coll
(1.15)
such that
This is the Boltzmann equation.
Once one solves the Boltzmann equation and finds the distribution function, one may obtain
various physical quantities. For example, the current density, J, is given by
J=e
X
k
3
v(k)g(k) .
(1.16)
One can also find the current density of the energy, namely, to analyze the heat transport.
We will discuss this quantity later.
The collision term of the Boltzmann equation
There are various sources contributing to the collision term of the Boltzmann equation. For
example, electrons may be scattered by randomly-distributed static impurities. This will give
rise to elastic scattering processes, during which the electrons can change just the direction
of the momentum, but not their energy. Another source is the electron-phonon interaction,
which gives rise to inelastic scattering processes, in which the electrons change their momenta
and their energies. Likewise, the electrons may be scattered off other electrons, or off
localized magnetic moments.
These various processes can all be described by assigning to each collision process a transition
probability per unit time (related to the scattering cross-section). This quantity gives the rate
for an electron to be scattered from a state k to another state, of k0 . (We use alternatively the
momenta or the energies to specify electronic states.) Denoting that transition probability
per unit time by Wkk0 , we have
∂g(k) X
Wk0 k g(k0 )[1 − g(k)] − Wkk0 g(k)[1 − g(k0 )] .
=
∂t
coll
k0
(1.17)
The first term on the right-hand-side of Eq. (1.17) is the scattering-in term: the transition
probability per unit time to go from any state k0 to the state k is multiplied by the distribution function of electrons having k0 (ensuring that there are electrons to be scattered in)
and by the probability to have available room at k, namely, by 1 − g(k). This is summed
over all states k0 . Similarly, the second term in Eq. (1.17) is the scattering out term. Note
that the right-hand-side of Eq. (1.17) vanishes when summed over k. This is a result of the
number (of electrons) conservation.
∗ ∗ ∗Exercise. At equilibrium, the distribution function is the Fermi function. Then the collision term, Eq. (1.17), vanishes. Derive the necessary condition on the transition probability
per unit time for this to happen.
Elastic impurity scattering
The explicit calculation of the collision term in the Boltzmann equation is usually done
up to second-order (in perturbation theory) in the scattering potential. (Remember that
the entire Boltzmann-equation approach is valid when the perturbations in the system are
slowly varying.)
4
When the impurity concentration, ni , is small, we may calculate the transition probability
per unit time due to a single impurity, and then multiply the result by the impurity concentration. Denoting the potential of each impurity by U (r), the transition probability per
unit time is given by the Fermi golden-rule (~ = 1),
Wkk0 = 2πni δ(Ek − Ek0 )|hk|U |k0 i|2 .
(1.18)
Wkk0 = Wk0 k .
(1.19)
We note that in this case
This is called ‘detailed balance’. When this is the case, the collision term (1.17) becomes
∂g(k) X
0
Wkk0 g(k ) − g(k) .
=
∂t
coll
k0
(1.20)
This vanishes at equilibrium, i.e., when g is replaced by the Fermi function, f [because of
the energy conserving delta function in Eq. (1.18)].
Linearization of the Boltzmann equation-the electrical conductivity
Let us consider the solution of the Boltzmann equation in a simple situation, in which the
only perturbation on the electrons is a constant (in space and time) electric field, and the
sole source of collisions is elastic impurity scattering.
Since the perturbation does not depend on time, the distribution function g will not depend
on t as well, i.e., ∂g/∂t = 0. Such a situation is called ‘steady-state’. In the present case, we
also have that ∂g/∂r = 0, since the field is uniform. Consequently, the distribution depends
on the momentum alone. We will also assume that the system is not too far away from
equilibrium, and will seek a solution for g up to first-order in the electric field.
Consider the field-driven term in the Boltzmann equation (1.15). Writing g(k) = f (Ek ) +
g (1) (k), where g (1) is linear in the field, we have (up to first-order in the electric field)
F·
∂g(k)
∂g(k)
∂f (Ek )
= −eE ·
' −eE ·
∂k
∂k
∂k
∂f dEk ∂f = eE ·
−
= eE · v(k) −
.
dk
∂E
∂E
(1.21)
We therefore conclude that the distribution function must have the form
g(k) = f (E) −
∂f ∂E
5
eE · v(k)τ (k) ,
(1.22)
where E ≡ Ek . The derivative of the Fermi function restricts the relevant energies to be at
about the Fermi energy. This is physically clear: the electric field affects only the electrons
lying close to the Fermi energy. We therefore may safely assume that the (yet unknown)
function τ (k) ' τ (EF ) ≡ τ.
We next consider the collision term, using Eqs. (1.17) and (1.18). We note that since the
scattering is elastic then necessarily k = k 0 , namely, the impurity scattering just changes
the direction of the wave vector. This means that the transition probability per unit time
(i.e., the scattering cross section) in the case of elastic scattering depends only on the angle
between k and k0 . Because of the energy-conserving delta function in Eq. (1.18), the
integral in Eq. (1.20) reduces to an integration over that angle. Putting all this together,
the Boltzmann equation of this case gives
Z
0
0
E · v(k) = τ dΩ D(cos θ) v(k) − v(k ) · E ,
(1.23)
where D is the scattering cross-section, θ is the angle between the directions of k and k0 ,
and dΩ0 marks the angular integration over the angles of k0 .
Generally one expects that the direction of v(k) is determined by k itself. Choosing the
polar direction along k, the vector E such that E = E(sin α, 0, cos α), and the vector k0 such
that k0 = k(sin θ cos φ, sin θ sin φ, cos θ), we find
k0 · E = Ek sin α sin θ cos φ + cos α cos θ) .
(1.24)
When this is inserted into Eq. (1.23), the angular integration kills the first term. Then Eq.
(1.23) yields
1
=
τ
Z
dΩ0 D(cos θ) 1 − cos θ .
(1.25)
We see that the collision rate 1/τ is dominated by back scattering, namely, by scattering
processes which change significantly the angle. Forward scattering (for which cos θ is close
to 1) contributes only little to the relaxation rate. Indeed, had Wkk0 dictated only forward
scattering, the collision rate would have vanished (namely, the collision term would have
disappeared from the Boltzmann equation). The relaxation time given by Eq. (1.25) is the
transport relaxation time. In fact, our result here is equivalent to replacing the collision term
by
∂g g(k) − f (Ek )
.
'−
∂t coll
τ
6
(1.26)
We have now determined all ingredients of our solution, Eq. (1.22), to the Boltzmann
equation in the present simple case. In order to find the (charge) current, it remains to
insert the solution into the expression for J, Eq. (1.16). Obviously, only the second term of
Eq. (1.22) contributes to the current (the first term does not contribute because the angular
integration vanishes). Hence we have
2
J=e
X
k
∂f (E ) k
E · v(k) v(k) .
τ (k) −
∂Ek
(1.27)
The derivative of the Fermi function restricts the relevant energies contributing to the sum
to be on the Fermi level. The angular integration is a bit more tricky. For a free electron
gas, v(k) = k/m, (m is the mass of the electron). Then we find
1
J = e2 N (0)vF2 τ E ,
3
(1.28)
where the factor 1/3 comes from the angular integration, and vF is the Fermi velocity. The
factor vF2 τ /3 is the diffusion coefficient (at three dimension). In general dimensions, d, the
diffusion coefficient, D, is
D=
vF2 τ
.
d
(1.29)
Comparing the result (1.28) with the one for the electronic density, Eq. (1.10), (omitting the
temperature-dependent correction term, which is usually small) we can re-write the above
equation in the form
J=E
e2 τ N m
≡ EσD ,
(1.30)
where
σD = e2 τ N/m = N (0)e2 D
(1.31)
is the electric conductivity. This expression for σ is termed ‘the Drude conductivity’.
∗ ∗ ∗Exercise. Repeat the above calculation for the case when the electric field is still uniform
(constant in space) but varies in time, such that E(t) = E cos ωt, and find the frequencydependent conductivity.
Linearization of the Boltzmann equation-transport coefficients
Let us now include in the previous treatment the effect of a temperature gradient. Namely,
the temperature is not fixed along the sample, but changes along a certain direction. When
7
this gradient is small enough, we may suppose that there is still a well defined temperature,
T (r), at each point, such that
f (Ek ) →
1
(Ek −µ(r))/(kB T (r))
e
+1
.
(1.32)
Here we have assign an r-dependence also to the chemical potential. This should be viewed
as adding to the (fixed) chemical potential an electric potential, which is created in the
system since under the effect of the temperature gradient the electron density is changed in
order to screen the effect of that gradient. (In fact, we should have done the same also in
the treatment above.) The term v · ∂g/∂r in the Boltzmann equation (1.15) now yields
v·
∂f E − µ
1
∂
k
[−∇T
]
−
∇µ
.
=
−
∂r e(Ek −µ(r))/(kB T (r)) + 1
∂E
T
(1.33)
Adding the effect of a uniform electric field E, the linearized Boltzmann equation takes the
form
∂f hE − µ
∂g ∇µ i
k
−
v(k) ·
(−∇T ) + e(E −
) =−
.
∂E
T
e
∂t coll
(1.34)
At this point we assume that the collision term is due to static impurity scattering, so that,
similarly to Eq. (1.22) above, the distribution function is given by
∂f hE − µ
∇µ i
k
g(k) = f (Ek ) + −
τ v(k) ·
(−∇T ) + e(E −
) .
∂E
T
e
(1.35)
(Here, it is implicitly assumed that both ∇T and ∇µ are uniform in space.)
Inserting the solution Eq. (1.35) into the expression for the (charge) current density, Eq.
(1.16), we find
∂f h
∇µ i
τ (k)v(k) −
v(k) · (E −
)
∂E
e
k
∂f E − µ h
i
X
k
+e
τ (k)v(k) −
v(k) · (−∇T ) .
∂E
T
k
J = e2
X
(1.36)
The first term here is similar to what we have found above [see Eq. (1.27)]. It gives the
electrical conductivity, describing the response of the electrons to both the electric field E
and the potential gradient included in ∇µ/e. The combination E − ∇µ/e is called the
‘electro-chemical potential’.
The second term in Eq. (1.36) shows that the effect of a temperature gradient on the
electrons is to produce an electric current. This is called ‘thermo-electric effect’. However,
8
there is another effect due to the temperature gradient: it produces heat current density.
The hear current density, denoted U, is defined as
U=
X
v(k) Ek − µ g(k) .
(1.37)
k
Since µ represents the free energy of the electron, the heat current is given by the current
density of the ‘internal energy’ (i.e., Ek ), minus the free energy (which is indeed the definition
of heat). Introducing the solution for the distribution function, Eq. (1.35), into Eq. (1.37),
we find
∂f h
∇µ i
v(k) · (E −
)
τ (k)(Ek − µ)v(k) −
∂E
e
k
∂f h
i
X
(Ek − µ)2
+
v(k) −
v(k) · (−∇T ) .
τ (k)
T
∂E
k
U=e
X
(1.38)
Inspecting Eqs. (1.36) and (1.38), we see that it is natural to define the following quantity
K(n) =
X
k
∂f τ (k)(Ek − µ)n −
v(k) ⊗ v(k) .
∂E
(1.39)
(K is a tensor.) Then, the general transport equations take the form
e (1)
K · (−∇T ) ,
T
1
∇µ
+ K(2) · (−∇T ) , Eef = E −
.
T
e
J = e2 K(0) · Eef +
U = eK(1) · Eef
(1.40)
For example, suppose that we apply a temperature gradient on an open system, so that
there is no electric current flowing through it. Then the first of Eqs. (1.40) tells us that an
electric field is set up on the system, given by
E=
1 (0)−1 (1)
K
K ∇T .
eT
(1.41)
Using this result in the second of Eqs. (1.40), we find that the heat current is
U = κ(−∇T ) ,
(1.42)
where κ (usually, a tensor) is the thermal conductivity, given by
κ=
1 (2)
(K − K(1) K(0)−1 K(1) ) .
T
9
(1.43)
How does one actually calculate the transport coefficients? To this end we use our trick
of calculating the sums which include the derivative of the Fermi function. We write Eq.
(1.39) in the form
Z
Z
Z
∂f ∂f dΩk
n
(n)
N (E)τ (E)(E − µ)
v ⊗ v ≡ dEΦ(E) −
,
K = dE −
∂E
4π
∂E
and use [see Eqs. (1.4), (1.5), and (1.6)] to obtain
Z
∂f π 2 kB2 T 2 ∂ 2 Φ dEΦ(E) −
= Φ(µ) +
.
∂E
6
∂E 2 E=µ
(1.44)
(1.45)
∗ ∗ ∗Exercise. Show that when only the leading orders are kept, one finds
K(2) =
π 2 kB2 T 2 (0)
K ,
3
(1.46)
and
(1)
K
π 2 kB2 T 2 ∂K(0) =
.
3
∂E E=µ
(1.47)
Thermo-electric effects
Let us consider a circuit (a simple ring, for concreteness), which is made of two materials.
The ring is connected to a battery, which provides the electric field E, and is kept at a fixed
temperature (namely, there is no temperature gradient). According to Eqs. (1.40),
J = e2 K (0) E , U = eK (1) E ,
(1.48)
1
U = K (0)−1 K (1) J ≡ ΠJ .
e
(1.49)
so that
Since the current driven in the ring, J, is the same in both metals forming the ring, then
the heat current is necessarily different, being ΠA J in metal A, and ΠB J in metal B. At the
two junctions, the balance is restored: one junction absorbs heat flux , say, (ΠA − ΠB )J, and
the other junction emits heat flux, so that the first one is heated, and the second is cooled.
This is the Peltier effect.
Transport in the presence of a (constant) magnetic field
Let us now consider the combined effect of a constant magnetic field and a constant electric
field on the electrons. Assuming that the collision integral is due only to elastic impurity
scattering [see Eq. (1.26)] the Boltzmann equation pertaining to this configuration reads
∂g(k)
1
g(k) − f (Ek )
−e E + v(k) × H ·
=
.
(1.50)
c
∂k
τ
10
We again seek for a solution for the distribution function which is just slightly away from
equilibrium,
g(k) = f (Ek ) + g (1) (k) ,
(1.51)
∂f (Ek ) v(k) · A(k) .
g (k) = eτ −
∂Ek
(1.52)
where in analogy Eq. (1.22),
(1)
(The function A(k) has yet to be found.) We will see below why such a form is dictated in
the present case.
Since
∂f (Ek ) ∂f (Ek ) ∂Ek ∂f (Ek ) =
=
v(k) ,
∂k
∂Ek
∂k
∂Ek
(1.53)
we see that the zeroth-order term of the distribution Eq. (1.51) does not contribute to the
magnetic field part of the Lorentz force (it includes the product v × H · v). However, in the
electric field part of the Lorentz force we may keep just the zeroth order term (as before).
As a result, we find that Eq. (1.50) becomes (to lowest possible order in the electric field)
∂f (E ) g(k) − f (Ek ) e
g(k) − f (Ek )
k
e −
v(k) · E =
+ v(k) × H ·
∂Ek
τ
c
∂k
(1)
(1)
g (k)
g (k) e
+ v(k) × H ·
.
(1.54)
≡
τ
c
∂k
Arranging terms, we find
∂f (E ) k
g (1) (k) = eτ −
v(k) · E −
∂Ek
∂f (E ) k
= eτ −
v(k) · E −
∂Ek
eτ
g (1) (k)
v(k) × H ·
c
∂k
2 2
eτ
∂f (Ek )
−
H × E · v(k) + . . .
mc
∂Ek
(1.55)
The last equality here is obtained upon solving by iterations, and assuming the free electron
gas relation
v(k) = k/m .
(1.56)
We see from the form of Eq. (1.55) that the general solution must have the form (1.52).
Moreover, it appears that the vector A there is independent of the momentum. Therefore,
using the solution Eqs. (1.51) and (1.52) in Eq. (1.54), we find
v·E=v·A+
eτ
eτ
v · H × A , i.e., E = A +
H×A .
mc
mc
11
(1.57)
Since the last equality gives
eτ H(H · A) − AH 2 , and H · A = H · E ,
mc
H×E=H×A+
(1.58)
we find
E−
A=
eτ
H
mc
eτ 2
× E + ( mc
) H(H · E)
.
eτ 2
1 + (H mc
)
(1.59)
eτ
must be dimensionless. Indeed,
Inspecting Eq. (1.59), we see that the quantity H mc
ωc =
eH
,
mc
(1.60)
is the cyclotron frequency. Moreover, from our discussion of the electrical conductivity above
[see Eq. (1.30)], and the formal analogy between Eq. (1.22) and Eq. (1.51), we find that
the electrical current is given by
J = σD A , where σD =
N e2 τ
is Drude conductivity .
m
(1.61)
Using our result (1.59) in Eq. (1.61), we finally obtain that the electrical current, when the
electrons experience both electric and magnetic field is given by
J = σD
E − ωc τ Ĥ × E + (ωc τ )2 Ĥ(Ĥ · E)
.
1 + (ωc τ )2
(1.62)
(Here, Ĥ = H/H.)
Without loss of generality, we may take the magnetic field to be along the ẑ direction. Then,
the z component of Eq. (1.62) is just
Jz = σD Ez ,
(1.63)
namely the motion along the magnetic field is un-changed. On the other hand, for the
motion in the plane perpendicular to the magnetic field, we have
Jx = σD
Ex + ωc τ Ey
Ey − ωc τ Ex
,
J
=
σ
,
y
D
1 + (ωc τ )2
1 + (ωc τ )2
which we can put in a matrix form
 



J
1 ωc τ
E
σD
 x=

  x  ⇒ J = σE .
2
1 + (ωc τ )
Jy
−ωc τ 1
Ey
12
(1.64)
(1.65)
Here σ is the conductivity tensor. Inverting this tensor, we find that the resistivity tensor ρ,
is such that

E = ρJ , with ρ =
1
1 
σD ω τ
c
−ωc τ
1

 .
(1.66)
Writing explicitly Eq. (1.66), we have
1
ωτ
Jx − c Jy ,
σD
σD
1
ωτ
Ey =
Jy + c Jx .
σD
σD
Ex =
(1.67)
It follows that the longitudinal resistivity (electric field and current along the same direction)
is unaffected by the magnetic field. On the other hand, the transverse resistivity is odd in
the magnetic field–this is one of the Onsager relations. In particular we see that when the
current is flowing along, say, the x− direction due to an electric field Ex , and the system is
open along the y− direction, such Jy = 0, then the magnetic field (via the Lorentz force)
causes the appearance of voltage along the y− direction, given by
Ey = Jx
eHτ m
1
ωc τ
= Jx
= Jx
.
2
σD
mc N e τ
N ec
(1.68)
This is the Hall voltage. One notes that the Hall coefficient, 1/N ec, is independent of the
relaxation time.
∗ ∗ ∗Exercise. We have found above that for the free electron gas, the longitudinal resistivity
is not affected at all by a constant magnetic field. Consider a system made up of two electron
species, which differ from one another by their masses and by their relaxation times. Find
the longitudinal resistivity in this case. Make sure that the result is even in the magnetic
field (this is another Onsager relation).
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