PHZ 5941 Condens Condensed ed Matter Matter I Proble Problem m Set Set 8 — Solut Solution ion
8.1 Problem 12.1, A&M, Pg. 239. For free electrons E ( ¯ 2 k) = h k2 /2m, surfaces of constant energy are spheres, and the the relevant area A is easily shown to be 2mE 2 − kz . A(E , kz ) = π( π (kx + k + ky ) = π( π ( k2 − kz2 ) = π 2 h ¯ 2
2
(1)
It immediately follows that h ¯ 2 ∂ m = A(E , kz ) = m. 2π ∂ E E ∗
(2)
8.2 Problem 12.2, A&M, Pg. 239. Near a band minimum at the point k = k0 the band dispersion has the form h ¯ 2 E (k) = E 0 + (k − k0 ) · M−1 · ( k − k0 ) 2 where
M−1
(3)
is the inve inverse rse of the effective effective mass tensor. tensor. The surfaces surfaces of constant constant energy energy are
now ellipsoids, and orbits in a uniform magnetic field (which we will take to point in the z ˆ direction) are ellipses. 1 To work out the area of the ellipse it is convenient to define the matrix M− made up of ⊥
the upper left 2 × 2 block of the full matrix M−1 , −1
M⊥
=
−1
−1
M xx M xy
−1 −1 M yx M yy
.
(4)
If we also define the vector q q = k − k0 and the two component vector q ⊥ = (q x , q y ) the equation describing the ellipse formed by the intersection of the surface E ( k) = E and a plane of constant kz has the form 1 · q q ⊥ · M− q q ⊥ + 2 2a · q ⊥ = C + + ⊥
2E h ¯2
(5)
−1 −1 1 where where a = q = q z (Mxz , Myz ) and C and C = −2E 0 /h ¯ 2 − q z2 M− both a and an d C are constant , i.e. zz (note: both
independent independent of q ⊥, and, as we will see, their actual values are irrelevant to the final answer.)
The next step is to complete the square. This can be done by shifting q ⊥ according to q ⊥
→
q ⊥ − M⊥ · a
(6)
which results in the equation 1 · q ⊥ = C + q ⊥ · M− ⊥
2 E h ¯2
(7)
where C = C + a · M⊥ · a. (Note: C is, of course, still just a constant). 1 Finally, if we rotate in the xy plane to the principal axes of M− (let’s label these axes ⊥ 1 1 and 2, and denote the corresponding eigenvalues of M− as M 1−1 and M 2− 1), the equation ⊥
describing the ellipse is q 12 q 22 2E + = C + 2 M 1 M 2 h ¯
(8)
The area of this ellipse is then 2 E A(E , kz ) = π M 1 M 2 C + 2 h ¯
(9)
Next (as discussed in class) note that the product of M 1−1 and M 2−1 is the determinant of −1
M⊥
which in turn is related to the zz component of the full matrix
M by
the following
relation 1 det M− ⊥ 1 M zz = = det M− det M. ⊥ −1 det M
(10)
(This follows from the usual expression for the inverse of a matrix). Thus we have M 1 M 2 =
1 1 det M− ⊥
=
det M
(11)
Mzz
and, putting everything together, h ¯ 2 ∂ ∗ m = A(E , kz ) = 2π ∂ E
det M
(12)
Mzz
(b) The low temperature specific heat of a Fermi gas has the form π2 2 cv = kB g(E F )T. 3
(13)
For free fermions with mass m and Fermi wave vector k F the density of states at the Fermi level has the form g(E F ) = 2
mkF h ¯ 2 π2
(14)
Thus, for a fixed number of particles (which determines k F ), the density of states is proportional to the mass m. More generally, we can parameterize the linear term in the specific heat in terms of a “specific heat effective mass.” For the quadratic dispersion relevant to this problem, the density of states has the form g(E ) =
d3k δ (E 4π3
2
− E 0 − h ¯
( k − k0) · M−1 · ( k − k0 )/2)
(15)
shifting the k integration k → k − k0
(16)
then yields g(E ) =
d3 k δ (E 4π 3
k · M−1 · k/2)
2
¯ − E 0 − h
(17)
Next, rotating to the principal axes of the matrix M−1 (labelled 1,2, 3) and rescaling the rotated components of k according to kα
→
M α1/2kα ;
α = 1, 2, 3
(18)
yields g(E ) = (M 1 M 2M 3 )
1/2
d3 k δ (E 4π3
k · k/2)
2
− E 0 − h ¯
(19)
Here M 1 , M 2 and M 3 are the eigenvalues of the matrix M. Thus M 1M 2 M 3 = det M. Up to this rescaling factor, this integral is the same as the integral for the density of states for free fermions with mass 1. The result is thus 1 g(E ) = det M1/2 2 2 h ¯ π (of course g(E ) = 0 for
2(E
− E 0 )
h ¯2
.
(20)
E < E 0 .)
Comparing this with the density of states for free fermions we see that the only difference is that the mass m has been replaced by the specific heat effective mass m∗ = (det M)1/3 .
(21)
It follows that g(E F ) = m∗ kF /(¯h2 π2 ) where kF is the Fermi wave vector corresponding to free fermions with the same number density as the actual system considered here. 3
8.3 Problem 12.3, A&M, Pg. 239. (a) For the quadratic dispersion E ( k)
= E 0 + ¯h2 ( k − k0 ) · M−1 · ( k − k0 )/2
(22)
the semiclassical equation of motion for v implies that v =
1 ∇k E (k) = h ¯ M−1 · ( k − k0 ). ¯ h
(23)
Taking the time derivative and multiplying both sides of this equation by the matrix
M
then yields ˙ h ¯ k = M · v˙ .
(24)
The remaining semiclassical equation of motion then implies that ˙ h ¯ k =
− M· v˙ = −eE
e
v c
×
B
(25)
In a uniform electric field this equation is then simply M· v˙ = −eE
(26)
This equation governs the dynamics of electrons in the absence of collisions. We can include collisions in the relaxation time approximation by writing down the following equation M· v (t + dt)
=
M· v (t) −
eEdt
1−
dt τ
(27)
Here the right hand side is a product of the average value of M · v (t + dt) in the absence of collisions (i.e. that given by the semiclassical equations of motion) and the probability that no collision occurred (1 − dt/tau). (We can ignore the fraction of electrons (dt/tau) which did experience collisions, since they will only have an average velocity of order dt and so contribute a term of order dt2). Dividing through by dt then yields the equation M· v˙
− = −eE
M· v
τ
(28)
For the steady state solution v˙ = 0 (remember that v here is the average velocity of all the electrons contributing to the current). This then implies − −eE
M· v
τ
=0
⇒
4
v =
−1
−eτ M
· E
(29)
Finally, expressing the current in terms of v we obtain j = −nev = ne 2 τ M−1 · E
(30)
; j = σ · E
(31)
Thus we can write σ = ne2 τ M−1 .
where the conductivity σ is now a tensor. the semiclassical equations of motion (for the quadratic (b) In a uniform magnetic field B dispersion we are considering here) has the form M· v = −e
Multiplying through by
M−1
v c
×
B
(32)
yields the equation e = 0 v˙ + M−1 · (v × B) c
(33)
This equation is linear, so we can look for a solution of the form v (t) = v (ω)e−iωt
(34)
Inserting this in the equation (and writing things out in component notation, with the usual convention that repeated indices are summed) we have e −1 −iωδ ij + Mik kjl Bl v j (ω) = 0 c
(35)
Writing the matrices out in all their glory we then have the equation,
−iω
1 0 0 0 1 0 0 0 1
+
e c
−1
−1
−1
M xx M xy M xz
−1 −1 −1 M yx M yy M yz −1 −1 −1 M zx M zy M zz
0
B 0
−B
0 0
0
0 0
· v (ω)
=0
(36)
that B = B zˆ ). (here we the zˆ direction to be parallel to B so After multiplying out the matrices, the requirement for a nontrivial solution (i.e. that the determinant of the matrix multiplying v (ω) vanishes) is found to be
det
−1
−BM xy −
ic ω/e
−1
BMxx
0
−1 −BM yy
−1 − ic ω/e BM yx
0
−1 −BM zy
−1 BM zx
0 − ic ω/e
5
=0
(37)
This yields the equation −1 −1 −1 −1 − ic ω/e)(BM yx − ic ω/e) − (BM xx )( −BM yy ) = 0 (−BM xy
(38)
for which the two roots are readily found to be ω± = ±
eB m∗ c
(39)
where ∗ −2
m
−1
−1
= M xx M yy
−
−1
−1
M xy M yx = det
−1 −1 M xx M xy −1 −1 M yx M yy
=
det M Mzz
.
(40)
Thus we confirm the result that the cyclotron effective mass is given by m∗ =
det M Mzz
8.4 Problem 12.7, A&M, Pg. 241. (a) Unoccupied. (b) Occupied.
6
.
(41)
