PHZ7427: Spring 2014
Problem set # 1: Solutions
due Monday, 01/27
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Office hours: TR 3 pm-4 pm
Please help your instructor by doing your work neatly. Every (algebraic) final result
must be supplemented by a check of units. Without such a check, no more
than 75% of the credit will be given even for an otherwise correct solution.
...
P1 20 points Bending mode in free-standing graphene
Read the attached article from Nature: Materials. The article refers to the fact that the bending
mode of an elastic membrane with dispersion ω = αq 2 and displacement h normal to the membrane
plane destroys long-range crystalline order. The prefactor p
α in the dispersion relation is related to
the bending rigidity κ, mentioned in the article, via α = κ/ρ, where ρ is mass density per unit
area. Derive Eq. (1) of the paper for the rms vertical displacement. (Notice that the ∝ sign, which
we normally use for “proportional to”, is used for “equal in order of magnitude” here; also kB is set
to unity, which shows that the authors are theorists ,.) Taking κ = 1 eV and assuming that the
membrane is at room temperature, estimate the lateral size of the membrane at which hhi1/2 becomes
equal to the atomic spacing and thus long-range crystalline order becomes impossible.
Solution:
The rms amplitude of atomic vibrations due to the bending mode is
Z
d2 q 1
1
~
n
+
,
h̄2 =
B
ρ
(2π)2 ω(q)
2
where nB is the Bose function. For T higher that ~ω/kB , nB +
h̄2 =
kB T
ρ
Z
1
2
kB T
d2 q
1
=
2
2
(2π) ω (q)
2πα2 ρ
≈
Z
(1)
kB T
~ω
dq
.
q3
and
(2)
The integral over q obviously diverges at q → 0. Now we need to recall that, in a finite size system,
the minimum value of q is 2π/L. Cutting the integral at this value and using α2 = κ/ρ, we obtain
h̄2 =
kB T 2
L .
8π 3 κ
(3)
The numerical coefficient in this formula has a “symbolic” meaning as when we cut the divergent
integral we can choose the cutoff with an order magnitude accuracy. If we keep this number anyway
and substitute T = 300 K and κ = 1 eV, we get
h̄2
1/2
≈ 0.01L
(4)
Given the bond length a = 1.4 Å, the membrane “melts” at L ≈ 140Å.
P2 20 points Zero point motion in solid hydrogen
Below 4 K, solid hydrogen has an fcc lattice with the mass density ρ = 88 kg/m3 . The speed of
longitudinal sound is s = 1, 140 m/s. Estimate the amplitude of zero-point displacements of atoms
and compare it to the lattice spacing. Is solid hydrogen a classical or quantum crystal?
To calculate the Debye frequency corresponding to the longitudinal sound mode, use Eqs. (23.22) and
(23.24) from Ashcroft & Mermin (notice that c in there is the speed of sound rather than of light! ,)
Solution
A conventional fcc cell of solid hydrogen contains four H2 molecules or eight H atoms (8 corner points,
each shared amongst 8 cells, and 6 face points, each shared between two cells: 8/8 + 6/2 = 4). The
side of the cube is found from
ρ=
8MH
→ a ≈ 5.3Å
a3
(5)
√
and the nearest neighbor distance is a/ 2 ≈ 3.7Å. The rms zero-point amplitude is
Z
d3 q 1
~
2
ū =
2ρ
(2π)3 ω(q)
(6)
In the Debye model, ω = sq for ≤ q ≤ qD thus
ū2 =
2
~qD
.
8π 2 ρs
(7)
The number density n = ρ/2MH = 2.6 × 1028 m−3 , thus qD = (6π 2 n)1/3 = 1.1 × 1010 m−1 . Then
1/2
ū2
= 0.6Å, which is still about 5 smaller than the n.n. distance. This estimate gives only a low
bound, however. In reality, there are also two transverse sound modes. Taking their contributions into
account would increase ū2 roughly by a factor of 3. There are also optical modes which are essentially
1/2
the vibrational modes of the H2 molecule. All of these factors increase the actual ratio of ū2
to
the nearest neighbor distance to about 0.5, which makes H2 quite a quantum crystal.
P3 20 points Grüneisen parameter
Find the Grüneisen parameter of a 1D chain of atoms with mass M connected by springs with spring
constant K and separated by lattice spacing a. Consider two limiting cases: kB T ~ωmax and
kB T ~ωmax , where ωmax is the maximal frequency of the phonon mode. You may need to use that
R π/2
dx ln(sin x) = −π ln 2/2.
0
Solution The Grüneisen parameter for a single-mode 1D system is given by
P
k γk Cv (k)
,
γ= P
k Cv (k)
(8)
∂nB
where γ(k) = − ∂~ω(k)
∂ ln L and Cv (k) = ~ω(k) ∂T . The phonon dispersion relation for a monoatomic chain
is given by ω = ωmax sin(ka/2) with ωmax = (4K/m)1/2 .
L
∂ sin(ka/2)
ka
∂ω
=
=
cot(ka/2)
∂ ln L
sin(ka/2)
∂L
2
ka
k 2 cot(ka/2)Cv (k)
P
R π/a
P
γ=−
k
Cv (k)
=−
0
dk ka
2 cot(ka/2)Cv (k)
.
R π/a
dkCv (k)
0
(9)
(10)
For kB T ~~ωmax , typical phonons have wavenumbers defined by the condition ~ωmax sin(ka/2) ∼
kB T or, replacing the sine function by its argument, k ∼ (1/a)(kB T /~ωmax ) 1/a. Therefore,
cot(ka/2) ≈ 2/ka and γ = −1 in this limit. In the opposite limit of kB T ~ω, we can replace the
Bose function by kB T /~ω(k) so that ∂nB /∂T = kB /~ω(k) and Cv (k) = kB . Then
γ(T ) = −
a2
2π
Z
π/a
k cot(ka/2) = −
0
2
π
Z
π/2
dxx cot(x) =
0
2
π
Z
π/2
dx ln sin x = − ln 2.
0
(11)
P4 Thomas-Fermi screening
8 points Prove Eq. (17.54) of Ashcroft & Mermin.
Solution:
With x ≡ cos θ and integrating over φ from 0 to 2π, we find
Z 1
Z
Z
2Q ∞
q2
q
Q ∞
iqrx
dxe
=
dq 2
dq 2
sin(qr)
φ(r) =
2
π 0
q + κ −1
πr 0
q + κ2
Z ∞
Z ∞
q
Q
q
Q
=
dq 2
sin(qr) =
exp(iqr)
Im
dq 2
2
πr −∞ q + κ
πr
q + κ2
−∞
Q
iκ
Q
=
Im 2πi
exp(−qr) = exp(−qr).
πr
2iκ
r
(12)
(13)
(14)
12 points In class, we showed that the Fourier transform of the potential of a point charge Q, embedded
into a 2D electron gas, is given by
φ̃(k) =
2πQ
,
k+κ
where k is a two-dimensional vector in the plane containing 2D electrons and κ = 2πe2 g(εF )is the
screening wavenumber1 in 2D (as usual, g(ε)is the electron density of states). Show that φ(r) falls
off only as a power of the in-plane distance r for r 1/κ (as opposed to the exponential fall off
in 3D). Determine the exponent of the power-law dependence. Most likely, you will need to use
R 2π
that 0 dθeia cos θ = 2πJ0 (a), where J0 (x) is the Bessel function, and perhaps also the Feynman
R∞
R∞
trick 1/x = 0 dy exp(−yx) (if you do, the integral 0 dxx exp(−ax)J0 (x) = a/(1 + a2 )3/2 will
also be handy).
Solution:
The inverse Fourier transform reads (here r is the in-plane radius-vector)
R 2πQ −ikr cos θ kdkdθ
φ(r) = k+κ
e
(2π)2
R ∞ kJ0 (kr)
Q
= 2π 0 dk k+κ
(15)
(16)
This integrand is nonanalytic when continued to the entire real line. The large-r behavior can be
obtained by integration by parts, using an identity
Z
xJ0 (x)dx = J1 (x).
With p ≡ kr, we find
Z ∞
Q
pJ0 (p)dp
φ(r) =
2πr 0
p + rκ
Z ∞
Q J1 (p) ∞
J1 (p)dp
=
+
.
2πr p + rκ 0
(p
+ rκ)2
0
(17)
(18)
The boundary terms cancel out because J1 (0) = J1 (∞) = 0. In the integral term, we can neglect p
compared to κr 1 in denominator, leaving us with
Z ∞
Q
1
J1 (p)dp
(19)
φ(r) = 2πr (rκ)2
}
| 0 {z
=
1
Q
1
2πr (rκ)2
(20)
and we see that φ(r) falls off as 1/r3 at r κ−1 .
The same result can also be obtained by using the Feynman trick:
Z ∞
Z ∞
Z ∞
pJ0 (p)dp
F (a) =
=
pJ0 (p)
e−(p+a)s dsdp
p+a
0
0
0
Z ∞
Z ∞
Z ∞
s ds
−ps
−as
e−as
pJ0 (p)e dpds =
e
.
=
(1
+
s2 )3/2
0
0
0
For a 1, typical s ∼ 1/a 1, and we thus can ignore the s2 term in the denominator which yields
Z ∞
1
F (a 1) =
se−as ds = 2 .
a
0
P5 20 points Plasmons in 2D
A 2D analog of Eq. (17.60) in Ashcroft & Mermin is
Z
2πe2
d2 k f (εk ) − f (εk+q )
(q, ω) = 1 +
.
q
(2π)2 εk+q − εk + ~ω
(21)
Find the dispersion relation ω(q) of a 2D plasmon for q kF .
Solution:
First of all, notice that the difference between the 3D and 2D is in the form of the Coulomb potential
in the Fourier space. In 3D, The Fourier transform of the point-charge potential is φ̃(q) = 4πQ/q 2 .
Consider now a point charge Q located in the plane z = 0. The Fourier transform of this potential
over the in-plane coordinates is given by (here, q is the in-plane vector)
Z ∞
Z ∞
1
2πQ
dqz
dqz
φ̃(q, qz ) = 4πQ
=
.
(22)
φ̄(q, z = 0) =
2 + q2
2π
2π
q
q
−∞
−∞
z
This explains why the prefactor 4π/q 2 in (17.60) is replaced by 2π/q. Expanding (21) for small q, we
obtain
Z
2πe2
d2 k
∂f
q · vk
(q, ω) = 1 +
.
(23)
−
2
q
(2π)
∂εk q · vk + ω
∂f
= δ(εk − εF ). Recalling the definition of the density of states
where vk = ~−1 ∂εk /∂k. At T = 0, − ∂ε
k
R
g(ε) = (1/2π 2 ) d2 kδ(εk − ε), we reduce the equation above to
(q, ω) = 1 +
πe2 g(εF )
q
Z
dφ q · vF
,
2π q · vF + ω
(24)
where φ is the azimuth of k measured from q. Plasmons are high-frequency excitations of the electron
liquid, so we can treat q · q/ω as a small parameter and expanding the integrand in Eq. (24) to 2nd
order in this parameter. This yields
Z
πe2 g(εF )
dφ 2 2
πe2 g(εF )q
(q, ω) = 1 −
q vF cos2 φ = 1 −
.
(25)
2
qω
2π
2ω 2
The dispersion of a plasmon is obtained from the equation = 0 which gives
q
ωp (q) = πg(εF )vF2 q/2.
In contrast to the 3D case, the plasmon dispersion vanishes at q = 0.
(26)
Bonus 20 points Correcting ω(q) for 3D plasmons
Show that, in contradistinction to statements in some textbooks, a 3D plasmon never runs into the
particle-hole continuum but comes very close to it (see Fig. 1).
Solution The Lindhard formula [Eq. (17.60)] in AM can be re-written as ( shifting the variables
k + q/2 → k + q and k − q/2 → k):
Z
4πe2
d3 k f (εk ) − f (εk+q )
(q, ω) = 1 + 2
.
(27)
q
(2π)2 εk+q − εk + ~ω
Split the integral into two terms and shift k + q → k in the second term. This gives
Z
4πe2
1
d3 k
1
(q, ω) = 1 + 2
fk
−
.
q
(2π)3
εk+q − εk + ~ω εk − εk−q + ~ω
(28)
For free electrons εk±q = εk ± (~k/m)q cos θ + ~2 q 2 /2m. We need to find the plasmon dispersion
at large q, when the order q term in the last equation is smaller than the order q 2 term. Then,
εk±q ≈ εk + ~2 q 2 /2m. The equilibrium Fermi function constraints the range of integration over k to
Rk
the interval 0 ≤ k ≤ kF . The angular integral gives 4π while the integral over k gives 0 F dkk 2 = kF3 /3.
Thus
"
#
(4π)2 e2 kF3
1
1
(q, ω) = 1 +
−
3q 2 (2π)3 ~2 q2 + ~ω − ~2 q2 + ~ω
2m
= 1−
(4π)2 e2 kF3
3q 2 (2π)3
2m
~2 q 2
m
(~ω)2 −
~2 q
2 = 1 −
2
2m
2e2 ~2 kF3
3πm
1
2 2 2
q
(~ω)2 − ~2m
The plasmon mode is obtained from the condition (q, ω) = 0 which gives
s
2
2e2 ~2 kF3
~2 q 2
+
.
~ω(q) =
2m
3πm
(29)
(30)
The upper boundary of the particle-hole continuum ~ωmax = ~vF q + ~2 q 2 /2m reduces to ~ωmax =
~2 q 2 /2m at larger q. Therefore ω(q) > ωmax and plasmon never crosses the continuum boundary.
1
Not to be confused with the bending rigidity!,
FIG. 1: A common misconception about the form of the plasmon dispersion in 3D. The correct behavior is shown by
the red curve.