(Oxford) Solutions for B6: Condensed Matter Physics,

B6: Uno
cial Past Paper Solutions (2011-2016)
Toby Adkins


November 21, 2017




Disclaimer: These solutions are produced for student use, and have no connection to the
department what-so-ever. As such, they cannot be guaranteed to be correct, nor have any
bearing on the actual mark schemes that were used in the exams. Furthermore, they are
almost literally word for word my practise solutions, so they are far from 'model' solu-
tions. The formatting may suer somewhat as a result. As to the question of why I typed
them, as supposed to just scanning my solutions; anyone that knows me personally will
attest to the dire state of my handwriting. For queries and corrections, please email me at


2011
Question 1
De
ne the direct lattice vectors a1 , a2 , and a3 . Then, the corresponding reciprocal lattice
vectors are given by
a2 × a3
b1 = 2π
a1 · (a2 × a3 )
a3 × a1
b2 = 2π
a1 · (a2 × a3 )
a1 × a2
b3 = 2π
a1 · (a2 × a3 )
These satisfy the orthogonality relationship:
ai · bj = 2πδij

Consider the Fourier decomposition of ρ(r) = ρ(r + R), such that
0
X
ρ(r) = ρk−k0 ei(k−k )·r
k0

Now perform a translation by a direct lattice vector R = n1 a1 + n2 a2 + n3 a3 :
! 0
X
ρ(r + R) = ρ(r) = ρk−k0 ei(k−k )·(r+R)
k0

In order for the
rst equality to hold, we require that
(k − k0 ) · R = 2π

which is only satis
ed by G = k − k0 . This means that we obtain:
X
ρ(r) = ρG eiG·r
G


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,Toby Adkins B6 Past Papers


as required. Consider Fermi's Golden Rule for the rate of a particular scattering process:
2π 2 dN
Γ= hk| V k0
~ | {z } dEf
scattering potential |{z}
density of
nal states

The scattering amplitude is given by
Z
0 1 0
hk| V k = 3 d3 r e−i(k−k )·r V (r)
L
V (r) must be periodic in the direct lattice, such that it satis
es V (r) = V (r + R). Then:
Z
1 X −i(k−k0 )·R 0
hk| V k0 = 3 e d3 x e−i(k−k )·x V (x)
L
R unit cell
where we have split the sum over the lattice points denoted by R, and the locations of
the scattering centres in the unit cell x, where r = x + R. The sum out the front of the
integral is made up of a series of rapidly oscillating terms that will cancel one another out
to zero unless k − k0 = G. This means that:
Z
hk| V k0 = d3 x e−iG·x V (x)
unit cell
This expression is known as the structure factor. We approximate that the scattering
potential can be written as
X
V (x) v fj δ(x − xj )
j

where the sum is over the atoms j in the unit cell, and fj is the form factor; the Fourier
transform of the scattering potential from a given atom j over all space, parametrising
the strength of the interaction with this scattering centre. Note that we have ignored the
dependence of the form factor on G. Then,
X
Shk` = fj e2πi(hxj +kyj +`zj )
j

where [xj , yj , zj ] are the coordinates of a given atom j within the unit cell.

For a conventional cubic cell, the lattice structure factor is constant. This means that we
can write the structure factor for the basis as:
(basis) π
Shk` = fGa + fAs ei 2 (h+k+`)

For the FCC lattice, we have lattice points located at
[0, 0, 0], [1/2, 1/2, 0], [1/2, 0, 1/2], [0, 1/2, 1/2]

so
(lattice)
Shk` = 1 + (−1)h+k + (−1)h+` + (−1)k+`

This means that the overall structure factor is given by the product
h ih π
i
Shk` = 1 + (−1)h+k + (−1)h+` + (−1)k+` fGa + fAs ei 2 (h+k+`)


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,Toby Adkins B6 Past Papers


according to the convolution theorem. For a re
ection to be allowed, we require that h, k, `
are all even, or all odd.

When we are considering powder diraction, we have to include multiplicity in our consid-
erations of intensity:
Ihk` ∝ Mhk` |Shk` |2

Then:
I111 23 |fGa + fAs (−1)|2
=
I200 2 × 3!/2! |fGa + fAs (−i)|2

We now approximate that fj v Zj , so
I111 8 |31 − 33i|2
= = 512.5
I200 2 × 3 |31 − 33|2

The small intensity of the 200 re
ection is due to the near complete cancellation in the
diraction (fGa v fAs ). For silicon, this would be identically equal to zero.

Question 2
Suppose that the nearest neighbour potential has a minimum at x = x0 . Then, consider
the Taylor series expansion of the potential around this minimum value:
1 ∂2V 1 ∂3V
V (x) ≈ V (x0 ) + (x − x0 )2 + (x − x0 )3 + . . .
2! ∂x2 x=x0 3! ∂x3 x=x0

In order to make the harmonic approximation, we require that the magnitude of the next
order term (the anharmonic term) is much smaller than that of the harmonic term, i.e:
1 ∂2V 1 ∂3V
(x − x0 )2 ≫ (x − x0 )3
2! ∂x2 x=x0 3! ∂x3 x=x0

Let
∂2V
κ=
∂x2 x=x0

Then, we can model the crystal as a series of atoms joined by springs of spring constant κ.




Using NII to write down the equation of motion of atom n:
mδ ẍn = κ(δxn+1 − δxn ) − κ(δxn − δxn−1 ) = κ(δxn+1 + δxn−1 − 2δxn )

We assume phononic (wave-like) solutions of the form:
δxn = Aei(kna−ωt)



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, Toby Adkins B6 Past Papers


Substituting this into the equation of motion:
 
−ω 2 m = κ eika + e−ika − 2
κ
ω 2 = (2 − 2 cos ka)
m  
4κ 2 ka
= sin
m 2

Thus, the dispersion relation for this oscillation is given by:
r  
κ ka
ω=2 sin
m 2

The sound velocity occurs for
r
dω κ
cs = lim vg = lim =a
k→0 k→0 dk m
Sketching the dispersion relation within the
rst Brillouin zone:




We need to
nd the maximum value of V (x), such that we can
nd
∂2V
κ=
∂x2 x=x0

so
∂V 12σ 12 σ6
=0: − + 6 =0
∂x x13 x7
6σ 6 x6 = 12σ 12
→ x0 = 21/6 σ

and
∂2V 12(−13)σ 12 6(−7)σ 6 156σ 12 42σ 6
   
= 4 − + = 4 − 8
∂x2 x14 x8 x14 x

Then:
∂2V
 
4 156 42
κ= = −
∂x2 x=x0 σ2 27/3 24/3


4