University of Kentucky
Department of Physics and Astronomy
PHY 525. Introduction to Solid State Physics II
Final Examination.
Date: Dec 12, 2001
Time: 8:00-10:00
Answer all questions.
1.
(25 points)
Consider a two dimensional square lattice of lattice parameter a. Each site
provides two conducting electrons.
(a)
Determine kF in terms of a. Under free electron model, at what value of ky
will the Fermi sphere cross the boundary at x=π/a?
(b) The electrons are actually only nearly free and a gap of 2U=0.1EF opens up at
the Brillouin zone boundary. At what value of ky will the Fermi surface in the
first Brillouin zone cross the boundary at x=π/a? How about the second
Brillouin zone?
You can express your answers in unit of π/a.
Solutions:
(a ) 2 ×
πk F
2
⎛ 2π ⎞
⎜ ⎟
⎝ L⎠
2
2
k
N
2
⇒ F = 2 = 2
2π L
a
=N
4π
a2
4π 3.54
π
⇒ kF =
=
= 1.128
a
a
a
⇒ kF =
2
π⎞
⎛π⎞
⎛
2
2
The sphere will cross the zone boundary at a value of k y so that ⎜ ⎟ + k y = k F = ⎜1.128 ⎟
a⎠
⎝a⎠
⎝
2
⎛π⎞
⇒ k y = 0.2732⎜ ⎟
⎝a⎠
π
⇒ k y = 0.523
a
2
2
h2k F
h2 ⎛
π⎞
=
⎜1.128 ⎟
2m
2m ⎝
a⎠
If the Fermi surface in the first Brillouin zone cross the zone boundary at k y
2
2
(b) The Fermi energy is
2
2
2
2
2
⎤ h2k F2
h2k F
h 2 ⎡⎛ π ⎞
h2 ⎛
π⎞
⎛ 0 .1 ⎞ h k F
2
∴
+ U = ⎜1 +
= 1.05
= 1.05 ×
⎟
⎜1.128 ⎟
⎢⎜ ⎟ + k y ⎥ =
2m ⎣⎢⎝ a ⎠
2m
2 ⎠ 2m
2m
2m ⎝
a⎠
⎝
⎦⎥
π⎞
⎛π⎞
⎛
⎛π⎞
2
⇒ ⎜ ⎟ + k y = 1.05 × ⎜1.128 ⎟ = 1.336⎜ ⎟
a⎠
⎝
⎝a⎠
⎝a⎠
2
⇒ ky
2
⎛π⎞
= 0.336⎜ ⎟
⎝a⎠
2
2
⇒ k y = 0.580
2
π
a
If the Fermi surface in the second Brillouin zone cross the zone boundary at k y
2
2
2
2
2
⎤ h2k F2
h 2k F
h 2 ⎡⎛ π ⎞
h2 ⎛
π⎞
⎛ 0 .1 ⎞ h k F
2
∴
− U = ⎜1 −
= 0.95
= 0.95 ×
⎟
⎜1.128 ⎟
⎢⎜ ⎟ + k y ⎥ =
2m ⎢⎣⎝ a ⎠
2m
2 ⎠ 2m
2m
2m ⎝
a⎠
⎝
⎥⎦
π⎞
⎛π⎞
⎛
⎛π⎞
2
⇒ ⎜ ⎟ + k y = 0.95 × ⎜1.128 ⎟ = 1.2088⎜ ⎟
a⎠
⎝a⎠
⎝
⎝a⎠
2
⇒ ky
2
⎛π⎞
= 0.2088⎜ ⎟
⎝a⎠
2
2
⇒ k y = 0.457
π
a
2
2
2.
(25 points)
(a)
Apply the Hund rules to find the ground state of an ion that has an outer
shell of 3d3. Write your answer in atomic notation.
(b)
Let the magnetic moment of the above ion be µ. Find the magnetization
as a function of magnetic field and temperature for a system formed by
these ions with a concentration of n.
(c)
Find the magnetization in the limit of µB<<kBT.
Solutions:
(a)
Lz:
2
1
0
S = 3/2
L=3
J=|L-S| =3/2
The ground state is 4F3/2.
(b)
-1
-2
If S = 1, S z = −1, 0, or + 1. In a magnetic field B, the energy level of these three states are :
⎧U +3 / 2 = − 32 µB
⎪
= − 12 µB
v v
⎪U
U = - µ ⋅ B = S z gµ B B = ⎨ +1 / 2
1
⎪ U −1 / 2 = 2 µB
⎪ U −3 / 2 = 3 µB
2
⎩
The relative population will be :
N +1/2
N +1/2
exp(-µB / 2k B T )
=
=
N
N +3/2 + N +1 / 2 + N −1 / 2 + N −3/2
N
=
exp(-µB / 2k B T )
N
Similarly,
N +3 / 2
exp(-3µB / 2k B T )
=
N
N
N −1/2 exp(µB / 2k B T)
=
N
N
N −3 / 2
exp(3µB / 2k B T )
=
N
N
3
1
( N -3/2 + 2 N -1/2 − 12 N1 / 2 − 32 N -3/2 )µ
∴M = 2
V
⎡ 3 exp(3µB / 2k B T) 1 exp(µB / 2k B T ) 1 exp(-µB / k B T ) 3 exp(-3µB / 2k B T ) ⎤ Nµ
=⎢
+
−
−
⎥ V
N
2
N
N
2
N
2
⎣2
⎦
⎡ 3 (exp(3µB / 2k B T) − exp(-3µB / 2k B T ) ) + 12 (exp(µB / 2k B T) − exp(-µB / 2k B T) ) ⎤
=⎢2
⎥ nµ
⎢⎣ exp(3µB / 2k B T ) + exp(µB / 2k B T ) + exp(-3µB / 2k B T) + exp(-µB / 2k B T ) ⎥⎦
⎡ 3sinh(3µB / 2k B T ) + sinh(µB / 2k B T) ⎤
=⎢
⎥ nµ
2
cosh
(3
B
/
2
k
T
)
2
cosh
(
B
/
2
k
T
)
µ
+
µ
B
B
⎣
⎦
(c)
In the limit µB >> k B T, sinh(µB / k B T ) →
3
∴ M→
µB
, and cosh (µB / k B T) → 1
k BT
3µB
µB
+
2k B T 2k B T
5nµ 2
nµ =
B
2+2
4k B T
3.
(25 points)
For an fcc lattice of magnetic spins it is impossible to find an antiferromagnetic
arrangement in which all the nearest neighbors of any spin are antiparallel to it.
The best that can be achieved, for example by having spins in alternate (2 0 0)
planes ↑ and ↓, is eight antiparallel and four parallel neighbors. (Miller indices
are referred to the conventional cubic unit cell).
(a)
Develop a Néel theory appropriate to the case of two sublattices when
only nearest neighbor exchange interactions are important; the effective field
acting on an ion on the A sublattice would then be BA=Ba-µMB-εMA with a
similar expression for BB. Show that the high temperature susceptibility is of the
form χ=C/(T+θ), where θ is related to the Néel temperature TN by
θ
µ+ε
=
.
TN µ − ε
(b)
If the same type of stoms occupy the two sublattices. Assume a particular
site in sublattice A has np parallel nearest neighbors in the same sublattice and na
antiparallel nearest neighbors in sublattice B. How will ε and µ depends on na
and np? Show that θ/ TN =3 for the fcc structure mentioned above.
.
Solutions:
(a)
⇒
⇒
⇒
⇒
⇒
⎧BA = Ba - µM B - εM A
⎨B = B - µM - εM
a
A
B
⎩ B
⎧ MAT
⎪ C = Ba - µM B - εM A
⎨ M T
B
⎪
= Ba - µM A - εM B
⎩ C
⎧ ⎛T
⎞
⎪⎪ ⎜⎝ C + ε⎟⎠ M A + µM B = Ba
⎨
T
⎪ µM A + ⎛⎜ + ε⎞⎟ = Ba
⎝C
⎠
⎪⎩
(Curie law ⇒ χ =
C
M
;B = )
χ
T
- - - (1)
⎧ ⎛T
⎞
⎪⎪ ⎜⎝ C + ε⎟⎠ M A + µM B = Ba
⎨
T
⎪ µM A + ⎛⎜ + ε⎞⎟ M B = Ba
⎝C
⎠
⎪⎩
2
⎤
⎡⎛ T
⎛T
⎞
⎞
⎜ + ε⎟ M A − µ 2 M A = ⎢⎜ + ε⎟ − µ ⎥ Ba
⎝C ⎠
⎠
⎣⎝ C
⎦
⎞
⎛T
⎜ + ε⎟ − µ
⎝C
⎠
1
=
MA =
B
BA
A
2
T
⎛
⎞
⎛T
⎞
⎜ + ε⎟ + µ
⎜ + ε⎟ − µ 2
⎝C
⎠
⎝C
⎠
M A and M B can be exchanged in the original simultaneous equatios, ∴ we expect
the same solution for M B :
1
BA
MB =
⎛T
⎞
⎜ + ε⎟ + µ
⎝C
⎠
2
∴ MA + MB =
BA
⎛T
⎞
⎜ + ε⎟ + µ
⎝C
⎠
MA + MB
2
2C
⇒ χ =
χ =
=
Ba
T + C( ε + µ )
⎛T
⎞
⎜ + ε⎟ + µ
⎝C ⎠
χ diverges at T = -θ
⇒ (-θ) + C(ε + µ ) = 0
⇒ θ = C( ε + µ )
- - - (2)
To evaluate C in terms of TN , we note that at T = TN , Ba can be neglected.
⎧⎛ T
⎞
⎪⎪⎜⎝ C + ε⎟⎠ M A + µM B = 0
∴ (1)
⇒ ⎨
T
⎪ µM A + ⎛⎜ ε + ⎞⎟ M B = 0
⎝
⎪⎩
C⎠
Non − trivial solution at TN ⇒
⎛ TN
⎞
+ ε⎟
⎜
⎝ C
⎠
µ
µ
⎛ TN
⎞
+ ε⎟
⎜
⎝ C
⎠
⎛ TN
⎞
+ ε⎟ = µ
⇒ ⎜
⎠
⎝ C
⇒ TN = C(µ - ε )
TN
⇒ C =
µ-ε
=0
- - - (3)
Substitute (3) into (2):
θ =
TN
(µ + ε)
µ-ε
⇒
θ
µ+ε
=
TN
µ-ε
(b)
If the same type of atoms occupy both sublattice, than the coupling must be
proportional to the number of nearest neighbors, i.e. n p / n a = ε / µ
As described in the question, for fcc structure, n p / n a = 4 / 8 = 1 / 2 = ε / µ
∴
θ
µ + ε 1 + ε / µ 1 + 1/ 2 3 / 2
=
=
=
= 3 ⇒ θ = 3TN
=
TN µ − ε 1 − ε / µ 1 − 1 / 2 1 / 2
4.
(25 points)
The penetration equation of a superconductivity may be written as λ2∇2B=B,
where λ is the penetration depth. (a) Show that B(x) inside a superconducting
plate perpendicular to the x axis and of thickness δ is given by
cosh( x / λ )
,
cosh(δ / 2λ)
Where Ba is the field outside the plate and parallel to it; here x=0 is at the center
of the plate. (b) The effective magnetization M(x) in the plate is defined by B(x)Ba = 4πM(x). Show that, in CGS, 4πM(x)= - Ba(1/8λ2 ) (δ2-4 x2), for δ<<λ. In SI
we replace the 4π by µ0.
B( x ) = B a
Solutions
(a)
2
2
λ ∇ B = B ⇒
∂2 B
1
B = 0
∂x 2
λ2
⇒ B = Ae x / λ + Ce -x / λ
At x = δ / 2 and x = - δ / 2, B = Ba
∴ Ae δ / 2 λ + Ce - δ / 2 λ = Ba
- - - (1)
− δ / 2λ
δ / 2λ
Ae
+ Ce
= Ba
- - - (2)
- δ / 2λ
-δ / λ
(1) ⇒ A = Ba e
- Ce
Substitute this into (2),
(Ba e - δ / 2 λ - Ce - δ / λ )e − δ / 2λ + Ce δ / 2 λ = Ba
⇒ C(e δ / 2 λ − e 3δ / 2 λ ) = Ba (1 − e − δ / λ )
⇒ C(1 − e − δ / λ )(e δ / 2 λ + e - δ / 2 λ ) = Ba (1 − e − δ / λ )
Ba
2Ba
⇒ C = δ / 2λ
=
δ
e
+ e - δ / 2λ
cosh 2 λ
∴ A = Ba e
- δ / 2λ
- Ce
-δ / λ
= Ba e
= Ba
=
=
∴ B(x) = Ae x / λ + Ce -x / λ =
- δ / 2λ
1+ e
-
-δ / λ
e -δ / λ B a
e δ / 2λ + e - δ / 2λ
- e -δ / λ
e δ / 2λ + e - δ / 2λ
Ba
e δ / 2 λ + e - δ / 2λ
Ba
δ
2 cosh 2 λ
Ba
ex/λ +
δ
δ
2 cosh 2 λ
−x/λ
2 cosh 2 λ
Ba
(e x / λ + e
=
δ
2 cosh 2 λ
x
= Ba
cosh λ
δ
cosh 2 λ
Ba
)
e− x/ λ
(b)
x
4 πM(x) = B(x) - Ba = Ba
cosh λ
δ
cosh 2 λ
− Ba
⎛ cosh x
⎞
λ
⎜
= Ba ⎜
− 1⎟⎟
δ
⎝ cosh 2 λ ⎠
⎛
⎞
1 x 2
⎜ 1+ 2 λ
⎟
1
≈ Ba ⎜
−
⎟
2
⎜1+ 1 δ
⎟
⎝
⎠
2 2λ
()
( )
()
(δ << λ )
( )
⎛
1 x 2 ⎡
1 δ 2 ⎤⎞
⎜ 1 + 2 λ − ⎢1 + 2 2 λ ⎥ ⎟
⎣
⎦⎟
= Ba ⎜
2
⎜
⎟
δ
1 + 2λ
⎜
⎟
⎝
⎠
2
2
⎡1 x
⎤
1 δ
≈ Ba ⎢ 2 λ − 2 2 λ ⎥
⎣
⎦
Ba 2
δ − 4x 2
= 2
8λ
()
(
( )
( )
)