Soluciones de problemas para preparar el examen (2014)
Ejercicio InglésUniversidad  Universidad de Barcelona (UB) 
Grado  Física  4º curso 
Asignatura  Fenómenos Colectivos y Transiciones de Fase 
Año del apunte  2014 
Páginas  14 
Fecha de subida  30/06/2014 
Descargas  13 
Descripción
Soluciones de problemas sobre exponentes críticos, aprox. Bragg Williams, modelo de Ising, campo medio, teoría de Landau...
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1
Solutions to problems for Part 3
Sample Quiz Problems
Quiz Problem 1. Draw the phase diagram of the Ising Ferromagnet in an applied magnetic field. Indicate the
critical point. Plot the magnetization as a function of the applied field for three temperatures T < Tc , T = Tc , T > Tc .
Solution. See lecture notes
—————–
Quiz Problem 2. Plot the behavior of the magnetization of the Ising ferromagnet as a function of the temperature,
for three applied field cases: h < 0, h = 0, h > 0. Indicate the critical point.
Solution. See lecture notes
—————–
Quiz Problem 3. Write down the definition of the critical exponents α, βe , γ, δ, η and ν. What values do these
exponents take within mean field theory.
Solution.
CV ∼ t−α ;
m ∼ tβ ;
χ ∼ t−γ ;
m(Tc ) ∼ h1/δ ;
c(r) ∼ e−r/ξ /rd−2+η
(1)
where ξ = t−ν , and t = T − Tc . Within mean field theory α = 0, β = 1/2, γ = 1, δ = 3, η = 0, ν = 1/2
—————–
Quiz Problem 4. Write down the mean field equation for the Ising ferromagnet in an applied field, on a lattice
with coordination number z and exchange constant J. From this equation find the critical exponent δ for the Ising
ferromagnet within mean field theory.
Solution.
1
m = T anh(βJzm + βh) ∼ (βJzm + βh) − (βJzm + βh)3
3
(2)
at the critical point βJz = 1, so m ∼ h1/3 and hence δ = 3.
—————–
Quiz Problem 5. Write down the van der Waals equation of state. Draw the P, v phase diagram of the van der
Waals gas and indicate the critical point.
Solution.
kb T
a
−
v − b v2
—————–
P =
(3)
Quiz Problem 6. Make plots of the van der Waals equation of state isotherms, for T > Tc , T < Tc and for
T = Tc . For the case T < Tc explain why the nonconvex part of the curve cannot occur at equilibrium and the
Maxwell construction to obtain a physical P, v isotherm.
Solution. The Maxwell construction requires that,
P ∗ (vg − vl ) =
vg
P dv
(4)
vl
The is required as the oscillation in the Van der Waals equations occur in a nonconvex region of the Helmholtz free
energy graph of this model. The nonconvex region is not an equilibrium or thermodynamic state as it is possible to
lower the free energy by constructing a tie line below the nonconvex region which corresponds to a mixture of the
two phases at the end points of the tie line.
2
—————–
Quiz Problem 7. Write down the Landau free energy for the Ising and fluidgas phase transitions. Explain the
correspondences between the quantities in the magnetic and classical gas problems.
Solution.
F = a(T − Tc )y 2 + by 4 + cy
(5)
For the Ising model y = m, c = h, for the van der Waals gas, y = vg − vl , c = P .
—————–
Quiz Problem 8. Explain the meaning of second quantization. Discuss the way that it can be used in position
space and in the basis of single particle wavefunctions. Write down the commutation relations for Bose and Fermi
second quantized creation and annihilation operators.
Solution.
Second quantization is a formulation of quantum mechanics and of quantum field theory that is expressed in
terms of creation and annihilation operators. In many body quantum physics creation and annihilation operators
create and destroy particles in many body basis sets constructed from single particle wave functions. In the case
of Fermions a many body basis function is a determinant, while for Bosons it is a permanent. The commutation
relations for Fermions and Bosons are similar, except that for fermions we have anticommutators and for Bosons we
have commutators. In many body quantum mechanics we have,
[ai , a†j ] = δij ;
f or bosons;
and
{ai , a†j } = δij ;
f or bosons;
and
{ψ(x), ψ † (x )} = δ(x − x );
f or f ermions;
(6)
while for quantum fields, we have,
[ψ(x), ψ † (x )] = δ(x − x );
f or f ermions;
(7)
These days, new theories are often formulated using creation and annihilation operators rather than the Heisenberg
or Schr¨
odinger formulations of quantum theory.
—————–
Quiz Problem 9. Write down the Hamiltonian for BCS theory, and the decoupling scheme used to reduce it to a
solvable form. Explain the physical reasoning for the decoupling scheme that is chosen.
Solution.
In the swave BCS theory a singlet state is assumed, so that,
Hpair − µN =
(
kσ
− µ) a† akσ +
kσ
where N =
kσ
Vkl a† a†
a a ,
k↑ −k,↓ −l↓ l↑
kσ
(8)
kl
nkσ is the number of electrons in the Fermi sea. We carry out an expansion in the fluctuations,
a−l↓ al↑ = bl + (a−l↓ al↑ − bl );
a† a†
k↑ −k↓
= b∗k + (a† a†
k↑ −k↓
− b∗k )
(9)
The mean field Hamiltonian keeps only the leading order term in the fluctuations so that,
HM F − µN =
(
kσ
− µ) a† akσ +
Vkl (a† a†
b
k↑ −k,↓ l
kσ
kσ
kl
This is the Hamiltonian that leads to the BSC solution.
—————–
+ b∗k a−l↓ al↑ − b∗k bl )
(10)
3
Quiz Problem 10. Consider the inverse BogoliubovValatin transformation,
γkσ = u∗k akσ − σvk a†
−k−σ
.
(11)
Show that if the operators a, a† obey standard fermion anticommutator relations, then the operators γ, γ † also obey
these relations, provided,
uk 2 + vk 2 = 1
(12)
Solution The anticommutator,
{γkσ , γlα } = {uk∗ akσ − σvk a†
−k−σ
, u∗l alα − αvl a†
−l−α
}.
(13)
Expanding the anticommutator gives,
{γkσ , γlα } = (uk∗ akσ − σvk a†
−k−σ
)(u∗l alα − αvl a†
−l−α
) + (u∗l alα − αvl a†
)(u∗k akσ − σvk a†
)
(14)
} − σvk u∗l {a−k−σ , a† }
(15)
−l−α
−k−σ
which reduces to,
{γkσ , γlα } = u∗k u∗l {akσ , alα } + σαvk vl {a†
−k−σ
, a†
−l−α
} − αu∗k vl {akσ , a†
−l−α
lα
The first two anticommutators are zero. The second two anticommutators are finite when the conditions δ(k, −l)δ(σ, −α)
hold. However when this condition holds, the two commutators are equal and opposite, so they sum to zero. Taking
the adjoint of Eq. (15) shows that {γkσ , γlα } = 0. Now consider,
{γkσ , γ † } = (u∗k akσ − σvk a†
−k−σ
lα
)(ul a† − αvl∗ a−l−α ) + (ul a† − αvl∗ a−l−α )(u∗k akσ − σvk a†
{γkσ , γ † } = u∗k ul {akσ , a† } + σαvk vl∗ {a†
lα
−k−σ
lα
−k−σ
lα
lα
, a−l−α } − αu∗k vl∗ {akσ , a−l−α } − σvk ul {a†
−k−σ
)
, a† } = 0
lα
(16)
(17)
which reduces to
{γkσ , γ † } = δkl uk 2 {akσ , a† } + σαvk 2 {a†
lα
−k−σ
lα
, a−l−α } = δkl δσα (uk 2 + vk 2 ).
(18)
This anticommutator thus requires Eq. (12) in order for the γ operators to obey Fermion anti commutator relations.
—————–
Quiz Problem 11. Given that the energy of quasiparticle excitations from the BCS ground state have the
spectrum,
E = [( −
F)
2
+ ∆2 ]1/2 ,
(19)
where ∆ is the superconducting gap and EF is the Fermi energy, show that the quasiparticle density of states if given
by,
D(E) =
N ( F )E
(E 2 − ∆2 )1/2
(20)
Solution We use the relation,
N(
F )d
= D(E)dE;
and
dE =
− F
d
[( − F )2 + ∆2 ]1/2
(21)
to find,
D(E) =
N(
F )[(
− F )2 + ∆2 ]1/2
− F
(22)
and using,
( −
F)
2
= E 2 − ∆2
(23)
N ( F )E
[E 2 − ∆2 ]1/2
(24)
yields,
D(E) =
4
—————–
Quiz Problem 12. Describe the physical meaning of the superconducting gap, and the way in which BCS theory
describes it.
Solution
The superconducting gap is the energy required to generate a quasiparticle excitation from the superconducting
ground state. In BCS theory, the quasiparticles behave like noninteracting fermions and the energy required to
generate a quasiparticle is at least 2∆(T ).
—————–
Quiz Problem 13. (Omit this question) BCS theory works very well even quite near the superconducting
transition, despite the fact that it is a mean field theory. Use the Ginzburg criterion to rationalize this result.
Solution
—————–
Quiz Problem 14. Given the general solutions to the BCS mean field theory
∆k = −
Vkl
l
∆l
,
2El
Ek = ((
k
− µ)2 + ∆k 2 )1/2
(25)
Describe the assumptions that are made in deducing that,
1=
N(
F )V
2
hωc
F −¯
h
¯ ωc /∆
N(
F )V
0
hωc
F +¯
(( −
F
dx
= N(
(1 + x2 )1/2
d
=
+ ∆2 )1/2
)2
F )V
Sinh−1 (
¯ ωc
h
)
∆
(26)
and hence,
∆ = 2¯
hωc Exp[
−1
]
N ( F )V
(27)
Solution
We assume an isotropic gap, and that the attractive coupling between electrons is constant −V , over the range
hωc < < F + ¯hωc . The density of states is assumed constant with value N ( F ).
F −¯
—————–
Assigned problems
Assigned Problem 1. Consider the Ising ferromagnet in zero field, in the case where the spin can take three values
Si = 0, ±1. a) Find equations for the mean field free energy and magnetization. b) Find the critical temperature and
the behavior near the critical point. Are the critical exponents (βe , γ, α, δ) the same as for the case S = ±1? Is the
critical point at higher or lower temperature than the spin ±1 case? c) Is the free energy for the the spin 0, ±1 case
higher or lower than the free energy of the ±1 case? Why?
Solution. The partition function and Helmholtz free energy are,
Z = [1 + 2Cosh(βJzm)]N ;
F = −kB T N ln[1 + 2Cosh(βJzm)]
(28)
The mean field equation is,
m=
2
1
2Sinh(βJzm)
≈ βJzm − (βJzm)3 + ...
1 + 2Cosh(βJzm)
3
3
(29)
The critical point is at βJz = 3/2, so the critical temperature is at kB Tc = 2Jz/3 which is lower than that for spin
1/2 due to the additional entropy of the spin one system. The critical exponents β and δ are clearly the same as for
the spin 1/2 case. The free energy is lower for the spin 1 case due to the higher entropy.
5
—————–
Assigned Problem 2. (i) Starting from Eq. (36) of the lecture notes, prove Eq. (39) of the lecture notes. (ii)
From Eq. (44) or (45) of the lecture notes prove Eq. (46) in three dimensions.
Solution. To fill in the details from Eq. (36) to (38), substitute the definitions of the Fourier transform and use
translational invariance, as illustrated in the notes. To prove equation (39), note that the matrix Jil is diagonalized
by the Fourier transform, and assuming only nearest neighbors, we then have,
J(k) = eikx x + e−ikx x + eiky y + e−iky y + ....
(30)
which reduces to Eq. (39). The approximate expression is found by expanding for small k.
To show Eq. (46) from Eq. (45), we use the integral,
∞
0
1/2 π
e−ax−b/x
dx = e−2(ab) ( )1/2 .
b
x3/2
(31)
with a = 1/ξ 2 , b = r2 /4, d = 3, we find,
1 −r/ξ
e
r
—————–
C(r) ≈
(32)
Assigned Problem 3. The Dieterici equation of state for a gas is,
P =
kB T −a/(kB T v)
e
v−b
(33)
where v = V /N . Find the critical point and the values of the exponents β, δ, γ.
Solution. The critical point is found by solving,
Pc =
kB Tc −a/(kB T vc )
e
;
vc − b
(34)
kB Tc −a/(kB T vc )
kB Tc
a
∂P
=0=−
e
+
e−a/(kB T vc )
∂v
(vc − b)2
vc − b kB T vc2
(35)
Which simplify to,
−
a
1
+
= 0;
vc − b kB Tc vc2
so
a
vc2
=
kB Tc
vc − b
(36)
The second derivative yields,
∂2P
kB Tc −a/(kB T vc )
kB T
a
=0=2
e
−2
e−a/(kB Tc vc )
∂v 2
(vc − b)3
(vc − b)2 kB Tc vc2
−2
kB Tc
a
kB Tc
a
(
e−a/(kB Tc vc ) +
)2 e−a/(kB Tc vc )
vc − b kB Tc vc3
vc − b kB Tc vc2
(37)
1
a
a
a
1
−2
−2
+(
)2 = 0
(vc − b)2
vc − b kB Tc vc2
kB Tc vc3
kB Tc vc2
(38)
so that,
2
and using Eq. (),
−2
1 1
1
+(
)2 = 0;
vc vc − b
vc − b
so
vc = 2b
(39)
6
so we find that,
vc = 2b;
kB Tc =
a
;
4b
Pc =
a
4e2 b2
(40)
To find the critical exponents we write v = vc + δv, T = Tc + δT, P = Pc + δP , so that,
Pc + δP =
kB (Tc + δT ) −a/[kB (Tc +δT )(vc +δv)]
kB Tc (1 +
e
=
vc + δv − b
b
1+
δT
Tc
δv
b
)
e−a/[kB Tc vc (1+δT /Tc )(1+δv/vc )]
(41)
This reduces to,
1 + p = e2
2
1+t
Exp[−
]
1 + 2x
(1 + t)(1 + x)
(42)
where p = δP/Pc , t = δT /Tc , x = δv/vc . To third order this expansion gives,
2
2
p = 3t + 2t2 − t3 + (−2t − 4t2 )x + 2tx2 − x3
3
3
(43)
Taking a derivative with respect to v at setting x = 0 leads to,
κT = (−V (
∂P
)T )−1 ≈ T − Tc −1
∂V
(44)
so γ = 1. The exponent δ is found by setting t to zero so that p ≈ x3 , so that δ = 3. To find β we assume that
pl = pg , xl = −xg , so that,
2
2
pl = 3t + 2t2 − t3 + (−2t − 4t2 )xl + 2tx2l − x3l
3
3
(45)
2
2
p − g = 3t + 2t2 − t3 − (−2t − 4t2 )xl + 2tx2l + x3l
3
3
(46)
Setting these equations to be equal yields,
2(−2t − 4t2 )xl =
2 3
x ,
3 l
so that
xl ≈ T − Tc 1/2
(47)
where we dropped the t2 term as it is higher order. In the above analysis the signs of the t and x are consistent but
have to be checked each time.
—————–
Assigned Problem 4. Consider the Landau free energy,
F =
1
1
1
am2 + bm4 + cm6
2
4
6
(48)
where c > 0 as required for stability. Sketch the possible behaviors for a, b positive and negative, and show that the
system undergoes a first order transition at some value a, b, c. Find the discontinuity in m at the transition.
Solution.
The mean field equation is,
δF
= am + bm3 + cm4 = 0;
δm
(49)
Note that in general we do a variation with respect to m, so when we add fluctuations later, we need to use the
EulerLagrange equation. Here the variation is the same as a partial derivative with respect to m. Solving the mean
field equation, we find five solutions.
m = 0,
m = ±m± ;
m2± =
−b ± (b2 − 4ac)1/2
2c
(50)
7
Though there are always five solutions, only the real solutions are physical. Analysis of the behavior of the model
reduces to identifying the real solutions, and finding which real solution has the lowest free energy. We can understand
the nature of the solutions by looking at the second derivative,
∂2F
= a + 3bm2 + 5cm4 ,
∂m2
(51)
which enables us to distinguish between maxima and minima. We also use the fact that F is symmetric in m and
that at large m, because c is positive, F is large and positive for large m. Finally, without loss of generality, we can
divide through by c, or equivalently set c = 1. Now consider the four cases for a, b.
(i) a > 0, b > 0. In this case b2 − 4ac < b, so m2± is always negative. Therefore the solutions m± are always
imaginary. The only real solution is m = 0, which is a minimum having F (0) = 0.
(ii) a < 0, b > 0. In this regime, b2 + 4ac > b2 , so m2+ > 0, so that m+ is real. However −b − (b2 + 4ac)1/2
remains negative, so m− remains imaginary. The real solutions are thus a maximum at m = 0 and two symmetric
minima at m2+ = ± 21 ([b2 + 4a]1/2 − b).
There is a phase transition between states (i) and (ii) that occurs at a = 0 where two new solutions emerge and the
extremum at m = 0 changes from a maximum for a < 0 to a minimum for a > 0. The nature of the transition is
found by making a small a expansion of the solutions m+ , which leads to m ≈ a1/2 . This is the Ising/Van der
Waals univesality class we have studied using mean field theory, where we found a ≈ T − Tc 
(iii) a < 0, b < 0. In this regime, b2 + 4ac > b2 , so m+ is real but m− remains imaginary. Therefore, as in case
(ii), there is a maximum at m = 0, and minima at ±m+ .
(iv) a > 0, b < 0. In this regime there are several things going on. First, the discriminant b2 − 4a is negative for
2
b < 2a, so in this regime there is only one real solution, a minimum at m = 0. For b2 > 2a, there are five real
solutions because b ± (b2 − 4a)1/2 > 0. Moreover, we know that the solution at m = 0 is a minimum, so we know
that ±m− are maxima, while ±m+ are minima.
The final issue we have to resolve is the behavior of the minima m+ as a function of a, b, in particular we need to
know if F (m+ ) is greater than or less than F (0). If the lowest free energy state changes it corresponds to a phase
transition. We can solve this problem by evaluating F (m+ ), or we can find solutions where F (m0 ) = 0 and then solve
m0 = m+ . The later leads to,
m20 =
3 b
b2
4a
[ ± ( − )1/2 ]
2 2
4
3
(52)
and setting m20 = m2+ , we find,
3 b
b2
4a
1
[ ± ( − )1/2 ] = [b + (b2 − 4a)1/2 ]
2 2
4
3
2
(53)
which has solution
a
b∗ = 4( )1/2 ;
3
with
a
m2∗ = m20 (b∗ , a) = 2( )1/2
3
(54)
m∗ is the magnetization on the phase boundary defined by b∗ . There is then a first order phase transition from
magnetization m∗ for b < b∗ , to magnetization m = 0 for b > b∗ . The behavior of the magnetization on the
phase boundary is m∗ ≈ a1/4 ≈ T − Tc 1/4 , which is the mean field result for the order parameter near a tricritical
point where a line of second order phase transitions meets a line of first order phase transitions.
—————–
Assigned Problem 5. The BCS pairing Hamiltonian is a simplified model in which only pairs with zero center of
mass momentum are included in the analysis. We also assume that the fermion pairing that leads to superconductivity
occurs in the singlet channel. The BCS Hamiltonian is then,
Hpair − µN =
(
kσ
where N =
kσ
kσ
− µ) a† akσ +
Vkl a† a†
a a ,
k↑ −k,↓ −l↓ l↑
kσ
(55)
kl
nkσ is the number of electrons in the Fermi sea. Defining,
bk =< a−k↓ ak↑ >, and b∗k =< a† a†
k↑ −k↓
>.
(56)
8
carry out a leading order expansion in fluctuations, leading to,
HM F − µN =
(
kσ
− µ) a† akσ +
Vkl (a† a†
b
k↑ −k,↓ l
kσ
kσ
+ b∗k a−l↓ al↑ − b∗k bl )
(57)
kl
This is the Hamiltonian that we will solve to find the thermodynamic behavior of superconductors, using an atomistic
model.
Solution. The meanfield Hamiltonian corresponding to Eq. (38) can be considered as a first order expansion in
the fluctuations i.e.
a−k↓ ak↑ =< a−k↓ ak↑ > +[a−k↓ ak↑ − < a−k↓ ak↑ >]
(58)
Substition of this into (38), along with the definitions (39) lead to HM F .
—————–
Assigned Problem 6. Using the BogoliubovValatin transformation (Eq. 107), show that the mean field BCS
Hamiltonian (Eq. (106)) reduces to Eq. (110), provided Equations (111) and (112) are true.
Solution. With this transformation (ie. plug (6) into (3)), the mean field Hamiltonian looks messy,
HM F − µN =
(
− µ) (a† ak↑ + a† ak↓ ) −
k
k↑
k
(
k
(∆k a† a†
k↑ −k,↓
k↓
+ ∆∗k a−k↓ ak↑ − b∗k ∆k ) =
k
k
− µ) ([u∗k γ † + vk γ−k↓ ][uk γk↑ + vk∗ γ †
−k↓
k↑
−∆k [u∗k γ † + vk γ−k↓ ][u∗k γ †
−k↓
k↑
] + [u∗k γ † − vk γ−k↑ ][uk γk↓ − vk∗ γ †
−k↑
k↓
− vk γk↑ ] − ∆∗k [uk γ−k↓ − vk∗ γ † ][uk γk↑ + vk∗ γ †
−k↓
k↑
])
] + b∗k ∆k
Expanding this yields,
(
k
− µ)[uk∗ uk γ † γk↑ + u∗k vk∗ γ † γ †
k↑ −k↓
k↑
+ vk uk γ−k↓ γk↑ + vk vk∗ γ−k↓ γ †
−k↓
]
k
+(
k
− µ)[u∗k uk γ † γk↓ − u∗k vk∗ γ † γ †
k↓
−∆k [u∗k u∗k γ † γ †
k↑ −k↓
k↓ −k↑
− vk uk γ−k↑ γk↓ + vk vk∗ γ−k↑ γ †
−k↑
− u∗k vk γ † γk↑ + vk u∗k γ−k↓ γ †
−k↓
k↑
−∆∗k [uk uk γ−k↓ γk↑ + uk vk∗ γ−k↓ γ †
−k↓
]
− vk vk γ−k↓ γk↑ ]
− vk∗ uk γ † γk↑ − vk∗ vk∗ γ † γ †
k↑ −k↓
k↑
] + b∗k ∆k
We now collect the terms in this expression into three catagories: Those which have no operators in them; those which
can be reduced to diagonal form ie. those which are of the form γk† γk and; those that are off diagonal (e.g. γ † γ † ).
The first stage is to collect together the terms which look to be in these three catagories. First the constant term,
b∗k ∆k
HM F − µN =
k
The following terms can be converted to diagonal form,
+(
k
− µ) (uk 2 γ † γk↑ + vk 2 γ−k↓ γ †
−k↓
k↑
−∆k [−u∗k vk γ † γk↑ + vk u∗k γ−k↓ γ †
k↑
−k↓
+ uk 2 γ † γk↓ + vk 2 γ−k↑ γ †
−k↑
k↓
] − ∆∗k [uk vk∗ γ−k↓ γ †
−k↓
)
− vk∗ uk γ † γk↑ ]
k↑
9
Finally the offdiagonal terms are,
+(
k
− µ) (u∗k vk∗ γ † γ †
k↑ −k↓
−∆k [(u∗k )2 γ † γ †
k↑ −k↓
− u∗k vk∗ γ † γ †
k↓ −k↑
+ vk uk γ−k↓ γk↑ − vk uk γ−k↑ γk↓ )
− vk2 γ−k↓ γk↑ ] − ∆∗k [u2k γ−k↓ γk↑ − (vk∗ )2 γ † γ †
k↑ −k↓
]
We have to rearrange the terms catagorized as diagonal above, as we need them in the form γ † γ. We do this using
the commutation relation, i.e. γγ † = 1 − γ † γ. This yields,
b∗k ∆k + 2(
HM F − µN =
k
− µ)vk 2 − ∆k u∗k vk − ∆∗k uk vk∗
k
+(
k
− µ) (uk 2 γ † γk↑ − vk 2 γ †
γ
−k↓ −k↓
k↑
+∆k [u∗k vk γ † γk↑ + vk u∗k γ †
γ
]
−k↓ −k↓
k↑
+ uk 2 γ † γk↓ − vk 2 γ †
γ
)
−k↑ −k↑
k↓
+ ∆∗k [uk vk∗ γ †
γ
−k↓ −k↓
+ vk∗ uk γ † γk↑ ]
k↑
Finally the offdiagonal terms are (as before),
+(
k
− µ) (u∗k vk∗ γ † γ †
k↑ −k↓
−∆k [(u∗k )2 γ † γ †
k↑ −k↓
− u∗k vk∗ γ † γ †
k↓ −k↑
+ vk uk γ−k↓ γk↑ − vk uk γ−k↑ γk↓ )
− vk2 γ−k↓ γk↑ ] − ∆∗k [u2k γ−k↓ γk↑ − (vk∗ )2 γ † γ †
k↑ −k↓
]
We note that nkσ = n−kσ , and that the expectation of other terms that transform into one another through the
transformation k → −k, are equivalent. Collecting terms then leads to Eqs. (110)(112) of the lecture notes.
—————–
Assigned Problem 7. We define
gk
(1 + gk 2 )1/2
(59)
− µ)(1 − vk 2 )1/2 vk + ∆k vk2 − ∆∗k (1 − vk 2 ) = 0,
(60)
vk =
show that Eq. (111) reduces to (117).
Solution Starting with,
2(
k
We have,
uk 2 = 1 − vk 2 =
1
;
1 + gk 2
uk =
1
.
(1 + gk 2 )1/2
(61)
Substitution into Eq. () leads to,
2(
k
− µ)gk + ∆k gk2 − ∆∗k = 0.
(62)
—————–
Assigned Problem 8. Show that Ek as defined in Eq. (118) is in agreement with Eq. (112).
Solution We need to show that the definitions,
Ek = (
k
− µ)(uk 2 − vk 2 ) + ∆k u∗k vk + ∆∗k vk∗ uk .
(63)
− µ)2 + ∆k 2 ]1/2
(64)
and
Ek = [(
k
are the same. To do this we use the results of Problem 9 in Eq. (53) to write,
Ek = (
k − µ)(
∆∗k
[Ek − ( k − µ)]2
∆
[Ek + ( k − µ)]2
−
)
+
∆
+ ∆∗k k .
2
2
k
4E
4E
2Ek
2Ek
k
which reduces to Eq. (54) as required.
k
(65)
10
—————–
Assigned Problem 9. Prove the relations Eq. (119121).
Solution We have,
gk =
Ek − ( k − µ)
;
∆k
Ek = [(
k
− µ)2 + ∆k 2 ]1/2
(66)
so that,
∆k 2 = Ek2 − (
k
− µ)2 ;
(67)
so that,
gk 2 =
Ek − ( k − µ)
[E − (
= k2
∆k
E −(
k
E −(
− µ)]2
= k
2
Ek + (
k − µ)
k
− µ)
k − µ)
(68)
Ek + ( k − µ)
2Ek
(69)
k
We then find,
vk 2 =
E − ( k − µ)
gk 2
= k
;
1 + gk 2
2Ek
uk 2 =
and hence,
uk vk∗ =
∆∗k
gk
2
=
g
u

=
k k
1 + gk 2
2Ek
(70)
(70)
—————–
Assigned Problem 10. Prove the relation (135).
Solution We start with,
bk =< a−k↓ ak↑ >=< (uk γ−k↓ − vk∗ γ † )(uk γk↑ + vk∗ γ †
k↑
−k↓
)>
(71)
To evaluate this expression at finite temperature, we must evaluate,
ˆ
< O >=
ˆ −β H )
tr(Oe
tr(e−β Hˆ )
(72)
The only terms that survive are the terms that are diagonal, such as nkσ . We thus find,
bk =< a−k↓ ak↑ >=< uk vk∗ γ−k↓ γ †
−k↓
− uk vk∗ γ † γk↑ >= uk vk∗ [1− < nk↑ > − < n−k↓ >]
k↑
(73)
Finally we use the fact that < nkσ >=< n−kσ > to find Eq. (135).
—————–
Assigned Problem 11. Starting from Eq. (137), prove the relation (145).
Solution In the case of the weakcoupling swave model Eq. (137) reduces to,
1
N(
F )V
h
¯ ωc
=
d
0
T anh( β2 ( 2 + ∆2 )1/2 )
( 2 + ∆2 )1/2
(74)
11
As shown in lectures, the critical point is found from the equation,
∞
1
N(
= ln(βc ¯
hωc /2)T anh(∞) + a1 ;
F )V
ln(x)Sech2 (x)dx = 0.8188.
a1 = −
where,
(75)
0
The critical temperature is then,
kB Tc = 1.13¯hωc Exp[−
1
N(
F )V
]
(76)
To find the behavior of the gap near Tc , we carry out a first order Taylor expansion of Eq. (64) using ∆2 as the small
quantity. This leads to,
2
T anh( β2 ( 2 + ∆2 )1/2 )
T anh( β2 (1 + 2∆ 2 ))
T anh( β2 ) β∆2 Sech2 ( β2 ) ∆2 T anh( β2 )
−
≈
≈
+
2
2
3
4
2
( 2 + ∆2 )1/2
(1 + 2∆ 2 )
(77)
Taking the gap to be real, we find,
1
N(
F )V
h
¯ ωc
=
d
T anh( β2 )
h
¯ ωc
+ ∆2
d [
0
0
β
β
β Sech2 ( 2 ) 1 T anh( 2 )
−
]
2
3
4
2
(78)
This expression may be written in the form,
1
β2
= ln( β¯hωc ) + a1 + ∆2 a2
N ( F )V
2
8
1
(79)
where,
∞
a1 = .8188 ;
and a2 =
dx[
0
Sech2 (x) T anh(x)
−
] = −0.853
x2
x3
(80)
We expand β about βc , but we only need to consider carry out this expansion in the first term on the right hand side,
so that,
1
N(
F )V
= ln[
¯ ωc
h
β2
] + a1 + ∆2 a2 ;
2kB Tc (1 − t)
8
where
t = (Tc − T )/Tc
(81)
We then find that,
ln(1 − t) ≈ −t = −∆2
β2
a2 ;
8
∆(T )
8 1/2 1/2
) t
≈(
kB Tc
a2 
(82)
T → Tc
(83)
so that
so that,
T
∆(T )
≈ 3.06(1 − )1/2 ;
kB Tc
Tc
This is correct near Tc . Also note that Eq. (127) and (141) immply that,
2∆(0)
4
≈
≈ 3.52,
kB Tc
1.13
(84)
—————–
Assigned Problem 12. Consider a ferromagnetic nearest neighbor, spin 1/2, square lattice Ising model where the
interactions along the xaxis, Jx , are different than those along the yaxis, Jy . Extend the low and high temperature
expansions Eq. (150) and Eq. (152) to this case. Does duality still hold? From your expansions, find the internal
energy and the specific heat.
Solution
eKij Si Sj = (Cosh(Kx ))N (Cosh(Ky ))N
Z=
{Si =±1} <ij>
(1 + tij Si Sj ).
{Si =±1} <ij>
(85)
12
tij = tx for horizontal bonds and tij = ty for vertical bonds, where tx = tanh(Kx ), ty = tanh(Ky ). The diagrams
used are similar, but now we have to treat subclasses with different numbers of hirizontal and vertical bonds, so that,
−βF = ln(Z) =
9
zN
ln(Cosh(K)) + N ln(2) + ln[1 + N t4 + 2N t6 + N (N + )t8 + 0(t10 )]
2
2
(86)
becomes
5
−βF = N [ln(2) + ln(Cosh(Kx )) + ln(Cosh(Kx )) + t2x t2y + t2x t2y (t2x + t2y ) + (t4x t4y ) + t2x t6y + t6x t2y + ....]
2
(87)
Similarly, the extension of the low temperature expansion,
Z=
{Si =±1}
9
eKSi Sj = eKzN [1 + N s4 + 2N s6 + N (N + )s8 + O(s10 )]
2
<ij>
(88)
to the anisotropic case leads to,
5
−βF = N [Kx + Ky + s2x s2y + s2x s2y (s2x + s2y ) + s4x s4y + s2x s6y + s6x s2y ...]
2
(89)
where sx = e−2Kx , sy = e−2Ky . Duality holds for both x and y directions.
—————–
Assigned Problem 13. Find the second virial coefficient for six cases: (i) the classical hard sphere gas; (ii)
Noninteracting Fermions; (iii) Noninteracting Bosons; (iv) The van der Waals gas.
Solution
Pv
=
kB T
∞
al (T )(
l=1
λ3 l−1
)
v
(90)
where the first virial coefficient a1 (T ) = 1, and the second virial coefficient is a2 (T ). The virial expansion is most
often carried out in the grand canonical ensemble, where we may write,
P
1
= 3
kB T
λ
∞
N
1
= 3
V
λ
bl z l ;
l=1
lbl z l
(91)
l=1
so that,
Pv
=
kB T
∞
l
l=1 bl z
∞
l
l=1 lbl z
∞
al (T )(
=
l=1
λ3 l−1
)
v
(92)
which gives relations between the quantities al (T ) and bl (T ). For the second virial coefficient the relationship is
a2 (T ) = −b2 (T ). We have already calculated the coefficients bl for the ideal Bose and Fermi gases, with the results,
bl =
(−1)l+1
;
l5/2
(Ideal F ermi)
bl =
1
l5/2
;
(Ideal Bose)
(93)
so we have,
1 λ3
Pv
= 1 − 5/2 ( ) + ....;
kB T
v
2
(Ideal Bose)
(94)
(Ideal F ermi)
(95)
and
1 λ3
Pv
= 1 + 5/2 ( ) + ....;
kB T
v
2
The van der Waals equation of state may be expanded to find the second virial coefficient,
P =
kB T
a2
− ,
v−b v
so that
Pv
1
a
b − a/kB T
=
−
≈1+
+ ...
kB T
(1 − b/v) kB T v
v
(96)
13
For the classical interacting gas, b2 is given by,
b2 =
1
2V λ3
dr1 d3 r2 [e−βu(r1 −r2 − 1] =
1
2λ3
d3 r[e−βu(r) − 1]
(97)
For the hard sphere problem, with a hard core radius of R, we then find that b2 = −2πR3 /(3λ3 ), so the virial
expansion for this case is,
Pv
2πR3
=1+
+ ....
kB T
3v
(98)
The behavior of the second virial coefficient as a function of temperature can be used to deduce the interaction
potential, and the importance of quantum effects as they have different temperature dependences.
—————–
PHY831  Quiz 5, Friday November 11, 2011
Answer all questions. Time for quiz  25 minutes
Name:
1. (i) Plot the behavior of the magnetization of the Ising ferromagnet as a function of the temperature, for three
applied field cases: h < 0, h = 0, h > 0. Indicate the critical point.
(ii) Write down the definition of the critical exponents α, β, γ, δ, η and ν for the Ising ferromagnet critical point.
What values do these exponents take within mean field theory?
(iii) Write down (or derive) the mean field equation for the spin 1/2 Ising ferromagnet in an applied field, on a
lattice with coordination number z and exchange constant J. From this equation find the critical exponent δ for the
Ising ferromagnet within mean field theory.
2. (i) Write down (or derive) the van der Waals equation of state. Make plots of the van der Waals equation of
state isotherms, for T > Tc , T < Tc and for T = Tc .
(ii) For the case T < Tc explain why the nonconvex part of the curve cannot occur at equilibrium and show
the Maxwell construction to obtain a physical P, v isotherm. Write the mathematical statement of the Maxwell
construction.
(iii) Write down the definition of the critical exponents α, β, γ, δ, η and ν for the liquidgas critical point. What
values do these exponents take within Van der Waals theory.
3. Write down the Hamiltonian for BCS theory, and the decoupling scheme used to reduce it to a solvable form.
Explain the physical reasoning for the decoupling scheme that is chosen.
—————–
PHY831  Quiz 6, Friday November 11, 2011
Answer all questions. Time for quiz  25 minutes
Name:
1. (i) Describe the physical meaning of the superconducting gap, and the way in which BCS theory describes it.
(ii) Given that the energy of quasiparticle excitations from the BCS ground state have the spectrum,
E = [( −
where ∆ is the superconducting gap and
by,
F
2
F)
+ ∆2 ]1/2 ,
(99)
is the Fermi energy, show that the quasiparticle density of states is given
D(E) =
N ( F )E
(E 2 − ∆2 )1/2
(100)
Draw this density of states and compare it to the density of states of the system in the absence of the pairing term
in the Hamilonian.
14
2. (i) Explain the importance of “linkedcluster” theorems in the perturbation theory of many particle systems.
(ii) Draw the low temperature series expansion diagrams to order s8 (where s = Exp[−2βJ]) for the square lattice,
nearest neighbor, spin half Ising ferromagnet partition function. What is the degeneracy of each of these diagrams?
Use these terms to write down an expansion for the Helmholtz free energy and give a physical reason why only the
terms of order N are kept.
3. (i) Write down the mathematical form of the virial expansion for many particle systems and explain why it is
important. What physical properties can be extracted from the second virial coefficient?
(ii) Given that,
bl =
1
(sum over all l − connected − cluster integrals)
l!λ3l−3 V
(101)
find the second virial coefficient for the classical gas with hard core repulsive interactions, where the hard core radius
is R.
...