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Solutions to problems for Part
Sample Quiz Problems
Quiz Problem . Draw the phase diagram of the Ising Ferromagnet in an applied magnetic eld.
Indicate the critical point. Plot the magnetization as a function of the applied eld for three
temperatures T lt Tc , T Tc , T gt Tc . Solution. See lecture notes Quiz Problem . Plot the
behavior of the magnetization of the Ising ferromagnet as a function of the temperature, for
three applied eld cases h lt , h , h gt . Indicate the critical point. Solution. See lecture notes
Quiz Problem . Write down the denition of the critical exponents , e , , , and . What values do
these exponents take within mean eld theory. Solution. CV t m t t mTc h/ cr er/ /rd
where t , and t T Tc . Within mean eld theory , /, , , , / Quiz Problem . Write down the mean
eld equation for the Ising ferromagnet in an applied eld, on a lattice with coordination number
z and exchange constant J. From this equation nd the critical exponent for the Ising
ferromagnet within mean eld theory. Solution. m T anhJzm h Jzm h Jzm h at the critical point
Jz , so m h/ and hence . Quiz Problem . 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. P kb T a v b v
Quiz Problem . Make plots of the van der Waals equation of state isotherms, for T gt Tc , T lt
Tc and for T Tc . For the case T lt 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
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.
Solution. and the decoupling scheme used to reduce it to a solvable form. Explain the
correspondences between the quantities in the magnetic and classical gas problems. c P . x
x x . and ai . c h. Write down the Landau free energy for the Ising and uidgas phase
transitions. Hpair N k k a ak k kl Vkl a a a a . We carry out an expansion in the uctuations. . x.
Write down the commutation relations for Bose and Fermi second quantized creation and
annihilation operators. y vg vl . Quiz Problem . Solution. for the van der Waals gas. al al bl al
al bl . F aT Tc y by cy For the Ising model y m. HM F N k k a ak k kl Vkl a a b k k. and x. we
have. Quiz Problem . j while for quantum elds. new theories are often formulated using
creation and annihilation operators rather than the Heisenberg or Schrdinger formulations of
quantum theory. j f or f ermions. Explain the physical reasoning for the decoupling scheme
that is chosen. f or bosons. while for Bosons it is a permanent. l b al al b bl k k This is the
Hamiltonian that leads to the BSC solution. These days. In the swave BCS theory a singlet
state is assumed. Explain the meaning of second quantization. a ij . a a k k b a a k k k b k
The mean eld Hamiltonian keeps only the leading order term in the uctuations so that. f or
bosons. Second quantization is a formulation of quantum mechanics and of quantum eld
theory that is expressed in terms of creation and annihilation operators. a ij . except that for
fermions we have anticommutators and for Bosons we have commutators. Write down the
Hamiltonian for BCS theory. x x x . k k. ai . The commutation relations for Fermions and
Bosons are similar. o Quiz Problem . Discuss the way that it can be used in position space
and in the basis of single particle wavefunctions. In many body quantum physics creation
and annihilation operators create and destroy particles in many body basis sets constructed
from single particle wave functions. Solution. so that. l l where N k nk is the number of
electrons in the Fermi sea. In the case of Fermions a many body basis function is a
determinant. f or f ermions. In many body quantum mechanics we have.
al vk vl a k l k l u vl ak . The second two anticommutators are nite when the conditions k. DE
Solution We use the relation. hold. a obey standard fermion anticommutator relations. show
that the quasiparticle density of states if given by. Taking the adjoint of Eq. l uk ak vk a k . DE
and using. al kl uk vk . k . l. so they sum to zero. DE N F E E / F F d N F E E / F d F / DEdE. l
uk ak vk a k u al vl a l . E F / . kl uk ak . k u ak vk a k k . Consider the inverse
BogoliubovValatin transformation. However when this condition holds. u al vl a l l . Show that
if the operators a. a l u al vl a l l u ak vk a k k which reduces to. Given that the energy of
quasiparticle excitations from the BCS ground state have the spectrum. N to nd. a l which
reduces to k . in order for the operators to obey Fermion anti commutator relations. shows
that k . also obey these relations. u ul ak . k . a k l vk u ak . al vk ul a k k k . yields. l u u ak . k
. Now consider. l . a vk vl a k l l k k . uk vk Solution The anticommutator. and dE N F F / F E .
This anticommutator thus requires Eq. Quiz Problem . the two commutators are equal and
opposite. then the operators . k . where is the superconducting gap and EF is the Fermi
energy. Quiz Problem . al u vl ak . a l l The rst two anticommutators are zero. a vk a l l k .
Expanding the anticommutator gives. u ak vk a k l k ul a vl al ul a vl al u ak vk a k l l .
provided.
N F V h F c h F c hc / F d / Sinh h c N and hence. . and the way in which BCS theory
describes it. m SinhJzm Jzm Jzm . Assigned problems Assigned Problem . despite the fact
that it is a mean eld theory. The density of states is assumed constant with value N F . The
critical exponents and are clearly the same as for the spin / case. CoshJzm F kB T N ln
CoshJzm The critical point is at Jz /. Omit this question BCS theory works very well even
quite near the superconducting transition. so the critical temperature is at kB Tc Jz/ which is
lower than that for spin / due to the additional entropy of the spin one system. case higher or
lower than the free energy of the case Why Solution. and that the attractive coupling between
electrons is constant V . over the range h h F c lt lt F c . Z CoshJzmN . In BCS theory. Are
the critical exponents e . the same as for the case S Is the critical point at higher or lower
temperature than the spin case c Is the free energy for the the spin ... F V dx N x / F V c Exp
h N F V Solution We assume an isotropic gap. Use the Ginzburg criterion to rationalize this
result. . The mean eld equation is. El Ek k k / Describe the assumptions that are made in
deducing that. . Quiz Problem . Quiz Problem . b Find the critical temperature and the
behavior near the critical point. the quasiparticles behave like noninteracting fermions and
the energy required to generate a quasiparticle is at least T . Describe the physical meaning
of the superconducting gap. in the case where the spin can take three values Si . Solution
Quiz Problem . The partition function and Helmholtz free energy are. a Find equations for the
mean eld free energy and magnetization. Consider the Ising ferromagnet in zero eld. The
free energy is lower for the spin case due to the higher entropy. Solution The
superconducting gap is the energy required to generate a quasiparticle excitation from the
superconducting ground state. Given the general solutions to the BCS mean eld theory k l
Vkl l . .
P kB Tc a/kB T vc kB T a e ea/kB Tc vc v vc b vc b kB Tc vc so that. vc b kB Tc vc so a vc
kB Tc vc b . . of the lecture notes. vc b kB Tc a/kB T vc kB Tc a P e ea/kB T vc v vc b vc b kB
T vc Which simplify to. The approximate expression is found by expanding for small k. Pc kB
Tc a/kB T vc e . Cr r/ e r Assigned Problem . The critical point is found by solving. . b r /. . /
eaxb/x dx eab / .. . To prove equation . we then have. substitute the denitions of the Fourier
transform and use translational invariance. in three dimensions. note that the matrix Jil is
diagonalized by the Fourier transform. or of the lecture notes prove Eq. Assigned Problem .
and using Eq. To show Eq. Find the critical point and the values of the exponents . prove Eq.
i Starting from Eq. from Eq. which reduces to Eq. as illustrated in the notes. The second
derivative yields. vc vc b vc b so vc b a a a vc b vc b kB Tc vc kB Tc vc kB Tc vc kB Tc a kB
Tc a ea/kB Tc vc ea/kB Tc vc vc b kB Tc vc vc b kB Tc vc a . b x/ with a / . To ll in the details
from Eq.. Solution. Jk eikx x eikx x eiky y eiky y . The Dieterici equation of state for a gas is.
of the lecture notes. and assuming only nearest neighbors. we nd. d . . we use the integral.. .
to . P kB T a/kB T v e vb where v V /N . Solution. ii From Eq.
b. T Tc T. pl t t t t t xl tx x l l p g t t t t t xl tx x l l Setting these equations to be equal yields.
Consider the Landau free energy. x v/vc . In the above analysis the signs of the t and x are
consistent but have to be checked each time. so that. The exponent is found by setting t to
zero so that p x . To third order this expansion gives. P Pc P . l so that xl T Tc / where we
dropped the t term as it is higher order. p e t Exp x t x kB Tc T a/kB Tc T vc v kB Tc e vc v b
b T Tc v b ea/kB Tc vc T /Tc v/vc where p P/Pc . T V P T T Tc V so . t T /Tc . xl xg . The
mean eld equation is. Pc P This reduces to. p t t t t t x tx x Taking a derivative with respect to
v at setting x leads to. Assigned Problem . To nd we assume that pl pg . m m . F am bm cm .
kB Tc a . so when we add uctuations later. Solving the mean eld equation. m Note that in
general we do a variation with respect to m. Here the variation is the same as a partial
derivative with respect to m. Find the discontinuity in m at the transition. Solution. b positive
and negative. vc b. t t xl x . F am bm cm where c gt as required for stability. Sketch the
possible behaviors for a. we need to use the EulerLagrange equation. and show that the
system undergoes a rst order transition at some value a. m b b ac/ c . so we nd that. b Pc a e
b To nd the critical exponents we write v vc v. c. so that . m . so that. we nd ve solutions.
to magnetization m for b gt b . l l where N k nk is the number of electrons in the Fermi sea.
there is a maximum at m . Therefore. we nd. with a m m b . F is large and positive for large
m. so m is always negative. b ac gt b . k k. The later leads to. b ac gt b . There is a phase
transition between states i and ii that occurs at a where two new solutions emerge and the
extremum at m changes from a maximum for a lt to a minimum for a gt . In this regime there
are several things going on. The real solutions are thus a maximum at m and two symmetric
minima at m b a/ b. so m gt . Therefore the solutions m are always imaginary. If the lowest
free energy state changes it corresponds to a phase transition. Assigned Problem . m and
setting m m . which leads to m a/ . Now consider the four cases for a. We also assume that
the fermion pairing that leads to superconductivity occurs in the singlet channel. In this
regime. iv a gt . We can solve this problem by evaluating F m . The nature of the transition is
found by making a small a expansion of the solutions m . i a gt . However b b ac/ remains
negative. so we know that m are maxima. The BCS Hamiltonian is then. ii a lt . we can divide
through by c. because c is positive. b. the discriminant b a is negative for b lt a. For b gt a. a
minimum at m . We can understand the nature of the solutions by looking at the second
derivative. so in this regime there is only one real solution. a / b b a / m is the magnetization
on the phase boundary dened by b . In this regime. there are ve real solutions because b b a/
gt . or we can nd solutions where F m and then solve m m . bk lt ak ak gt. Hpair N k k a ak k
kl Vkl a a a a . Moreover. which is a minimum having F . Though there are always ve
solutions. The nal issue we have to resolve is the behavior of the minima m as a function of
a. so that m is real. b lt . we know that the solution at m is a minimum. as in case ii. or
equivalently set c . only the real solutions are physical. where we found a T Tc iii a lt . so m
remains imaginary. The only real solution is m . . There is then a rst order phase transition
from magnetization m for b lt b . and b lt a a k k k gt. F a bm cm . The behavior of the
magnetization on the phase boundary is m a/ T Tc / . m which enables us to distinguish
between maxima and minima. Dening. and minima at m . In this case b ac lt b. b. Analysis of
the behavior of the model reduces to identifying the real solutions. First. The BCS pairing
Hamiltonian is a simplied model in which only pairs with zero center of mass momentum are
included in the analysis. b b a / b b a/ which has solution a b / . which is the mean eld result
for the order parameter near a tricritical point where a line of second order phase transitions
meets a line of rst order phase transitions. so m is real but m remains imaginary. while m are
minima. We also use the fact that F is symmetric in m and that at large m. b lt . in particular
we need to know if F m is greater than or less than F . Finally. b gt . and nding which real
solution has the lowest free energy. b gt . without loss of generality. This is the Ising/Van der
Waals univesality class we have studied using mean eld theory.
those which are of the form k k and. First the constant term. reduces to Eq. Assigned
Problem . those which can be reduced to diagonal form ie. the mean eld Hamiltonian looks
messy. carry out a leading order expansion in uctuations. plug into .g. provided Equations
and are true. those that are o diagonal e. show that the mean eld BCS Hamiltonian Eq. . With
this transformation ie. The rst stage is to collect together the terms which look to be in these
three catagories. . . leading to. ak ak lt ak ak gt ak ak lt ak ak gt Substition of this into . k k uk
uk k u vk k k k k vk uk k k vk vk k k k u uk k u vk k k k k k vk uk k k vk vk k k k u u k k k k u vk
k vk u k k k k k vk vk k k b k k uk uk k k uk vk k k k vk uk k vk vk k k k We now collect the
terms in this expression into three catagories Those which have no operators in them. k uk k
vk k k k uk k vk k k k k u vk k vk u k k k k k uk vk k k k vk uk k k . The meaneld Hamiltonian
corresponding to Eq. using an atomistic model. can be considered as a rst order expansion
in the uctuations i. ak ak b k k k k k u vk k uk k vk k k k u vk k uk k vk k k k k u vk k u k k k k
vk k uk k vk uk k vk k k k b k k Expanding this yields. Using the BogoliubovValatin
transformation Eq. along with the denitions lead to HM F .e. HM F N k b k k The following
terms can be converted to diagonal form. Solution. l b al al b bl k k This is the Hamiltonian
that we will solve to nd the thermodynamic behavior of superconductors. HM F N k k a ak k kl
Vkl a a b k k. HM F N k k a ak a ak k k k k a a k k. Solution.
. is in agreement with Eq. as we need them in the form . k k k / are the same. k k vk / vk k vk
vk . k gk k gk . We dene vk show that Eq. uk vk Substitution into Eq. as required. and that the
expectation of other terms that transform into one another through the transformation k k. HM
F N k b k k k vk k u vk uk vk k k k uk k vk k k k uk k vk k k k k k u vk k vk u k k k k k uk vk k k
k vk uk k Finally the odiagonal terms are as before. . to write. reduces to . . Show that Ek as
dened in Eq. leads to.e. Solution Starting with. We do this using the commutation relation. Ek
and Ek k k uk vk k u vk vk uk . We have. Finally the odiagonal terms are. Assigned Problem .
This yields. Collecting terms then leads to Eqs. k u vk k k k u vk k k k vk uk k k vk uk k k k k k
u k k k vk k k u k k vk k k We have to rearrange the terms catagorized as diagonal above.
Solution We need to show that the denitions. Ek k E k Ek k k k k k . k u vk k k k u vk k k k vk
uk k k vk uk k k k k k u k k k vk k k u k k vk k k We note that nk nk . are equivalent. k E E E
Ek k k k which reduces to Eq. gk uk . k gk gk / . To do this we use the results of Problem in
Eq. i. gk / Assigned Problem . of the lecture notes.
N F V hc d T anh / / . bk lt ak ak gtlt uk k vk uk k vk k k gt To evaluate this expression at nite
temperature. . Assigned Problem . Solution In the case of the weakcoupling swave model
Eq. so that. k Ek k Ek k . Prove the relations Eq. we must evaluate. gk We then nd. Prove the
relation . k Ek k k / . Solution We have. lt O gt trOe H tre H The only terms that survive are
the terms that are diagonal. Starting from Eq. Assigned Problem . reduces to. bk lt ak ak gtlt
uk vk k k uk vk k gt uk vk lt nk gt lt nk gt k Finally we use the fact that lt nk gtlt nk gt to nd Eq.
. uk v k Ek k E k k E k k E k Ek k k k E k gk k . such as nk . gk Ek uk Ek k Ek gk gk uk k gk
Ek Assigned Problem . vk and hence. gk so that. Solution We start with. We thus nd. prove
the relation . .
using as the small quantity. a . . x x We expand about c . This leads to. . As shown in
lectures. so that. and immply that. kB Tc . N F V hc d T anh hc d Sech T anh This expression
may be written in the form. we nd. c Exp h N F V To nd the behavior of the gap near Tc .. N F
V lnc c /T anh a .. the critical point is found from the equation. but we only need to consider
carry out this expansion in the rst term on the right hand side. T T . spin /. Jy . square lattice
Ising model where the interactions along the xaxis. nd the internal energy and the specic
heat. T anh / T anh T anh Sech T anh / Taking the gap to be real. Assigned Problem . we
carry out a rst order Taylor expansion of Eq. ln c a a h N F V where. to this case. Does
duality still hold From your expansions. and a dx Sech x T anhx . Solution Z Si ltijgt eKij Si Sj
CoshKx N CoshKy N tij Si Sj . kB Tc Tc T Tc a . Also note that Eq. so that T / / t kB Tc a F V
ln h c a a . and Eq. ln t t so that. / . kB Tc t where t Tc T /Tc This is correct near Tc . h where.
are dierent than those along the yaxis. N We then nd that. Consider a ferromagnetic nearest
neighbor. kB Tc . Si ltijgt . a lnxSech xdx . The critical temperature is then. Jx . Extend the
low and high temperature expansions Eq.
For the second virial coecient the relationship is a T b T . Find the second virial coecient for
six cases i the classical hard sphere gas. iv The van der Waals gas. F lnZ becomes F N ln
lnCoshKx lnCoshKx t t t t t t t t t t t t . where tx tanhKx . with the results... but now we have to
treat subclasses with dierent numbers of hirizontal and vertical bonds. so that.. Assigned
Problem . Pv kB T l l bl z l l lbl z bl z l . The virial expansion is most often carried out in the
grand canonical ensemble. Duality holds for both x and y directions.. where we may write. sy
eKy .. kB T v and Pv / . l/ Ideal F ermi bl l/ . kB T v Ideal F ermi Ideal Bose l . ii Noninteracting
Fermions.... tij tx for horizontal bonds and tij ty for vertical bonds.. and the second virial
coecient is a T .. iii Noninteracting Bosons. F N Kx Ky s s s s s s s s s s s s . bl so we have..
Pv / . vb v so that Pv a b a/kB T . l N V lbl z l l l al T l v which gives relations between the
quantities al T and bl T . Solution Pv kB T al T l l v where the rst virial coecient a T .. kB T b/v
kB T v v . Z Si zN lnCoshK N ln ln N t N t N N t t eKSi Sj eKzN N s N s N N s Os ltijgt to the
anisotropic case leads to. P kB T a .. ty tanhKy .. We have already calculated the coecients
bl for the ideal Bose and Fermi gases.. x y x y x y x y x y x y where sx eKx . P kB T so that.
the extension of the low temperature expansion. Ideal Bose The van der Waals equation of
state may be expanded to nd the second virial coecient. x y x y x y x y x y x y Similarly. The
diagrams used are similar.
. E where is the superconducting gap and by.Quiz . . Explain the physical reasoning for the
decoupling scheme that is chosen. kB T v The behavior of the second virial coecient as a
function of temperature can be used to deduce the interaction potential. on a lattice with
coordination number z and exchange constant J. . i Write down or derive the van der Waals
equation of state. . show that the quasiparticle density of states is given N F E E / DE 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. T lt Tc and for T Tc . so the virial expansion for this case
is. ii For the case T lt Tc explain why the nonconvex part of the curve cannot occur at
equilibrium and show the Maxwell construction to obtain a physical P. and the way in which
BCS theory describes it. v isotherm. for three applied eld cases h lt . and for the liquidgas
critical point.. and the importance of quantum eects as they have dierent temperature
dependences. Time for quiz . h gt . ii Write down the denition of the critical exponents . and
for the Ising ferromagnet critical point. Answer all questions. minutes Name . . i Describe the
physical meaning of the superconducting gap. Pv R . i Plot the behavior of the magnetization
of the Ising ferromagnet as a function of the temperature. .. Friday November . From this
equation nd the critical exponent for the Ising ferromagnet within mean eld theory. h . for T gt
Tc . minutes Name . Answer all questions. and the decoupling scheme used to reduce it to a
solvable form. we then nd that b R / . iii Write down the denition of the critical exponents .
What values do these exponents take within Van der Waals theory. . Indicate the critical
point. . PHY . Time for quiz . Write down the Hamiltonian for BCS theory. is the Fermi
energy. b V dr d r eur r d reur For the hard sphere problem. Write the mathematical
statement of the Maxwell construction. ii Given that the energy of quasiparticle excitations
from the BCS ground state have the spectrum.. F F / . with a hard core radius of R. Friday
November . .Quiz . What values do these exponents take within mean eld theory iii Write
down or derive the mean eld equation for the spin / Ising ferromagnet in an applied eld. Make
plots of the van der Waals equation of state isotherms. b is given by. For the classical
interacting gas. PHY .
nearest neighbor. where the hard core radius is R. spin half Ising ferromagnet partition
function. . i Explain the importance of linkedcluster theorems in the perturbation theory of
many particle systems. What physical properties can be extracted from the second virial
coecient ii Given that. . . ii Draw the low temperature series expansion diagrams to order s
where s ExpJ for the square lattice. 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. i Write down the mathematical form of the
virial expansion for many particle systems and explain why it is important. bl sum over all l
connected cluster integrals ll V nd the second virial coecient for the classical gas with hard
core repulsive interactions.