Physics 212: Statistical mechanics II
Lecture IX
Let us work out a simple example of the linear-response formula from the last lecture. Suppose
that the starting Hamiltonian describes a single spin-half in a magnetic field along the ẑ axis:
∆
Sz ,
h̄
H0 =
(1)
which has two eigenstates: spin down with energy −∆/2, and spin up with energy ∆/2. We know
from the derivation last time that a perturbation Hamiltonian proportional to Sz will not induce
any transitions and hence no linear response. Instead let us try
H1 (t) = λ(t)Sx ,
(2)
where λ(t) has units of frequency (inverse time) Then the response in a measurement of Sx at time
t to the perturbation is
Z t
λ(t0 )
dt0 h Sx (t − t0 ), Sx
δhSx i =
i.
(3)
h̄
−∞
We can use the unperturbed Hamiltonian to obtain
0
0
Sx (t − t0 ) = eiH0 (t−t )/h̄
Sx e−iH0 (t−t )/h̄
−i∆(t−t0 )/(2h̄)
0 )/(2h̄)
i∆(t−t
e
e
0
Sx
=
0 )/(2h̄)
−i∆(t−t
0
0
e
0
0
ei∆(t−t )/(2h̄)
.
(4)
Then the matrix multiplication gives
h̄
Sx (t − t ) =
2
0
0
ei∆(t−t )/h̄
0
0
0
e−i∆(t−t )/h̄
where we have used
Sx =
h̄
2
0
1
1
0
= cos(∆(t − t0 )/h̄)Sx − sin(∆(t − t0 )/h̄)Sy ,
,
Sy =
h̄
2
0
i
−i
0
(5)
(6)
which satisfy [Sx , Sy ] = ih̄Sz .
Now only the Sy term in Sx (t − t0 ) gives a nonzero result in the commutator with Sx . Since
(Sy , Sx ) = −i[Sy , Sx ] = −h̄Sz , we obtain the simple form
δhSx i =
Z t
−∞
0
0
0
dt hSz sin(∆(t − t )/h̄)λ(t )i =
Z t
−∞
dt0 sin(∆(t − t0 )/h̄)λ(t0 )hSz i.
(7)
The statistical average of Sz is time-independent, and is just
hSz i =
h̄ eβ∆/2 − e−β∆/2
h̄
= tanh(β∆/2).
β∆/2
−β∆/2
2e
2
−e
(8)
Combining everything, the response is
δhSx i = tanh(β∆/2)
Z t
dt0 sin(∆(t − t0 )/h̄)λ(t0 ).
−∞
1
(9)
The fact that in this example temperature just gives a prefactor that is independent of the time
dependence of λ(t0 ) is special: it results from the simple commutators of the spin operators. We see
that the response goes to zero at high temperatures kT ∆. The periodic form of the integrand
makes sense in a picture of classical precession of the spin, and is consistent with what we could
have derived with ordinary time-dependent perturbation theory. Clearly a weak constant-in-time
perturbation will have less effect than an equally weak perturbation at the resonant frequency ∆/h̄.
Not covered in class: The frequency-space version of the linear-response formula can be used
to write a compact relationship known as the fluctuation-dissipation theorem between response
functions and correlation functions (cf. problem set). We now introduce a quick rewriting of the
Kubo formula from the last lecture, then move on to scaling theories of equilibrium systems. The
Kubo formula derived last time was
δhÔ2 i = λ
Z t
−∞
dt0 h Ô2 (t − t0 ), Ô1 (t0 ) i,
(10)
where (A, B) = −i[A, B], and remember that here Ô1 (t0 ) carries a physical time dependence, while
Ô2 (t − t0 ) is defined using the unperturbed Hamiltonian and the time-independent operator Ô2 .
Consider the case where Ô1 couples a force F to a density b: by this we mean
1 X
F (q, t)b(−q),
V q
H1 =
(11)
where the sum is over spatial wavevectors. As an example, F could be a chemical potential and b a
number density, or F could be a Zeeman magnetic field and b a spin density. We are going to simplify
the Kubo formula under the assumption that the equilibrium Hamiltonian H0 is translationally
invariant.
The the Kubo formula for the response in a density a(q0 ) is
−i
δha(q )it =
V
0
Z t
dt0 F (q, t0 )h a(q0 , t − t0 ), b(−q) i.
(12)
−∞
When can this commutator be nonzero in a translation-invariant system? Translation invariance
implies
ha(q0 )b(−q)i = ha(q)b(−q)iδq,q0 .
(13)
This result can be derived as follows: translation invariance implies that hã(r1 )b̃(r2 )i = hã(r1 +
r0 )b̃(r2 + r0 )i for any r0 . Now
ha(q0 )b(−q)i =
Z
Z
dr1
V
0
dr2 eiq ·r1 e−iq·r2 hã(r1 )b̃(r2 )i =
V
Z
Z
dr1
V
0
dr2 eiq ·r1 e−iq·r2 hã(r1 − r2 )b̃(0)i.
V
(14)
Now change coordinates in the integral to r1 ± r2 . The positive sign gives a delta-function that
forces q = q0 , and the negative sign gives the desired result.
Applying this to the Kubo formula gives
δha(q0 )it =
Z ∞
−∞
dt0 F (q, t0 )χab (q, t − t0 ),
2
(15)
where the response function χab (q, t) is defined as
χab (q, t) = iV −1 θ(t)h[a(q, t), b(−q, 0)]i.
(16)
and θ(t) is the step function that vanishes if the argument is negative, which allowed the upper
limit of the t0 integral to be taken to +∞.
We will not do any detailed calculations using this formula, as methods for such calculations
are the main subject of a course in many-body theory in condensed matter (e.g., Physics 216).
However, we can say one thing about this formula simply by inspection. If the correlations never
die out in time, then the response must be infinite. A simple example is a free-electron system
with translational invariance: an excitation of nonzero momentum cannot ever decay back to the
ground state, so there is no dissipation, and the resulting conductivity is infinite. The connection
between symmetries and long-lived modes is something that we will return to after learning more
about equilibrium systems.
Here is where class resumed: The next few lectures will consider phase transitions in
classical equilibrium systems. The free energy F = E − T S of a statistical-mechanical system can
develop singularities as the size of a system becomes infinite. Even this simple fact was somewhat
controversial until Onsager’s solution of the 2D Ising model, in the 1940s. At a phase transition
there are two phases which have the same free energy. We can divide phase transitions into two
classes: those where the first derivatives of the free energy, such as pressure or magnetization, have
singularities (first-order transitions), and those where only the second derivatives of the free
energy, like susceptibility, have singularities (second-order transitions).
Many systems have a line of first-order transitions terminated by a second-order critical point.
An example is the uniaxial (Ising) magnet in dimensions greater than or equal to two, as will
be shown. To start understanding phase transitions, let us consider the Ising model as a simple
representative example.
The Ising Hamiltonian is
H = −J
X
σiz σjz .
(17)
hiji
Here the notation hiji means that i and j are nearest neighbors. Assume J > 0, so that the ground
state is ferromagnetic, and write σ = σ z since the spins in this model are just classical variables
taking values ±1. Also assume that the model lives on the cubic lattice in d dimensions.
Statistical averages with this Hamiltonian take the form
P
Ae−βH
=
−βH
σe
hAi = Pσ
P
σ
Ae−βH
.
Z
(18)
Here A will typically be some product of the spin operators. For example, the spin-spin correlation
is one quantity we will be quite interested in, as it can distinguish between long-range ordered
and disordered phases.
hσi σj i →
m2
0
ordered phase
disordered phase
as |i − j| → ∞.
Here m is some number that measures the strength of the long-range order.
3
(19)
At zero temperature, the free energy is just the internal energy H, and the system is entirely
in its ground state, which consists of all spins aligned. It is conceivable that as the temperature
is increased, this perfect alignment is destroyed by thermal fluctuations, since there are only two
fully ordered configurations but many disordered configurations. The existence of phase transitions
in classical systems from an ordered ground state at zero temperature will turn out to depend
sensitively on the dimensionality and symmetry of a system.
The Ising model in three dimensions, the most physically relevant case, has not been exactly
solved. To attempt to understand its behavior, the “mean-field” approximation is now used to
obtain a qualitative picture. Consider one spin in the Ising model, and suppose that the equilibrium
magnetization is hσi i = m, or
P
−βH
σ σi e
= m.
(20)
Z
Consider one spin in the lattice. This spin is only coupled to its 2d nearest neighbors in d dimensions
(i.e., the coordination number of the lattice is z = 2d). A logical self-consistency condition is that
the expected value of the magnetization of this spin should come out to be m, if the magnetization
of each of its neighbors is m. That is, we are replacing the sum over all spin configurations in the
above by sums over a single spin on the lattice; the error involved in this approximation will be
discussed later.
Mathematically, this self-consistency equation becomes (letting K = βJ)
ezKm − e−zKm
= tanh(zKm).
(21)
ezKm + ezKm
This form occurred because the energy for σi = +1 is zKm, assuming each neighbor has mean spin
+m. Now let’s understand the self-consistency equation
hσi i =
m = hσi i = tanh(zKm).
(22)
The number of solutions of this transcendental equation for m depends on the value of the parameter
zK. If zK < 1, the only solution is m = 0. This can be seen graphically by plotting the tanh
function. If zK > 1, there are three solutions: m = 0 and m = ±m∗ , where 0 < m∗ ≤ 1 is the
positive root. The existence of multiple solutions is connected with the ordered phase of the model,
as now reviewed. First note that if K → ∞, as at zero temperature, then the right side becomes
tanh ±∞ = 1, so m∗ = 1 corresponding to complete order. This suggests physically that for large
K, the m = 0 solution is not actually realized: it is “unstable” to the other two solutions (and
in fact can be shown to correspond to a local maximum of the free energy). As K decreases with
increasing temperature, m∗ increases until K = z −1 , when the three solutions collapse into one.
The Ising model can be solved exactly in one dimension. The solution will help us understand
when phase transitions exist. (It can also be solved in two dimensions, but that is quite technical
and beyond the scope of this course. For a discussion, see Feynman’s book on statistical mechanics.)
The first way to solve it in one dimension is via a transfer matrix method. This just means that
the partition function can be written in terms of matrices.
The partition function of one spin in a magnetic field is eβH + e−βH . The partition function of
two spins is e2βH+K + 2e−K + eK−2βH , which can be written as Z2 = Z2↑ + Z2↓ , with
Z2↑
Z2↓
!
=
eK+βH
e−K−βH
e−K+βH
eK−βH
eβH
e−βH
4
=T
eβH
e−βH
=T
Z1↑
Z1↓
!
.
(23)
The idea of this multiplication is that new terms in the partition function for ZN +1 are generated
by adding a spin, with energy depending on the last spin of the previous N -spin chain. The top
left element of the matrix corresponds to added spin up and previous spin up; the top right corner
to added spin up and previous spin down (thus it gets multipled by the second component of the
↑
↓
input vector). In general, the partition function of the N -spin chain is just ZN = ZN
+ ZN
, where
↑
ZN
↓
ZN
!
= T N −1
eβH
e−βH
.
(24)
This transfer matrix expression for the partition function is quite useful. Remember that the
free energy is F = −kT log Z. In the large-N limit, the free energy per site will be dominated by
the larger of the two eigenvalues of the transfer matrix:
F
≈ −kT log λ+ .
N
(25)
The eigenvalues of the transfer matrix are found, from solving a quadratic, to be
K
cosh βH ±
λ± = e
q
2
sinh βH +
e−4K
.
(26)
λ± = eK 1 ± e−2K = (2 cosh K, 2 sinh K).
(27)
As a check, assume H = 0. Then this becomes
h
i
These are indeed the eigenvalues of
T =
eK
e−K
e−K
eK
.
(28)
Hence the free energy is given in the thermodynamic limit, with zero magnetic field, by
F
≈ −kT log λ+ = −kT log(2 cosh K).
N
(29)
For large K, or low temperature, this becomes just −kT log(eJ/kT ) = −J. We can interpret this
simply as an energy of J per bond for the ground state.
The idea of the mean field theory is that it includes a simplifying assumption about fluctuations.
It becomes more correct as the number of spins connected to a given spin increases (that is, as the
coordination number increases), as then the average field felt by a single spin has smaller relative
fluctuations. In fact the mean field theory above is exact if every spin is connected with equal
strength to every other spin. Since the mean field theory seems to get better with increasing
coordination number, one might expect it to be correct in the limit of infinite dimensionality.
Actually we’ll see later that it does quite a bit better than one might expect!
Reviewing the one-dimensional case, it is actually quite easy to see that there can never be
an ordered phase at finite temperature in a model with only short-ranged interactions. (Here and
elsewhere “finite” means neither zero nor infinite.) Consider flipping all the spins after site i. In a
system of L sites, there are of order L states which have one boundary between up spins and down
5
spins. Then the entropy increase is log L, while the energy cost is that of flipping a single bond
(2J). So the free energy difference from the ordered state to that with one flipped bond is
∆F = 2J − T log L < 0,
(30)
so that the disordered state is favored for a large enough system no matter how low the temperature.
Essentially the energy cost of a “domain wall” between one spin orientation and another is so low
in one dimension that the entropy contribution wins at any finite temperature. In two dimensions,
the energy cost to flip a domain goes with the perimeter of the domain, or linearly in its radius.
This increase in the domain wall energy turns out to be enough, in the Ising model, to stabilize the
ordered low temperature phase.
The general statement of the above argument is that a system with a discrete order parameter,
like the Ising model, cannot have a finite-temperature phase with long-range order in one dimension. It can in two dimensions, as the exact solution of the Ising model shows. For a system with
a continuous order parameter, like a Heisenberg or XY model, the “Mermin-Wagner-HohenbergColeman” theorem states that there cannot be a phase with true long-range order in dimensions
one or two. We will see how this happens in a later lecture.
6