4The variational method

III Theoretical Physics of Soft Condensed Matter



4.2 Smectic liquid crystals
We use this to talk about the isotropic to smectic transition in liquid crystals.
The molecules involved often have two distinct segments. For example, we may
have soap molecules that look like this:
The key property of soap molecules is that the tail hates water while the head
likes water. So we expect these molecules to group together like
In general, we can imagine our molecules look like
and like attracts like. As in the binary fluid, we shall assume the two heads are
symmetric, so
U
AA
=
U
BB
6
=
U
AB
. If we simply want the different parts to stay
away from each other, then we can have a configuration that looks like
z
In general, we expect that there is such an order along the
z
direction, as
indicated, while there is no restriction on the alignments in the other directions.
So the system is a lattice in the
z
direction, and a fluid in the remaining two
directions. This is known as a smectic liquid crystal, and is also known as the
lamellar phase. This is an example of microphase separation.
As before, we let
φ
be a coarse grained relative density. The above ordered
phase would then look like
φ(x) = cos q
0
z.
for some
q
0
that comes from the molecular length. If our system is not perfectly
ordered, then we may expect it to look roughly like A cos q
0
z for some A.
We again use Landau–Ginzburg model, which, in our old notation, has
βF =
Z
a
2
φ
2
+
b
4
φ
4
+
κ
2
(φ)
2
+
γ
2
(
2
φ)
2
dr.
If we write this in Fourier space, then we get
βF =
1
2
X
q
(a + κq
2
+ γq
4
)φ
q
φ
q
+
b
4V
X
q
1
,q
2
,q
3
φ
q
1
φ
q
2
φ
q
3
φ
(q
1
+q
2
+q
3
)
.
Notice that the quartic term results in the rather messy sum at the end. For the
iso-smectic transition, we choose κ < 0, γ > 0.
Again for simplicity, we first consider the case where
b
= 0. Then this is a
Gaussian model with
G(q) = a + κq
2
+ γq
4
= τ + α(q q
0
)
2
,
Varying
a
gives a linear shift in
G
(
q
). As we change
a
, we get multiple different
curves.
q
G(q)
a = a
c
q
0
Thus, as a decreases, S(q) = h|φ
q
|
2
i =
1
G(q)
blows up at some finite
q = q
0
=
r
κ
2γ
, a
c
=
κ
2
4γ
.
We should take this blowing up as saying that the
|q|
=
q
0
states are highly
desired, and this results in an ordered phase. Note that any
q
with
|q|
=
q
0
is
highly desired. When the system actually settles to an ordered state, it has to
pick one such
q
and let the system align in that direction. This is spontaneous
symmetry breaking.
It is convenient to complete to square, and expand
G
about
q
=
q
0
and
a = a
c
. Then we have
G(q) = τ + α(q q
0
)
2
,
where
τ = a a
c
, α =
1
2
G
00
(q
0
) = 2κ.
Then the transition we saw above happens when τ = 0.
We now put back the quartic term. We first do this with mean field theory,
and then later return to the variational method.
Mean Field Theory
In mean field theory, it is easier to work in real space. We look for a single field
configuration that minimizes
F
. As suggested before, we try a solution of the
form
φ = A cos q
0
z,
which is smectic along
z
. We can then evaluate the free energy (per unit volume)
to be
βF
V
= βF[φ] =
a
2
φ
2
+
κ
2
(φ)
2
+
γ
2
(
2
φ)
2
+
b
4
φ
4
where the bar means we average over one period of the periodic structure. It is
an exercise to directly compute
φ
2
=
1
2
A
2
, (φ)
2
=
1
2
A
2
q
2
0
, (
2
φ)
2
=
1
2
A
2
q
4
0
, φ
4
=
3
8
A
4
.
This gives
βF
V
=
1
4
aA
2
+ κA
2
q
2
0
+ γA
2
q
4
0
+
3b
8
A
4
=
1
4
A
2
(a a
c
)
| {z }
τ
+
3b
8
A
4
.
Note that
q
0
is fixed by the system as we originally had, while
A
is the amplitude
of the fluctuation which we get to choose. Plotting this, we get a familiar graph
A
βF/V
τ > 0
τ < 0
τ > 0
τ < 0
If
τ >
0, then the optimum solution is given by
A
= 0. Otherwise, we should
pick
A 6
= 0. Observe that as we slowly reduce
τ
across 0, the minimum varies
continuously with τ:
a
|A|
a
c
Variational method
We now consider the variational theory. In the notation we were using previously,
we have
H =
X
q
+
φ
q
φ
q
G(q) +
b
4V
X
q
φ
q
1
φ
q
2
φ
q
3
φ
(q
1
+q
2
+q
3
)
.
Our trial H
0
is
H
0
=
X
q
φ
q
φ
q
J(q).
Since this is a Gaussian model, we know that
F
0
=
X
q
+
log
J(q)
π
.
To use our inequality, we need to evaluate our other two bits. We have
hH
0
i
0
=
X
q
+
hφ
q
φ
q
i
0
J(q).
We already calculated
hφ
q
φ
q
i
0
=
1
J(q)
.
Thus, we have
hH
0
i
0
=
X
q
+
1.
Here it is clear that we must impose a cutoff on
q
. We can think of this 1 as the
equipartition theorem.
We can also compute
hHi
0
=
X
q
+
1
J(q)
G(q) +
b
4V
X
q
1
,q
2
,q
3
hφ
q
1
φ
q
2
φ
q
3
φ
(q
1
q
2
q
3
)
i
0
| {z }
.
In the Gaussian model, each
φ
q
is a zero mean Gaussian random variables, which
have certain nice properties. Wick’s theorem tells us we have
habcdi
0
= habi
0
hcdi
0
+ haci
0
hbdi
0
+ hadi
0
hbci
0
.
Applying this, together with the result
hφ
q
1
φ
q
0
i
0
= h|φ
q
|
2
i
0
δ
q
1
,q
2
,
we obtain
U = 3
"
X
q
h|φ
q
|
2
i
0
#
2
= 12
"
X
q
+
1
J(q)
#
2
.
Thus, we have
hH
0
i
0
=
X
q
+
G(q)
J(q)
+
3b
V
X
q
+
1
J(q)
!
2
.
We can then compute
˜
F =
X
q
+
log
J(q)
π
1 +
G(q)
J(q)
+
3b
V
X
q
+
1
J(q)
!
2
.
We minimize over J(q) by solving
˜
F
J(q)
= 0
for all q. Differentiating, we obtain
1
J(q)
G(q)
J(q)
2
6b
V J(q)
2
X
q
0
+
1
J(q
0
)
= 0.
Multiplying through by J
2
, we get
J(q) = G(q) +
6b
V
X
q
0
+
1
J(q
0
)
.
For large V , we can replace the sum by an integral, and we have
J(q) = G(q) +
3b
(2π)
d
Z
dq
0
J(q
0
)
.
It is very important that once we have fixed
J
(
q
), the second term is a constant.
Writing
C =
2
V
X
q
0
+
1
J(q
0
)
=
1
(2π)
d
Z
dq
0
J(q
0
)
,
we can then say the minimum is given by
J
(
q
) =
G
(
q
) + 3
bC
. Thus, solving for
J(q) is equivalent to finding a C such that
C =
1
(2π)
d
Z
dq
0
G(q) + 3bC
.
This is a self-consistency equation for C (and hence J).
There is a very concrete interpretation of
C
. Recall that
hφ
2
i
is the average
value of φ
2
at some point r (which is independent of r. So we can compute
hφ(r)
2
i =
1
V
Z
hφ(r)
2
i dr =
1
V
X
q
h|φ
q
|
2
i =
1
V
X
q
1
J(q)
.
Thus, what our computation above amounts to is that we replaced
G
(
q
) by
G(q) + 3bhφ
2
i.
A good way to think about this is that we are approximating the
φ
4
term in
the free energy by
b
4
φ
4
3b
2
hφ
2
iφ
2
.
We have to pick a value so that this approximation is consistent. We view the
hφ
2
i
above as just another constant, so that the free energy is now a Gaussian.
We can compute the expectation of
φ
2
using this free energy, and then we find
that
hφ
2
i =
1
(2π)
d
Z
dq
0
G(q) + 3bhφ
2
i
.
This is the self-consistency equation as above.
We could have done the approximation above without having to through the
variational method, but the factor of
3b
2
is then no longer obvious.
To solve this self-consistency equation, or at least understand it, it is conve-
nient to think in terms of τ instead. If we write our G(q) as
G(q) = τ + α(q q
0
)
2
,
then we have
J(q) = ¯τ + α(q q
0
)
2
, ¯τ = 3bhφ
2
i.
The self-consistency equation now states
¯τ = τ +
3b
(2π)
d
Z
d
d
q
¯τ + α(q q
0
)
2
.
We are interested in what happens near the critical point. In this case, we expect
¯τ
to be small, and hence the integral is highly peaked near 0. In
d
= 3, we can
make the approximation.
3b
(2π)
3
Z
d
3
q
¯τ + α(q q
0
)
2
=
3b
2π
2
Z
0
q
2
dq
¯τ + α(q q
0
)
2
3bq
2
0
2π
2
Z
0
dq
¯τ + α(q q
0
)
2
.
While there is a nasty integral to evaluate, we can make the substitution
q 7→ ¯τq
to bring the dependency on ¯τ outside the integral, and obtain
¯τ = τ +
sb
¯τ
, s =
3q
2
0
2π
2
1
α
Z
0
dy
1 + y
2
q
2
0
α
.
The precise value of
s
does not matter. The point is that it is constant indepen-
dent of
¯τ
. From this, we see that
¯τ
can never reach zero! This means we can
never have a continuous transition. At small
¯τ
, we will have large fluctuations,
and the quantity
h|φ
q
|
2
i = S(q) =
1
¯τ + α(q q
0
)
2
becomes large but finite. Note that since
¯τ
is finite, this sets some sort of length
scale where all q with |q q
0
| ¯τ have large amplitudes.
We can think of the above computation as looking at neighbourhoods of the
φ
= 0 vacuum, as that is what the variational methods see. So
¯τ
never vanishes
in this regime, we would expect a discontinuous isotropic-smectic transition that
happens “far away” from this vacuum.
Consider fluctuations about an ordered state, φ = φ
0
+ δφ, where
φ
0
= A cos q
0
z.
We can do computations similar to what we did above by varying
δφ
, and obtain
a
˜
F
(
A
). Then the global minimum over all
A
then gives us the “true” ground
state. Instead of doing that, which is quite messy, we use the heuristic version
instead. For A = 0, we had
τ = ¯τ + 3bhφ(r)
2
i.
For finite A, the quartic term now has an extra contribution, and we bet
¯τ = τ +
sb
¯τ
+
3bA
2
2
.
Compare this with mean field theory, where we have
F
MF T
(A) =
V
4
τA
2
+
3b
8
A
4
.
We see that for small
A
, the fluctuations are large, and mean field theory is
quite off. For large
A
, the fluctuation terms are irrelevant and mean field theory
is a good approximation. Thus, we get
A
F
MFT
We see that as
τ
decreases, the minimum discontinuously jumps from
A
= 0
to a finite value of A. Mean field theory is approached at large A.
We can then plot the minimum value of A as a function of τ :
MFT
τ
|A|
τ
c
A
c
τ
H
We have not calculated
A
c
or
τ
c
, but it can be done. We shall not do this, but
we can state the result (Brazovskii, 1975):
τ
c
' (sb)
2/3
, A
c
' s
1/3
b
1/6
.
It turns out the variational approach finally breaks down for
τ < τ
H
s
3/4
b
1/2
.
We have
τ
H
τ
c
if
b
s
. The reason this breaks down is that at low enough
temperatures, the fluctuations from the quartic terms become significant, and
our Gaussian approximation falls apart.
To figure out what
τ
H
is, we need to find leading corrections to
˜
F
, as
Brazovskii did. In general, there is no reason to believe
τ
H
is large or small. For
example, this method breaks down completely for the Ising model, and is correct
in no regime at all. In general, the self-consistent approach here is ad hoc, and
one has to do some explicit error analysis to see if it is actually good.
Brazovskii transition with cubic term
What happens when we have a cubic term? This would give an
A
3
term at
mean field level, which gives a discontinuous transition, but in a completely
different way. We shall just state the results here, using a phase diagram with
two parameters τ and c. In mean field theory, we get
c
τ
I
S
H
where H is a hexagonal phase, which looks like
where each cylinder contains a high concentration of a fixed end of the molecule.
This is another liquid crystal, with two crystal directions and one fluid directions.
The self-consistent version instead looks like
c
τ
I
S
H
˜c
Here c only matters above a threshold ˜c.