See ABH, Phys. Rev. B 76, 054447 (2007).
1

H
  ij
J ij

S i


S j
w
F

U

T
S
U

( z / 2 ) J

S
 i
  
S
 j
 S  k ln W
W
2

S
(S) 
S
( 0 )
 aS
2  b S
4  
T = T c
+1
F
 a ( T

T c
) S
2  u S
4  
S

0
S
 c ( T c

T )
1 / 2
T

T c
T

T c
3
According to the renormalization group, fluctuations (neglected by mean field theory) cause this exponent to assume a modified value often close to 1/3.
T = T c
-1
Spin S
T = T c
-2
T/T c

*
F

J
 n
S n
S n

1
 ckT
 n
S n
2  
S ( q )

N

1
 n e i q
 x n S n
F
  q
S ( q )

S ( q ) [ J cos ( qa )
 c k T ]
S n
  q e
 iq x n

1
2
 q

S ( q )
( q )

1
S ( q )

S ( q )
4
FERRO
J < 0 aq/ p
Note: the paramagnetic phase becomes unstable with respect to the formation of long-range order. For
FERRO this happens at q=0. For ANTIFERRO it happens at q = p
/a, as expected.
J > 0
ANTI-
FERRO aq/ p

H

  [

1 ( q ) n

J
1 ckT
S n

J
1
S n

1
 cos( qa )
J
2

S n
J
2
S n

2
] cos ( 2 qa )
T = T c
+ 1 T = T c
+ 2   
1
( q ) /
 q
| q
 q *

0
 a q
*  cos

1
(

J
1
/ 4 J
2
)
5
T = T c
Lifshitz criterion: if q at T=T c is incommensurate (q=q*), then q is temperature dependent in the ordered phase. q at
T<T c can be commensurate only if it is a high symmetry wave vector.
How can we understand this principle?
Note that the value of q depends on the specific interactions in the system. That is to say, if we were to add a third neighbor interaction that would modify the extremum condition that determined q*. If nothing else, the J’s will be very slightly temperature-dependent because of the thermal expansivity. In contrast, return to the previous slide: there you see that the extremum is at a symmetry point which is stable against addition of weak further neighbor interactions.

In the above discussion it was not said whether several values of wave vector could be simultaneously selected. This brings up the whole question of accidental degeneracy as it becomes relevant in its many guises. For this discussion it is useful to consider the “vacuum” as being the disordered phase when all the order parameters are zero.
The symmetry of the vacuum, is the symmetry of this paramagnetic state. For any excitation relative to the vacuum, i. e. for any set of values of the order parameters, there will be a family of states with identical free energy which are generated by applying all the symmetry operators of the vacuum to that excitation.
A simple application of this principle is that if the free energy has an instability at a wave vector q, it will also have symmetry-related instabilities at all wave vectors in the STAR of q. (The STAR of q consists of all wave vectors which are generated by symmetry operators of the vacuum providing that we do not accept wave vectors which are equivalent in the sense that their difference is a reciprocal lattice vector, a RLV.) An important STAR includes q and –q if q and –q do not differ by an RLV.
This is not all!!
6

Can I possibly find more to say under this heading?? YES.
You may ask: is it possible that two wave vectors not related by symmetry could simultaneously become unstable? This is an accidental degeneracy of the type theoreticians excluded two centuries ago in the development of thermodynamics.
I will give an example of this discussion later. But for now
I call your attention to the phase rule. Consider a system
(like water, see right) that has gas, liquid, and solid phases all of which are characterized by specifying two intensive variables, the pressure P and the temperature T. If you vary one of them, say, the temperature, you can reach exceptional points where two phases coexist. This does not happen at a generic point in phase space: it requires adjusting one parameter to its special value. Is it possible that by varying the temperature you can reach the triple point where all three phases coexist? Absolutely not: to reach that point you need to adjust the temperature AND pressure.
p
S
L
G
T
7

LOCK-IN: Consider the competition between this q* near aq/ p
=1/4 and aq/ p
=1/4:
8
F

( 1 / 2 )
 
1
( q
*
) | S ( q
*
) |
2  u | S ( q
*
) |
4 
( 1 / 2 )
 
1
( 1 / 4 ) S ( 1 / 4 ) S ( 1 / 4 )
*
 u [ S ( 1 / 4 ) S ( 1 / 4 )
*
]
2 
( only if q

1 / 4 )
 d
[ (
S ( 1 / 4 )

)
4 
S ( 1 / 4 )
4 ]
If d were zero, then F would be minimal for q=q*. However, by proper choice of f in S(1/4)=|S(1/4)|exp(i f
), the term in d can be made negative.
Then, when S 4 is large enough, the Umklapp term in d can lock in aq=1/4 .
So, as the temperature is lowered further into the ordered phase, the range over q for which phase locking occurs grows.
I should have mentioned that once on order parameter condenses, entropic effects prevent another order parameter of the same symmetry from condensing.

When J
1 and J
2 compete
(both antiferromagnetic)
AF
Experimental trajectory
J
1
/J
2
If there are several J’s, then when such a system is mapped onto the model system with two J’s, the ratio
J
1
/J
2 can be temperature dependent.
(Nature 2003)
Usually, one ignores phase locking. It is very difficult to really establish experimentally discontinuities in slope. 
9
T lock

When we consider a system with several vector spins per unit cell, we need a more formal approach. One introduces the susceptibility i. e. the spin correlation function:

( R ,

,

; R

,
 
,

)
 
S

( R
 
) S

( R
   
)
 where R labels the unit cell and
 labels sites 1, 2, … n in the unit cell.

( q ;

,

;
 
,

)
 
R

( R ,

,

; R

,
 
,

) e iq

( R
  
R
   
)
This is a 3n x 3n matrix. This matrix is a function of wave vector q.
Stat mech tells us that the leading term in the Landau expansion is
F

1
2
  q
   
 
1
( q ;

,

;
 
,

) S


( q ,

) S

( q ,
 
) where S

( q ,

)

1
N

R
S

( R
 
) e iq

( R
 
)
Warning: many people use e iqR
10

To repeat :
F

1
2
  q
   
 
1
( q ;

,

;
 
,

) S


( q ,

) S

( q ,
 
)
You can think of the inverse susceptibility matrix as being
 
1 
V int
( q ;

,

;
 
,

)
 kT


,



,
 
This is the generalization of temperature independent interaction
F

1
2
( T

T c
) S
2
Imagine diagonalizing the quadratic form F. In terms of its eigenvectors
 n
 

U n ;

( q ) S

( q ,

) the free energy is
F

1
2
 n k ( T

T n
) |
 n
|
2
The mode
 with the highest T n is the critical mode and its eigenvalue k(T-T n
) is the critical eigenvalue. This mode determines the structure of the ordered phase. Its amplitude is fixed by the terms of order S 4
(which we have not shown explicitly.)
11

To repeat:
F

1
2
  q
   
 
1 ( q ;

,

;
 
,

) S


( q ,

) S

( q ,
 
)
F

1
2
 n k ( T

T n
) |
 n
| 2
This is reminiscent of phonons, where
 1 plays the role of the dynamical matrix. The eigenvectors
 of this matrix are characterized by symmetry labels. The critical eigenvector gives the distribution of spin components over the unit cell when the system orders and within mean field theory the amplitude of the ordering is governed by the terms of order S
4
.
NOTE: if the ground state does not have the same symmetry as the state just below the ordering temperature T c
, then we can be sure that there is an additional phase transition below T c
. As the temperature decreases the eigenvector will have admixtures from other eigenvectors
of the same symmetry. If Q is the critical eigenvector and X another eigenvector of the same symmetry as Q, then

F
 a Q
3
X
 a

Q
 2
QX
 cQX
12

F

1
2
  q
   
 
1
( q ;

,

;
 
,

) S


( q ,

) S

( q ,
 
)
Here I am not going to discuss F from the formal point of view of group theory. As it happens, there are canned programs (which
I will refer to later) which do the group theory analysis. My aim here is to treat some simple examples where a knowledge of group theory is not necessary. The point of this discussion is to be sure that we understand the basic language of group theory so that we can check that we know how to use the output of such canned programs.
13
In these simple examples we will invoke the well-known principle for the eigenvectors of a matrix M. Namely, if we have a set of mutually commuting operators which commute with M, the eigenvectors of M are also simultaneously eigenvectors of the commuting operators.
(Group theory was invented to treat the case when the operators which commute with M do not commute with each other.) The point is that we definitely do not want to try to construct the inverse susceptibility and then diagonalize it. However, we will rely on the unambiguous experimental determination of the selected wave vector q.

14
Kenzelmann et al, PRB 76, 014429 (’06) Kenzelmann et al, PRL 95, 087206 (’05)
F
 cQ
 
2 n

1
Q ([ 2 n

1 ] q )
*  c
*
[ Q ( q )
*
]
2 n

1
Q ([ 2 n

1 ] q )
Time reversal says that the total number of powers of Q is even. We have wave vector conservation because Q(q)* = Q(-q).

Look at the phase diagram for an antiferromagnet with the z-axis being the easy axis in a magnetic field H along the z axis. If we lower the temperature ordering will be into either the low field state or high field state. Only if the field is adjusted to be exactly at the critical value will one access the bicritical (magenta) point.
“SPIN FLOP”
15
This example indicates that if we consider a phase transition when the temperature is lowered, we ought not allow an accidental degeneracy in which two different irreducible representations condense. In simpler language: we are only allowed to break one symmetry at a time .
If, unusually, two irreps condense simultaneously in a continuous
Transition, then what? The, the most probable explanation is that
There is a hidden symmetry that you did not identify .
Case in point: in a Coulomb potential the 2s and 2p orbitals (which have different symmetry) have the same energy. This is explained by identifying the hidden Laplace-Runge-Lenz symmetry vector. (Wikipedia)

Finding the eigenvectors of the commuting operators which leave

-1 invariant will reproduce the results of group theory.
Accordingly, let us consider the magnetic ordering of TbMnO
3
(TMO).
The ordering wave vector is along b which is called y here. q=(0.52
p
/b) y
16
= Mn = Tb
There are 8 spins per unit cell.

At right I show a schematic phase diagram of TMO.
Paramagnetic
Paraelectric
The structure of the HTI phase is a predominantly magnetically ordered collinear incommensurate phase described by irrep
G
3
(explained below).
P = spontaneous polarization
The low field LTI phase has additional contributions from
G
2
, so that a magnetic spiral is formed which gives rise to a ferroelectric phase.
=
LTI low
temperature incommensurate phase
17

18
irrep
If
 is an eigenvector, m x
  l
( m x
)
 m z
  l
( m z
)

S(1) S(2) S(3) S(4)
G l
(m x
) l
(m z
) x y z x y z x y z x y z
G
1
G
2
G
3
G
4
L
1 a b c a b c a b c a b c
-
L
-1 a b c a b c a b c a b c
-
L
1 a b c a b c a b c a b c
L
-1 a b c a b c a b c a b c
S(5) S(6) S(7) S(8)
G l
(m x
) l
(m z
) x y z x y z x y z x y z spin component labels x denotes -x
G
1
G
2
G
3
G
4
L
-
L
-
L
L
1 0 0 C 0 0 F 0 0 F 0 0 C
-1 A B 0 C D 0 C D 0 A B 0
1 0 0 C 0 0 F 0 0 F 0 0 C
-1 A B 0 D E 0 D E 0 A B 0
NOTES: All parameters are complex valued. The constants for different irreps are unrelated. So for denotes a
2
, etc.
G
1 has 5 constants:
G
G
1
1 a denotes a
1
, for
G
2 is contained
5 times in the original reducible representation.
G contained 7, 5, and 7 times respectively.
L
= exp(i p
2
, G q).
3
, and
G a
4 are

What about spatial inversion? This operation is not in the group of the wave vector because inversion takes q into –q. However, the free energy is invariant under inversion. It is amazing that the world wide neutron scattering community (including many famous people!) didn’t know how to take this into account. As a relative novice I uncovered this problem. In hindsight this problem had been solved long ago. But this important work was either forgotten or overlooked. There are several ways to take account of inversion. One way is to use the rather arcane formalism of corepresentations. Another is to use the full subgroups of the entire space group (rather than the group of the wave vector). This latter method has been used to develop a canned program which I will refer to in a moment. Meanwhile
I continue with the “poor man’s” approach to symmetry which does not invoke corepresentations.
To illustrate the simple approach, let us consider irrep
G
3
. I consider this irrep because it is the one that experiment identifies as the first magnetic irrep to condense as the temerature is lowered.
19

For notational simplicity, let a, b, c, C, and F be denoted x
1
, x
2
, x
3
, x
4
, and x
5
, respectively. Because we are considering a wave vector having only a y component, we only need keep track of how the y coordinate of position transforms. (Because spin is a pseudo vector its orientation is invariant under inversion: the only effect of inversion is to relocate the spin.) If we invert spin #1 (or #2, #3, or #4), we change the sign of its y component of position which takes exp(iqy) into its complex conjugate.
Thus
I x
1
 x
1
* , I x
2
 x
2
* , Ix
3
 x
3
*
If we invert spin #5, it goes into spin #7 and #6 goes into #8. So
I x
4
 x
5
* , Ix
5
 x
4
*
Invariance under inversion yields the relation
F
  nm
G nm x n
* x m
  nm
G nm
[ Ix n
*
] [ Ix m
]
This has to be true no matter what we choose for the x’s. Suppose only x
1 and x
2 are nonzero. Then
( G
12

G
21
) x
*
1 x
2

( G
21

G
12
) x
1 x
*
2

0

G
12

G
21
20

Note that G is Hermitian: G
21
= G
12
*. So if G
12
= G
Proceeding similarly we find that G is of the form
21
, then G
21
=G
12 is real.
a b c
   b d e
   c e f g g
    g g

  g   g
Where a, b, c, d, e, f, and g are real valued and the
Greek letters are complex valued. The eigenvectors of this form of matrix must be of the form
  e i

[ r , s , t ;

,

*
] where r, s, and t are real and
 is complex .
We included a phase factor because we are working in a complex vector space. In comparison with what we had before inversion: a, b, and c were complex quantities. But now we see that they must all have the same phase. The Tb amplitudes C and F, which were arbitrary complex numbers, now must be complex conjugates of one another. If one ignores the consequences of inversion symmetry, one has these phases to determine from the data and that proved to be a very difficult task, so invariably some guesswork was performed. Furthermore, when these constraints are ignored, the symmetry of the structure is wrong: a single irrep would then possibly induce ferroelectricity, a result we know to be incorrect.
21

1. Search for ISODISTORT and click on the entry
2. Construct crystallographic input file (ISODISTORT will prompt you) I upload the cif file prepared previously.
3. space-group preferences: I click OK, but maybe you can do better.
4. Types of distortions to be considered. Magnetic Tb and Mn.
Then OK.
5. Method 2: SM (0,a,0) a=0.52. I think # of inc. mod. = 1. Then OK
6. Pick IR. As Wills says, we may not be sure of the convention.
I took mSM3 because I thought that Kenzelmann took the best reference. Then click OK.
7. Finish selecting: I chose P1Z. The OK .
8. Click on Complete modes details. Then OK
9. Then “Magnetic mode definitions” gives the IR basis vectors.
22

23
4
1
2
3
4
1
2
3 a tom z
2
3
4
1
M x
M y
ME s s s s t t r t t r r r

G
3
5
8
7
6
5
8
7
6
ME
Mag. mom.
z x y z x y








24

25
We argued against having an accidental degeneracy so that the disordered phase would have simulataneous instabilities to ordering. In what I call
“the Green Bible” all `possible’ subgroups which can arise are tabulated.* (This is actually too bold a claim!
If you have a first order transition, I think anything is possible, at least in principle.) Here I give a simple mechanism for having two different irreps simultaneously appear at a first order phase transition.
A first order transition arises from
F
 a ( T

T
0
)

2  b

4  c

6
At right I show curves for F vs
 for a sequence of temperatures (black highest, blue lowest)
* H. T. Stokes and D. M. Hatch “Isotropy Subgroups of the 230 Crystallographic Space Groups”

Now suppose we have two different symmetry order parameters,
Q
1 and Q
2
, each of which, if they were uncoupled, would have a first order phase transition. With T
1
> T
2 and all parameters >0
F
 a
1
( T

T
1
) Q
1
2
 a
2
( T
 w Q

T
2
) Q
2
2
1
2
Q
2
2
 u
1
Q
1
4
 u
2
Q
2
4
 v
1
Q
1
6
 v
2
Q
2
6
It can be shown rigorously that if w is large enough this model has a single phase transition in which both order parameters simultaneously condense. To see that this is reasonable, suppose that Q
1
If w is large enough, then the coefficient of Q
2
2 , which is a
2
( T

T
2
)
 w

Q
1
 2 condenses.
26 can become negative, so that Q
2 is sure to also order.

In TMO we have an incommensurate colinear phase at 42K and then a spiral at 32K at which point one has a ferroelectric. In the Nature paper this was attributed to the hypothetical lock-in phenomena. Nearly at the
Same time we similarly analyzed
Ni
3
V
2
O
8
.
To discuss the symmetry of this situation it is helpful to introduce order parameters.
27

Landau theory for the higher T transition is as we have described it above.
For the lower T transition the vacuum is no longer the trivial state with zero ordering. In principle we need to discuss the symmetry of the second ordering in the presence of the first ordering. This is a nasty project!!
Consider the hypothetical phase diagram shown below.
T
<
T
>
T l
LTI
HTI
HTI =
G
3
LTI =
G
3
+
G
2
PARA
Conclusion: we can learn everything that we need by considering the
Landau expansion in the two order parameters: one for the HTI type of magnetic order and the second one to describe the additional ordering that comes in at the lower T.
28

The spin wave function for the unit cell (which is just the Fourier coefficient
At the wave vector q), has the form of the basis function whose coefficients may or may not be determined from a refinement of scattering data. What we can say is that
S ( q )

Q
1
( q )

1
( q )

Q
2
( q )

2
( q ) and S (
 q )

S ( q )
* where
 n with n=1,2 are the basis vector for the two irreps. Note that of course the Q’s are complex because their phase regulates where the origins of the waves are placed. If we had only one wave, this phase
would be unobservable. So really what is important is the relative phase.
Initially we do not treat the interactions between irreps and write
F

( T


T ) | Q

|
2  u | Q

|
4 
( T


T ) | Q

|
2  u

| Q

|
4  
F
There is a problem with this formulation: why should the same value of the wave vector be selected for the two irreps (which predominantly describe ordering in different directions??? Surely the exchange interactions for, say, S x are different, if only slightly, than those for S y
. Solution: only if q
>
=q
<

F
 aQ

( q

)
2
Q

(
 q

)
2  c.
c .
 q

 q

29


F
 a
{
Q

( q )
2
[ Q

( q )
*
]
2 
[ Q

( q )
*
]
2
Q

( q )
2 }

2 a | Q

( q ) Q

( q ) |
2 cos [ 2 (


 

)]
 

 


( n

1 / 2 ) p
The two irreps are out of phase – this is what is needed for a spiral.
But this actually follows from simple hand-waving arguments: As the temperature is lowered, the spins more and more want to have fixed length (as they do at T=0). Thus when spin ordering from one irrep is large, the spin ordering from the other irrep must be small.
In phase
30
Out of phase

If
 is an eigenvector, m m z x



 l l
( m x
( m z
)
)


G l
(m
G
G
1
G
2
G
3
4 x
) l
(m
L
1
-
L
-1
-
L
1
L
-1 z
)
Let Q
3 then Q
3 be the complex amplitude (order parameter) for
* will be that for
G be that for
G
2
G
3 and Q
2 that for
G
2
3
* and Q
*. Note also
I
Q n
= Q
2 n
.
* will
*.
L
We want to see how the Q’s can be combine with the spontaneous polarization P to induce a nonzero value of P. To induce a nonzero value of P the interaction should be linear in P. For time reversal invariance we must have an even number of Q’s To conserve wave vector one will be Q and the other will be Q*=Q(-q). We can not have Q n
Q n
* because this transforms like unity. So

F
 
 c

P

Q
2
*
Q
3
 d

P

Q
2
Q
3
*

F
 i

 r

P

[ Q
*
2
Q
3

Q
2
Q
*
3
]
F is real, so c

=d

*. Under inversion
Q
2
*Q
3
-> Q
2
Q
3
* and P -> -P, so c

= -d

= i r

, where r
 is real.
The square bracket is even under m x
, odd under m z
Therefore only r z and odd under inversion.
can be nonzero. The polarization is found to lie along z!!
31

1. Landau theory provides a natural way to understand the phenomenology of phase transitions.
2.
An alternative to group theory can provide a convenient route to categorizing the symmetry of system whose symmetry is not too complicated. Of course, group theory provides the definitive answers concerning symmetry.
3. Where Landau theory becomes essential is when one
Investigates coupling between several different symmetries.
4. The principal of “no accidental degeneracy” has wide application.
5 A corollary: “if it allowed by symmetry, it happens.”