Symmetry Analysis of Multiferroics

A. B. Harris (NIST, 2007)

The objective of this set of lectures is to describe in detail how to apply symmetry considerations to the magnetic and ferroelectric states of multiferroics which have a combined magnetic and ferroelectric phase transition. The aim is to do this is as transparent a way as possible.

1

OUTLINEI. Setting the Stage

II. Symmetry of the Quadratic Free Energy

A. Physics of this Free Energy

B. Eigenvector Condition (TbMnO3)

C. Group of the Wavevector (TbMn2O5 , YMn2O5)

D. Impose Inversion Symmetry

E. Introduce and Analyze Order Parameters

III. Consider Quartic Terms in the Free Energy

IV. Construct the Magnetoelectric Coupling

using the Symmetry of the Order Parameters

V. Discussion

2

HISTORICAL REMARKS

Prior to our 2005 Physical Review Letter1

it was not recognized that, in addition to the

symmetry operations which leave the wave-

ector invariant, one could exploit inversion

symmetry. (This may have been recognized

in some quarters, but certainly none of the

multiferroic community knew it!) This is docu-

mented in Ref. 2, on which these notes are

based and to which citation may be made.

A reader who understands representation theory can skip to slide #27!!

3

INVERSE SUSCEPTIBILITYThe susceptibility is a key quantity in continuoustransitions like those weare about to study. One isused to saying that the susceptibility diverges at acontinuous transition. Alternatively, one can plotthe free energy, F, (see the appendix) of a ferromagnet versus the magnetization, M. By time reversal only even powers of M can appear.The actual value of M is theone which minimizes F.

Free energy F = (T-Tc)M2 + uM4

for a sequence of temperatures

T< < Tc < T1 < T2. Note that the

free energy is unstable relativeto the formation of long-range

order for T < Tc

4

INVERSE SUSCEPTIBILITYMHuMAMF 42)2/1(

11 AHAM

21)2/1( MF

For small H the magnetization M is

The coefficient of M2 is the inverse susceptibility:

21 |)(|)(2

1qMqF

q

In the Appendix we allow for all Fourier components:

Ferromagnet

For a ferromagnet the instability first occurs at q=0 as one reduces the temperature.

T2 > T1 > Tc

When the first Fourier componentcondenses, the others are inhibitedfrom condensing because there isa finite amount of spin to go around.

5

INVERSE SUSCEPTIBILITY

The instabilityoccurs at wave-

vector q=/a

AntiferromagnetFor the antiferromagnet the instability leads toup-down-up-down spins, i. e. aq = . Observethat if you have a plotof the susceptibilityversus wavevector,you don’t have to knowwhether the system isferromagnetic or anti-ferromagnetic: thelocation of the minimumin inverse chi tells youwhich it is.

6

WAVEVECTOR SELECTION:INCOMMENSURATE SPINS

For competing interactions, the minimum in inverse chi can be anywhere in the zone, as at right.

Incommensurate Magnet

The minimum in inverse chi locates the wavevector of the ordered state.

This is often referred to as “wavevector selection.”

So far we have treated M as a scalar, as it is for the Ising model. What happens if M is a vector?

T = Tc

T = T1 > TcT = T2 > T1

7

INVERSE SUSCEPTIBILITY

,)(O][||][2

1 4222 MMMKMTTF yxc

Suppose we have vector spins with an easy axisalong z. Then

Where K>0 is the anisotropy energy. Here as T

approaches Tc from above, only Mz becomes

unstable relative to long-range order. Asanother example suppose the free energy is

MMKMTTFzyx

,,

20 ||)(

2

1

(Continued next slide)

8

How do we deal with

?||)(2

1

,,

20

MMKMTTF

zyx

Now we have an anisotropy tensor, which we can diagonalize to find the anisotropies associated with each principal axis. Keeping only the order para-meter associated with the easiest axis, , we haven̂

)(O]ˆ][[2

1 42 MnMTTF c

This example teaches us that if the free energyis a quadratic form, ordering takes place in thechannel defined by the eigenvector associatedwith the eigenvalue that is vanishing.

INVERSE SUSCEPTIBILITY9

INCOMMENSURATE CHI

,)]([exp),(),( RiqqsRSq

Now we consider the free energy of a systemwhich has n>1 vector spins in the unit cell:

Let S(R,) denote the -component of the

spin at position within the unit cell at R. Then

where s(q,) is complex with s(-q,)=s(q,)*.The quadratic free energy is the Hermitian form

.),(),()( *;

qsqsqFF

This quadratic form has complex-valued eigen-vectors which we can normalize, but they havean arbitrary phase factor (which I will discusslater). Obviously we focus on the critical eigen-value, i. e. the one that first becomes unstable.

10

INTERPRETATIONFor each value of the wavevector we have 3n eigen- ectors and 3n eigenvalues. As the temperature is lowered we have wavevector selection: at some q, one of these eigenvalues will become unstable. This determines the wavevector of the ordered state (at

least for temperature just below Tc). Now consider

the eigenvector. Its normalization is not fixed by the quadratic free energy. The quartic terms fix the nor- malization and give the square root temperature dependence characteristic of mean field theory.

Reminder: all the coefficients of the critical eigen-vector (which represents the Fourier transform ofthe spin distribution) are complex-valued. Note:we reject the possibility of accidental degeneracy.Degeneracy can only occur if symmetry requires it.

11

INTERPRETATIONSo we write the eigenvector as

1,],,;

,,;,,[

;2;1;;

2;1;,2;1;

nzzzny

yynxxxi

ssss

sssssXe

where the components inside the square bracketare normalized so that the sum of the squares oftheir magnitudes is unity. I will discuss the phase

later. The amplitude X varies as (Tc-T)1/2.

s is the complex amplitude of the wave

of the -component of the sublattice. Sothe eigenvector tells us the pattern into whichthe spins order, as we will see in a moment.

12

APPLICATIONOne might attempt to write down and anlyze the matrix of coefficients F. That is a hopeless task because ourmicroscopic models are not reliable. Instead, we usethe symmetry of the crystal to limit the possible eigen- vectors which are determined by fitting the diffraction data. The standard approach is to invoke group theory.I will avoid that as much as possible. Instead I willinvoke the principle that first-year graduate students

learn: if a set G1, G2, …of mutually commuting opera-

tors commute with the Hamiltonian, then the eigen-vectors of the Hamiltonian can be classified according

to their eigenvalues g1, g2, … In the language of group

theory, this case is the case when all the irreducible representations are one dimensional.

13

CALCULATIONThe essential input we require is the symmetryof the incommensurate wavevector (usuallyas determined by neutron diffraction) i. e. thesymmetry operations of the crystal which alsoleave this wavevector invariant. Since we areconcerned with phase transitions in whichferroelectricity appears in coincidence withmagnetic ordering, spatial inversion I is asignificant symmetry operation, although it doesnot usually leave the wavevector invariant. So I’m afraid we have to actually confront the crystal structure of the paramagnetic phase.

14

EXAMPLE A: TbMnO3

The space group opera-tions3 are defined in Table I and the ion positions are listed in Table II (x=0.9836,y=0.0810)4. The figure below shows the mirror plane

(z=1/4) for mz and the glide

plane (x=3/4) for mx with

associated translation

uy = b/2.

From neutron diffraction5,6

the wavevector lies along b (i. e. y) and in rlu is q= 0.28. The wavevector is invariant

under mx and mz.

Table II: Mn and Tb positions.4

= Mn

= Tb

15

Table I: General positions forPbnm (TMO).3,4 Here x means –x.

Eigenfunctions for TbMnO3 (TMO)

In Table III I give the eigenfunctions for TMO. First I explain how to read this table. In each column headed by the symmetry label (n) I list the components of eigenfunctions on the sites 1 – 8 as labeled in Table II. The three numbers in each box are forthe x, y, and z components. These eigenfunctions are eigenfunctions of mx and mz with respective

eigenvalues (mx) and (mz). Here= exp(iq), where q is the incommensurate wavevector in reciprocal lattice units (rlu), soin ``real” units, it is 2q/b.

Table III: Eigen-vectors. = exp(iq).

16

EIGENFUNCTIONS

The eigenfunction for 1 has

five parameters xM, yM, zM, zT1,

and zT2 indicating that there

are five independent eigenfunctions of this symmetry. So far, there is no restriction on these five complex parameters. As we will see below, inversion symmetry places important constraints on these parameters.

On the next slides we will show how the eigenvectors are easily constructed.

Table III (repeated)

17

To construct eigenfunctions of mx and mz it is convenient to write in terms of subvectors as

)2(,],,,,,[ )()()()()()( TMTMTM ZZYYXX

where each symbol is a vector with four components.

X(M) has the x-components of the four Mn sublattices,

Y(M) and Z(M) the y and z components, respectively, of

the four Mn sublattices. Similarly, X(T), Y(T) , and Z(M)

have the x, y, and z components, respectively, of the four Tb sublattices. The representation of Eq. (2) isconvenient because the symmetry operations onlyinterrelate the four components within a given symbol.Consequently, each eigenfunction has components only with in a single four-component subvector.

EIGENFUNCTIONS 18

EIGENFUNCTIONS The main ideas involved in determining how spinfunctions transform are: 1) The transformed

spin at r is obtained by OS acting on the spin

which was at the original location [OR]-1 r, 2)

exp(iqr) is a number and O does not act on it,and 3) the subscripts ``i” and ``f” mean ``before”and ``after” transformation. In Table IV we listthe initial and final sites under the operationswe will invoke.

Table IV (at right): initial site, i,

sites, f, after mx, mz, and I. The

sites are numbered according to Table II.

19

EIGENFUNCTIONS

To construct the eigenfunctions of mx and mz,

we need to discuss how the symmetry opera-tions act on Fourier transforms of spin functions

like sx(q,1). First of all, spin is a pseudovector,

so the mirror mdoes’t reverse the sign of the

-component of spin but does reverse the signof the other components. Let O denote an

operator acting on spin and space, OR (OS) an

operator acting only on space (spin), and lowercase denote a Fourier transform. On the nextslide we indicate the effect of a symmetry

operator O=OR OS on a Fourier transform.

20

EIGENFUNCTIONS

)],([)]([exp

]),([)([exp

)],([)]([exp),(

1

111

,1

,

iSiRP

fRRSfR

iiiffR

ff

PSOPOiqN

OROSORiqN

RSORiqNqsf

OR = a point group operation GO + a translation uO:

so that (since GO leaves q invariant)

Oi

OiOOiOiR

uqPq

uqPqGuqPGqPOq

)(

)(][)()( 1

Thus, the final result is

OiOiR uPGPO )()(

)3(,),()]([exp)(exp),( 1iSi

RO RSORiqNuqiqOs

21

EIGENFUNCTIONS The result of the previous slide was

For mz, u0=0, and for mx, q.u0=q (recall rlu!).

Also, OS gives a plus or minus sign. Thus

where the sign depends on O and and (’)is the initial (final) sublattice. Thus for thepseudovector spins, with = exp(iq),

)',()exp(),( qsuiqqOs O

where = -1, except that = +1, so that

mz2 = 1 and mx

2 = 2 . Note: ( ’)’=.

)5(,)',(),(

)4()',(),(

,

,

qsqsm

qsqsm

zz

xx

22

)3(,),()]([exp)(exp),( 1iSi

RO RSORiqNuqiqOs

EIGENFUNCTIONS

.)(1,/)(1 2zzzxxx mmPmmP

Now we show how the eigenvectors are constructed. For instance, to construct the eigenvector scaled by

xM, we assign m1x the value 1 and then deduce the

other components according to the symmetry ( ‘s). So we introduce projection operators

To check the effect of Px, use mx2 = 2 = (mx)2:

d.e.q.])[(

]/)(1)[(

)](/1)[(

]/)([

]/)(1[][

2

22

2

xx

xxx

xxx

xx

xxxxx

Pm

mmm

mmm

mm

mmmPm

23

EIGENFUNCTIONS

Thus P PxPz projects an eigenfunction of both mx

and mz with respective eigenvalues (mx) and (mz).

We extend the notation of Eq. (2) so that (a,b,c,d)sT

represents a subvector S(T), whose four componentsare a, b, c, and d, and where s=x, y, or z (tolabel components) and T=M or T (for Mn or Tb).We now illustrate the use of these projection opera-

tors to construct eigenfunctions of mx and mz. To

get the eigenfunction scaled by xM , we let the

projection operator act on X(M) = [1,0,0,0]xM in the

notation of Eq. (2):

24

EIGENFUNCTIONS

.]/)()(),(,/)(,1[

]/)()(),(,/)(,1[

]0),(,0,1][/)(1[]0),(,0,1[

]0,)(,0,1[]0,0,0,1][)(1[]0,0,0,1[

22

2

Mxzxzx

Mxxxzxzxxx

Mxzx

Mxzx

Mxzxzx

Mxzzx

Mxzx

mmmm

mmmm

mmmP

mPmmPPP

In the first line we used Table IV which tells us that mz

takes site #1 into site #3 and Eq. (5) with zx=-1. In

the third line Table IV tells us that mx takes site #1 into

site #2 and #3 into #4 and we used Eq. (4) with xx=1.

The above eigenvector regulates the x-components on

the Mn sites. To get the amplitudes for 1 set (mz)=1

and (mx)=. This leads to =[1,1,-1,-1]xM in agree-

ment with the eigenvector of Table IV scaled by xM.

25

EIGENFUNCTIONS

To further illustrate the construction of the eigen-vectors we apply the projection operator to the

Tb wavefunction [1,0,0,0]xT :

Txxzxxxz

Txxxz

Txzx

Txzxzx

Txzzx

Txzx

mmmm

mmm

mPmP

mmPPP

]/)(,0,0,1)][(1[]/)(,0,0,1)][(1[

]0,0,0,1][/)(1)][(1[

]0,0,0,1)][(1[]0,0,0,1][)(1[

]0,0,0,1][)(1[]0,0,0,1[

2

2

So for (mz)=+1, there are no eigenvectors of the

x-component. Otherwise the eigenfunction coversonly sites #5 and #8. One can verify that theabove result agrees with Table III. Enough checks!

26

EIGENFUNCTIONS: SUMMARYFor convenience I repeat Table III at right. For those rejoining here, each column contains the most general eigenvector having the symmetry specified either by the eigenvalues of the symmetry operators, or by the irreducible representation . At this stage of the development the 5 (or 7) free parameters may assume arbitrary complex values. The guiding principle is that we do not admit the possibility of accidental degeneracy. When ordering takes place only one symmetry eigen-vector can condense. The Fouriertransform of the spin distribution isproportional to this ``critical” eigen-vector. The symmetry () and theamplitudes (x, y, z) are determined tobest fit to diffraction data.

27

INVERSION SYMMETRY Up to now our results are absolutely standard. It is often said (in reviews7) that the only symmetries that can be exploited are those that leave the wavevector invariant. This is not so because when one has inversion symmetry, the free energy clearly must be invariant with respect to inversion.1,2,8 We nowanalyze the consequences of that. We assume that only a single representation iscritical and write the free energy as an Hermitianform in in the amplitudes (which we here denote

Qi) of one of the columns () of Table III:

.)()( *

,

)( nn

mnnm QQFF

28

So Q1() = xM, Q2() = yM, etc.

INVERSION Clearly, we now have to determine how inversion Itransforms the coordinates Q. For that purposewe modify the derivation of Eq. (3) so that itapplies to I. Replace ``leaves q invariant” by``changes the sign of q.” Thereby we get

),()]([exp),( 1iSi

Rf RSIRiqNqIs

)6(),()],()]([exp[ **1iii

R

qsRSRiqN

What does this mean? The initial and final sitesare given in Table IV. For the Mn sites, they are at

centers of symmetry so f = i. So Is(q,n)=s(q,n)*

for Mn sites (n=1,2,3,4). For the Tb sites the pairsof initial and final sites are 5 and 7 and 6 and 8, so

Is(q,5)=s(q,7)* and Is(q,6)=s(q,8)*.

29

INVERSIONNow we deduce the effect of inversion on the sym-

metry adapted coordinates, denoted Qn(), which

are the amplitudes in terms of the eigenfunctionsof Table III. As an illustration we consider the

coordinates of 1 from column 1 of Table III:

Mxxxx xqsqsqsqsQ 22/)]4,()3,()2,()1,([)( 11

Myyyy yqsqsqsqsQ 22/)]4,()3,()2,()1,([)( 12

Mzzzz zqsqsqsqsQ 22/)]4,()3,()2,()1,([)( 13

114 22/)]8,()5,([)( Tzz xqsqsQ

215 22/)]6,()7,([)( Tzz xqsqsQ

30

INVERSION

spins)(Mn1,2,3n)()( * nn QQI

.)()(;)()( *1415

*1514 QIQQQI

From the discussion of Eq. (6) one sees that

)7()](][)([)()()( 1*

1,

1*

1,

1 mnmn

nmmnmn

nm IQIQFQQFF

Now write the free energy (for ) in terms of

the Q’s and require invariance under inversion:

Here and below we invoke the convention thatRoman letters are real and Greek ones complex. Thus

31

INVERSION32

To repeat:

We now need to study how inversion affects the

matrix. For example, IQ1* Q4 = Q5* Q1. After inversion

the 5,1 element of the matrix is the old 1,4 element.

A simpler example: IQ1* Q2 = Q2

* Q1, so after inversion

the 2,1 element is the old 1,2 element.

INVERSIONSince these two forms have to be equal no matterhow we choose the Q’s, the matrices have to beequal. Thus and must be real and

.,,, *** de

Thus the free energy is of the form (with Romanletters real and Greek ones complex).

The critical eigenvector is an eigenvector of theabove matrix.

33

INVERSIONTo repeat: we found the free energy to be

Because of the special form of this matrix (due toinvoking inversion symmetry) the eigenvector isshown in the Appendix to be of the form6

,],,,,[)exp()( *1 QcbaiA

where the subscript Q says that the components are

Q1, Q2, Q3, … and we normalize by a2+b2+c2+2||2=1.

Since the eigenvalue problem takes place in acomplex vector space, we can not omit the arbitrary phase . We discuss this result below.

34

INVERSIONWe now discuss the result of the last slide:

)a8(,],,,,[)exp()( *1 QcbaiA

Previously (in Table III) we showed that symmetry

led to an eigenvector of symmetry 1 which had

five complex coefficients. Now we see that apartfrom an overall phase factor, three of these arereal, and the other two (for Tb), rather than beingindependent complex numbers are the complexconjugates of one another: So instead of having5 complex amplitudes to fit the diffraction data,we have three real and one complex parameter.

The result for = 3 is similar. The result for

24) is (see the Appendix)

35

)b8(,],,,,,,[)exp()( **2 QcbaiA

INVERSIONSo at right we have updatedTable III to take account ofinversion symmetry. The eigen-

functions of each symmetry n

should be assigned an inde-

pendent phase factor n and

amplitude An. If only one

symmetry is present, the phaseis usually irrelevant. But if twosymmetries are simultaneouslypresent, the phases matterbecause the relative phase hassignificance, as we shall see.Reminder: Roman letters arereal and Greek ones complex.

As Table III, but withInversion symmetry.

36

ORDER PARAMETERSWe now discuss how the magnetic structure can bedescribed by order parameters.2,8 Suppose that the

structure has n symmetry and the spin wavefunction

is put into the form of Eq. (8). How do we think aboutan order parameter? Here the order parameter has tobe complex because any phase is allowed in Eq. (8).As a function of temperature the normalized eigen-vector will be essentially constant, and only themagnitude A will vary. So we identify the complexorder parameter as

37

,)exp()( nnn iAq

where An is real.

ORDER PARAMETERSThe order parameter inherits the symmetry of the

representation n. Thus3,2for;1,4 for nmnm nnxnnx

4,2for;1,3for nmnm nnznnz )9(.allfor,)((-q), ** nqI nnnn

At right is the phase diagram9 of TMO.

As one cools at H=0 throughT> = 40K,

order in 3 appears6 in the HTI

(high-temperature incommensurate)phase.8 Then at T=28K additional order

in 2 appears6 in the low temperature

(LTI) phase. Cooling at H>10T leadsto a different phase. The polarizationis nonzero only in the LTI phases.

38

Para

P=0

TMO

MAGNETIC STRUCTUREHere I discuss qualitatively the magnetic structure.Neutron diffraction6 indicates that in the HTI phase

the ordering with 3 (i. e. 3) involves the spins

pointing along the b (i. e. y) axis. Since the a, b,and c directions are inequivalent, one axis mustbe easiest, and obviously this is the b axis. Notethat the pattern of sublattice spins is (++--). Thequartic terms in the free energy favor fixed spinlength8 which is not consistent with a collinearsinusoidal state. Thus, it is not surprising8,10 thata phase transition occurs at lower temperaturewhere transverse order appears along the c axis

scaled by the order parameter 2, for 2. The

pattern of these transverse moments is again(++--) indicating nearly isotropic interactions.

39

PHENOMENOLOGY

412

223 )())(2/1())(2/1( iSNTTTTF

This behavior is consistent with the followingsimple phenomenological Landau expansion:

40

and is consistent with the famous review of incommensurate systems by Nagamiya.10

It is not obvious that the above free energyensures that the wavevectors of the two orderparameters are equal. If the interactions arenearly isotropic, one would expect this to bethe case. Within Landau theory the term thatlocks to two order parameters is of the form8

,])()()()([ 222

233

222

233 qqqquV

where u<0 and this term only exists if q2 = q3.

MAGNETOELECTRIC COUPLING41

Now we discuss the Landau theory of magneto-electric (ME) coupling V.1,2,8 Order B is induced byorder A if there is a term in the free energy linearin B times 2 (or more) powers of A. Time reversalinvariance ensures that such a term linear in themagnetization can not occur, so the polarization Pcan not induce magnetic order. But magnetizationcan induce P. The interaction we seek is thereforeof the form V = ss P and the ME free energy is then

where the electric susceptibility E is finite since

we assume that without magnetism there islittle tendency to have a spontaneous polarization.

,/)2/1(2

ME VPF E

ME COUPLING

,/)2/1(2

ME cssPPF E

The ME coupling is schematically of the form

42

Where c is a constant. Now minimize with

respect to P

,sscP

so this interaction does just what we want:it induces a polarization only when ss isnonzero.

ME COUPLING43

We drop the Umklapp term with G nonzero (forwhich see Ref. 11). Then we have

)10()()(

PqqcV mnnmnm

,)'()()',( ,''

GqqmnnmqqnmG

PqqqqcV

where G is a reciprocal lattice vector and delta isunity if q+q’=G and is zero otherwise.

As we argued, the magnetoelectric interaction Vis linear in P and quadratic in the magnetic orderparameters. To conserve wavevector V is

ME COUPLING44

PqqaqqaV srsr )]()()()([ *

Warning: the discussion that follows applies toTMO. For different symmetry systems the MEInteraction need not assume the form found be-low.2,12 In TMO, V must be invariant with respect

to mx, mz, and I. To handle the cases that will be

of interest to us, let us assume that there aretwo irreducible representations (irreps) present.

r and s. In Eq. (10) the term with n=m is zero

because it changes sign under I. Physically thisreflects the fact that a single wave has a centerof symmetry which, for an incommensurate q,can be as close to an inversion center of thelattice as you like. Thus

ME COUPLING

One can show, using Eq. (9) that V is invariant under mx.

PqqaqqaV )]()()()([ 23*

23

45

This has to be invariant under mz. Using Eq. (9) we have

])]}[()][([)]()][([{

]}[)]()][([)]([)]([{

23*

23

23*

23

Pmqqaqqa

PmqmqmaqmqmaVm

z

zzzzzz

This equalsV only if mzP=-P, i. e. for =z (i. e. c). So

zzz PqqaqqaV )]()()()([ 23*

23

To repeat:

ME COUPLING

zzz

zzz

Pqqaqqa

PqIqIaqIqIaVI

)]()()()([

])]}[()][([)]()][([{][

2323*

23*

23

We have V as

zzz PqqaqqaV )]()()()([ 23*

23

This has to be invariant under inversion. We have

To have IV = V, az must be imaginary: az=ir, with r real.

We set n(q) = |n| exp (in). Then

46

)11(.sin||2 2332 zPrV

The polarization in TMO (at low H) is only found8 inthe LTI phase where it lies along c (i. e. z) as in (11).

ME COUPLINGThe appearance of the relative phase is easilyunderstood. The phase of the irrep regulates the location of origin of the cosine wave. The wave has inversion symmetry about this point. So, ifthe two irreps have their origin at the same point,

i. e. if 2 = 3, then the entire structure has

inversion symmetry about a lattice site.

The fact that P is proportional to |s2 s3| also agrees

with experiment8 in that the temperature depend-

ence of Pz looks like that of an order parameter

and not (order parameter)2. This happens because

the HTI order parameter s3 is more or less

saturated by the time one gets into the LTI phase.

47

HIGH FIELD PHASE48

In the high field phase (see the figure on slide #38)the polarization lies alog a (i. e. x) in the analog ofthe LTI phase. Thus the magnetic structure must

be 3 + n such that their product transforms like

Pa. For the product to be odd under mx, we need

n=1 or 4. For it to be even under mz, we need

n=1 or 3. So this analysis indicates that n=3.(This was first predicted in Ref. 13.)It should be noted that having a different irrepdoes not necessarily mean a totally unrelatedstructure. It may simply mean that the structurehas been rotated. Because different spincomponents transform different under symmetryoperations, this causes a change in the irrep.

TbMn2O5 (T25) 49

T25’s space group is Pbam,14,15

whose symmetry operationsare in Table VI and whose ionpositions are in Table VII.

Table VI. Pbam spacegroup operations.3

Table VII. Positions of spins in T25: x=0.09,y=-0.15, z=0.25,14

X=0.14 and Y=0.17.15

Since the irreps for T25 are twodimensional, the simple methodused above is clumsy.2 An alternate way, used here, is tosupplement the standard analysisby including inversion symmetry.

Below T=43K, T25 has a commen-surate phase with q=(1/2,0,1/4)and an incommensurate phasewith q=(0.48, 0, 0.32).16

T25The standard analysis (withoutinversion symmetry) for the

wavevector (0.50, 0, qz) leads to

Table VIII, which is read just asTable III, except that the twocolumns do not represent dif-ferent symmetry irreps, but rathertransform like two columns of thesame irrep. Since the two columnsare degenerate in free energy,the actual spin function is alinear combination (determinedby a fit to diffraction data) ofthe two columns. The complex

coefficients 1 and 2 are the

order parameters.Table VIII for T25.

50

T25 TRANSFORMATIONSBefore discussing inversion symmetry let usconsider the transformation properties of theeigenfunctions of Table VIII. To do that we needto know how the spin functions transform underthe symmetry operations. Here it is convenientto redefine the spin Fourier transforms so that

the exponential is exp (iq . R) rather than

exp [iq.(R+)]. With this change, a transformation1) changes the orientation of the spin, 2) takesthe spin from an initial sublattice to a final sub-lattice, and 3) introduces a phase factor (givenIn columns 4, 6, and 8 in the Table IX, below)depending on the position change. (See Ref.2.)

51

T25 TRANSFORMATIONS

Table IX. As Table IV.For exp(i) and exp(i’)see the text. Inversion

takes q into – q. ab is

not included.

In Table IX we give the way thesublattices transform. The spin

transformation factor (for

the symmetry operation m) are

the same as before: =1 if and is -1 otherwise. Thus

)1,()2,(

)2,()2,()3,(

)3,()3,()1,(

qsqsI

qsqsqsm

qsqsqsm

yy

zzyzzy

xxxxxx

52

In the first equation, xx=1,

and in the second, yz=-1.

One needs the phase factorsin Table IX, taken from Ref. 2. I does not carry a factor.

INVERSION SYMMETRY

.),(),(),;()2/1( 1 mqsnqsmnqFnm

Now we discuss the effect of inversion symmetry.To do this we will study the free energy when it iswritten in terms of the symmetry adapted coordi-nates of Table VIII. If we take the 12 symmetrycoordinates of column #1, then the free energy is a

quadratic form with a 12 by 12 matrix Fnm. In terms

of Fourier components the free energy is written as

53

For convenience I will absorb the ½ into -1.Now we introduce symmetry adapted coordinatesfor the first column of Table IX as we did in slide#30. These are listed on the next two slides.

SYMMETRY COORDINATES

2/)]8,()7,([,2/)]6,()5,([

2/)]12,()10,([,2/)]11,()9,([

2/)]4,()2,([,2/)]3,()1,([

65

43

21

qsqsQqsqsQ

qsqsQqsqsQ

qsqsQqsqsQ

xxxx

xxxx

xxxx

2/)]8,()7,([,2/)]6,()5,([

2/)]12,()10,([,2/)]11,()9,([

2/)]4,()2,([,2/)]3,()1,([

1211

109

87

qsqsQqsqsQ

qsqsQqsqsQ

qsqsQqsqsQ

yyyy

yyyy

yyyy

aqsqsQqsqsQ

qsqsQqsqsQ

qsqsQqsqsQ

zzzz

zzzz

zzzz

12.2/)]7,()8,([,2/)]6,()5,([

2/)]10,()12,([,2/)]11,()9,([

2/)]4,()2,([,2/)]3,()1,([

1817

1615

1413

Below we give the Q’s which transform like thefirst column of the two-dimensional irrep.

54

SYMMETRY COORDINATES

2/)]7,()8,([,2/)]5,()6,([

2/)]11,()9,([,2/)]12,()10,([

2/)]3,()1,([,2/)]4,()2,([

65

43

21

qsqsRqsqsR

qsqsRqsqsR

qsqsRqsqsR

xxxx

xxxx

xxxx

2/)]8,()7,([,2/)]6,()5,([

2/)]11,()9,([,2/)]12,()10,([

2/)]3,()1,([,2/)]4,()2,([

1211

109

87

qsqsRqsqsR

qsqsRqsqsR

qsqsRqsqsR

yyyy

yyyy

yyyy

bqsqsRqsqsR

qsqsRqsqsR

qsqsRqsqsR

zzzz

zzzz

zzzz

12.2/)]8,()7,([,2/)]6,()5,([

2/)]11,()9,([,2/)]12,()10,([

2/)]3,()1,([,2/)]4,()2,([

1817

1615

1413

55

Below we give the Q’s which transform like thesecond column of the two-dimensional irrep.

SYMMETRY OPERATIONS

)13(.,

,,

nnynny

nnxnnx

QRmRQm

RRmQQm

Now we explicitly display the effect of thesymmetry operations using Table IX and

= 1 if but is -1 otherwise:

56

Using Table IX one can determine the effect ofinversion on the Q’s and R’s. We find that

,*

fii nnn RIQ where the necessary data is in Table X.

Table X. Transformation parameters for inversion.

DOUBLE DEGENERACY

mnmyyn

mynymn

QFQQmFmQ

QmFQmRFR

|||][|

||||1

The meaning of a two-dimensional irrep is thatthe free energy matrix F (shown on the nextslide) for the R’s is exactly the same as thatfor the Q’s:

This guarantees that all eigenvalues are doublydegenerate. What this means is that symmetrydictates that these two modes (or any linearcombination of them) become unstable(relative to ordering) simultaneously. Thus thefree energy in terms of their order parameters is

)O(]||||[))(2/1( 422

21 cTTF

57

F =

Here Sn is a 4 x 4 real

symmetric matrix, Rn

a 4 x 4 real matrix.The other matricesassume the formsshown below.

where a is real.

58

Below we will verify this form. Meanwhile, see that F acting on , given below, yields the a vector ofthe same form. Roman = real, i2=-1.= [a,b,c,d; *; e, f, g, h; *; im, in, io, ip; *]

VERIFYING THE FORM OF F

6115161*6115

*6

515*1515

*1 ][][

HHRHRRHR

IQHIQFIQHQF

2112121*2112

*2

212*1212

*1 ][][

HHRHRRHR

IQHIQFIQHQF

7117171*7117

*7

717*1717

*1 ][][

HHRHRRHR

IQHIQFIQHQF

1,1817,111,17*17117,1

*18

1717,1*11717,1

*1 ][][

HHRHRRHR

IQHIQFIQHQF

5656656*5656

*5

656*5656

*5 ][][

HHRHRRHR

IQHIQFIQHQF

6655166*6655

*6

555*5555

*5 ][][

HHRHRRHR

IQHIQFIQHQF

1,1313,111,13*13113,1

*13

1313,1*11313,1

*1 ][][

HHRHRRHR

IQHIQFIQHQF

(a)

(b)

(c)

(d)

(e)

(f)

(g)

Now we use invarianceunder inversion I:

NOTES: When we have a result of the form RnXRm, then

we can conclude (in the second line of these equations)

that X=Hnm. In (a) H12 is real. In (b) H15=H16*. In (d)

H1,13 is imaginary. In (e) H1,17=-H1,18*. No info in (g).

59

T25 SPIN STRUCTURENote that the spin Fourier transform comes from alinear combination of the Q’s and the R’s. Sowe parametrize the eigenvector as (see bottom ofslide #57)

60

= 1 [a,b,c,d; *; e, f, g, h; *; im, in, io, ip; *]Q

Here the subscript Q indicates that these are the

values of Q1, Q2, … Q18 and we will normalize the

vector so that the sum of the magnitudes squaredof its components is unity. The matrix for the R’sis identical to that for the Q’s. This guaranteesthat the Q mode and the R mode become unstablesimultaneously and that the critical R eigenvectorhas the same form as .

T25 SPIN STRUCTURE

= 2 [a,b,c,d; *; e, f, g, h; *; im, in, io, ip; *]R ,

61

Since the variables and matrices are degenerate,the eigenvector for the R’s is the same as thatfor the Q’s:

where the subscript R indicates that these are thecomponents of R. We want the spin Fourier com-ponents, which we get by writing

)14(.|),(|),(),( 0201 12 qsqsqs

When 2=0, then all the R’s on slide #55 are zero,

which gives relations between pairs of s(q,)’s.

Likewise, when 1=0, all the Q’s are zero and we

have other relations between pairs of s(q,)’s.

T25 SPIN STRUCTURE

etc.,)10,()12,(,)9,()11,(,)7,()8,(

)5,()6,(,)1,()3,(,)2,()4,(

qsqsqsqsqsqs

qsqsqsqsqsqs

xxxxxx

xxxxxx

62

To illustrate: when 2=0, then

so that

.],,,;,,,;,,,[),(

],,,;,,,;,,,[),(

],,,;,,,;,,,[),(

**1

**1

**1

ioipioipiminiminnqs

hghgfefenqs

dcdcbabanqs

z

y

x

These results give the first column in Table X, below.The second column can be obtained similarly, or

maybe more easily by applying my to the left column.

In any event, the final result is shown in Table XI.

T25 SPIN STRUCTURE63

In any event, the result for the spinstructure of T25 after taking ac-count of inversion is given in TableXI, where the Roman letters are realand the Greek ones complex. NoteThe effect of inversion: the two Tborbits (sites 5-6 and 7-8) are nolonger independent, but now arecomplex conjugates of one another.Also the others are now restrictedto be real. Of course, we have anoverall undetermined phase whichwe put into the order parameters.

Table XI. As Table III.

SYMMETRY OF THE ORDER PARAMETERS64

Up to now we have studied how the spin comp-onents transform under symmetry operations. Wecan summarize these results by instead attributingthe transformation to the transformation of theorder parameters. This is very convenient becausethis enables us to analyze the symmetry ofLandau expansions in terms of these order para-meters. Applying the transformation law of TableIX to the eigenvectors of Table XI, we see that

This information is crucial to constructing theME interaction, as we shall see.

(15)

YMn2O5 (YMO)65

The results for YMO are included in those forT25 because the only difference is that inYMO there is no analog of the magnetic Tbsites. It is interesting to express thestructure determined by Chapon et al.,17

in terms of order parameters n.

The simplest way to identify the order parameters

is to note the symmetry of the spin structure: mx

takes it into the negative of itself and my takes it

into an orthogonal state. Thus only s2 is nonzero.

ME INTERACTIONHere I discuss the ME interaction. Of course the Landau expansion will be the same for both T25and YMO. We start with the general trilinear form:

66

.)()(

PqqcV mnnm

nm

Eq. (15) says that I[1 2*] = 1 2*. So only terms

with n=m in V can be inversion invariant. So.]|)(||)(|[ 2

22

1

PqqaV Eq. (15) says that the factor in square brackets is

Odd under my. Therefore only =y contributes:

.]|)(||)(|[ 22

21 yPqqaV

ME INTERACTION67

,]|)(||)(|[ 22

21 yPqqaV

To repeat: the final result for the ME interaction is

which gives the spontaneous polarization (alongy, as observed18,19) as

.]|)(||)(|[const. 22

21 qqP

This relation is also obtained for

RbFe(MoO4)2,12 where its prediction

for the temperature dependencewas confirmed by comparing P(T)with the neutron diffraction

measurement of ||2.

CONCLUSION

The invocation of inversion symmetry usuallycuts down to half the number of fitting parametersneeded for magnetic structure determinations in incommensurate systems.

The characterization of the magnetic orderingby the introduction of order parameters has many advantages. The most important one isthat the symmetry of the unit cell (importantin these magnetically frustrated systems) isproperly taken into account, which is not thecase20 with continuum theories.21

68

APPENDIX A: FREE ENERGY

)1A(,]ln [Tr kTHF

To help understand the free energy, F, I now give asimple approximation for it. F is the minimum of

where H is the Hamiltonian and is the trial densitymatrix (normalized to have unit trace) and which is

varied to minimize F. Take H to be -JSiSj for nearest

neighbors on a cubic lattice and let each S assumethe values +1 and -1 (Ising spins). Take the densitymatrix to be such that the value +1 occurs withprobability (1+ )/2 and -1 with probability (1-)/2.

Then Tr [Si] = = M is the order parameter and ,))()(2/( TSNzJF

Where z=6 is the coordination number of the cubiclattice of N sites and -TS is the second term inEq. (A1) for F. (Continued next slide.)

69

We now evaluate –TS:

.]2ln12/2/[

]2/)1[(ln]2/)1[(

]2/)1[(ln]2/)1[(

]ln[Tr

42

NkT

NkT

NkT

kTTS

So up to quartic order the free energy is

.2/)(

]6/][2/[)2/(/42

422

uTTk

kTJzNF

c

For T -> Tc-, ~ (Tc – T)1/2 .

70

I now describe the result when we allow the orderingto be according to a wavevector q, so that instead ofthe probability p of a spin being up being uniform,now it is that of a spin-density wave given by

Of course, this kind of state for incommensurate qis not favored by a nearest neighbor interaction whichgives either a ferromagnet or an antiferromagnet,depending on the sign of J. Therefore we allow for competing interactions and we introduce

4/)]exp()()exp()([)2/1()( rqiqrqiqrp

)exp()()( rqirJqJr

(Continued next slide.)

71

)O()()()]([2

1 4 qqqJkTFq

For this system the free energy is

This formula holds for a ferrr- or antiferro-magnet.It is useful to identify the wavevector-dependentsusceptibility (q) as

)()(1 qJkTq

For nearest neighbor interactions on a cubiclattice this gives

)O(q6

)]cos()cos()[cos(2)(422

1

qJaJkT

qaqaqJkTq zyx

For a ferromagnet J is negative and the inversesusceptibility is minimal at q=0, as expected.

72

FORM OF EIGENVECTORThe matrix under consideration for irreps

3 and 1 is

G =

One can show that when G operates on a vectorof the form [a,b,c,d,x+iy, x-iy], where all theletters represent real quantities, it gives backa vector of this same form. That is enough tosay that any eigenvector is of this form.

(continued on the next slide)

73

G =

The matrix we are analyzing is

Now we form U-1GU, where

Since U-1GU is a real symmetric matrix, its eigen-

vectors n are real valued. Then the eigenvector

of G, namely Un has the form asserted.

74

The analysis for the 2 and 4 eigenvectors is similar.

The free energy is now a function of 7 coordiates:

H(2)

=

By equating H and IH we get relation, such as a, b, and are real, *, ’’*, etc. Thus, inversion sym-metry leads to the matrix H assuming the formIts eigenvectors are of the form [a, b, c,

75

A FEW REFERENCES

1. G. Lawes, et al., Phys. Rev. Lett. 95, 087205 (2005).

2. A. B. Harris, cond-mat/0610241 (submitted to PRB).

3. A. J. Wilson, International Tables for Crystallography (Kluwer Academic, Dordrecht, 1995) Vol. A.

4. E. E. Sauerbrei et al., Acta Crystallogr. B 29, 2304 (1975).

5. R. Kajimoto et al., Phys. Rev. B 70, 012401 (2004).

6. M. Kenzelmann et al., Phys. Rev. Lett. 96, 087206 (2005).

7. E. F. Bertaut, J. de Physique, Colloque C1, 32, 462 (1971).

8. M. Kenzelmann et al., Phys. Rev. B 74, 014429 (2006).

9. T. Kimura et al., Nature 426, 55 (2003).

10. T. Nagamiya, in Solid State Physics, ed. F. Seitz and D.

Turnbull (Academic, New York, 1967), Vol. 20, p346.

Continued on the next slide

76

11. J. J. Betouras et al., Phys. Rev. Lett. 98, 257602 (2007).12. M. Kenzelmann et al., Phys. Rev. Lett. 98, 267205 (2007).13. A. B. Harris and G. Lawes, “Ferroelectricity in Incomm- mensurate Magnets,” in The Handbook of Magnetism and Advanced Magnetic Materials,” Ed. H. Kronmuller and S. Parkin, (Wiley, 2008); cond-mat/0508617.14. G. Buisson, Phys. Stat. Sol. 16, 533 (1973).15. G. Buisson, Phys. Stat. Sol. 17, 191 (1973).16 L. Chapon et al., Phys. Rev. Lett. 93, 177402 (2004).17. L. Chapon et al., Phys. Rev. Lett. 96, 097601 (2006).18. A. Inomata and K. Kohn, J. Phys. Cond. Mat. 8, 2673 (1996). Data for T25 and YMO.19. I. Kagomiya et al., Ferroelectrics 286, 167 (2003). Data for T25.20. M. Kenzelmann and A. B. Harris, cond-mat/0610471 accepted as a comment (re Ref. 21) by Phys. Rev. Lett.21. M. Mostovoy, Phys. Rev. Lett. 96, 067601 (2006).

77