Presentation on theme: "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."— Presentation transcript:
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
OUTLINE I.Setting the Stage II.Symmetry of the Quadratic Free Energy A. Physics of this Free Energy B. Eigenvector Condition (TbMnO 3 ) C. Group of the Wavevector (TbMn 2 O 5, YMn 2 O 5 ) 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 Letter 1 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 SUSCEPTIBILITY The susceptibility is a key quantity in continuous transitions like those we are about to study. One is used to saying that the susceptibility diverges at a continuous transition. Alternatively, one can plot the 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 the one which minimizes F. Free energy F = (T-T c )M 2 + uM 4 for a sequence of temperatures T < < T c < T 1 < T 2. Note that the free energy is unstable relative to the formation of long-range order for T < T c 4
INVERSE SUSCEPTIBILITY For small H the magnetization M is The coefficient of M 2 is the inverse susceptibility: 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. T 2 > T 1 > T c When the first Fourier component condenses, the others are inhibited from condensing because there is a finite amount of spin to go around. 5
INVERSE SUSCEPTIBILITY The instability occurs at wave- vector q= /a Antiferromagnet For the antiferromagnet the instability leads to up-down-up-down spins, i. e. aq = . Observe that if you have a plot of the susceptibility versus wavevector, you don’t have to know whether the system is ferromagnetic or anti- ferromagnetic: the location of the minimum in inverse chi tells you which 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 = T c T = T 1 > T c T = T 2 > T 1 7
INVERSE SUSCEPTIBILITY Suppose we have vector spins with an easy axis along z. Then Where K>0 is the anisotropy energy. Here as T approaches T c from above, only M z becomes unstable relative to long-range order. As another example suppose the free energy is (Continued next slide) 8
How do we deal with 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 have This example teaches us that if the free energy is a quadratic form, ordering takes place in the channel defined by the eigenvector associated with the eigenvalue that is vanishing. INVERSE SUSCEPTIBILITY 9
INCOMMENSURATE CHI Now we consider the free energy of a system which 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 This quadratic form has complex-valued eigen- vectors which we can normalize, but they have an arbitrary phase factor (which I will discuss later). Obviously we focus on the critical eigen- value, i. e. the one that first becomes unstable. 10
INTERPRETATION For 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 T c ). 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 of the spin distribution) are complex-valued. Note: we reject the possibility of accidental degeneracy. Degeneracy can only occur if symmetry requires it. 11
INTERPRETATION So we write the eigenvector as where the components inside the square bracket are normalized so that the sum of the squares of their magnitudes is unity. I will discuss the phase later. The amplitude X varies as (T c -T) 1/2. s is the complex amplitude of the wave of the -component of the sublattice. So the eigenvector tells us the pattern into which the spins order, as we will see in a moment. 12
APPLICATION One might attempt to write down and anlyze the matrix of coefficients F. That is a hopeless task because our microscopic models are not reliable. Instead, we use the 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 will invoke the principle that first-year graduate students learn: if a set G 1, G 2, …of mutually commuting opera- tors commute with the Hamiltonian, then the eigen- vectors of the Hamiltonian can be classified according to their eigenvalues g 1, g 2, … In the language of group theory, this case is the case when all the irreducible representations are one dimensional. 13
CALCULATION The essential input we require is the symmetry of the incommensurate wavevector (usually as determined by neutron diffraction) i. e. the symmetry operations of the crystal which also leave this wavevector invariant. Since we are concerned with phase transitions in which ferroelectricity appears in coincidence with magnetic ordering, spatial inversion I is a significant symmetry operation, although it does not usually leave the wavevector invariant. So I’m afraid we have to actually confront the crystal structure of the paramagnetic phase. 14
EXAMPLE A: TbMnO 3 The space group opera- tions 3 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 m z and the glide plane (x=3/4) for m x with associated translation u y = b/2. From neutron diffraction 5,6 the wavevector lies along b (i. e. y) and in rlu is q= 0.28. The wavevector is invariant under m x and m z. Table II: Mn and Tb positions. 4 = Mn = Tb 15 Table I: General positions for Pbnm (TMO). 3,4 Here x means –x.
Eigenfunctions for TbMnO 3 (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 for the x, y, and z components. These eigenfunctions are eigenfunctions of m x and m z with respective eigenvalues (m x ) and (m z ). Here = exp(i q), where q is the incommensurate wavevector in reciprocal lattice units (rlu), so in ``real” units, it is 2 q/b. Table III: Eigen- vectors. = exp(i q). 16
EIGENFUNCTIONS The eigenfunction for 1 has five parameters x M, y M, z M, z T1, and z T2 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 m x and m z it is convenient to write in terms of subvectors as 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) is convenient because the symmetry operations only interrelate 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 spin functions transform are: 1) The transformed spin at r is obtained by O S acting on the spin which was at the original location [O R ] -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 list the initial and final sites under the operations we will invoke. Table IV (at right): initial site, i, sites, f, after m x, m z, and I. The sites are numbered according to Table II. 19
EIGENFUNCTIONS To construct the eigenfunctions of m x and m z, we need to discuss how the symmetry opera- tions act on Fourier transforms of spin functions like s x (q,1). First of all, spin is a pseudovector, so the mirror m does’t reverse the sign of the -component of spin but does reverse the sign of the other components. Let O denote an operator acting on spin and space, O R (O S ) an operator acting only on space (spin), and lower case denote a Fourier transform. On the next slide we indicate the effect of a symmetry operator O=O R O S on a Fourier transform. 20
EIGENFUNCTIONS O R = a point group operation G O + a translation u O : so that (since G O leaves q invariant) Thus, the final result is 21
EIGENFUNCTIONS The result of the previous slide was For m z, u 0 =0, and for m x, q. u 0 = q (recall rlu!). Also, O S gives a plus or minus sign. Thus where the sign depends on O and and ( ’) is the initial (final) sublattice. Thus for the pseudovector spins, with = exp(i q), where = -1, except that = +1, so that m z 2 = 1 and m x 2 = 2. Note: ( ’)’= . 22
EIGENFUNCTIONS Now we show how the eigenvectors are constructed. For instance, to construct the eigenvector scaled by x M, we assign m 1x the value 1 and then deduce the other components according to the symmetry ( ‘s). So we introduce projection operators To check the effect of P x, use m x 2 = 2 = (m x ) 2 : 23
EIGENFUNCTIONS Thus P P x P z projects an eigenfunction of both m x and m z with respective eigenvalues (m x ) and (m z ). We extend the notation of Eq. (2) so that (a,b,c,d) s T represents a subvector S (T), whose four components are a, b, c, and d, and where s=x, y, or z (to label components) and T=M or T (for Mn or Tb). We now illustrate the use of these projection opera- tors to construct eigenfunctions of m x and m z. To get the eigenfunction scaled by x M, we let the projection operator act on X (M) = [1,0,0,0] x M in the notation of Eq. (2): 24
EIGENFUNCTIONS In the first line we used Table IV which tells us that m z takes site #1 into site #3 and Eq. (5) with zx =-1. In the third line Table IV tells us that m x 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 (m z )=1 and (m x )= . This leads to =[1,1,-1,-1] x M in agree- ment with the eigenvector of Table IV scaled by x M. 25
EIGENFUNCTIONS To further illustrate the construction of the eigen- vectors we apply the projection operator to the Tb wavefunction [1,0,0,0] x T : So for (m z )=+1, there are no eigenvectors of the x-component. Otherwise the eigenfunction covers only sites #5 and #8. One can verify that the above result agrees with Table III. Enough checks! 26
EIGENFUNCTIONS: SUMMARY For 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 Fourier transform of the spin distribution is proportional to this ``critical” eigen- vector. The symmetry ( ) and the amplitudes (x, y, z) are determined to best fit to diffraction data. 27
INVERSION SYMMETRY Up to now our results are absolutely standard. It is often said (in reviews 7 ) 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 now analyze the consequences of that. We assume that only a single representation is critical and write the free energy as an Hermitian form in in the amplitudes (which we here denote Q i ) of one of the columns ( ) of Table III: 28 So Q 1 ( ) = x M, Q 2 ( ) = y M, etc.
INVERSION Clearly, we now have to determine how inversion I transforms the coordinates Q. For that purpose we modify the derivation of Eq. (3) so that it applies to I. Replace ``leaves q invariant” by ``changes the sign of q.” Thereby we get What does this mean? The initial and final sites are 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 pairs of 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
INVERSION Now we deduce the effect of inversion on the sym- metry adapted coordinates, denoted Q n ( ), which are the amplitudes in terms of the eigenfunctions of Table III. As an illustration we consider the coordinates of 1 from column 1 of Table III: 30
INVERSION From the discussion of Eq. (6) one sees that Now write the free energy (for ) in terms of the Q’s and require invariance under inversion: Here and below we invoke the convention that Roman letters are real and Greek ones complex. Thus 31
INVERSION 32 To repeat: We now need to study how inversion affects the matrix. For example, IQ 1 * Q 4 = Q 5 * Q 1. After inversion the 5,1 element of the matrix is the old 1,4 element. A simpler example: IQ 1 * Q 2 = Q 2 * Q 1, so after inversion the 2,1 element is the old 1,2 element.
INVERSION Since these two forms have to be equal no matter how we choose the Q’s, the matrices have to be equal. Thus and must be real and Thus the free energy is of the form (with Roman letters real and Greek ones complex). The critical eigenvector is an eigenvector of the above matrix. 33
INVERSION To repeat: we found the free energy to be Because of the special form of this matrix (due to invoking inversion symmetry) the eigenvector is shown in the Appendix to be of the form 6 where the subscript Q says that the components are Q 1, Q 2, Q 3, … and we normalize by a 2 +b 2 +c 2 +2| | 2 =1. Since the eigenvalue problem takes place in a complex vector space, we can not omit the arbitrary phase . We discuss this result below. 34
INVERSION We now discuss the result of the last slide: Previously (in Table III) we showed that symmetry led to an eigenvector of symmetry 1 which had five complex coefficients. Now we see that apart from an overall phase factor, three of these are real, and the other two (for Tb), rather than being independent complex numbers are the complex conjugates of one another: So instead of having 5 complex amplitudes to fit the diffraction data, we have three real and one complex parameter. The result for = 3 is similar. The result for 2 4 ) is (see the Appendix) 35
INVERSION So at right we have updated Table III to take account of inversion symmetry. The eigen- functions of each symmetry n should be assigned an inde- pendent phase factor n and amplitude A n. If only one symmetry is present, the phase is usually irrelevant. But if two symmetries are simultaneously present, the phases matter because the relative phase has significance, as we shall see. Reminder: Roman letters are real and Greek ones complex. As Table III, but with Inversion symmetry. 36
ORDER PARAMETERS We now discuss how the magnetic structure can be described 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 about an order parameter? Here the order parameter has to be complex because any phase is allowed in Eq. (8). As a function of temperature the normalized eigen- vector will be essentially constant, and only the magnitude A will vary. So we identify the complex order parameter as 37 where A n is real.
ORDER PARAMETERS The order parameter inherits the symmetry of the representation n. Thus At right is the phase diagram 9 of TMO. As one cools at H=0 throughT > = 40K, order in 3 appears 6 in the HTI (high-temperature incommensurate) phase. 8 Then at T=28K additional order in 2 appears 6 in the low temperature (LTI) phase. Cooling at H>10T leads to a different phase. The polarization is nonzero only in the LTI phases. 38 Para P=0 TMO
MAGNETIC STRUCTURE Here I discuss qualitatively the magnetic structure. Neutron diffraction 6 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 must be easiest, and obviously this is the b axis. Note that the pattern of sublattice spins is (++--). The quartic terms in the free energy favor fixed spin length 8 which is not consistent with a collinear sinusoidal state. Thus, it is not surprising 8,10 that a phase transition occurs at lower temperature where 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 This behavior is consistent with the following simple 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 energy ensures that the wavevectors of the two order parameters are equal. If the interactions are nearly isotropic, one would expect this to be the case. Within Landau theory the term that locks to two order parameters is of the form 8 where u<0 and this term only exists if q 2 = q 3.
MAGNETOELECTRIC COUPLING 41 Now we discuss the Landau theory of magneto- electric (ME) coupling V. 1,2,8 Order B is induced by order A if there is a term in the free energy linear in B times 2 (or more) powers of A. Time reversal invariance ensures that such a term linear in the magnetization can not occur, so the polarization P can not induce magnetic order. But magnetization can induce P. The interaction we seek is therefore of 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 is little tendency to have a spontaneous polarization.
ME COUPLING The ME coupling is schematically of the form 42 Where c is a constant. Now minimize with respect to P so this interaction does just what we want: it induces a polarization only when ss is nonzero.
ME COUPLING 43 We drop the Umklapp term with G nonzero (for which see Ref. 11). Then we have where G is a reciprocal lattice vector and delta is unity if q+q’=G and is zero otherwise. As we argued, the magnetoelectric interaction V is linear in P and quadratic in the magnetic order parameters. To conserve wavevector V is
ME COUPLING 44 Warning: the discussion that follows applies to TMO. For different symmetry systems the ME Interaction need not assume the form found be- low. 2,12 In TMO, V must be invariant with respect to m x, m z, and I. To handle the cases that will be of interest to us, let us assume that there are two irreducible representations (irreps) present. r and s. In Eq. (10) the term with n=m is zero because it changes sign under I. Physically this reflects the fact that a single wave has a center of symmetry which, for an incommensurate q, can be as close to an inversion center of the lattice as you like. Thus
ME COUPLING One can show, using Eq. (9) that V is invariant under m x. 45 This has to be invariant under m z. Using Eq. (9) we have This equalsV only if m z P =-P , i. e. for =z (i. e. c). So To repeat:
ME COUPLING We have V as This has to be invariant under inversion. We have To have IV = V, a z must be imaginary: a z =ir, with r real. We set n (q) = | n | exp (i n ). Then 46 The polarization in TMO (at low H) is only found 8 in the LTI phase where it lies along c (i. e. z) as in (11).
ME COUPLING The appearance of the relative phase is easily understood. The phase of the irrep regulates the location of origin of the cosine wave. The wave has inversion symmetry about this point. So, if the 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 |s 2 s 3 | also agrees with experiment 8 in that the temperature depend- ence of P z looks like that of an order parameter and not (order parameter) 2. This happens because the HTI order parameter s 3 is more or less saturated by the time one gets into the LTI phase. 47
HIGH FIELD PHASE 48 In the high field phase (see the figure on slide #38) the polarization lies alog a (i. e. x) in the analog of the LTI phase. Thus the magnetic structure must be 3 + n such that their product transforms like P a. For the product to be odd under m x, we need n=1 or 4. For it to be even under m z, 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 irrep does not necessarily mean a totally unrelated structure. It may simply mean that the structure has been rotated. Because different spin components transform different under symmetry operations, this causes a change in the irrep.
TbMn 2 O 5 (T25) 49 T25’s space group is Pbam, 14,15 whose symmetry operations are in Table VI and whose ion positions are in Table VII. Table VI. Pbam space group 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 two dimensional, the simple method used above is clumsy. 2 An alternate way, used here, is to supplement the standard analysis by including inversion symmetry. Below T=43K, T25 has a commen- surate phase with q=(1/2,0,1/4) and an incommensurate phase with q=(0.48, 0, 0.32). 16
T25 The standard analysis (without inversion symmetry) for the wavevector (0.50, 0, q z ) leads to Table VIII, which is read just as Table III, except that the two columns do not represent dif- ferent symmetry irreps, but rather transform like two columns of the same irrep. Since the two columns are degenerate in free energy, the actual spin function is a linear combination (determined by a fit to diffraction data) of the two columns. The complex coefficients 1 and 2 are the order parameters. Table VIII for T25. 50
T25 TRANSFORMATIONS Before discussing inversion symmetry let us consider the transformation properties of the eigenfunctions of Table VIII. To do that we need to know how the spin functions transform under the symmetry operations. Here it is convenient to redefine the spin Fourier transforms so that the exponential is exp (iq. R) rather than exp [iq. (R+ )]. With this change, a transformation 1)changes the orientation of the spin, 2) takes the spin from an initial sublattice to a final sub- lattice, and 3) introduces a phase factor (given In 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 the sublattices transform. The spin transformation factor (for the symmetry operation m ) are the same as before: =1 if and is -1 otherwise. Thus 52 In the first equation, xx =1, and in the second, yz =-1. One needs the phase factors in Table IX, taken from Ref. 2. I does not carry a factor.
INVERSION SYMMETRY Now we discuss the effect of inversion symmetry. To do this we will study the free energy when it is written in terms of the symmetry adapted coordi- nates of Table VIII. If we take the 12 symmetry coordinates of column #1, then the free energy is a quadratic form with a 12 by 12 matrix F nm. 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 coordinates for the first column of Table IX as we did in slide #30. These are listed on the next two slides.
SYMMETRY COORDINATES Below we give the Q’s which transform like the first column of the two-dimensional irrep. 54
SYMMETRY COORDINATES 55 Below we give the Q’s which transform like the second column of the two-dimensional irrep.
SYMMETRY OPERATIONS Now we explicitly display the effect of the symmetry operations using Table IX and = 1 if but is -1 otherwise: 56 Using Table IX one can determine the effect of inversion on the Q’s and R’s. We find that where the necessary data is in Table X. Table X. Transformation parameters for inversion.
DOUBLE DEGENERACY The meaning of a two-dimensional irrep is that the free energy matrix F (shown on the next slide) for the R’s is exactly the same as that for the Q’s: This guarantees that all eigenvalues are doubly degenerate. What this means is that symmetry dictates that these two modes (or any linear combination of them) become unstable (relative to ordering) simultaneously. Thus the free energy in terms of their order parameters is 57
F = Here S n is a 4 x 4 real symmetric matrix, R n a 4 x 4 real matrix. The other matrices assume the forms shown below. where a is real. 58 Below we will verify this form. Meanwhile, see that F acting on , given below, yields the a vector of the same form. Roman = real, i 2 =-1. = [a,b,c,d; *; e, f, g, h; *; im, in, io, ip; *]
VERIFYING THE FORM OF F (a) (b) (c) (d) (e) (f) (g) Now we use invariance under inversion I: NOTES: When we have a result of the form R n XR m, then we can conclude (in the second line of these equations) that X=H nm. In (a) H 12 is real. In (b) H 15 =H 16 *. In (d) H 1,13 is imaginary. In (e) H 1,17 =-H 1,18 *. No info in (g). 59
T25 SPIN STRUCTURE Note that the spin Fourier transform comes from a linear combination of the Q’s and the R’s. So we parametrize the eigenvector as (see bottom of slide #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 Q 1, Q 2, … Q 18 and we will normalize the vector so that the sum of the magnitudes squared of its components is unity. The matrix for the R’s is identical to that for the Q’s. This guarantees that the Q mode and the R mode become unstable simultaneously and that the critical R eigenvector has 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 that for the Q’s: where the subscript R indicates that these are the components of R. We want the spin Fourier com- ponents, which we get by writing 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 62 To illustrate: when 2 =0, then so that These results give the first column in Table X, below. The second column can be obtained similarly, or maybe more easily by applying m y to the left column. In any event, the final result is shown in Table XI.
T25 SPIN STRUCTURE 63 In any event, the result for the spin structure of T25 after taking ac- count of inversion is given in Table XI, where the Roman letters are real and the Greek ones complex. Note The effect of inversion: the two Tb orbits (sites 5-6 and 7-8) are no longer independent, but now are complex conjugates of one another. Also the others are now restricted to be real. Of course, we have an overall undetermined phase which we put into the order parameters. Table XI. As Table III.
SYMMETRY OF THE ORDER PARAMETERS 64 Up to now we have studied how the spin comp- onents transform under symmetry operations. We can summarize these results by instead attributing the transformation to the transformation of the order parameters. This is very convenient because this enables us to analyze the symmetry of Landau expansions in terms of these order para- meters. Applying the transformation law of Table IX to the eigenvectors of Table XI, we see that This information is crucial to constructing the ME interaction, as we shall see. (15)
YMn 2 O 5 (YMO) 65 The results for YMO are included in those for T25 because the only difference is that in YMO there is no analog of the magnetic Tb sites. It is interesting to express the structure 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: m x takes it into the negative of itself and m y takes it into an orthogonal state. Thus only s 2 is nonzero.
ME INTERACTION Here I discuss the ME interaction. Of course the Landau expansion will be the same for both T25 and YMO. We start with the general trilinear form: 66 Eq. (15) says that I[ 1 2 *] = 1 2 *. So only terms with n=m in V can be inversion invariant. So Eq. (15) says that the factor in square brackets is Odd under m y. Therefore only =y contributes:
ME INTERACTION 67 To repeat: the final result for the ME interaction is which gives the spontaneous polarization (along y, as observed 18,19 ) as This relation is also obtained for RbFe(MoO 4 ) 2, 12 where its prediction for the temperature dependence was confirmed by comparing P(T) with the neutron diffraction measurement of | | 2.
CONCLUSION The invocation of inversion symmetry usually cuts down to half the number of fitting parameters needed for magnetic structure determinations in incommensurate systems. The characterization of the magnetic ordering by the introduction of order parameters has many advantages. The most important one is that the symmetry of the unit cell (important in these magnetically frustrated systems) is properly taken into account, which is not the case 20 with continuum theories. 21 68
APPENDIX A: FREE ENERGY To help understand the free energy, F, I now give a simple approximation for it. F is the minimum of where H is the Hamiltonian and is the trial density matrix (normalized to have unit trace) and which is varied to minimize F. Take H to be -JS i S j for nearest neighbors on a cubic lattice and let each S assume the values +1 and -1 (Ising spins). Take the density matrix to be such that the value +1 occurs with probability (1+ )/2 and -1 with probability (1- )/2. Then Tr [S i ] = = M is the order parameter and Where z=6 is the coordination number of the cubic lattice of N sites and -TS is the second term in Eq. (A1) for F. (Continued next slide.) 69
We now evaluate –TS: So up to quartic order the free energy is For T -> T c -, ~ (T c – T) 1/2. 70
I now describe the result when we allow the ordering to be according to a wavevector q, so that instead of the 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 q is not favored by a nearest neighbor interaction which gives either a ferromagnet or an antiferromagnet, depending on the sign of J. Therefore we allow for competing interactions and we introduce (Continued next slide.) 71
For this system the free energy is This formula holds for a ferrr- or antiferro-magnet. It is useful to identify the wavevector-dependent susceptibility (q) as For nearest neighbor interactions on a cubic lattice this gives For a ferromagnet J is negative and the inverse susceptibility is minimal at q=0, as expected. 72
FORM OF EIGENVECTOR The matrix under consideration for irreps 3 and 1 is G = One can show that when G operates on a vector of the form [a,b,c,d,x+iy, x-iy], where all the letters represent real quantities, it gives back a vector of this same form. That is enough to say that any eigenvector is of this form. (continued on the next slide) 73
G = The matrix we are analyzing is Now we form U -1 GU, where Since U -1 GU is a real symmetric matrix, its eigen- vectors n are real valued. Then the eigenvector of G, namely U n 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 form Its eigenvectors are of the form [a, b, c, 75
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