Magnetism of the Localized Electrons on the Atom
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1 Magnetism of the Localized Electrons on the Atom 1. The hydrogenic atom and angular momentum 2. The many-electron atom. Spin-orbit coupling 4. The Zeeman interaction 5. Ions in solids 6. Paramagnetism Comments and corrections please: Dublin January 27 1
2 1. The hydrogenic atom and angular momentum Consider a single electron in a central potential. A hydrogenic atom is composed of a nucleus of charge Ze at the origin and an electron at r,,. First, consider a single electron in a central potential e = Ze/4 e r z = - ( 2 /2m) 2 - Ze 2 /4 e r l -e In polar coordinates: r 2 = 2 / r 2 +(2/r) / r + 1/r 2 { 2 / 2 + cot / + (1/sin 2 ) 2 / 2 } Ze l The term in parentheses is -l 2. Schrödinger s equation is = E x The wave function means that the probability of finding the electron in a small volume dv ar r is *(r) (r)dv. ( * is the complex conjugate of ). Eigenfunctions of the Schrödinger equation are of the form (r,, ) = R(r) ( ) ( ). The angular part ( ) ( ) is written as Y l ml (, ). The spherical harmonics Y l ml (, ) depend on two integers l, m l, where l is and m l l. ( ) = exp(im l ) where m l =, ±1, ±2... The z-component of orbital angular momentum, represented by the operator l z = -i /, has eigenvalues < l z > = m l. ( ) = P l ml (cos ), are the associated Legendre polynomials with l m l, so m l =, ±1, ±2,...±l. Dublin January 27 2
3 z The square of the orbital angular momentum l 2 has eigenvalues l(l+1). The orbital angular momentum has magnitude [l(l+1)] and its projection along z can have any value from -l to +l. The quantities l z and l 2 can be measured simultaneously (the operators commute). In the vector model, The total angular momentum is a vector which precesses around z. Spherical harmonics. s Y = (1/4 ) m l [l(l+1)] p Y 1 = (/4 ) cos Y 1 ±1 = ± (/8 ) sin e ±i d Y 2 = (5/16 )(cos 2-1) Y 2 ±1 = ± (15/8 ) sin cos e ±i Y 2 ±2 = (15/2 ) sin 2 e ±2i f Y = (7/16 )(5cos - cos ) Y ±1 = ± (21/64 )(5cos 2-1)sin e ±i Y ±2 = (15/2 ) sin 2 cos e ±2i Y ± = ± (5/64 ) sin e ±i Dublin January 27
4 Spherical harmonics. n =1 l= s n =2 l=1 ml = p ml =±1 n = l=2 ml = ml =±1 d ml =±2 Dublin January 27 4
5 The radial part of the wavefunction R (r) The radial part R(r) depends on l and also on n, the total quantum number; n > l; hence l =, 1,...(n-1) R(r) = V nl (Zr/na )exp[-(zr/na )].4 V 1 = 1. Here a = 4 2 /me 2 = 52.9 pm is the first Bohr radius, the basic length scale in atomic physics. 2 Rnl ( ) The energy levels of the 1-electron atom are = r / a E = -Zme 4 /8h 2 2 n 2 = -ZR /n 2 The quantity R = me 4 /8h 2 = 1.6 ev is the Rydberg, the basic energy in atomic physics. For the central Coulomb potential e, the potential energy V(r) depends only on r, not on or. E depends only on n. Dublin January 27 5
6 The three quantum numbers n, l, m l denote an orbital, a spatial distribution of electronic charge. Orbitals are denoted nx, x = s, p, d, f for l =, 1, 2,. Each orbital can accommodate up to two electrons with spin m s = ±1/2. No two electrons can be in a state with the same four quantum numbers (Pauli exclusion principle). The hydrogenic orbitals are listed in the table n l m l m s No of states 1s 1 ±1/2 2 2s 2 ±1/2 2 2p 2 1,±1 ±1/2 6 s ±1/2 2 p 1,±1 ±1/2 6 d 2,±1,±2 ±1/2 1 4s 4 ±1/2 2 4p 4 1,±1 ±1/2 6 4d 4 2,±1,±2 ±1/2 1 4f 4,±1,±2,± ±1/2 14 The Pauli principle states that no two electrons can have the same four quantum numbers. Each orbital can be occupied by at most two electrons with opposite spin. Dublin January 27 6
7 2. The many-electron atom Hartree-Foch approximation, each electron sees a different self-consistent potential V l (r). Coulomb interactions between electrons E n n = s 4s s 6p 4p p 6d d 6f 4 1s 2s s 4s 5s 6s 2p p 4p 5p 6p 2s 2p 2 d 4d 5d 6d 4f 5f 6f 1s s p d f l 5g 6g Dublin January 27 7
8 Dublin January 27 8
9 Example: The carbon atom 1s 2 2s 2 2p 2 There are no options for the first four electrons. There must be a pair of them with opposite spin in each s orbital. However, there are 15 ways of accomodating two electrons in the p orbitals. The 15 states fall into three groups (terms). Notation: States with L =,1,2,,4,5,6 are S,P,D,F,G,H 2S+1 X Dublin January 27 9
10 Term L S (ML, MS) 1 S (,) P 1 1 (1,1)(1,)(1,-1)(,1)(,)(.-1)(-1,1)(-1,)(-1,-1) 1 D 2 (2,)(1,) (2,1)(2,)(,)(-1,)(-2,) The terms are widely separated in energy; one is the lowest Finally we need to couple the spin and orbital angular momentum to form a resultant J. J = L + S L-S J L+S Addition of L and S in the vector model Hund s rules; A prescription to give the ground state of a multi-electron atom 1) First maximize S for the configuration 2) Then maximize L consistent with that S ) Finally couple L and S; J = L - S if shell is < half-full; J = L + S if shell is > half-full. In the example, S = 1, L = 1, J =. The ground state of carbon is P, which is nonmagnetic (J = ). General notation for multiplets is 2S+1 X J where X = S, P, D... for L =, 1, 2... J L S In spectroscopy, the energy unit cm -1 is used. Handy conversions are:1 ev 1165 K and 1 cm K Dublin January 27 1
11 Some examples: Fe + d 5 ooooo S = 5/2 L = J = 5/2 6 S 5/2 Ni 2+ d 8 oo S = 1 L = J = 4 F 4 Nd + 4f oooo ooooooo S = /2 L = 6 J = 9/2 4 I 9/2 Dy + 4f 9 ooooo S = 5/2 L = 5 J = 15/2 6 H 15/2 Dublin January 27 11
12 . Spin-orbit coupling This relatively-weak relativistic interaction is responsible for Hund's third rule. In the multi-electron atom, the spin-orbit term in the Hamiltonian can be written as so = L.S is > for the first half of the d or 4f series and < for the second half. It becomes large in heavy elements. is related to the one-electron spin-orbit coupling constant by = ± /2S for the first and second halves of the series. The resultant angular momentum (see above) is J = L ± S The identity J 2 = L 2 + S L.S is used to evaluate H so.the eigenvalues of J 2 are J(J + 1) 2 etc, hence L.S can be calculated. 4f 1 Ce + Spin-orbit coupling constants in the d and 4f series ion L d 1 Ti d 2 Ti d V d 4 Cr d 6 Fe d 7 Co d 8 Ni f 2 54 Pr f Nd + 4 4f 5 Sm + 5 4f 8 Tb f 9 Dy f 1 Ho f 11 Er f 12 Tm f 1 Yb Dublin January 27 12
13 9 8 7 L S J Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Application of Hund s rules to the trivalent rare-earth ions. Dublin January 27 1
14 Magnetic properties of free atoms: only the elements marked in bold are nonmagnetic, w ith J = Dublin January 27 14
15 4. The Zeeman interaction The magnetic moment of an ion is represented by the term m = (µ B / ) (L + 2S) The Zeeman Hamiltonian for the magnetic moment in a field B applied along z is m.b Z = (µ B / )(L + 2S).B z The vector model of the atom, including magnetic moments. First project m onto J. J then precesses around z. We define the g-factor for the atom or ion as the ratio of the component of magnetic moment along J in units of µ B to the magnitude of the angular momentum in units of. g = -(m.j/µ B )/(J 2 / ) = m.j/j(j + 1) µ B.e but m.j = (µ B / ){(L + 2S).(L + S)} (µ B / ){(L 2 + L.S + 2S 2 )} (µ B / ){(L 2 + 2S 2 + (/2)(J 2 - L 2 - S 2 )} (µ B / ){((/2)J 2 (1/2)L 2 + (1/2)S 2 )} J 2 = J(J + 1) 2 ; J z = M J (µ B / ){((/2)J(J + 1) (1/2)L(L + 1) + (1/2)S(S + 1)} hence g = /2 + {S(S+1) - L(L+1)}/2J(J+1) Dublin January m J L S S
16 The g-factor is also the ratio of the z-component of magnetic moment (in units of µ B ) to the z component of angular momentum (in units of ) m z /J z = m.j/j 2 = gµ B / The magnetic Zeeman energy is E Z = m z B. This is (m Z /J z ).(J z B) = (gµ B / )J z B Hence E Z = -gµ B M J B M J -5 / 2 Note the magnitudes of the energies involved, with g - / 2 = 2 and µ B = J T -1. The splitting of two J = 5 / 2-1 / 2 adjacent energy levels is gµ B B. For B = 1 T, this is 1 / 2 only J, equivalent to 1.4 K. [k B = J K -1 ] / 2 5 / 2 In a large field at low temperature the moment is gµ B B saturated at the value -gµ B J Bohr magnetons. The effect of applying a magnetic field on an ion with J = 5/2. Susceptibility gives m eff 2 = g 2 µ B2 J(J+1) Dublin January 27 16
17 Co 2+ free ion Energy levels of Co 2+ ion, d 7. Note that the Zeeman splitting is not to scale. Dublin January 27 17
18 Data on rare-earth ions Dublin January 27 18
19 Data on d ions Dublin January 27 19
20 Summary: For free ions: Filled electronic shells are not magnetic. A and a electron is paired in each orbital Only partly-filled shells may possess a magnetic moment The magnetic moment is related to the total angular momentum by m = g(µ B / )J, where J is the total quantum number given by Hund s rules. The third point has to be modified for ions in solids Orbital angular momentum for d ions is quenched. The spin-only moment is m = gµ B S, with g =2. Certain crystallographic directions become easy axes of magnetization magnetocrystalline anisotropy. Dublin January 27 2
21 5. Ions in solids When an ion is embedded in a solid, the Coulomb interaction of the charge distribution of the ion with the potential cf created by the surrounding charges is the crystal-field interaction. cf = cf (r) (r) d r Dublin January 27 21
22 The Hamiltonian now has four terms = + so + cf + Z Typical magnitudes of energy terms (in K) so cf Z in 1 T d f so must be considered before cf for 4f ions, and the converse for d ions. Hence J is a good quantum number for 4f ions, but S is a good quantum number for d ions. The 4f electrons are generally localized, and d electrons are localized in oxides and other ionic compounds. Dublin January 27 22
23 The most common coordination for d ions is 6-fold (octahedral) or 4-fold (tetrahedral). Both have cubic symmetry, if undistorted. The crystal field can be estimated from a point-charge sum. q 5.1 One-electron states Octahedral and tetrahedral sites. To demonstrate quenching of orbital angular momentum, we consider the l = 1 states, 1, -1 corresponding to m l =, ±1. = R(r) cos ±1 = R(r) sin exp {± } The functions are eigenstates in the central potential V(r) but they are not eigenstates of cf. Suppose the oxygens can be represented by point charges q at their centres, then for the octahedron, cf = ev cf = Dq(x 4 +y 4 +z 4 - y 2 z 2 -z 2 x 2 -x 2 y 2 ) where D e/4pe o a 6. But ±1 are not eigenfunctions of V cf, e.g. i * V cf j dv ij, where i,j = -1,, 1. We seek linear combinations that are eigenfunctions, namely = R(r) cos = zr(r) = p z (1/ 2)( )= R (r)sin cos = xr(r) = p x (1/ 2)( )= R (r)sin sin = yr(r) = p y Dublin January 27 2
24 Note that the z-component of angular momentum; l z = i / is zero for these wavefunctions. Hence the orbital angular momentum is quenched. The same applies to d orbitals; the eigenfunctions there are d xy = (1/ 2)( ) = R (r)sin 2 sin2 xyr(r) d yz = (1/ 2)( ) = R (r)sin cos sin yzr(r) t 2g orbitals d zx = (1/ 2)( ) = R (r)sin cos cos zxr(r) d 2 2 x -y = (1/ 2)( ) = R (r)sin 2 cos2 (x 2 -y 2 )R(r) e g orbitals d 2 2 z -r = = R (r)(cos 2 1) (z 2 -r 2 )R(r) The three p-orbitals are degenerate in a cubic crystal field, whether octahedral or tetrahedral, whereas the five d-orbitals split into a group of three t 2g and a group of two e g orbitals p x,, p y, p z dx 2 -y 2, d z 2, e g t2g d xy, d yz, d zx d x 2 -y 2, d z 2, oct / tet d xy, d yz, d zx oct tet Notation; a or b denote a nondegenerate single-electron orbital, e a twofold degenerate orbital and t a threefold degenerate orbital. Capital letters refer to multi-electron states. a, A are nondegenerate and symmetric with respect to the principal axis of symmetry (the sign of the wavefunction is unchanged), b. B are antisymmetric with respect to the principal axis (the sign of the wavefunction changes). Subscripts g and u indicate whether the wavefunction is symmetric or antisymmetric under inversion. 1 refers to mirror planes parallel to a symmetry axis, 2 refers to diagonal mirror planes. 1Dq 6Dq Dublin January t 2 e
25 s, p and d orbitals in the crystal field Dublin January 27 25
26 As the site symmetry is reduced, the degeneracy of the one-electron energy levels is raised. For example, a tetragonal extension of the octahedron along the z-axis will lower p z and raise p x and p y. The effect on the d-states is shown below. The degeneracy of the d-levels in different symmetry is shown in the table. P x, p y p z The effect of a tetragonal distortion of octahedral symmetry on the one-electron energy levels. The splitting of the 1-electron levels in different symmetry. s p l 1 2 Cubic 1 Tetragonal 1 1,2 Trigonal 1 1,2 Rhombohedral 1 1,1,1 d 2, 1,1,1,2 1,2,2 1,1,1,1,1 f 4 1,, 1,1,1,2,2 1,1,1,2,2 1,1,1,1,1,1,1 Dublin January 27 26
27 One-electron energy diagrams b 1g e 1/2 1/2 a 1g /5 b 2g 2/5 e 1/ e g 2/ e a 1 field-free octahedral tetragonal trigonal monoclinic ion O h D 4h C v C 2 Dublin January 27 27
28 z z Jahn Teller Effect y x x x x x z z z E D cfse x y A system with a single electron (or hole) in a degenerate level will tend to distort spontaneously. The effect is particularly strong for d 4 and d 9 ions in octahedral symmetry (Mn +, Cu 2+ ) which can lower their energy by distorting the crystal environment. This is the Jahn-Teller effect. If the local strain is, the energy change E = -A +B 2, where the first term is the crystal-field stabilization energy D cfse and the second term is the increased elastic energy. The J-T distorsion may be static or dynamic. Dublin January 27 28
29 d x2-y2, d z2 e g d xy, d yz, d xz t 2g d xy, d yz, d xz /5 o 2/5 c t 2 E 2/5 t o t c /5 t 2/5 o /5 c e d x2-y2, d z2 t 2g d xy, d yz, d xz e g d x2-y2, d z2 cubic tetrahedral spherical octahedral Dublin January 27 29
30 5.2 Many-electron states In insulators, the electrons in an unfilled shell interact strongly with each other giving rise to a series of sharp energy levels which are determined by the action of the crystal field on the orbital terms of the free atom. The spacing of theses levels may be determined by spectroscopy, and the crystal-field determined. T 1g T 1g T 2g P F A 2g T 1g T 2g T 1g Orgel Diagrams These diagrams show the effect of a cubic crystal field on the Hund s rule ground state term. Since a half-filled shell has spherical symmetry, the cases d n and d 5+n are equivalent. Also, since a hole is the absence of an electron, the cases d n and d 1-n are related. A 2g T 2g d, d 8 octahedral (d 2, d 7 tetrahedral) E E d 2, d 7 octahedral (d, d 8 tetrahedral) E g S D A 1 A 1 T 2g E g D o D t d 5 octahedral or tetrahedral d 4, d 9 octahedral (d 1, d 6 tetrahedral) Dublin January 27 d 1, d 6 octahedral (d 4, d 9 tetrahedral)
31 High-spin and low-spin states An ion is in a high-spin state or a low spin state, depending on whether the Coulomb interaction U leading to Hund s first rule (maximize S) is greater or less than the the crystal-field splitting D. Consider a d 6 ion such as Fe +. e g e g e g t 2g t 2g D e g U D U t 2g t 2g U > D, gives a high-spin state, S = 2 e.g. FeCl 2 U < D, gives a low-spin state, S = e.g. Pyrite FeS 2 Dublin January 27 1
32 Tanabe-Sugano diagrams These show the splitting of the ground state and higher terms by the crystal field. The high-spin lowspin crossover is seen. Diagrams shown are for d-ions octahedral environments. d 6 1 I G F P H E T 2g A 1g T 1g E g g; ( t2g) ( eg) T 2g T 1g E 5 D T 2g T 1g 1 A g t 6 1 ( 2 g ) 5 2 T2g; ( t2g)( eg) crystal field splitting Dublin January 27 2
33 d 5 Matching the optical absorption spectrum of Fe + - doped Al 2 O with the calculated Tanabe-Sugano energy-leve diagram to determine the cubic crystal field splitting at octahedral sites. Dublin January 27
34 d 2 d 7 Note the similarities between the Tanabe-Sugano diagrams for d 2 and d 7. The differences are associated with the possible low-spin states for d 7 (e.g. Co 2+ ) Dublin January 27 4
35 d For Cr + in Al 2 O, the cf parameter Dq/B is 2.8 Dublin January 27 5
36 d For Cr + in Al 2 O, the cf parameter Dq/B is 2.8 Dublin January 27 6
37 5. Single-ion anisotropy The electrostatic interaction of the ionic charge distribution (r) with the potential cf created by the rest of the crysta gives rise to the crystal field splittings. It is also the source, via spin-orbit coupling, of magnetocrystalline anisotropy. where E = cf (r)dr The anisotropy energy is therefore cf (r) = -(e/4 ) { (R) / R - r } dr E a (r) = -(e/4 ) { (r, ) (R ) / R - r }dr dr Both the charge distribution (r) and the potential cf (r) can be expanded in spherical harmonics. Using the Wigner-Eckart theorem, it is possible to write the corresponding crystal-field Hamiltonian in terms of angular momentum operators J x, J xy J z J 2 which is a particularly useful way to find the energy-levels (eigenvalues). The Hamiltonian matrix is written in an M L or M J basis for the d transition elements or 4f rare earths respectively. In concise form cf = n=2,4,6 m=-n n B n m Ô n m Crystal field parameters n r n A m n Stevens operators Stevens coefficients Crystal field coefficients In a site with uniaxial anisotropy, the leading term is cf = B 2 Ô 2 The Stevens operator Ô 2 is {J 2 z - J(J+1)} cf = B 2 {J 2 z - J(J+1)} Dublin January 27 7
38 9/2 Charge distributions of the rare-earth ions. Those with a positive quadrupole moment ( 2 > ), italic type are distinguished from those with a negative quadrupole moment ( 2 < ) bold type. Note the quarter-shell changes, ±1/2 ±9/2 e.g. Nd + J = 9/2 ±/2 ±5/2 ±7/2 ±9/2 ±1/2 B 2 < B 2 > Dublin January 27 8
39 6 Paramagnetism 6.1 Langevin theory z m 2 sin d H O Dublin January 27 9
40 6.2 Brillouin theory The general quantum case was treated by Brillouin; m is gµ B J, and x is defined as x = µ mh/k B T. There are 2J+1 energy levels E i = -µ gµ B M J H, with moment m i =gµ B M J where M J = J. J-1, J-2, -J The sums over the energy levels have 2J+1 terms. Their populations are proportional to exp(-e i /kt) a) Susceptibility To calculate the susceptibility, we can take x << 1, because the susceptibility is defined as the initial slope of the magnetization curve. We expand the exponential as exp(x) = 1 + x +.., <m> = -JJ gµ B M J (1 + µ gµ B M J H/k B T)/ -JJ (1 + µ gµ B M J H/k B T) M J Recall -JJ 1 = 2J + 1-5/2 -JJ M J = -/2 -JJ M 2 J=5/2-1/2 J = J(J + 1)(2J + 1)/ Hence <m> = µ g 2 1/2 µ B2 HJ(J + 1)(2J + 1)/(2J + 1)k B T /2 The relative susceptibility is N<m>/H, where N is the 5/2 number of atoms/m -5/2. r = µ Ng 2 µ B2 J(J + 1)/k B T This is the general form of the Curie law. Again it can be written r = C/T where the Curie constant C = µ Ng 2 /2 µ B2 J(J+1)/k B or C = µ Nm eff2 /k B where m eff = gµ B [J(J+1)]. A typical value of C for J = 1, N = m - 5/2 is.5 K. Note that results for the classical limit and S = 1/2 are obtained when J (m = gµ B J) and J = 1/2, g = 2. (m = µ B ). H Dublin January /2-1/2 1/2
41 Magnetization Curve To calculate the complete magnetization curve, set y = µ gµ B H/k B T, then <m> = gµ B / y[ln J -J exp{m J y}] [d(ln z)/dy = (1/z) dz/dy] The sum over the energy levels must be evaluated; it can be written as exp(jy) {1 + r + r r 2J } where r = exp{-y} The sum of a geometric progression (1 + r + r r n ) = (r n+1-1)/(r - 1) J -J exp{m J y} = (exp{-(2j+1)y} - 1)exp{Jy}/(exp{-y}-1) multiply top and bottom by exp{y/2} = [sinh(2j+1)y/2]/[sinh y/2] <m> = gµ B ( / y)ln{[sinh(2j+1)y/2]/[sinh y/2]} = gµ B /2 {(2J+1)coth(2J+1)y/2 - coth y/2} setting x = Jy, we obtain <m> = mb J (x) where the Brillouin function B J (x) = {(2J+1)/2J}coth(2J+1)x/2J - (1/2J)coth(x/2J). Again, this reduces to the previous equations in the limits J (m = gµ B J) and J = 1/2, g = 2. 1.,.8 1/ Comparison of the Langevin function and the Brillouin functions for J = 1/2 and J = Dublin January x
42 7 6 Magnetic moment, µ B per ion S = 7/2 (Gd + ) S = 5/2 (Fe + ) S = /2 (Cr + ) 1.K 2.K.K 4.21K Brillouin functions B J /T (TK -1 ) Reduced magnetization curves for three paramagnetic salts, with Brillouin-theory predictions The theory of localized magnetism gives a good account of magnetically-dilute d and 4f salts where the magnetic moments do not interact with each other. Except in large fields or very low temperatures, the M(H) response is linear. Fields > 1 T would be needed to approach saturation at room temperature. The excellence of the theory is illustrated by the fact that data for quite different temperatures superpose on a single Brillouin curve plotted as a function of x H/T Dublin January 27 42
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