Magnetism I: from the atom to the solid state

Transcription

1 Magnetism I: from the atom to the solid state

2 Preface The Lecture on Magnetism I: from the Atom to the solid state is an introduction to the fundamental concepts in magnetism. It deals with magnetic effects in atoms (diamagnetism and paramagnetism) and with the occurrence of magnetic order in the solid state in virtue of the exchange interaction. The lecture will emphasize the purely quantum mechanical origin of magnetism in matter, which cannot be derived nor from the Maxwell equations nor from classical mechanics in general. A very extended introduction in modern magnetism can be found in the book by J. Stöhr and H.C. Siegmann Magnetism: from fundamentals to the Nanoscale dynamics, Springer-Verlag, Berlin Heidelberg 006. This lecture is dedicated to H.C. Siegmann for his inspiration and for his great contributions to establishing modern magnetism. Zürich, September 007 D. Pescia ii

3 Contents Preface ii Magnetism in atoms. The spin operator in quantum mechanics Stern-Gerlach Experiment and the non-integer spin values Compounding angular momentum Diamagnetism and paramagnetism The formation of the magnetic moment in atoms The energy levels in a He-Atom Magnetism in solids 0. The H molecule Friedel-Oscillations: why is Fe magnetic? Band structure and magnetism The quenching of angular momentum Non-integer magneton numbers Electronic structure of ferromagnetic bulk Ni, Fe and Co Magnetism at surfaces and magnetic obverlayers iii

4 Chapter Magnetism in atoms. The spin operator in quantum mechanics The angular momentum operator appearing in the Schrödinger equation for the Hydrogen atom is defined as L op = r op p op = i h [ r ]. e x e y e z i h x y z x y z y z z y = i h z x x z x y y x It can be transformed into spherical coordinates by considering that x = r sin ϑ cosϕ y = r sin ϑ sin ϕ z = r cosϑ and r = x + y + z cos ϑ = z x + y + z tan ϕ = y x We transform, as an example L z = i h [x y y x ] = i h { r sin ϑ cosϕ [ r y r sin ϑ sin ϕ [ r To proceed further we need the table + ϑ + ] ϕ r y ϑ y ϕ + ϑ + ]} ϕ x r x ϑ x ϕ r/ z = cosϑ r/ y = sin ϑ sin ϕ r/ x = sin ϑ cosϕ ϑ/ z = sin ϑ/r ϑ/ y = cosϑsin ϕ/r ϑ/ x = cosϑcosϕ/r ϕ/ z = 0 ϕ/ y = cosϕ/(r sin ϑ) ϕ/ x = sin ϕ/(r sin ϑ)

5 CHAPTER. MAGNETISM IN ATOMS As an example of the application of this table we perform ϕ y yϕ y ϕ = y tanϕ ϕ tanϕ = x cos ϕ = r sin ϑ sin ϕ cos ϕ = cosϕ r sin ϑ As a result, L z = i h. The other components can be transformed similarly ϕ. ( L x = i h sin ϕ ϑ + cosϑcos ϕ ) ( L y = i h cosϕ ϕ ϑ cosϑsin ϕ ) ϕ In these coordinates, it is easy to solve the eigenvalue equations: i h ϕ Y m l (ϑϕ) = hmy m l (ϑ, ϕ) [ L = h sin ϑ ϑ (sin ϑ ϑ ) + ] Y m sin ϑ ϑ l (ϑ, ϕ = h l(l + )Y m l (ϑ, ϕ) Accordingly, we see that the angular momentum in quantum mechanics has discrete eigenvalues and a set of eigenfunctions which build a VONS on the unit sphere in there dimensions. This means that the angular momentum operator within each eigenspace can be represented by matrices. To each quantum number the three components of the angular momentum are represented each by a (l + ) (l + ) matrix. The matrix elements can be computed by taking the integrals over the spherical harmonics Y l,m (ϑ, ϕ), m = l,..., +l. We give the matrix elements following the convention of Condon and Shortley: ( Y j,m, L ) z h Y j,m = m ( Y j,m±, L ) x h Y j,m ( Y j,m±, L ) y h Y j,m = i = / (j m)(j ± m + ) (j m)(j ± m + )

6 CHAPTER. MAGNETISM IN ATOMS 3 As an example: with Y = 3 sin 8π ϑeiϕ, Y 0 = 3 3 sin 8π ϑe iϕ we obtain ( h0) (Y, L xy ) = i (Y, sin ϕ ϑ Y ) }{{} =0 +i (Y cosϑ 4π sin ϑ cosϕ ϕ Y ) }{{} =0 cos ϑ and Y = = 0 ( ( ) Y 0, L x Y = i Y 0, sin ϕ ) ( ϑ Y +i Y 0, cotϑcosϕ ) ϕ Y }{{}}{{} ( ) ( ) i i = i + i = In this way, the matrices for L x and L y (l = ) are: [L x ] = [L y ] = 0 i i 0 i 0 i Stern-Gerlach Experiment and the non-integer spin values In 9 Stern and Gerlach did the first experimental observation of the quantization of the angular momentum: A beam of Ag-atoms (configuration 4d 0 5s ) is injected into a Stern-Gerlach apparatus, where a strong gradient of the magnetic field act with a force F = ( µ B) = ( µ ) B db (in our case F z = µ z z ) upon the magnetic moment associated with the dz s-electron (we know that it exists because of Dirac equation µ z = gµ B S z with g = and µ B = Bohr Magneton). The force depend on the sign of the magnetic moment. For continuous values of the spin we would expect a continuous distribution of atoms leaving the apparatus. A double peak structure is observed, showing that the spin is quantized. The experiment of Stern and Gerlach showed that. The spin of the electron is quantized and can assume only discrete values. The spin can only assume two values, so that from S + = it follows the unusual value S = /, suggesting an eigenvalue S z = ± h for the angular momentum of the electron, which therefore is not comparable with the angular momentum observed in the Schrödinger theory.

7 CHAPTER. MAGNETISM IN ATOMS 4 Figure.: The Stern-Gerlach experiment with the measured intensity (right). The hypothesis of a two-value electron spin angular momentum was introduced in 95 by Uhlenbeck and Goudsmit, based on the Stern Gerlach experiment and on the spectrum of Na (the Na doublet, a concept the fine structure introduced by A. Sommerfeld). The existence of non-integer Figure.: Thermal scheme of the Na-Atom with optical transitions in the visible spectral range values for the angular momentum can be proved within the group theoretical treatment SO(3). The matrices for the non-integer values can be obtained by generalizing the matrix elements computed for integer values to non-integer

8 CHAPTER. MAGNETISM IN ATOMS 5 values. as an example, the matrices for spin /-particles are ( S z / 0 h = 0 / ) ( Sx 0 / h = / 0 ) Sy h = ( 0 i i 0 ) One writes S = h σ σ z = ( 0 0 ) σ x = ( 0 0 ) σ y = ( 0 0 ) σ: Pauli-matrices Die eigenfunction of S z cannot be represented in conventional 3d-space, only in C : u /,/ u /, / ( ) 0 ( ) 0 A general quantum mechanical state for a spin /-particle is therefore ( ) ( ) ( ) ( ) 0 c/ ( r, t).= c( r, t, /) c / ( r, t) + c /( r, t) = 0 c /( r, t) c( r, t, /) Basis states to a general spin S are called spinors. c( r, t, s) c( r, t, s ). c( r, t, s) According to Born c( r, t, m) dv is the probability to find a particle at the time t, at the location r and with S z = m S m= S dv c( r, t, m) =.. Compounding angular momentum For determining the electronic structure in atoms one needs to know how the orbital angular momentum L couples with the spin angular momentum S. In addition, when more electrons are involved, one needs to know how their orbital and spin angular momentum couples. Mathematically, the question

9 CHAPTER. MAGNETISM IN ATOMS 6 in to find all the common eigenspaces and all the possible eigenvalues of the matrices L z E + E S z and ( L E + E S), (respectively J z and J, where the operation means the Kronecker (or tensor) product of the two matrices. After this problem is solved, one need some rules on how to determine which value belong to the ground state of the system. The first problem is solved for the group SO 3 by the Clebsch-Gordan development (possible eigenvalues) and the Clebsch-Gordan coefficients, which produces the symmetry adapted wave function transforming according to irreducible representations of J z and J. The second problem is solved by the Hund s rules, as we will see later. We now elucidate the group theoretical results that are involved in the the coupling of two angular momenta S and S. The Clebsch-Gordan development leads to possible values of the total angular momentum J amounting to S +S, S +S,..., S S. To each value of J we have J + eigenvalues of J z. Notice that the possible eigenvalues of J also labels the eigenspaces of an Hamiltonian that contains the spin orbit S L interaction, as both J and the spin orbit interaction are scalar under rotation and therefore have, according to the Wigner-Eckart theorem, the same eigenspaces. Notice that, again because of the W.-E. theorem, J z, L z and S z have also common eigenspaces, as both are the z-components of a vector operator under rotations. At the appropriate place we will make good use of these group theoretical results.. Diamagnetism and paramagnetism To understand the occurrence of the most elementary magnetic behavior in matter it is enough to consider the response to B of the energy levels of one single electron in the field of a positive charged ion with spherically symmetric potential. In order to establish the Hamilton operator determining the interaction of such an electron, its spins and an applied magnetic field B we need some informations form classical electrodynamics. In order to find the classical Hamilton function of an electron in a potential V ( r) and an applied magnetic field B we write E and B according to Ansatz B = A; E = φ A t This Ansatz satisfies the Maxwell equation (ME) div B = 0 and rot E = B t. Inserting this Ansatz in the remaining ME we obtain equations for A and φ. Notice that A and φ are defined up to Gauge transformation A = A + χ; φ = φ χ t

10 CHAPTER. MAGNETISM IN ATOMS 7 Theorem: The classical Hamilton function is H = m ( p e A) + eφ Proof : Assuming this Hamilton function we obtain the following Hamilton equations: r = H p ; p = H r From ẋ i = H p i = m (p i ea i ) and ṗ i = m ( p e A)e A x i e φ x i = j ẋ j e A j x i e φ x i mẍ i = ṗ i e A i t e j A i x j ẋ j = e( φ x i + A i t ) + e j we can transform the Hamilton equations to m d r dt = e E + e( r B) ẋ j ( A j x i A i x j ) which is nothing else that the Newton equation with the Lorentz force driving the motion (as it should be). In transforming the Hamilton function to a QM Hamilton operator one must consider that p and A do not commute: ( p e A) = p e( p A + A p) + e A = p e A p e A + e A Accordingly, the SE (Schrödinger equation) writes i h u t = h i he u + A m m u e m Au + e A m u + V u Note. As the H-operator is only defined up to a gauge transformation, the choice of different gauge fields leads to different eigenfunctions. However, one can show that a gauge transformation only leads to a multiplication of a wave function u( x) with a phase factor e iē h χ, so that the sought for eigenvalues are not affected.

11 CHAPTER. MAGNETISM IN ATOMS 8 We now consider a magnetic field which is uniform along the z-direction: A = ( r B) and From i h u t = [ h u + V u] m + i he A m u (a) + e m A u (b) A u = /( r B) u = /( r u) B we obtain (a) = e ( B L)u, where L is the orbital momentum. We define m µ = e L as the operator of the (orbital) magnetic Moment, so that m (a) = ( B µ) u The second term (b) can be written as e m A u = e 8m ( r B) u = e 8m ( r B ( r B) ) The final Hamilton operator of a single (for the moment) spinless electron in a magnetic field becomes H op = [ h m + V ( r)]u ( B µ)u + e 8m ( r B ( r B) )u We now estimate the relative strength of the magnetic field induced correction to the eigenvalues of the atom with Coulomb interaction by computing the first order correction arising form the presence of a magnetic field over the s-orbital wave function of Hydrogen. We take V ( r) = e 4πǫ 0 and use the r atomic units a 0 = h 4πǫ 0 and E e m 0 = e 4πǫ 0 a 0 as units for length and energy, i.e. ρ = r/a 0, h = H/E 0. Accordingly h = ρ ρ + e h B me 0 b l + e a 0 B 8mE 0 ( ρ b ( ρ b) ) l and b being dimensionless vectors. Further manipulations lead to h = ρ ρ + ( a 0 a B ) ( b l) + 8 ( a 0 a B ) 4 ( ρ b ( ρ b) )

12 CHAPTER. MAGNETISM IN ATOMS 9. a B = h =.5 eb 0 8 m being a characteristic magnetic length. From B/T this equations we notice that (a) is of the order ( a 0 a B ) B/T smaller than the value of a s-level in Hydrogen. In addition, (b) is factor ( a 0 a B ) 4 smaller that the typical s-level eigenvalue arising form the Coulomb interaction and a factor ( a 0 a B ) smaller that (a). The term (a) is responsible for the atomic Paramagnetism, (b) produces atomic Diamagnetism. Both are very small corrections of the atomic energy levels calculated including the Coulomb interaction and the spin -orbit coupling, as long as B is within the range of values available in laboratories ( T). Diamagnetism is only observable is the magnetic moment is exactly zero, so that term a vanishes exactly. From the expression for the diamagnetic energy change one can define an effective. diamagnetic moment per atom through E D = µd B. In the present case we obtain, assuming that < s ( ρ b ( ρ b) ) >= /3, E D = +/3 E 0 0 (B/T) (.) µ D = /3 E 0 0 B/(T) (.) Thus, the diamagnetic moment per electron points anti parallel to the applied magnetic field. Notice that in a field of B T µ D 0 ev/t. The diamagnetic susceptibility defined as µ o µd amounts to about a 3 0 B 0 6. As the correction to the energy levels due to term (a) is order of magnitude larger, we expect (a) to dominate as soon as some magnetic moment is built up in the atom. We now consider the effect of (a) the so called Zeeman term on the fine structure of an atom. As electrons do have both orbital and spin angular momentum, the magnetic coupling interaction with a magnetic field must be extended to include the magnetic moment arising from the spin with µ S = g e S. For orbital moments g =. For the spin moment the Dirac m equation shows g =.Quantum electrodynamic corrects this value to g =.003, which is extremely close to the experimental value. Thus, the operator describing the interaction of a magnetic moment with a magnetic field becomes H Z = e m ( L + g S) B. = µ B ( L + g S) B µ B = e h m is the Bohr magneton. L and S angular momenta in units of h. To find the first order correction we consider a level with quantum number J and solve the eigenvalue problem of H Z within the J + dimensional space containing the symmetry adapted wave functions to this J-value: u n,l,j,mj. We seek the first order correction, i.e. the magnetic field is small enough so

13 CHAPTER. MAGNETISM IN ATOMS 0 that the Zeeman splitting is smaller than the level splitting arising form the spin-orbit coupling. We choose B = (0, 0, B): H Z = µ B B (J z + (g ) S z ) The eigenvalue problem of J z is simply solved, because (u n,l,j,mj, J z u n,l,j,mj = m j, m j = J, J,..., J. The eigenvalue problem of S z is simplified by the Wigner-Eckart theorem, which states that (u n,l,j,mj, J z u n,l,j,mj ) = τ LSJ m j. τ LSJ is common to all m j and QM shows that τ LSJ = J(J+)+S(S+) L(L+). J(J+) Finally we obtain (u n,l,j,mj, H z u n,l,j,mj ) = µ B B m j g LSJ with the Lande factor g LSJ = +(g ) J(J+)+S(S+) L(L+). Accordingly, a J(J+) magnetic field lifts the J +-degeneracy of the fine structure level completely. The Zeeman splitting is symmetric around the unperturbed level E nj. The distance between two consecutive Zeeman levels is E Z = µ B B g LSJ, i.e. it is proportional to B These results are known as the anomalous Zeeman effect, to be compared with the normal Zeeman effect, with g = and g LSJ =. The statistical mechanics of an ensemble of non interacting N identical atoms each carrying a magnetic moment J allows to compute the partition function as Z N = [ exp( g LSJ µ B m j B ] N (.3) m j k B T the total free energy f(t, B) per atom as k B T ln [ m j exp( g ) LSJ µ B m j B k B T ] and the and the mean magnetic moment per atom as [ f(t, B) m j ( m j ) exp( g ] LSJ µ B m j B k < µ z > = = g LSJ µ B T B B [ m j exp( g ] LSJ µ B m j B k [ B T m j (m j ) exp( g ] LSJ µ B m j B = g LSJ µ B k B T [ m j exp( g LSJ µ B m j B k B T ] (.4) To perform the sum explicitly we define α =. g LSJ µ B J B k B T and compute [ m j (m j ) exp( α m ] j d J dα [ m j exp( α m ] j J With [ m j exp( α m j J ] = J [ m j exp( α m j J ] (.5) exp( m j α ) = sinh[j+α] J m J J sinh α ] (.6) J

14 CHAPTER. MAGNETISM IN ATOMS we finally obtain B J (α) = J + J < µ z >= g LSJ µ B J B J (α) coth[ J + J α] J coth α J α =. g LSJ µ B J B k B T (.7) For the particular case J = / (a spin /, s-state Atom) we obtain the simple equation For small arguments α, B J (α) J+ α J 3 < µ z >= µ B tanh[ µ B B k B T ] (.8) and we obtain the Curie law < µ (P) z > (g LSJ) J(J + ) µ B 3k B T B (.9) which contains the purely quantum mechanical quantity (g LSJ ) J(J + ). The paramagnetic susceptibility in this limit amounts to with the Curie constant χ P µ 0 C T (.0) C = N V (g LSJ ) J(J + ) µ B 3k B (.) At room temperature is χ P = O(0 3 ). Notice that the determination of the Curie constant is a key experiment to access the quantities J and g LSM determining the ground state electronic configuration of atoms and thus provide a reliable test of our quantum mechanical approach to the ground state configuration in atoms..3 The formation of the magnetic moment in atoms We need now to understand the formation of magnetic moments in atoms,i.e. the origin of the quantum number J, which means considering a manyelectron problem in the central field of a positively charged ion. In classical physics similar particles can be easily distinguished: they can be localized by

15 CHAPTER. MAGNETISM IN ATOMS specifying the initial condition of the motions, identified and their trajectory can be followed exactly by solving the equation of motion. Later, it is possible to exactly tell where each particle is: the classical trajectory is used to identify the particle itself. Quantum mechanically is the situation completely different. A particle, which is localized at one position at the time t = 0 is in a state where the position is specified, let us say x >. The amplitude that the particle in this state has a momentum p is < p x >= L e( ī h p x) which means that any momentum p is equally probable (a particular version of Heisenberg uncertainty principle). Accordingly, a particle, which is localized at the initial time, can be anywhere later. A number of identical particles localized at some point in time will appear in an unpredictable way somewhere else at a later time: their origin can no longer be reconstructed and thus the particles cannot be distinguished. In QM the uncertainty principle makes identical particle indistinguishible. We would like illustrate this principle to two identical particles (to be concrete let us consider two electrons with spin /). The two particles are described by coordinates r, r in Euclidean space and by coordinates m, m is spin space. The wave functions are of the type ψ( r, r )χ(m, m ). The Hamiltonian H is assumed to be invariant with respect to the permutation of the two coordinates, i.e. H(, ) = H(, ). We define an Hermitic operator P, P, b Ps, that permutate the coordinates in both Euclidean and spin space, P b, P s,ψ b ( r, r )χ s (m, m ) = ψ b ( r, r ) χ s (m, m ) (the index b (s) referring to the Orbital (spin) space. As [P,, H] = 0, P, and H have common eigenspaces. The eigenfunctions of P, must obey the equation P, ψ(, ) = λψ(, ) where λ is a real eigenvalues. Applying P, on both side of this equation leads to P, ψ(, ) = λ ψ(, ) =!ψ(, ) From this it is evident that λ = ±. The eigenfunctions belonging to are symmetric with respect to interchange of particle coordinates, i.e. ψ(, ) = ψ(, ). The eigenfunctions belonging to are antisymmetric, i.e. ψ(, ) = ψ(, ). An important theorem proved by Pauli within Quantum field theory states that

16 CHAPTER. MAGNETISM IN ATOMS 3 (a) Only fully symmetrical wave functions are allowed for a system of identical particles with integer spin (so called Boson). (b) Only fully antisymmetric states are allowed for Fermions (identical particles with non-integer spins) (c) The symmetry properties involve the whole wave functions, including the spin variables, even if the Hamilton operator does not contains spin dependent interactions. (d) Sates with other symmetries or mixed symmetries, which appear in the representation theory of the permutation group S N are forbidden in nature We now investigate the consequence of the requirement of the antisymmetric wave functions for the energy spectrum of two non-interacting spin -particles moving in a one dimensional space, for simplicity. The Hamilton operator writes H = m p E + E m p The spectrum of H i is given (in units of h (π) a m by ǫ n i = n i, n i = 0, ±, ±,... The eigenfunctions are u ni (x i ) u(m i ), m i = ±/, u + = >,u = >. The eigenfunctions of the total operator are with the eigenvalues u n (x )u(m ) u n (x )um ǫ n,n = n + n Each eigenvalue is more or less degenerate, because the permutation of the number n and n leads to the same eigenvalue and each eigenfunction with well defined orbital wave function appears with 4 different spin basis functions. This situation changes dramatically because of the requirement that only antisymmetric functions should be allowed with each eigenspace. Following a theorem by Weyl, we construct the suitable antisymmetric wave functions as follows: (a) in a first step, one find all possible values of ( S). = ( i S i ), i.e. all possible values of the operator for the total spin. (b) In a second step, one search, within the eigenspace to each S, for the eigenfunction of S z. H. Weyl has proven that all such eigenfunctions have the same symmetry with respect to particle permutation: the symmetry property of the eigenfunction of S z under permutation are called Spinrasse.

17 CHAPTER. MAGNETISM IN ATOMS 4 (c) Having determined the spin functions, one must choose the corresponding orbital wave functions so that the total wave function is antisymmetric. For our original spin / particles this Weyl procedure means the following: We need to find the eigenvalues and eigenvectors of the matrix S = ( S E + E S ) = The eigenvalues are λ = (3-times degenerate) and λ = 0 (non-degenerate), i.e. the possible values for S are, 0. The 3 eigenfunctions to S = are u / u / (u / u / + u / u / ) = u / u / = > = > > ( > > + > >) These three states are called a triplet(χ t ) and are clearly symmetric with respect to particle interchange. The eigenfunction to S = 0 is (u / u / u / u / ) = ( > > > >) It is a singlet-(χ s ) and antisymmetric. Thus, the energy spectrum of the two non interacting particle will consist of two completely separated schemes, one with singlet spin functions (S = 0), which tolerate only symmetric orbital wave functions and one with triplet wave functions (S = ), which tolerate only antisymmetric orbital wave functions. The S = 0 antisymmetric orbital Figure.3: Thermal scheme of a system with two non-interacting spin /- particles(one-dimensional). functions will be (u n (x )u n (x ) + u n (x )u n (x ))

18 CHAPTER. MAGNETISM IN ATOMS 5 The S = orbital function will be (u n (x )u n (x ) u n (x )u n (x )) Consider now the implication of this result: all eigenvalues with n = n - and in particular the ground state n = n = 0 disappears from the S = eigenvalues scheme. In other words: (a) Although H is spin independent and in the present case the particles are not interacting the non distinguishibility postulate of QM remove the degeneracy of some states and even exclude a well defined spin state from some eigenvalues. The Pauli principle acts as some sort of effective interaction that remove some degeneracy and distinguish between spin states. This interaction is know in the literature as exchange interaction. It is a purely QM effect and completely disappears together with the spin upon transition to classical mechanics. In the present case the exchange interaction dictates that the ground state of identical spin /-particles is a singlet. This is a powerful result that must be taken carefully into account when trying to create a magnetic moment in atoms. (b) The exchange interaction is responsible that two identical spin / particles can only be in the same orbital state (n = n ) if they form a singlet. In other words: more than two electrons cannot be in the same orbital state: if they are, they (sloppy) must have opposite spin. This statement is the most famous version of the Pauli principle and is the key for the construction of the electronic structure in atoms with many electrons, for determining their magnetic moment and, ultimately, for the stability of matter..4 The energy levels in a He-Atom The Pauli principle in the version (b) provides a very powerful rule for filling the various hydrogen-like orbitals with electrons in order to construct the electronic configuration of complex atoms. However, it does not provide all rules required. There is a set of rules known as Hund s rules which are extremely important and ultimately allows determining the ground state magnetic moment of any atom. In order to construct these rules we need to increase the degree of complexity, allowing the two electrons of the previous example, to interact, via the Coulomb interaction. Doing this with plane waves would introduce difficult integrals, which are not very realistic. This is because we embed the two electrons in the strong attractive potential of a ion core and compute the electronic structure of He (or He-similar ions with Z protons and only two electrons). For an elementary understanding of

19 CHAPTER. MAGNETISM IN ATOMS 6 Hund s rules we do not need to work on the fine structure, so that we neglect spin-orbit coupling. The Hamiltonian of the two electrons in an attractive potential originating form the nucleus is H = H 0 (, ) + V (, ) with H 0 (, ) = h m ( + ) Ze Ze r r V (, ) = e r (.) Let us first neglect V. The energy levels of H 0 (i) are E ni = Z e, a being a n i Figure.4: The vector system in the He-atom the Bohr radius The eigenfunctions of E ni are ϕ ni l i m i = R n i,l i Y r li,mi(ϑ, ϕ). The ground state of H 0 corresponds to the state in which both electrons are in a s-orbital and has energy E 0 = E = Z e. Its wave function is a ψ 0 = ϕ s ()ϕ s () χ(, ) ϕ s ()ϕ s () = π (Z a )3 e Z a (r +r ) and χ(, ) = (u / u / u / u / ) The wave function ϕ s ()ϕ s () is symmetric with respect to change of the coordinate vectors and the ground state of He is a singlet, as required by Pauli principle. Again, we do not have any net magnetic moment: S = is prohibited in the ground state. Let us now introduce V as a perturbation. The energy of the ground state is modified to E G = E 0 + Q with Q = dv ()dv ()ϕ s () e r ϕ s ()

20 CHAPTER. MAGNETISM IN ATOMS 7 In order to explicitly compute this last integral, one develops /r in spherical harmonics: r r r r = 4π r l + (r ) l Y r l.m(ϑ, ϕ )Y l,m (ϑ, ϕ ) r > r l,m = 4π r l + (r ) l Yl,m r (ϑ, ϕ )Y l,m (ϑ, ϕ ) r < r 0 l,m Inserting this development in Q and using the orthogonality of spherical harmonics we obtain Q = 4e π (Z a )6 dr r Zr e r a [ dr r r e Zr a + dr r e Zr a ] 0 Partial integration leads to Q = 5Ze 8a and positive. Notice that the correction Q is of the same order of magnitude as E 0. Summarizing : the two He-electrons have a ground state configuration (s) with energy E G = Ze a (Z 5 8 ). The first excited state of the H 0 -operator corresponds to the electronic configuration (s) (s), with antisymmetric wave function r ψ S=0 = [ϕ s ()ϕ s () + ϕ s ()ϕ s ()] χ s ψ S= = [ϕ s ()ϕ s () ϕ s ()ϕ s ()] χ t the S = 0-wave functions belong to the parahelium thermal scheme, the S = -wave functions belong to the orthohelium thermal scheme. Parahelium shows diamagnetism, orthohelium is a paramagnet. Without consideration of V para and ortho (s)(s) states are degenerate. The Coulomb interaction lifts this degeneracy, leading to a very important result for the formation of magnetic moment: the triplet states have lower energy than the singlet states, i.e. in contrast to the ground state the S = state is favored. We would like to show this important fact with an explicit calculation. We consider the four states eigenspace of H 0 to (s)(s) and solve the eigenvalue problem of H 0 + V within this space. The Hamiltonian matrix reads mit E s + E s + Q + J E s + E s + Q J E s + E s + Q J E s + E s + Q J Q = J = dv dv ϕ e s ()ϕ s () r dv dv ϕ s ()ϕ s () e r ϕ s ()ϕ s ()

21 CHAPTER. MAGNETISM IN ATOMS 8 from which the sougth for eigenvalues can be read out immediately. The integral Q is the Coulomb energy. The integral J is the result of the exchange interaction and is called exchange integral. It provides the exchange energy contribution that arises form the correlation of the two electron as a consequence of symmetrizing the wave functions according to the Pauli principle. The Coulomb interaction produces, via Pauli principle, a splitting of the initial degeneracy of the (s)(s) configuration: the singlet state has the energy E s +E s +Q+J, the triplet state has the energy E s +E s +Q J. These results can be summarized as follows: There are two distinct thermal schemes for He, consisting of para (singlet) - and ortho (triplet) states. Levels with the same quantum numbers of the orbital wave functions (e.g. (s)(s)) are prohibited in ortho helium the splitting between ortho and para states depends on the relative sign and strength of Q(n, l) and J(n, l). For example, both integrals Q n,s and J n,s are positive, so that the t-states level is lower than the S = 0-level. There is an intuitive explanation for this result. the Coulomb energy is large when the two electrons are closer to each other. In an antisymmetric orbital wave functions the two electrons are far away form each other, an this reduces the Coulomb repulsion with respect to the symmetric orbital wave function, where the two electrons are closer to each other. This last point is an example of the empirical first Hund rule. the lowest energy state corresponds to a state of maximum spin number compatible with the Pauli exclusion principle. The second Hund rule states that, after having established the maximum S, the lowest energy correspond to maximum L 3. The third rule states that the total angular momentum quantum number J minimizing the energy is L + S if the shell is more than half full, L S is the shell is less than half full. These rules, together with the Pauli principle that two electrons can only be in the same orbital state if they have opposite spins determine completely the total magnetic moment of the ground sate of an atom. The spectroscopic label for the ground state atomic configuration writes S+ L J. In the absence of spin-orbit coupling, optical transitions between the two thermal schemes are absolutely forbidden. Should one be able to pump a He

22 CHAPTER. MAGNETISM IN ATOMS 9 atom gas (by electrons excitation or other mechanism) into the triplet excited state, then it will practically stay forever (many months) in that state. Notice that because of spin orbit splitting the triplet states degeneracy is lifted and a fine structure appear in the excitation spectrum of ortho helium Note: Heisenberg-Dirac-Van Vleck found an effective operator, formally acting in spin space, caricaturing the t s splitting by the exchange interaction (which of course is acting on orbital wave functions and not in spin space:the Heisenberg-Dirac-Van Vleck operator. Dirac defined the operator H Spin = (E s + E s + Q) E J P with the exchange operator P P + > + >= + > + > P + > >= > + > P > + >= + > > P > >= > > The eigenvalues of the operator H spin when restricted to the ss eigenspace, are identical with the eigenvalues of the physical operator. A useful way of writing P is P = E + σ σ = [E + 4 S S ] This last form is known as the Heisenberg-Dirac-Van Vleck Operator, which represents a very useful caricature of the exchange interaction. It can be formally generalized to many electrons: H Spin = r E r + Q(r, s) J(r, s) [ r,s 4 + S r S s ] r s where r, s is a set of quantum numbers. Dirac has shown with quite general arguments that the eigenvalues of this effective spin Hamilton operator are correct within first order perturbation theory. The difficulty is one of computing the various coupling constants forthcoming in the operator, which requires knowledge of the orbital wave functions.

23 Chapter Magnetism in solids In the first chapter we have shown how atomic magnetic moments are produced. On these grounds, diamagnetism and paramagnetism are well understood. However, we still need to understand why Fe is ferromagnetic, that is we need to understand the mechanisms that couple atomic magnetic moments when they are put together to form a solid. For this purpose we seek, as a way of capturing the essential ingredients, the electronic structure of the H -molecule.. The H molecule We use the Born-Oppenheimer approximation where the two nuclei A, B are kept fixed at a distance R which enters the problem as a parameter. The Hamilton operator neglecting both motion of the nuclei and spin-orbit Figure.: The geometry of the H -calculation coupling writes H 0 = h m ( + ) e [ r A + r A + r B + r B r R ] 0

24 CHAPTER. MAGNETISM IN SOLIDS The indices, refer to the electrons, the indices A, B to the nuclei. we want to solve the eigenvalue equation [H 0 ǫ(r)]ϕ(r,, ) = 0 following the method introduced by Heitler and London. Accordingly, the eigenvalue problem is solved with the subspace determined by the four trial functions with ψ A ()ψ A () ψ B ()ψ B () ψ A ()ψ B () ψ B ()ψ A () ψ A () = (πa 3 ) / e ( r A a ) ψ A () = (πa 3 ) / e ( r A a ) ψ B () = (πa 3 ) / e ( r B ) a ψ B () = (πa 3 ) / e ( r B a ) being derived from the basis states of the separated atoms. The polar states where both electrons are close to one nucleus are excluded, as they are thought to have a too large Coulomb repulsion. Of core, only symmetrized trial functions need to be considered, corresponding to the singlet and triplet spin functions: ϕ s = [( + S )] / (ψ A ()ψ B () + ψ A ()ψ B ()) ϕ t = [( + S )] / (ψ A ()ψ B () ψ A ()ψ B ()) with S = dv ψ A ()ψ B () = dv πa 3 e r A +r B a being the overlap integral. Notice that the choice of H-atomic wave function is physically plausible, but the wave functions ψ A () and ψ B () are not orthogonal. This must be taken into account in order to normalized the symmetrized trial wave functions to. One must notice however that ϕ s χ s is orthogonal to ϕ t χ t. Within the subspace defined by the singlet state ϕ s χ s and the three triplet states ϕ t χ t the matrix of H 0 is diagonal: ( (ϕt, H 0 ϕ t ) 0 0 (ϕ s, H 0 ϕ s ) and there is one threefold degenerate eigenvalue ǫ t = (ϕ t, H 0 ϕ t ) and one nondegenerate eigenvalue ǫ s = (ϕ s, H 0 ϕ s ). For calculating the matrix elements )

25 CHAPTER. MAGNETISM IN SOLIDS one must consider that We obtain Let us now discuss Q = = [ h m e r A ]ψ A () = E s ψ A () ǫ t = E s + Q J S ǫ s = E s + Q + J + S dv dv ψa()ψ B()[ e + e + e + e r B r A r R ] dv ψa() e dv ψ r B() e + B r A dv dv ψa()ψ B() e + e r R The first integral is the Coulomb energy of the first electron with the nucleus B, the second in the Coulomb energy of the second electron with nucleus A, the third is Coulomb interaction of the two electrons and the fourth is the Coulomb repulsion of the two nuclei. Q is called Coulomb-integral. The integral J = e S R + dv dv ψ A ()ψ B () e ψ A ()ψ B () r S dv ψ A () e ψ B () S dv dψ B () e ψ A () r B r A represents the exchange energy, arising form the requirement of symmetrizing the wave functions. Both Q and J depends on the distance R. In the figure the R dependence of ǫ s E s and ǫ t E s is plotted. The energy of the s states decrease up to R 0.5a, where it increases because of the Coulomb repulsion of the nuclei. Only the s state can produce a chemical bonding, the t state is energetically unfavored. R 0 is the equilibrium distance of the nuclei in the the H -molecule. The experimental value is A. The R- dependence of the energy is determined by the R dependence of Q(R) and J(R). Both can be computed exactly within the subspace chosen in the H-L method, as shown in the figure. Q(R) is a small positive quantity, being small and negative only for some R. J is instead negative for R > R 0 : Accordingly, Q + J is negative for R > R 0, while Q J is positive. Notice the a negative sign of J is the essential ingredient for a stable chemical bonding, which then occurs in the singlet state. However, a negative J leads to a preferred

26 CHAPTER. MAGNETISM IN SOLIDS 3 Figure.: Plot of the eigenvalues as a function of R. The singlet state has a minimum at a well defined distance R 0, at which the chemical binding of the two H-atoms is realized antiparallel coupling of the two spins and is a simple but very clear and robust demonstration that anti parallel spin alignment is the key coupling mechanism when atoms are assembled to form a solid. We thus have a very complicated situation where the Pauli principle, essential for the chemical bonding, works against parallel spin alignment. This result, obtained by a simple calculation of the H molecule, is robust and there exist a very strong theorem by Lieb and Mattis that states that in a linear arrangement of atoms the non-magnetic state, i.e. the state with lowest total spin, is the ground state. One needs to go to higher than one dimension to escape this theorem, because in higher dimensions electrons states with different symmetry atomic orbitals with different quantum numbers can hybridize and provide a route to escape the strong Pauli principle. Notice that the in atoms, when the electronic states involved have different orbital quantum numbers, the exchange energy is positive and leads to the first Hund rule.. Friedel-Oscillations: why is Fe magnetic? The long sought explanation for the origin of ferromagnet was provided after years of research in a review article by M.B. Stearn, Physics Today, April 978, p.34. The first condition for ferro magnetism is that we have some localized magnetic moments, and this conditions is met in Fe by the localized

27 CHAPTER. MAGNETISM IN SOLIDS 4 d-electrons, which keep a part of their atomic magnetic moment produced by the Hund-rule intra atomic exchange. The second condition for ferromagnetism it that all these moments line up parallel to each other, in apparent contrast to Pauli principle and to the negative inter atomic exchange intervening during formation of the chemical bond. This second condition requires a novel mechanism for exchange other than direct exchange between the d-electrons provided within the HL method. This alternative mechanism is provided, according to Stearns, by the indirect exchange between localized d-electrons through RKKY coupling with the delocalized part of the d wave functions. The RKKY coupling mechanism is a central one in modern research on magnetism as well and we want to introduce it with a simple, computable exact model which is also relevant in thin films coupling phenomena. We follow the paper by K. Yosida, A. Okiji, Phys. Rev. Lett. 4, 30 (965). We put two infinite, one dimensional metals F and N in contact at x = 0. We assume that V (x) = 0, x 0 V (x) = V 0 h α /m, x > 0 We fill the metals with non-interacting electrons. The filling is done according Figure.3: Potential step to the Pauli principle, that requires the only two electrons can occupy a state with the same k-value. The filling is stopped at the Fermi level E f. The question we would like to answer is the following. In the case α = 0 the electron density ρ(x) is uniform over the entire space. The electron density is perturbed by the appearance of a finite α in metal N: the electrons stumble at the interface x = 0. We now compute the electron densityρ N (x). The SE of this one-dimensional problem is ( d dx + m h E)ψ F(x) = 0(x 0) [ d dx + m h (E V 0)]ψ N (x) = 0(x > 0)

28 CHAPTER. MAGNETISM IN SOLIDS 5 The boundary conditions are the continuity of the wave function and of its derivative at x = 0. In order to deal with normalized wave functions, we introduce a wall at x = ±L/, and assume periodic boundary conditions at these walls. To obtain a physical result we will let finally L. The allowed k-values are k = π n, n = 0, ±, ±,... We let the wave function L vanish at the external walls for simplicity. To find the solutions, we need to distinguish between two cases: E < V 0 Ansatz in F: ψ F (x) = A L sin k(x + L/)x, with k = me/ h. Ansatz in N: ψ N (x) = B e k x. This produces an algebraic equation for k : k + m(e V h 0 ) = 0 i.e. k = ± α k is a real number and the solution are exponentially decreasing or increasing functions. The exponentially increasing wave function must be neglected, as it is unphysical for x. The physical solution is therefore ψ N = k e k x. Matching of the wave function lead to an algebraic equation for A and B: A A k sin k(x + L/) x=0 = B L k e k x x=0 cosk(x + L/) x=0 = B L k ( k )e ik x x=0 (normalization condition: A + B = ). The solution is B = 4 A L k. The charge density in N, caused by the exponentially decreasing wave function is B A = 4 k x k e k x L α e k The total electron density by such bound (b) electrons is obtained by integration form 0 to α, the density of k values being L. Accordingly π k α ρ(x) b = L 4 π L 8 = π α α 0 α 0 k α e k x dk k α k e kx dk We now consider E V 0 With the Ansatz ψ F = A ψ N = B L sin k(x + L/) L sin k (x L/)

29 CHAPTER. MAGNETISM IN SOLIDS 6 Figure.4: ρ(x) b /8 as function of k F x. we obtain the algebraic equations with solution A sin kl/ = B sin k L/ A k coskl/ = B k cos k L/ B A = k α sin k L/ + k The total electron density is ( f meaning free ): ρ f (x) = 8 kf π α k α sin k L/ + k sin k (x L/) dk with k = k α. With the substitution q. = k α we obtain ρ f (x) = 8 kf α k k sin k(x L/) π + α 0 k + α sin kl/ dk We would like to evaluate this integral for large values of L. With sin (x L/) = / / cos(kx) cos(kl/) / sin(kx) sin(kl/) we can transform the integral into a sum of three terms. All integrands are rapidly oscillating function of k, as shown in the figure. Physically relevant is the means value of the integrand within the interval [L, L ], with L L very small and L, L very large. In this way the integrands become

30 CHAPTER. MAGNETISM IN SOLIDS 7 Figure.5: Integrand ofρ f as function of q. It is a strongly oscillating function with frequency L. Physically relevant in the envelope of the graphs, which shows small oscillations superposed on a background (L L ) L L (L L ) cos(kx) L L L (L L ) L The result of the averaging procedure is. k k + α α sin (kl/) + k dl cos(kl/)k k + α dl α sin (kl/) + k sin(kl/)k k + α dl α sin (kl/) + k (L L ) k arctan[ α + k tan(kl) k L / L /

31 CHAPTER. MAGNETISM IN SOLIDS (L L ) cos(kx) k k + α [ L L + k + α α ) tan(kl) α α k α + k arctan[( k L/ (k + α ) α 4 (L L ) ( kα ln(k + α α cos(kl) L / L / L / ]] As we obtain α + k arctan[ tan(kl) k arctan[ ( k α ) tan(kl) L/ (k + α ) α 4 L / L / L / ln(k + α α cos(kl) L L / = 0 = (L L ) k ]] = (L L ) k ρ f (x) = 8 kf α dk π kf α cos(kx)[k k π α + α 0 (k + α )]dk

32 CHAPTER. MAGNETISM IN SOLIDS 9 Figure.6: Graphic representation of (L L ) arctan[ α +k tan(kl) k L/ L / as a function of k. We call that part of ρ f which is independent of x ρ 0 and write the remaining part as ρ(x) 0 k + ρ(x) ( k F α k ) with ρ(x) 0 k = 8 π α ρ(x) ( k F α k ) = 8 π The integrals α 0 cos(kx)[k k + α (k + α )]dk k cos(kx)[k k + α (k + α )]dk F α 0 0 cos(kx)dk k cos(kx)dk vanishes. The calculation of these integrals can be performed by allowing a small uncertainty of x: 0 dk x+ x cos(ky)dy = 0 dk cos(k x )sin k k as x >. with complex integration it is possible to show that 8 dk cos(kx)[k k π α + α ] = ρ(x) b 0 Thus, the contribution of the bound states cancels out exactly from the final result. For small α we get kf α k F ( α ) [k k α (k + α )] α4 8k k F = 0

33 CHAPTER. MAGNETISM IN SOLIDS 30 so that we can write ρ(x) ( k F α k ) as 8 π α k cos(kx)( α4 α 8k)dk π F α This last integral can be evaluated as and for large x k F cos(kx) k α π [π x + cos( k F x)/(k F ) + x Si(k F x)] (.) ρ(x) ( k F α k ) α π sin(k F x) 4k Fx + O( x ) The charge density close to a potential step acquires an oscillatory contribution with period π/k F. The total charge density behave, close to the boundary ρ N k F π α x + O(x ) (.) being continuous across the boundary. The long range oscillatory contribution is called Friedel oscillations or, within the magnetism community, the Rudermann-Kittel-Kasuya-Yosida (RKKY) interaction. In the case that the boundary a plane is, the oscillatory component falls like x. The RKKY interaction arising form a point defect falls like r 3 ab, where r the distance form the defect. dk Figure.7: Numerical ρ f (x) (left) and ρ f (x)+ρ b (x) (right). Only the oscillatory contribution remains. The application of this result to magnetism is performed by filling the semi-infinite segment F with a ferromagnet and the segment N with a nonmagnetic metal. The exchange energy J in F produces a splitting between

34 CHAPTER. MAGNETISM IN SOLIDS 3 majority and minority electrons. This splitting is absent in the N segment, so that only the majority spin electrons feel a potential step at zero. We make the contact to magnetism by taking the potential step to be J. = h α /m. As Figure.8: Energy as a function of k and k for majority (+) and minority (-) spins. a consequence of the potential step we expect that minority spins propagate without change across the boundary. The majority spins instead acquire an oscillating density. Close to the interface P = < S z > h ρ k F π O(x) ρ k F π α πk F (x) = ρ+ ρ ρ + + ρ α k F π /α x (.3) i.e. the spin polarization in then segment N starts positive, decreases and propagates further in an oscillatory manner. At the location x in N there is a spin imbalance: Because of exchange coupling, a spin at a location x in N would lower his energy by aligning along the direction of P. In this way, the exchange interaction can propagate, oscillating between positive and negative depending on the position x..3 Band structure and magnetism We have found an explanation for parallel alignment in solid, but we still do not have all essential elements of magnetism in solids. One puzzle is the quenching of angular momentum observed typically in the transition atom metals which are the most known example for ferromagnetism. The second puzzle is that the remaining angular momentum arising form the spin is in general not an integer multiple of µ B as one could expect from atomic Hund s rules. We will now treat these puzzles in a qualitative way, as the quantitative solution needs the Spin density functional theory developed by Kohn and Sham, which is outside the scope of this lecture.

35 CHAPTER. MAGNETISM IN SOLIDS 3 Figure.9: a): the left-hand side shows a typical hysteresis curve (M versus magnetic field H) recorded for exchange-coupled Co films. At the shift field H = H j the magnetizations of the individual films are aligned to the direction specified by the external magnetic field. The critical field H j is measured as a function of the Cu spacer thickness τ by scanning a focused laser beam over a wedge-like multi layered structure, shown schematically on the right-hand side. b): H j versus τ for a room temperature grown wedge-like multi layered structure. A finite shift field means AFM coupling in the ground state. A vanishing shift field means FM coupling. The thickness of the Co films are 3. ML and 5.8 mono layers, respectively. Inset, the Fourier transform, the two peaks corresponding to the two periodicities.4 Ml and 5.4 ML. The long period dominates. c): as b) but with the Cu wedge and the final Co film deposited and measured at 60 K. The short period now dominates, see Fourier transform).

36 CHAPTER. MAGNETISM IN SOLIDS The quenching of angular momentum When an electron is placed into a rotationally invariant potential, the eigenvalues of the Hamilton operator carry eigenspace which are degenerate with respect to the eigenfunction of L z. This means that, within each eigenspace, basis functions which are eigenfunction of L z can be found and that L z is a good quantum number. The orbital angular momentum is well defined and produces a magnetic moment which must be accounted for when finding the total magnetic moment of the atomic configuration. In a solid, the crystal potential is lo longer invariant against the full rotation group but only with respect to a small subgroup, called point group, and the the translational group. The eigenstates of the Hamilton operator can be labeled by a k-vector. For any given k vector, they are one dimensional except at point and lines of higher symmetry, which form a discrete set in the Brillouin zone. Accordingly, they carry real eigenfunctions (up to a phase factor). This is because the complex conjugate of a wave function is also an eigenvector to the same eigenvalue, so that a non-degenerate eigenvalue cannot carry, as an eigenstate, a function which produces a linear independent function upon complex conjugation. A real wave function ψ have the very important property that (ψ, L z ψ) cc = (ψ, L z ψ) = 0, i.e. the value of the angular momentum is vanishing for almost all states of a crystal. At high symmetry points and lines, essential degeneracy might produce, for some levels, some angular momentum for those specific energy levels, while others, despite degeneracy, carry no angular momentum. An exact quenching vanishing of angular momentum cannot be observed. However, the band structure calculation of most transitions metals do indeed show that the crystal potential effectively produces an almost quenching of orbital momentum, as the weight of states with zero angular momentum dominates..3. Non-integer magneton numbers In angular momentum quenched metals, magnetic moments should be an integer multiple of µ B, i.e. their magneton number is an integer. This would lead, for example, to atomic magnetic moments of,3, respective 4 µ B for Ni, Co and Fe. The experimentally measured value in bulk Fe is 0.66,.75 and.6 µ B. To explain these values we introduce Stoner-Wohlfahrt-Slater model of magnetism in a solid. While the formation of a magnetic moment, because of the first Hund rule, is almost the rule in atoms, it is a very rare event in solid, where the electrons can be considered as delocalized and are therefore better described by the a band structure. To appreciate this we consider a model of free elec-