Magnetism III [Lln22]

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Magnetism III [lln22]

The description of magnetism in matter calls for tools from quantum mechanics and statistical mechanics. This module explores topics that employ such tools – topics traditionally covered (or meant to be covered) in solid-state physics courses.1

Angular momentum of electrons: Magnetism in matter is dominated by orbital and spin angular momenta of electrons, which require a quantum mechanical description. The net angular momentum of atoms originates in incomplete electronic shells. Many atoms thus carry spin and/or angular momentum. Conduction electrons, which are free to roam between atoms, leave behind the orbital angular momentum of incomplete shells and carry along their spin angular momentum.

Orbital angular momentum operator: L = (Lx, ly,Lz). Eigenvalue equations for atomic orbital angular momentum:2

L2|l, mi = l(l + 1)|l, mi, l = 0, 1, 2,...

Lz|l, mi = m|l, mi, m = −l, −l + 1, . . . , l.

3 p Atomic orbital magnetic moment: |µ| = l(l + 1) µB, µz = −mµB. Eigenvalue equations for atomic spin angular momentum:

2 1 3 S |l, msi = s(s + 1)|s, msi, s = 2 , 1, 2 ,...

Sz|l, msi = ms|l, msi, ms = −s, −s + 1, . . . , s.

4 p (z) Atomic spin magnetic moment: |µs| = s(s + 1)gµB, µs = −msgµB. ˆ Zeeman energy-level splitting in a magnetic ﬁeld B = Bzk:

E = E0 + msgµBBz

1To a large extent, this module adapts materials from Blundell 2011. 2All angular momenta are in units of ~, which is often suppressed (as here) in the notation. 3 . All electronic magnetic moments are units of the Bohr magneton µB = e~/2me. 4The g-factor for electrons is g ' 2.

1 Two-electron wave function: Electrons are fermions. The two-electrons wave function must be permuta- tion antisymmetric. It can be expressed as the product of a spatial part ψ and a spin part χ. The spatial part can be written as a linear combination of products of one- electron (spatial) wave functions φa, φb,...:

1 h i ψ(r1, r2) = √ φa(r1)φb(r2) ± φb(r2)φa(r1) . 2 The plus (minus) sign makes it symmetric (antisymmetric) under permuta- tion of positions. The function has parity (−1)p with even or odd p. The spin part is constructed from the basis, | ↑↑i, | ↑↓i, | ↓↑i, | ↓↓i, as linear combinations with a definite parity: 1 1 χ(m1, m2) = | ↑↑i, √ | ↑↓i + | ↓↑i , | ↓↓i, √ | ↑↓i − | ↓↑i . 2 2 The first three are symmetric (even p) and constitute a triplet (s = 1), whereas last is antisymmetric (odd p) and constitues a singlet (s = 0). To make the two-electrons wave function antisymmetric, its spatial and spin parts must have opposite parity. The Pauli conclusion principle is thus enforced. If both electrons are in the same quantum state, either the spatial part or the spin part of the two- electron wave function vanishes identically, whichever has odd parity. When the electron pair experiences an interaction (e.g. Coulomb repulsion) that energetically favors on spatial parity over the other, this has the effect that the spin degeneracy between singlet and triplet is lifted. ˆ When the electron pair is subject to a magnetic field B = Bz k, the triplet splits up energetically, whereas the singlet is unaffected.

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2 Atomic electrons in a magnetic ﬁeld: Consider an atom with Z electrons positioned in a uniform magnetic ﬁeld B.

In a semi-classical description each electron is instantaneously at position ri moving with instantaneous velocity vi. 1 Vector potential associated with uniform magnetic ﬁeld: A = B × r. 2

Kinetic momentum of electron: pi = mevi.

Canonical momentum of electron (with charge −e): pi + eA(ri).

Z X Total spin angular momentum: ~S = ~si. i=1

Z Z X X Total orbital angular momentum: ~L = ~Li = ri × pi. i=1 i=1

Scalar potential (due to nuclear charge) experienced by electrons: Φi. Hamiltonian:

Z 2 X [pi + eA(ri)] H = + Φ + gµ B · S 2m i B i=1 e Z Z X p2 e2 X = i + Φ +µ (L + gS) · B + (B × r )2 2m i B 8m i i=1 e e i=1 | {z } H0

– The first term does not depend on B. It describes the kinetic energy and potential energy of each electron as it moves in the static electric field of the nucleus, possibly modified summarily by the other electrons. – The second term is linear in B. It is the dominant magnetic effect and describe the phenomenon of paramagnetism. It is absent for atoms whose total angular momentum vanishes. – The third term is quadratic in B. It is present for all atoms and becomes the dominant magnetic effect for atoms that are not paramagnetic. It describes the phenomenon of diamagnetism. – Neglected in this Hamiltonian are the interaction effects between electrons. They give rise to couplings between the total orbital and spin angular momenta (LS-coupling) and couplings between the total angular momenta, Ji = Li + si, of electrons (JJ-coupling).

3 Diamagnetism: Diamagnetic response, like dielectric response, is universal in atomic matter. Diamagnetic materials are those that do not exhibit stronger magnetic response of other types. A classical plausibility argument for diamagnetism invokes Lenz’s rule of neg- ative feedback – a consequence of Faraday’s law for magnetic fields generated by induced currents. A diamagnetic response can be derived for a single charge carrier moving in a magnetic field [lex121]. However, the effect is wiped out by classical statistical averaging. The Bohr-van Leeuwen theorem states that on the basis of classical statistical mechanics of charge carriers moving in a magnetic field, neither diamagnetism nor paramagnetism exist. Quantum mechanics is the basis of both. A quantum theory of diamagnetism starts from the third term of the Hamil- tonian constructed on the previous page, rewritten here for an ensemble of N identical electron orbitals in a macroscopic system of volume V :

Z Ne2 X h0|H |0i = h0|(B × r )2|0i. dia 8m V i e i=1 Diamagnetism is empirically known to be only very weakly T -dependent. Taking a ground-state expectation value ignores any T -dependence in this simple ﬁrst crack at a theory. ˆ 2 2 2 2 2 2 2 Setting B = B k we can write (B × ri) = B (xi + yi ) ' 3 B ri .

In a diamagnet, the expectation value h0|Hdia|0i carries the leading magnetic- ﬁeld dependence of the Helmholtz free energy F . The diamagnetic response thus follows from a second derivative as follows:

2 2 Z . ∂ F Ne µ0 X χ = −µ = − h0|r2|0i. 0 ∂B2 6m V i e i=1

Among the ﬁrst 60 elements, 31 are diamagnetic. A more reﬁned theory of diamagnetism will be presented elsewhere. In solid materials, diamagnetism can be strongly anisotropic.

4 Paramagnetism: The dominant form of paramagnetism is caused by unpaired atomic electrons. Magnetic atoms have a nonzero total angular momentum, J = L + S, a combination, in general, of orbital and spin angular momentum. The paramagnetisms of N independent atomic magnetic moments is analyzed in textbook applications of statistical mechanics of canonical ensembles. The canonical partition function factorizes. Semi-classical theory [tex84]:5

– The magnetic moment is treated as a three-component vector µ. – Energy function for each moment: H = −µ · B. – Partition function evaluated in spherical coordinates: N sinh(βµB) . 1 ZN = 4π , β = . βµB kBT

– Gibbs free energy: G(T,B,N) = −kBT ln ZN . . ∂G 1 – Magnetization: M = − = Nµ coth(βµB) − . ∂B T,N βµB . 1 y – Langevin function: L(y) = coth(y) − = + O(y3). y 3 . ∂M 2 1 β – Susceptibility: χ = µ0 = Nµ0µ 2 2 − 2 . ∂B T,N βµ B sinh (βµB) Nµ µ2 Curie’s law: χ = 0 + ··· (dominant term at low T ). 3kBT . ∂2 β2µ2B2 – Heat capacity: C = k β2 = Nk 1 − . H B ∂β2 B sinh2(βµB)

Flaw of semi-classical model: lim CH = NkB > 0. T →0 – Universal magnetization curve:

1.0

0.5

0.0 M / N μ

-0.5

-1.0 -4 -2 0 2 4

μB/kBT

5These are exercises of a diﬀerent course (PHY525: Statistical Physics I).

5 Two-level system [tex85]:

1 – Magnetic moment of an electron spin: S = 2 , g = 2.

– Two energy levels: E± = ±µBB. N µBB – Partition function: ZN = 2 cosh . kBT

– Gibbs free energy: G(T,B,N) = −kBT ln ZN . . ∂G µBB – Magnetization: M = − = NµB tanh . ∂B T,N kBT 2 . ∂M Nµ0µB 2 µBB – Susceptibility: χ = µ0 = sech . ∂B T,N kBT kBT Nµ µ2 Curie’s law: χ = 0 B + ··· (dominant term at low T ). kBT 2 2 . 2 ∂ µBB 2 µBB – Heat capacity: CH = kBβ 2 = NkB sech . ∂β kBT kBT – Universal magnetization curve and heat capacity:

1.0 1.0

0.5 0.8 B B 0.6

0.0 / Nk H M / N μ C 0.4 -0.5 0.2

-1.0 0.0 -4 -2 0 2 4 0.0 0.5 1.0 1.5 2.0 2.5 3.0

μBB/kBT kBT/μBB

– The magnetization curve is similar in structure but not identical, apart from the diﬀerent scale. – The heat capacity of the quantum result is, as expected, consistent with the third law of thermodynamics.

– The broad peak of CH is known as Schottky anomaly – one of several contributions to the heat capacity of a solid magnetic material.

Brillouin paramagnetism [tex86]:

– Atomic angular momentum J with z-component m = −J, −J+1,..., +J.

– Energy levels: Em = mgµBB (2J + 1 in number). – Solution use geometric series.

6 N N " +J # " 1 # X mx sinh (J + 2 )x . gµB – Partition function: ZN = e = 1 , x = . sinh x kBT m=−J 2

– Gibbs free energy: G(T,B,N) = −kBT ln ZN . . ∂G – Magnetization: M = − = MsatBJ (y), y = Jx. ∂B T,N

Saturation value: Msat = NgµBJ. . 2J + 1 2J + 1 1 y – Brillouin function: B (y) = coth y − coth . J 2J 2J 2J 2J

Quantum limit: B 1 (y) = tanh(y). 2 1 Classical limit: B (y) = coth(y) − . ∞ y 1 3 5 – One-parameter family of magnetization curves (J = 2 , 1, 2 , 2, 2 , ∞):

1.0

0.5 sat 0.0 M /

-0.5

-1.0 -4 -2 0 2 4

JgμBB/kBT

Van Vleck paramagnetism: If the atomic ground state has J = 0, then the dominant form of paramagnetism is absent. However, excited states with J 6= 0 produce a paramagnetic response as a second-order perturbation:

2 2 Z ! N X |h0|Lz + gSz|ni| e µ0 X χ = 2µ2 − h0|r2|0i . V B E − E 6m i n n 0 e i=1 P P The sum n is over excited atomic states and the sum i over electronic oribitals in the ground state. The van Vleck paramagnetism (ﬁrst term) is comparable in magnitude to the diamagnetism (second term), but oppositie in sign. Both are largely independent of temperature (unlike the dominant form of paramagnetism).

7 Fine structure: The ground state of an atom with incomplete occupation of electronic shells has either a net orbital angular momentum L or a net spin angular momentum S or both for a total angular momentum J = L + S. In the absence of any interactions, such an electronic ground state has a degeneracy (2L + 1)(2S + 1).

In the presence of a spin-orbit coupling, HLS = λL · S, L and S are not separately conserved, whereas J = L + S is. Even then, the magnitudes, L2 = L(L + 1) and S2 = S(S + 1) are conserved (nonrelativistically). States with given L and S exist for a range of J: |L − S| ≤ J ≤ L + S. The spin-orbit coupling reduces the degeneracy to multiplets with ﬁxed J: λ J2 = L2 + S2 + 2L · S ⇒ hH i = J(J + 1) − L(L + 1) − S(S + 1). LS 2 . Land´einterval rule: ∆EJ = E(J) − E(J − 1) = λJ.

-2 -1 ,ITL7 0

Visualization of degeneracy reduction 1 due to spin-orbit coupling: 2 3 3 Example with L = 3 and S = 2 . 3 9 ⇒ Jmin = 2 ,Jmax = 2 .

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Predicting the nature of the electronic ground state of an atom requires knowledge of the values of L, S, and J produced by unpaired electrons.

8 Hund’s rules: Empirical rules (of limited and ranked scope) regarding the values of L, S, and J of the electronic ground-state conﬁguration in a single incomplete subshell (speciﬁed in the periodic table of [lln10]).

#1 Total spin: S = Smax.

#2 Total orbital angular momentum: L = Lmax. |L − S| : shell less than half full, #3 Total angular momentum: J = L + S : shell more than half full.

Rule #1 is argued on the fact that electrons with the same spin orientation are forced into diﬀerent l orbitals, which lowers the Coulomb repulsion. Rule #2 is argued (more dubiously) on the fact that electrons that maximize L move in the same direction and can thus avoid each other more easily. Rule #3, supposed to minimize the spin-orbit coupling energy, is the most restricted in scope, given that other couplings or sometimes stronger. Example: rare-earth ion Dy3+ (dysprosium).

– Electronic conﬁguration of atom: 4f106s2. – Dy3+ lacks the two 6s elcetrons and one of the 4f electrons. – The incomplete 4f shell has 9 electrons. – Electrons in an f shell have l = 3. – The 4f shell accommodates a maximum 2l + 1 = 7 electrons with the same spin orientation. – Two electrons are thus spin-paired, which leaves 5 electrons that are 5 not. Hence S = 2 (rule #1), implying 2S + 1 = 6. – The net orbital angular momentum of the 7 electrons with the same spin orientation vanishes. – The maximum orbital angular momentum of the two remaining electrons (with opposite spin orientation) is L = 3 + 2 = 5 (rule #2). 5 15 – The 4f shell is more than half full. Hence J = 5 + 2 = 2 (rule #3). 6 – Symbolic representation: H15/2.

Hund’s rules are applicable to incomplete 3d shells in transition metals (elements 21 to 30) and to incomplete 4f shells in lanthanides (57 to 71). They are more accurate for the latter set for reasons discussed later.

9 Russell-Saunders coupling: The coupling between the spin S and the orbital angular momentum L of an electron is interpreted as the electron spin alignment with the magnetic ﬁeld produced by (relative) orbital motion of the nucleus.

– Orbital angular momentum: ~L = mer × v. – Nuclear charge: q = Ze. q – Electric potential: Φ(r) = . 4π0r q r dΦ r – Electric ﬁeld: E = − 2 = − . 4π0r r dr r µ0 qv × r E × v dΦ ~ – Magnetic ﬁeld: B = 3 = 2 = − 2 L. 4π r c dr merc ge – Electron spin magnetic moment: m = ~ S. 2me 2 6 1 e~ g 1 dΦ – Potential energy: HLS = − m · B = 2 2 S · L. 2 4mec r dr 1 dΦ Ze – If Coulomb potential is applicable: = 3 . r dr 4π0r – Atomic physics yields estimates for hr−3i for estimates of hS · Li.

Land´eg-factor:

– Electron magnetic moment: m = µB(gLL + gSS) = µBgJ J

with gL = 1, gS = 2, and gJ to be determined. 2 – Use m · J = µBgJ J = µB(gLL · J + gSS · J). 1 1 – Use L · J = J2 + L2 − S2 S · J = J2 + S2 − L2. 2 2 – Use J2 = J(J + 1), L2 = L(L + 1), S2 = S(S + 1). ⇒ 2gJ J(J + 1) = gL J(J + 1) + L(L + 1) − S(S + 1) + gS J(J + 1) − L(L + 1) + S(S + 1) .

3 S(S + 1) − L(L + 1) – Land´e g-factor: g = + . J 2 2J(J + 1)

6 1 The relativistic Thomas factor 2 looks artiﬁcial, but appears naturally in a fully relativistic calculation.

10 Nuclear spins:

1 Protons and neutrons are fermions with spin I = 2 . Their magnetic moments are much smaller than that of the electron.

. e~ −27 2 −4 Nuclear magneton: µN = = 5.0508 × 10 Am ' 5.4 × 10 µB. 2mp

Nuclear magnetic moment: µ = gI µNI.

The nuclear g-factors gI are hard to predict, even for the neutron and the proton, due to their complex quark composition.

Some common nuclear spins: Nucleus Z N I pltry gr n: neutron p: proton, 1H 2 d: deuteron, H 0tl - 1.913 -3.826 t: triton, 3H 1,, 101, 2.793 5.586 N: # of neutrons d:2H 111 0.857 0.857 't1I Z: atomic number (# of protons) t--3H LL2 -2.128 4.255 A = Z + N: mass number 12c 660 0 0 l3c X: element in AX 67+ 0.702 1.404 laN I: nuclear spin (in units of ) 771 0.404 0.404 ~ 160 µ: nuclear magnetic moment 8 8 0 0 0 17o 8 9 5 1.893 57 2 - -0.7 µN: nuclear magneton leF 10 1 9 2 2.628 5.257 gI : nuclear g-factor 3lP 1 2.263 l5 t6 2 r.r32 33s t6 l7 3 4.643 0.429 [table from Blundell 2011] 2

Identical nucleons tend to form singlet pairs. Nuclei with an odd number of protons and/or an odd number of neutrons have a nonzero nuclear spin.

Hyperfine structure: The orbital angular momentum L and spin S of atomic electrons produce a magnetic field at the nuclear position and thus interact with nuclear spin I. µ h i Magnetic dipole field: B = 0 3(m · ˆr)ˆr − m . 4πr3 Magnetic dipole interaction between electron spin and nuclear spin: µ h i H = 0 µ · µ − 3(µ · ˆr)(µ · ˆr) dip 4πr3 S I S I µ g g µ µ h i = 0 S I B N S · I − 3(S · ˆr)(I · ˆr) . 4πr3

11 This dipolar spin-spin interaction averages to zero for electronic s-orbitals. However, the overlap of the s-orbital wave function with nuclear position produces the Fermi contact interaction:

2µ H = 0 g g µ µ S · I. con 3 S I B N

There also exist a dipolar interaction between the orbital angular momentum and the nuclear spin. A more complete analysis starts from the Hamiltonian,

1 2 H = (p − eA) + 2µBS · (∇ × A) + V (r), 2me µ A = 0 µ × r, µ = g µ I. 4πr3 I N The hyperﬁne interaction is distilled from this expression in two steps [lex123]:

2 0 . p µ0gI µBµN h 1 2 1 L · Ii H = H − − V (r) = (S · ∇)(I · ∇) − (S · I)∇ + 3 , 2me 2π r r r µ g µ µ L − S + 3(S · ˆr)ˆr 8π = 0 I B N I · + S δ(r) . 2π r3 3

The hyperﬁne splitting in the ground state of atomic hydrogen with electronic 1 1 L = 0,S = 2 and nuclear I = 2 involves transitions between a singlet combination and a triplet combination. The transition energy corresponds to the famous 21cm HI emission detected by astronomers. For comparison, the average wavelength of the cosmic mi- crowave background is less than 1cm.