Chapter11 Paramagnetism and Diamagnetism
Chapter Eleven Diamagnetism and Paramagnetism
The story of magnetism begins with a mineral called magnetite “Fe3O4”. The first truly scientific study of magnetism was made by William Gilbert. His book “On the magnet” was published in 1600. In 1820 Hans Christian Oersted discovered that an electric current produces a magnetic field. In 1825 the first electromagnet was made by Sturgeon.
Each stores the same magnetic energy ~0.4J. 1 paramagnetism ferromagnetism ) ) 3 3
diamagnetism M (emu/cm M M (emu/cm M
H (Oe) H (Oe) Magnetic susceptibility per unit volume M χ ≡ M : magnetization, magnetic moment per unit volume H B : macroscopic magnetic field intensity CGS
In general, paramagnetic material has a positive susceptibility (χ >0). diamagnetic material has a negative susceptibility (χ<0).
2 P. Curie 1859~1906 1895 Curie proposed the Curie law 1903 1905 Langevin proposed the theory of diamagnetism and paramagnetism 1906 Wiess proposed the theory of ferromagnetism 1920’s The physics of magnetism was developed with theories involving electron spins and exchange interactions – the beginnings of QM
Three principal sources for the magnetic moment of a free atom : Electrons’ spin Electrons’ orbital angular momentum about the nucleus Change in the orbital moment induced by an applied magnetic field
First two sources give paramagnetic contributions.
The last source gives a diamagnetic contribution. 3 Reviews, in CGS B = H + 4π M = µ H N M = m m : magnetic dipole moment V [emu/cm3] e e m = − ∑ ri × pi = − L = −µBL Angular 2m i 2m momentum m = µBgo ∑ si = µBgoS Spin i e Bohr magneton µ = = 9.27×10-24 J/T; 1T =1Vs/m2 B 2m Electronic g-factor go=2.0023 µ electron I v
L and S are quantum operators with no basis in classical physics. 4 Theory of diamagnetism was first worked out by Paul Langevin in 1905. Negative magnetism present in an applied field even though the material is composed of atoms that have no net magnetic moment.
Motion of the electron is equivalent to current in the loop. Faraday’s law dΦ Emf = − B Paul Langevin [volt] dt 1872~1946 Consider a circular loop with radius r, the effect of applied field H is to change µ by an amount ∆µ in negative direction.
2 NA ρ e Z 2 I r χ = − R A 2 µ 6mc [emu/Oe-cm3] 5 H The change in moment of a single orbit : Emf A dH π r 2 dH r dH E = = − = − = − dt 2π r dt 2 dt dv eE er dH er a = = = − dv = − dH dt mc 2mc dt ∫ ∫ 2mc er ∆v = − H 2mc er e2r 2 ∆µ = ∆v = − H the plane of the orbit is perpendicular 2c 4mc 2 to the applied field
All possible orientations in the hemisphere θ sin 2 θ dA R r 2 = R2 sin 2 θ = R2 ∫ A R2 π / 2 2 = sin 2 θ (2πRsin θ )(Rdθ ) = R2 2 ∫ 2πR 0 3 6 A single electron with consideration of all possible oriented orbits, eR e2 R2 ∆µ = ∆v = − H 2c 6mc2
For an atom that contains Z electrons, z e2 R2 e2H z ∆µ = − i = − 2 R is the radius of the ith orbit ∑ 2 H 2 ∑ Ri i i=1 6mc 6mc i=1 Volume density of mass The number of atoms per unit volume : N ρ/A A Atomic mass Avogadro’s number
On the volume basis, 2 2 z NA ρ e Z 2 NA ρ e 2 χ = − ∆µ = − R H Susceptibility, 2 R 2 ∑ i A 6mc A 6mc i=1 Classical Langevin result Carbon (radius =0.7Å) 23 3 −10 2 6.02×10 (2.22g/cm ) (4.80×10 esu) 6 −8 2 χ = − 2 (0.7×10 cm) 12.01g 6(9.11×10−28g)(3×1010cm/s ) -6 3 7 = −1.5×10 emu/Oe − cm Experimental value : -1.1×10-6 emu/Oe-cm3 The agreement between calculated and experimental values for other diamagnetic materials is generally not that good, but it is at least within an order of magnitude.
This classical theory of Langevin is a good example of the use of a simple atomic model to quantitatively explain the bulk properties of a material.
Negative susceptibility magnitude : 10-5 ~10-6 cm3/mole
The other consistency is that χ is independent of temperature for most diamagnetic materials.
Quantum theory of diamagnetism results in the same expression. 8 Diamagnetic materials Pierre Weiss Closed-shell electronic structures result in no net moment The monatomic rare gases He, Ne, Ar, …
Polyatomic gases, H2, N2, … Some ionic solids, NaCl, … Some covalent solids, C, Si, Ge, …
Superconductors : Meissner effect 1865~1940
The first systematic measurements of susceptibility of a large number of substances over an extended range of temperature were made by Pierre Curie and reported by him in 1895. -- mass susceptibility is independent of temperature for diamagnetics, but that it vary inversely with temperature for paramagnetics. C C' χ = χ = in general T T −θ Curie law Curie-Weiss law (1907) 9 Quantum theory of diamagnetism 2 1 Q Hamiltonian H = p − A + QV 2m c Terms due to the applied field B = ∇× A ie e2 H '= (∇ • A + A•∇)+ A2 2mc 2mc 2 In the uniform magnetic field B = Bkˆ ∂ ∂ ∂ ˆ ˆ ˆ ˆ ˆ ˆ ∇× A = i + j+ k× (Ax i + Ay j+ Azk) ∂x ∂y ∂z ∂ ∂ ∂ ∂ ∂A ∂ Az Ax ˆ Ax Az ˆ y Ax ˆ = − i + − j+ − k ∂y ∂z ∂z ∂x ∂x ∂y yB xB One solution A = − ˆi + ˆj+ 0kˆ 2 2 ieB ∂ ∂ e2B2 = − + 2 + 2 H ' x y 2 (x y ) 4mc ∂y ∂x 8mc ~ 10 Lz z-component of angular momentum ~ ∂ ∂ paramagnetism Lz = −i x − y ∂y ∂x For diamagnetic materials with all electrons in filled core shells, the sum ∑ L z + 2S z over all electrons vanishes 2 For a spherically symmetric system x 2 + y2 = r 2 3
2 2 e2B2 e r the additional energy E'= x 2 + y2 = B2 8mc 2 12mc 2 2 2 ∂E' e r the associated magnetic moment µ = − = − B ∂B 6mc 2 in agreement with the classical Langevin result
11 Vibrating Sample Magnetometer (VSM)
→→ measuring sample’s magnetic moment oscillate sample up and down measure induced emf in coils A and B compare with emf in coils C and D from known magnetic moment12 13 Classical theory of paramagnetism In 1905, Langevin also tried to explain paramagnetism qualitatively. He assumed that paramagnetic materials have molecules or atoms with the same non-zero net magnetic moment µ. In the absence of magnetic field, these atomic moments point at random and cancel one another. M=0 When a field is applied, each atomic moment tends to align with the field direction; if no opposing force acts, complete alignment would be produced and the material demonstrates a very large moment. But thermal agitation of atoms tends keep the atomic moments pointed at random. Partial alignment and small positive χ. As temperature increases, the randomizing effect of thermal agitation increases and hence, χ decreases. This theory leads naturally into the theory of ferromagnetism.
14 A unit volume material contains n atoms with a magnetic moment µ each. H E = −µ • H = −µHcosθ Let the direction of each moment be presented by a vector, θ and let all vectors be drawn thru. the center of the sphere. µ E µH cosθ − dn = KdA e kBT = K 2π sinθdθ e kBT # at angle θ θ+dθ µH cosθ π k T n = ∫dn = K 2π ∫dθ sinθ e B total # 0 µH cosθ n π M = ∫dnµ cosθ = K 2πµ∫dθ cosθ sinθ e kBT 0 0 π nµ ∫ dθ cosθ sinθ ea cosθ 0 Let a = µH/kBT, M = π ∫ dθ sinθ ea cosθ 0 −1 ax nµ ∫ dx xe a −a e + e 1 = 1 = µ − 1 −1 n a −a = nµcoth(a)− ax e − e a 15 ∫ dx e a 1 Langevin function L(a)=coth(a)-1/a L(a) can be expressed in series, a a3 2a5 L(a) = − + −... o 3 45 945
M/M When a is small (<0.5), L(a) is a straight line w/. a slope 1/3.
When a is large, L(a) → 1.
a=µH/kBT Two conclusions : (1) Saturation will occurs at large a implying that at large H and low T the alignment tendency of field is going to overcome the randomizing effect of thermal agitation. (2) At small a, L(a) depends linearly on a implying that M varies linearly with H. 16 The Langevin theory also leads to the Curie law. nµa nµ µH For small a, M = = 3 3 k BT 2 2 nµ M nµ C C = χ = = = 3k B H 3k BT T Oxygen is paramagnetic and obeys the Curie law w/. χ=1.08×10-4emu/gOe.
3k T χ 3(32g/mole )(1.38×10−16erg/K )(1.08×10−4 emu/gOe ) µ = B = n 6.02×1023 = 2.64×10-20erg/Oe − mole
= 2.85µB Even in heavy atoms or molecules containing many electrons, each with orbital and spin moments, most of the moments cancel out and leave a magnetic moment of only a few Bohr magnetons. µH (2.64×10−20erg/Oe )(10000Oe) = = = Typically, a −16 0.0064 <0.5 k BT (1.38×10 erg/K )(300K) smaller enough L(a)=a/3 17 The Langevin theory of paramagnetism which leads to the Curie law, is based on the assumption that individual carrier of magnetic moment do not interact with one another, but are acted on only by applied field and thermal agitation. However, many paramagnetic materials obey the more general law, C χ = Curie-Weiss law (1907) T−θ considering the interaction between elementary moments. Weiss suggested that this interaction could be expressed in terms of
a fictitious internal field, “molecular field HM”. In addition to H.
The total field acting on material H + HM= H + γM M M C χ = = = (T − Cγ ) M = C H Htot H + γM T M C C χ = = = H T − Cγ T −θ
where θ (=Cγ) is a measure of the strength of the interaction. 18 1 T −θ 1 θ = = T − Theoretical predictions χ C C C
θ Intercept reveals information of θ
θ (FeSO4)<0, θ (MnCl2)>0 θ>0 indicates that the molecular field aids the applied field, makes elementary moments align with H, and tends to increase χ. θ<0 indicates that the molecular field opposes the applied field, makes elementary moments disalign with H, and tends to decrease χ. 19 Precession of atomic magnetic moments about H, τ = µ × H like gyroscope, angular momentum in the presence of gravitation. dL τ = r × mg = dt
If the atom was isolated, the only effect of an increase in H would be an increase in the rate of precession, but no change in θ. For a system with many atoms, all subjected to thermal agitation, there is an exchange of energy among atoms. When H is applied, the exchange of energy disturbs the precessional motion enough so that the value of θ for each atom decreases slightly, until the distribution of θ values becomes appropriate to the existing values of field and temperature. 20 Quantum theory of Paramagnetism The central postulate of QM is that the energy of a system is discrete. quanta of energy improving the quantitative agreement between theory and experiment.
Classical case Quantum case J=1/2 J=2 H H θ
θ is continuous variable. θ is restricted to certain variables. Space Quantization
21 The orbital and spin magnetic moments for an electron eh e h e µ = = = − p orbit 4πmc 2mc 2π 2mc orbit eh e 1 h e µ = = = − p spin 4πmc mc 2 2π mc spin In general e µ = −g p where g is spectroscopic splitting factor, or g-factor 2mc g=1 for orbital motion and g=2 for spin In a multiple-electron atom, the resultant orbital angular momentum and the resultant spin momentum are characterized by quantum numbers L and S. Both are combined to give the total angular momentum that is described by the quantum number J. H e µ = + = + µ g J(J 1) g J(J 1) B θ 2mc µ µ e H In the direction of H, µ = g m = gm µ H 2mc J J B There are 2J+1 numbers of allowed azimuthal quantum number mJ : 22 J, J-1, … , -(J-1), -J. For a free atom, the g factor is given by the Landé equation J(J +1) + S(S +1) − L(L +1) g =1+ 2J(J +1) The energy levels of the system in a magnetic field
ms µs 1/2 -µ 2µB -1/2 µ E = −µ • H = mJgµBH
For a single spin with no orbital moment, mJ=±1/2 and g=2. The equilibrium population for the two level system : N exp( µH/k T) 1 = B lower level Total number of atoms µ + −µ N exp( H/k BT) exp( H/k BT) is conserved.
N2 exp( −µH/k BT) = upper level N = N1+N2 = constant N exp( µH/k BT) + exp( −µH/k BT) 23 The resultant magnetization for N atoms per unit volume ex − e−x µH = − µ = µ = µ = M (N1 N2 ) N x −x N tanh (x) where x e + e k BT µ 2H For x<<1, tanh(x)~x and M ≅ Nµ x = N k BT For a system with known g and J,
Potential energy of each moment in H U = −mJgµBH −U/k T m gµ H/k T The probability of an atom w/. energy U e B = e J B B
mJgµBH/k BT ∑mJgµBe Boltzmann statistics m Therefore, M = N J mJgµBH/k BT ∑e mJ : J, J-1, … , -(J-1), -J mJ 2J +1 (2J +1)x 1 x M = NgJµB coth − coth 2J 2J 2J 2J µ gJ BH24 = NgJµBBJ (x) where x = Brillouin function k BT The saturation magnetization Mo = N µH = N gJµB Number of atoms The maximum moment of each per unit volume atom in the direction of H M 2J +1 (2J +1)x 1 x = BJ (x) = coth − coth Brillouin function, 1927 Mo 2J 2J 2J 2J
When J=∞, the classical distribution, the Brillouin function reduces to the Langevin function. M 1 = coth(x)− Mo x When J=1/2, the magnetic moment consists of one spin per atom. M = tanh (x) x<<1 x Mo 2 2 2 2 µBH µ H M ≅ NµH x = Ng J = N k BT k BT 25 KCr(SO4)12H2O
Henry, Phys. Rev. 88, 559 (1952)
The Brillouin function, like the Langevin, is zero for x equals to zero and tends to unity as x becomes large. The course of the curve in between depends on the value of J for the atom involved. µH gJµ H µ H classical x = ≠ quantum x = B = H k BT k BT k BT µ : net magnetic moment in the atom 26 When H is large enough and T low enough, a paramagnetic can be saturated.
For example, KCr(SO4)12H2O at H=5Tesla and T=4.2K only magnetic ion Cr3+, J=3/2, L=3, and S=3/2 g=2/5 the lower curve shown in the figure → data does not follow this predicted curve
The data is well described by the quantum model with J=S=3/2 and g=2 . L=0 no orbital component
3 3 µ = gJμ = g J(J +1)μ = 2 ( +1)μ B = 3.87μ B B B 2 2
“Spin-only” moment
27 metals
Gd(C2H3SO4)3•9H2O
28 What magnetic moment a metal should have? incorporate coupling of magnetic moment to determine Hund’s rule the magnetic state
(1) Maximize S = ∑ m s subject to the Pauli exclusion principle
ms=±1/2 put spins (in different states) parallel
(2) Maximize L = ∑ m subject to the Pauli exclusion principle
m= -, -+1, …, -1, fill higher m state preferentially
L −S when shell is less than half full (3) J = L + S when shell is more than half full J= L-S, L-S+1, …, L+S-1, L+S allowed values Whether it is a minimum or maximum in the ground state depends on filling function.
29 L : 0, 1, 2, 3, 4 Magnetic state 2S+1 LJ S, P, D, F, G Examples (1) Full shells : L=0, S=0, and hence, J=0
(2) Half filled shells : L=0 and S=(2+1)(1/2)
Eg. Gd3+ 4f75S25P6 (n=4,=3) 3 2 1 8 0 State S7/2 -1 -2 -3 L=0 and S=7/2 J=L+S=7/2
30 Eg. Cs3+ 4f15S25P6 Eg. Pr3+ 4f25S25P6 (n=4, =3) (n=4, =3) 3 3 2 2 1 2 1 3 0 State F5/2 0 State H4 -1 -1 -2 -2 -3 -3 L=3 and S=1/2 L=5 and S=1 J=L-S=5/2 J=L-S=4 Eg. Dy3+ 4f95S25P6 (n=4, =3) 3 2 1 6 0 State H15/2 -1 -2 -3 L=5 and S=5/2 31 J=L+S=15/2 Rare earth ions
p = g J(J +1) For Sm3+ and Eu3+, there are discrepancies between experimental p and theoretically calculated value.
High states of the L-S multiplet. 32 Quenching of the orbital angular momentum : Iron group ions
Effective number of Bohr magneton p = 2 S(S +1) p = g J(J +1)
In the iron group ions the 3d shell, the outermost shell, is responsible for the paramagnetism and experiences the intense inhomogeneous electric field produced by neighboring ions. called “the crystal field” breaking L-S coupling (J is meaningless) splitting 2L+1 sublevels (diminishing the 33 contribution of orbital motion to magnetic moment) An atom w/. L=1 is placed in the uniaxial crystalline electric field of two positive ions along the z-axis. Take S=0 to simplify the problem.
In a free atom, a single electron in a p-state. x + iy R n (r)Y11(θφ) ∝ f(r) 2
R n (r)Y10(θφ) ∝ f(r) z Three degenerate p states x − iy R n (r)Y1−1(θφ) ∝ f(r) 2 34 In a crystal of orthorhombic symmetry the charges on neighboring ions will produce an electrostatic potential ϕ about the nucleus, eϕ = Ax 2 + By 2 − (A + B)z 2
Reconstructing unperturbed ground state
Ux = xf(r), U y = yf(r), Uz = zf(r) + perturbed term eϕ
Non-diagonal elements Ux eϕ U y = U y eϕ Uz = Uz eϕ Ux = 0
Diagonal elements ϕ = 2 2 2 + 2 − + 2 Ux e Ux ∫ f(r) x [Ax By (A B)z ]dxdydz 2 = ∫ f(r) [Ax 4 − Ax 2 y2 ]dxdydz
= A(I1 − I2 ) 35 ϕ = 2 2 2 + 2 − + 2 U y e U y ∫ f(r) y [Ax By (A B)z ]dxdydz 2 = ∫ f(r) [By 4 − By 2z2 ]dxdydz
= B(I1 − I2 ) ϕ = 2 2 2 + 2 − + 2 Uz e Uz ∫ f(r) z [Ax By (A B)z ]dxdydz 2 = ∫ f(r) (A + B)[y 2z2 − z4 ]dxdydz
= (A + B)(I 2 − I1)
The orbital moment of each of the states
Ux Lz Ux = U y Lz U y = Uz Lz Uz = 0
Crystal field splits the original degenerate states into non-magnetic states separated by energies >> µH. quenching of the orbital angular momentum 36 Paramagnetic susceptibility of conduction electrons • Conduction electrons are not bound to atoms so that they are not included in atomic magnetization.
• Each electron has µ=±µB. • Curie-type paramagnetic contribution (electron spins align w/. H) nµ 2H M = B where n is # of conduction electrons k BT This requires each electron spin align w/. H, independent of all others.
Fermi-Dirac distribution • Pauli pointed out that only electrons near EF (±kBT) can flip over. 1.1 1.0 kBT H=0 H≠0 0.9 0.8
) 0.7 0.6 T=0 ,T
ε εε 0.5 T≠0 FF f( 0.4 0.3 0.2 0.1 ε 37 0.0 F ε
k BT T Fraction of electron participating = ε F TF 2 2 T µBH nµBH M = n = independent of temperature TF k BT k BTF
Consider the density of states D(ε) in the presence of H Spin ↑ electrons lowered in energy by µH in H ↑ ε Spin ↓ electrons raised in energy by µH in H ↑ flip spins ε ε µH ε F ↓ ↓ H ↑ ↑ 2µH µH DOS DOS 38 The concentration of electrons w/. magnetic moment parallel to H ε ε 1 F 1 F 1 = ε + µ ≅ ε + ε µ N↑ ∫ dεD( H) ∫ dεD( ) D( F ) H H↑, 2 −µH 2 0 2 T=0K The concentration of electrons w/. magnetic moment antiparallel to H ε ε 1 F 1 F 1 = ε − µ ≅ ε − ε µ N↓ ∫ dεD( H) ∫ dεD( ) D( F ) H 2 µH 2 0 2 Pauli spin magnetization
3N 2 3N 2 M = (N↑ − N↓ )µ = µD(ε F )µH = µ H = µ H 2ε F 2k BTF independent of temperature valid when kBT<<εF
39 References : Kittel, Introduction to Solid State Physics, 8th Ed., Ch.11-12, 2005.
Aschroft and Mermin, Solid State Physics, Harcourt, Ch.31-33, 1976.
Mattis, Theory of Magnetism, Springer, 1985.
Cullity, Introduction to Magnetic Materials, Addison-Wesley, 1972.
Comstock, Introduction to Magnetism and Magnetic Recording, Wiley-Interscience, 1999.
Spin-dependent Transport in Magnetic Nanostructures, Edited by Maekawa and Shinjo, Taylor & Francis, 2002. 40 Cooling by isentropic demagnetization Entropy : A measure of the disorder of a system The greater the disorder, the higher is the entropy. In the magnetic field the moments will be partly lined up (partly ordered), so that the entropy is lowered by the field. The entropy is lowered if the temperature is lowered, as more of the moments line up. When the material is demagnetized at constant entropy, entropy can flow from the system of lattice vibrations into the spin system.
S = k B ln Z where Z is the partition function N m=I ε M Z = ∑exp m=-I k BT
41 a
c b
B/T Entropy for a spin ½ system depends only for the population distribution. a→b : the system is magnetized isothermally. b→c : the magnetic field is turned off adiabatically.
H∆ T = T where H∆ is the effective field f i H µ H T ~10-7K If we start at H=5T and Ti=10mK ≈ 0.5 f k BT -6 The first nuclear cooling was done on Cu, Ti=20mK Tf ~1.2×10 K 42 Cu nuclei in the metal
3.1 T = T f i B where B∆=3.1Oe.
1.2 µ K
Rd The present record for a spin temperature is about 280pK. PRL70, 2818 (1993).43