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IEEE Mag Society summer school 2018 (USFQ)

Summer School of the IEEE Magnetics Society 2018

USFQ Quito

Fundamentals of Magnetism

Antonio Azevedo

Department of Physics, UFPE

Recife PE, Brazil

[email protected]

Carnival in Recife: Rooster of dawn parade

http://www.google.com.br/url?sa=i&rct=j&q=recife+carnaval+galo+madrugada+2013&source=images&cd=&cad=rja&docid=cgRnRWcAYJUT1M&tbnid=rC-Yb3xalX7cZM:&ved=0CAUQjRw&url=http://acaoambientalsp.blogspot.com/2013/01/dj-leandro-som-e-iluminacao-casamento.html&ei=NcIXUeqHGZL8qQGIjoGYCQ&bvm=bv.42080656,d.aWc&psig=AFQjCNGT_WvZcAdcwDDLx_aeJTiIXgkJfg&ust=1360597890661028


• Soshin Chikazumi, Physics of Ferromagnetism. Oxford University Press, 2nd Edition (1997), 668 pp.

• J. M. D. Coey, Magnetism and Magnetic Magnetic Materials. Cambridge University Press (2010) 633 pp.

• S. Blundell, Magnetism in Condensed Matter. Oxford (2001), 251 pp.

• Mathias Getzlaff, Fundamentals of Magnetism. Springer-Verlag (2008), 384 pp.

• Nicola A. Spaldin, Magnetic Materials - Fundamentals and Applications. Cambridge University Press,

(2011), 290 pp.

• B. D. Cullity and C. D. Graham, Introduction to Magnetic Materials. John Wiley & Sons, Inc., 2nd edition

(2009) 550 pp.

• Robert M.White, Quantum Theory of Magnetism - Magnetic Properties of Materials. Springer-Verlag Berlin

Heidelberg 2007, 3rd Edition (2007), 366 pp.

• D. C. Jilles, An Introduction to Magnetism and Magnetic Magnetic Materials, Magnetic Sensors and

Magnetometers, 3rd edition CRC Press, (2014) 480 pp.

• R. C. O’Handley, Modern Magnetic Magnetic Materials. John Wiley & Sons (2000), 767 pp.

Books

Fundamentals of Magnetism1. Introduction: Basic concepts

2. Magnetostatic Phenomena

3. Microscopic origin of magnetism

4. Magnetic properties of materials

5. Magnetism of Atoms: free electrons; localized electrons

6. Ferromagnetism

7. Exchange interaction

8. Ferrimagnetism

9. Antiferromagnetism

10.Magnetism of itinerant electron systems

11.Domain wall and magnetization curves

12.Magnetic anisotropies

13.Spin orbit interaction and Rashba effect



The seven ages of magnetism (J.M.D. Coey, Magnetism and Magnetic Materials)

C. Chien coined the term “Golden Age of Magnetism” to the series of astonishing discoveries that gave birth

the area of Spintronics

De magnete

(Gilbert)

Oersted,

Faraday,

Maxwell

Weiss,

Heisenberg,

Néel

NMR, EPR,

radar, etc

Magnetic

recording

MR &

Spintronics


Basic concepts: magnetic fields

1. The magnetic field H (H-field )of a solenoid that consists of a coil of

length l with n windings through which a current I flows is:

𝐻 =𝑛𝐼

𝑙. 𝐻 = Τ𝐴 𝑚

2. The magnetic field strength H and the magnetic induction B (B-field) in vacuum are related by:

𝑩 = 𝜇0𝑯. 𝐵 = Τ𝑉 ∙ 𝑠 𝑚2 = Τ𝑊𝑏 𝑚2 = 𝑇 (Tesla)

(𝜇0 = 4𝜋 × 10−7 Τ𝑉 ∙ 𝑠 𝐴 ∙ 𝑚 is the vacuum permeability)

3. Inside a material:

𝑩 = 𝜇𝑯 = 𝜇0𝜇𝑟𝑯,

where 𝜇𝑟 = 1 + 𝜒; 𝜇𝑟 = relative permeability; 𝜒 = susceptibility, (𝜇𝑟 & 𝜒 are dimensionless, in free space 𝜇𝑟 = 1and 𝜒 =0).

𝜇𝑟 𝑚𝑒𝑡𝑔𝑙𝑎𝑠𝑠 ~107; 𝜇𝑟(𝑖𝑟𝑜𝑛)~10

6


4. Inside a magnetic material with magnetization M, the B-field consists of two contributions: 𝜇0𝑯 plus a

contribution due to the presence of the material 𝜇0𝑴.

𝑩 = 𝜇0(𝑯 + 𝑴).

As 𝑴 = 𝜒𝑯, the magnetic induction 𝑩 = 𝜇0 1 + 𝜒 𝑯 = 𝜇0𝜇𝑟𝑯.

Basic concepts: magnetization, diamagnetism and paramagnetism

5. In general, the susceptibility of isolated atoms consists of a diamagnetic and a paramagnetic part

(𝜒𝑑𝑖𝑎 and 𝜒𝑝𝑎𝑟𝑎, respectively).

6. Diamagnetism is due to the fact that an external magnetic

field induces a change in the current of the orbiting electrons →

magnetic moment is opposite to direction of the external field

7. Paramagnetism arises from alignment of existing magnetic

moments (resulting from electron spins and orbiting electrons).

𝒎𝒊 Definition of magnetization M

𝑴 =σ 𝒎𝑖

∆𝑉.

𝑀 ≡ Τ𝐴 𝑚; ( 𝑀 ≡ Τ𝑒𝑚𝑢 𝑐𝑚3)


Basic concepts: different types of magnetic ordering

http://www.aacg.bham.ac.uk/magnetic_materials/type.htm

Suceptibility: χ =𝑀

𝐻


Basic concepts: Ampère and Biot-Savart laws

𝐼 = 1.0 𝐴; 𝑟 = 1.0 𝑐𝑚; 𝜇0 = 4𝜋 × 10−7 Τ𝑇 ∙ 𝑚 𝐴

𝐵 = 2 × 10−5 𝑇 = 0.2 𝐺𝐵𝑒𝑎𝑟𝑡ℎ~0.5 𝐺

𝑩 𝒓 =𝜇04𝜋

ර𝐶

𝐼𝑑𝒍′ × (𝒓 − 𝒓′)

𝒓 − 𝒓′ 3

Biot-Savart law

ර𝐶

𝑩 ∙ 𝑑𝒍 = 𝜇0𝐼 → 𝑩 =𝜇0𝐼

2𝜋𝑟ො𝜑

Oersted (1820)

Ampère’s law

B-field is created by: (i) electric currents.

(ii) magnetization (due to microscopic sources (electron spin)).

Current carrying loop


න

𝑉

J(r’) × (𝒓 − 𝒓′)

𝒓 − 𝒓′ 3dV′

Volume distribution of current density


𝑩 𝒓 =𝜇0

4𝜋𝑉

J(r’)×(𝒓−𝒓′)𝒓−𝒓′ 3

dV′

Where, 𝛻𝟏

𝒓−𝒓′= −

(𝒓−𝒓′)

𝒓−𝒓′ 3

𝑩 𝒓 = −𝜇0

4𝜋𝑉 J(r’) × 𝛁

1

𝒓−𝒓′dV′

𝑩 𝒓 =𝜇0

4𝜋𝑉 𝛻 ×

J(r’)𝒓−𝒓′

dV′

𝑩 𝒓 = 𝛻 ×𝜇0

4𝜋𝑉


dV′

• There are no sources or sinks of the magnetic induction B, i.e.,

there are no magnetic charges

• B is a solenoidal field

• A is the vector potential: 𝑨(𝒓) =𝜇0

4𝜋𝑉


dV′

By using the vector identity:𝛻 ∙ 𝛻 × 𝑨 = 0,we obtain:

𝛻 ∙ 𝑩(𝒓) = 0

We may expand the vector potential 𝑨(𝒓) up to 2nd order (𝒓 ≫ 𝒓′)

𝑨 𝒓 =𝜇0

4𝜋

1

𝒓𝑉 J(r’)dV

′ +𝜇0

4𝜋𝑉 J(r’)

𝒓∙𝒓′

𝒓 3dV′ + ⋯

The first term is null. 𝑉 J𝑥 (r’)dV′ = 𝑉 𝛻 ∙ 𝑥J(r’) dV′ = ׯ 𝑥J(r’)𝑑𝑆 = 0

There is no magnetic monopole

The first nonzero term is the magnetic dipole!

Basic concepts: 𝛻 ∙ 𝑩(𝒓) = 0


The first nonzero term of the multipole expansion is given by: 𝑨 𝒓 =𝜇0

4𝜋 𝒓 3𝒓 ∙ 𝑉 𝒓′J(r’) dV′

It can be written as (see classical books on electrodynamics),

𝑨 𝒓 = −𝜇0

4𝜋 𝒓 3𝒓 ×

𝒓′ × J(r’) dV′

2=

𝜇04𝜋

𝒓′ × J(r’) dV′

2×

𝒓

𝒓 3=

𝜇04𝜋

𝒎 × 𝒓

𝒓 3

𝒎 =𝟏

𝟐න 𝒓′ × J(r’) dV′

is the dipole magnetic moment of the current distribution J(r’).

From the above equation we can calculate the B-field of a magnetic dipole m by means of 𝑩 𝒓 = 𝛻 × 𝑨 𝒓 =

𝜇0

4𝜋

1

𝒓 3𝛻 × 𝒎 × 𝒓 − 𝒎 × 𝒓 × 𝛻

1

𝒓 3. Remember that: 𝒎 ∙ 𝛻 𝒓 = 𝒎; 𝛻 ∙ 𝒓 = 3;𝛻

1

𝒓 3= −

3𝒓

𝒓 5. Thus we obtain,


3ො𝒓 Ƹ𝑟 ∙ 𝒎 − 𝒎

𝑟3B-field of a dipole m at r

Basic concepts: Magnetic dipole and dipole magnetic field


Basic concepts: Dipole-dipole interaction

Classical electromagnetism𝛻 ∙ 𝑩 = 0 ∴ There is no magnetic monopole! The 1st nonzero term of the multipole expansion is a dipole 𝒎

Magnetic dipole

𝑩 𝒓 =𝜇0

4𝜋

3ො𝒓 Ƹ𝑟∙𝒎 −𝒎

𝑟3; 𝒎 = 𝑖𝑨

Dipole-dipole interaction

𝒎𝟏

𝒎𝟐

Ԧ𝑟𝜃1

𝜃2𝑩 𝒓 =

𝜇0𝑚1

4𝜋𝑟32 cos 𝜃ො𝒓 + sin 𝜽 𝜽

𝑈 = −𝒎𝟐 ∙ 𝑩 𝒓

𝑈 = −𝜇0𝑚1𝑚2

4𝜋𝑟32 cos 𝜃1 cos 𝜃2 − sin 𝜃1 sin 𝜃2If 𝑚1 = 𝑚2 = 𝑚 and 𝜃1 = 𝜃2 = 𝜃

𝑈 = −3𝜇0𝑚

2

4𝜋𝑟3cos2 𝜃 −

1

3

Torque: 𝝉 = 𝒎 × 𝑩

Pot. Energy: 𝑈 = −𝒎 ∙ 𝑩

Force: 𝑭 = −𝛻𝑈 = 𝛻 𝒎 ∙ 𝑩 = 𝒎 ∙ 𝛁 𝑩

m

m

(𝑎)(a) 𝜃 = 0 (stable), 𝑈 = 𝑈𝑚𝑖𝑛. m m

(𝑏)(b) 𝜃 = Τ𝜋 2 (unstable), 𝑈 = 𝑈𝑚𝑎𝑥.

We can estimate the critical temperature Tc, below which the order sets in.

𝑘𝐵𝑇𝐶~𝑧𝜇04𝜋

𝜇𝐵2

𝑎3= 8

(𝑒ℏ)2

4𝑚2𝑎3

𝜇𝐵 is the Bohr magneton, a = 2.49Å (for bcc iron).

Iron becomes ferromagnetic at 𝑇𝐶 =1043 K!

Therefore, the dipole-dipole interaction is too weak to explain the ferromagnetic ordering of iron.

→ 𝑇𝐶~ 0.3 𝐾

Critical temperature for some ferromagnetic

substance

Iron (Fe) 1043 K

Cobalt (Co) 1394 K

Nickel (Ni) 631 K

Gadolinium (Gd) 293 K

Manganese arsenide

(MnAs)

318 K

Basic concepts: Can the dipole-dipole interaction explain magnetic order?

iron


Basic concepts: magnetic scalar potential and magnetic charges

Ampère’s law applied to H-field: ׯ 𝑯 ∙ 𝑑𝒍 = 𝐼

(integral form)

𝜙𝑚 𝒓 = −1

4𝜋න

𝑉

𝛻 ∙ 𝑴 𝒓′

𝒓 − 𝒓′𝑑𝑉′ +

1

4𝜋ර

𝑆

𝑴 𝒓′ ∙ ො𝑛

𝒓 − 𝒓′𝑑𝑆′

𝜌𝑚 = −𝛻 ∙ 𝑴 (Volume charges)𝜎𝑚 = 𝑴 ∙ ො𝑛 (Surface charges)

As 𝛻 ∙ 𝑩 = 0 and 𝑯 = −𝛻𝜙𝑚, then

𝛻 ∙ 𝜇0 𝑯 + 𝑴 = 0 ∴ 𝛻 ∙ 𝑯 = −𝛻 ∙ 𝑴 ≡ 𝜌𝑚 (magnetic charge density)

𝛻 ∙ −𝛻𝜙𝑚 = −𝛻 ∙ 𝑴 ∴ 𝛻2𝜙𝑚 = 𝛻 ∙ 𝑴 (Poisson’s law for magnetostatic)

Using the Stokes th. it is transformed in the differential form:

ׯ 𝑯 ∙ 𝑑𝒍 = 𝑆

(𝛻 × 𝑯) ∙ 𝑑𝑺

𝐼 = 𝑆

𝑱 ∙ 𝑑𝑺൞→ 𝑆 (𝛻 × 𝑯) ∙ 𝑑𝑺 = 𝑆 𝑱 ∙ 𝑑𝑺 ∴ 𝛻 × 𝑯 = 𝑱

With no current distribution (𝑱 = 𝟎): 𝛻 × 𝑯 = 0 → 𝑯 = −𝛻𝜙𝑚,

where 𝜙𝑚 is the magnetic scalar potential.


Textbook example: a uniformly magnetized sphere

𝜌𝑚 = −𝛻 ∙ 𝑴 = 0𝜎𝑚 = 𝑴 ∙ ො𝑛 = 𝑀 cos 𝜃

𝜙𝑚 𝒓 = −𝑀

4𝜋ׯ

𝑆


𝒓−𝒓′𝑑𝑆′ → ൞

𝜙𝑚 𝒓 =𝑀

3𝑟 cos 𝜃 =

𝑀

3𝑧, for 𝑟 ≤ 𝑅 inside

𝜙𝑚 𝒓 =𝑀

3

𝑅3

𝑟2cos 𝜃, for 𝑟 ≥ 𝑅 outside

Inside: 𝑯 = −𝛻𝜙𝑚 → 𝑯 = −𝑴

3and 𝑩 = 𝜇0 H+M =

𝟐

3𝜇0𝑴. The H-field is uniform!

Outide: 𝑯 = −𝛻𝜙𝑚 → 𝑩 = −𝜇0𝛻𝜙𝑚 =𝜇0𝑅

3

3𝑟3(𝟑𝑴 ∙ ෝ𝒏 − 𝑴). H-field is the field of a magnetic dipole!

Remember that: 𝑴 = Τ3 4𝜋𝑅3 𝒎

• Both the H and B fields are uniform inside the sphere

• The magnetic field H is oppositely directed to the magnetization

• The H-field acts to demagnetize the sphere

• B-field is continuous and H-field is discontinuous @ the surface

• 𝑯𝑖𝑛𝑠 in opposite direction to 𝑴

• North pole acts as “source”, south pole acts as “sink”


Basic concepts: demagnetizing field

uniformly

magnetized

block

J M D Coey 2014 J. Phys.: Condens. Matter 26 064211

• The B-field and H-field outside the magnet are useful quantities for magnet applications.

• Inside the material 𝑯 is oppositely directed to 𝑴, hence it is known as demagnetizing field 𝑯𝑑.

• For ellipsoidal shaped samples 𝐻𝑑 ≅ 𝑁𝑴, where the demagnetizing factor N varies from 0 to 1.


Demagnetizing field: effect of sample shape

How to calculate 𝑯𝒅?

The magnetization at all points must be known (including the surface)

Then, we find 𝜙𝑚 𝒓 = −1

4𝜋𝑉

𝛻∙𝑴 𝒓′

𝒓−𝒓′𝑑𝑉′ +

1

4𝜋𝑆ׯ


𝒓−𝒓′𝑑𝑆′ and 𝑯𝑑(𝒓) = −𝛻𝜙𝑚 𝒓

𝑯𝑑 𝒓 = −𝑁𝑴

where N is the demagnetizing tensor!

In particular, ellipsoidal samples, exhibit simple expressions for N.

Same

material


Basic concepts: demagnetizing factors

𝑯𝑖𝑛𝑡 = 𝑯𝑒𝑥𝑡 − 𝑁𝑴


Technique for magnetic field generation

Generate magnetic field in a controlled way is very important


𝑩 = Magnetic induction [𝑩] = Testa (T); 𝑯 = Magnetic field [𝑯] = A/m. Obs: in air 𝑩 = 𝜇0𝑯; 𝜇0 = 4π × 10

−7 𝐻/𝑚 or [N/A2 or T.m/A or V.s/A.m]

Helmholtz coils

𝐵 𝑧 = 0 ≅ 9 × 10−7𝑁𝐼

𝑅B = Tesla

N = # of turns in each coil

I = Coil current in A

R = Coil radius in m

N=400; R = 0.1 m; I = 2,0 A

𝐵 𝑧 = 0 ~70 × 10−4 𝑇 = 70 Oe

Very convenient for AC

modulation field generation!

https://www.didaktik.physik.uni-muenchen.de

𝐻 𝑥 = 𝐶1𝑁𝐼

𝐿

𝐿 + 2𝑥

2 𝐷2 + (𝐿 + 2𝑥)2+

𝐿 − 2𝑥

2 𝐷2 + (𝐿 − 2𝑥)2

Single layer solenoid

For 𝐻 ≥ 1 𝑘𝑂𝑒, cooling is necessary

Magnetic field generation: low value fields


Toroidal electromagnet

𝐵 = 𝜇0𝐻 =𝜇0𝑁𝐼

𝐿, Air core

𝐵 =𝜇𝑟𝜇0𝑁𝐼

𝐿−𝑑 +𝜇𝑟𝑑, Iron core

𝜇𝑟 = Τ𝜇 𝜇0

If 𝜇𝑟 = 2000, 𝐵 ~ 2𝑇

Magnetic field generation: high value fields

Configurations commonly used in labs

By using resistive electromagnets, fieldsup to 2 T can be easily generated

(Needs water cooling !)


Example: C-core electromagnet

ׯ 𝑯 ∙ 𝑑𝒍 = 𝑁𝑖 ∴ 𝐻𝑚𝑙𝑚 + 𝐻𝑔𝑙𝑔 = 𝑁𝑖 (1)𝐵𝑔

𝐴𝑔=

𝐵𝑚

𝐴𝑚∴ 𝐵𝑔 ≈ 𝐵𝑚 (2).

2 → 1 ,𝐵𝑔

𝜇0𝑙𝑔 +

𝑙𝑚

𝜇𝑟= 𝑁𝑖

for 𝜇𝑟~104, field at the gap is: 𝑯𝒈 ≅

𝑵𝒊

𝒍𝒈

𝐻𝑚 = magnetic field in the coreHg = magnetic field in the gap

𝑙𝑚 = length of the magnetic circuitN = number of windings

I = electrical current

𝑙𝑔 = gap length

Example

𝑙𝑔 = 1′′ = 0.025𝑚

N = 500

I = 100 A

𝐻 = 2𝑀 Τ𝐴 𝑚 ≅ 2.5 𝑇!

(Needs cooling)


Magnetic susceptibility (𝜒), Magnetic permeability (𝜇) and Magnetization curves

• Susceptibility of the material (𝜒)

𝜒 =response

excitation=

𝑴

𝑯→ 𝑴 = 𝜒𝑯 (linear media)

Obs. 𝜒 is dimensionless

• Permeability of the material (𝜇)

𝑩 = 𝜇0 𝑯 + 𝑴 = 𝜇0 1 + 𝜒 𝑯 = 𝜇𝑯

where, 𝜇 = 𝜇0 1 + 𝜒 = 𝜇0𝜇𝑟

M vs H curves are called magnetization curves (how do they look like?)

• Linear M-H curves

• Retains no M when

H is removed

• Nonlinear M-H curves

• M saturates

• Hysteresis (irreversibility)

• We come back later

Excitation

H

Response

M

Material


Microscopic magnetic dipole momentsOrbital and spin angular moments

Classical and quantum treatment


Orbital motion: connection between angular and magnetic moments

𝒍 = 𝒓 × 𝒑 = −𝑚𝑒𝑟𝑣 Ƹ𝑧𝒎 = 𝑖𝐴 Ƹ𝑧 =𝑒

𝑇× 𝜋𝑟2 Ƹ𝑧

𝒎 =𝑒𝑣

2𝜋𝑟× 𝜋𝑟2 Ƹ𝑧 =

𝑒𝑣𝑟

2Ƹ𝑧

𝒎 = −𝑒

2𝑚𝑒𝒍 = −𝛾𝒍 𝛾 = gyromagnetic ratio


Connection between magnetization and mechanical moment

An electron in an orbital motion is

equivalent to a current loop.

𝒎 = −𝑒

2𝑚𝑒𝒍 = −𝛾𝒍

𝛾 = gyromagnetic ratio

Einstein – de Haas effect (1915)

• There is the inverse effect (Barnett effect) in which

magnetization is induced by rotation.

• Both effects show that magnetic moments are

associated with angular momentum.

Ԧ𝒍


Orbital angular moment: quantum treatment

𝐻 = Τ𝑝2 2𝑚𝑒 + 𝑉 Ԧ𝑟 = Τℏ2𝛻2 2𝑚𝑒 + 𝑉 Ԧ𝑟

Suppose an electron in a spherically symmetrical potential

The eigenfunctions are: ൯𝜓𝑛𝑙𝑚𝑙(𝑟, 𝜃, 𝜑 = 𝑓(𝑟)𝑌𝑙𝑚(𝜃, 𝜑)

n = principal quantum number (n=1, 2, 3, …)

l = azimuthal quantum number (l = 0, 1, 2, 3, …)

𝑚𝑙 = magnetic quantum number (𝑚𝑙= 0, ±1, … ± 𝑙)

If the electron is submitted to magnetic induction 𝑩 = 𝐵 Ƹ𝑧

Ԧ𝑝 → Ԧ𝑝 − 𝑞 Ԧ𝐴, where Ԧ𝐴 = Τ1 2 (𝐵 × Ԧ𝑟)

𝐻 =𝑝2

2𝑚𝑒+ 𝑉 Ԧ𝑟 −

𝑞

2𝑚𝑒( Ԧ𝑟 × Ԧ𝑝) ∙ 𝐵 +

𝑞2

8𝑚𝑒𝐵 × Ԧ𝑟

2

The eigenfunctions of the 𝐿𝑧 = Ԧ𝑟 × Ԧ𝑝 𝑧 operator are 𝑌𝑙𝑚 𝜃, 𝜑

Ԧ𝑟 × Ԧ𝑝 𝑧𝑌𝑙𝑚 𝜃, 𝜑 = 𝑚ℏ𝑌𝑙

𝑚 𝜃, 𝜑

The additional term of energy, dependent on B is given by

𝐸𝑧𝑙 = −𝑚𝑙𝑒ℏ

2𝑚𝑒𝐵 = −𝑚𝑙𝜇𝐵𝐵

𝜇𝐵 = 9.274 × 10−24𝐴𝑚2 (Bohr magneton)

𝑌𝑙𝑚 𝜃, 𝜑 = spherical harmonics

The magnitude of 𝑳 and 𝑙𝑧, are respectivelly

𝑳 = 𝑙(𝑙 + 1)ℏ and 𝑙𝑧 = 𝑚𝑙ℏ, where 𝑚𝑙 = −𝑙 … .0 … 𝑙.

• For d-electrons (l = 2 e ml = 2, 1, 0, -1, -2) 𝑳 = 6ℏ, being 𝑚𝑙ℏthe projection along z axis

• The angular vector L never aligns with B

• Classical interpretation: L precesses around B

𝝎 = 𝛾𝑩Larmor frequency


Spin angular moment: pure quantum mechanical concept

•In addition to 𝑳, electrons have an intrinsic spin angular moment 𝑺

•The associated magnetic moment is Ԧ𝜇𝑠 = − Τ𝑔𝑠𝜇𝐵 ℏ Ԧ𝑆, where 𝑔𝑠~2

•The magnitude of S is Ԧ𝑆 = 𝑠(𝑠 + 1)ℏ and the z-component is 𝑠𝑧 = 𝑚𝑠ℏ

•As 𝑠 = Τ1 2, then 𝑆 = 3 Τℏ 2 and 𝑚𝑠 = ± Τ1 2

• 𝜇𝑠 = 𝑔𝑠𝜇𝐵 Τ3 2 and 𝜇𝑠𝑧 = −𝑔𝑠𝜇𝐵𝑚𝑠

•The angular vector S never aligns with B

•Classical interpretation: S precesses around B


Equation of motion of the spin magnetic moment

The spin magnetic moment is given by Ԧ𝜇𝑠 = −𝛾 Ԧ𝑆, where 𝛾 = Τ𝑔𝑠𝜇𝐵 ℏ =𝑔𝑠

ℏ

𝑒 ℏ

2𝑚𝑒≅

𝑒

𝑚𝑒is the gyromagnetic ratio

If it is submitted to a magnetic field 𝐵0 = 𝐵0 Ƹ𝑧

As previously mentioned, there appears a torque given by:

𝑇 =𝑑 Ԧ𝑆

𝑑𝑡(1)

or 𝑇 = −𝛾 Ԧ𝑆 × 𝐵0 (2). As (1)= (2): 𝑑𝜇𝑠

𝑑𝑡= −𝛾 Ԧ𝜇𝑠 × 𝐵0 = −𝜇0𝛾 Ԧ𝜇𝑠 × 𝐻0

In a piece of material, the magnetization is given 𝑴 = 𝑛𝝁, where n is the # of unbalanced spins/volume, we obtain:

𝑑𝑀

𝑑𝑡= −𝛾𝜇0𝑀 × 𝐻0

By adding damping to the above equation we obtain the equation of Landau-Lifschitz-Gilbert equation of

motion!


Magnetic properties of materialsMacroscopic properties


The magnetic behavior of the materials can characterized by their corresponding values of 𝜒 and 𝜇:

1. Empty space; 𝜒 = 0, since there is no matter to magnetize, and 𝜇 = 𝜇0.

2. Diamagnetic; 𝜒 is small and negative, and 𝜇 slightly less than 1.

3. Para- and antiferromagnetic; 𝜒 is small and positive, and 𝜇 slightly greater than 1.

4. Ferro- and ferrimagnetic; 𝜒 and 𝜇 are large and positive, and both are functions of H.

• Molar magnetic susceptibility (𝜒𝑚): 𝜒𝑚 = 𝜒𝑉𝑚

(𝑉𝑚 is the molar volume, [𝜒𝑚]= Τ𝑚3 𝑚𝑜𝑙)

• Mass susceptibility (𝜒𝑔): 𝜒𝑔 = Τ𝜒 𝜌

(𝜌 is the material density, [𝜒𝑔]= Τ𝑚3 𝑘𝑔)


Diamagnetism: classical interpreation

• Diamagnetic materials do not have any net magnetic moment on their atoms because all bound electrons are paired with spins antiparallel (i.e.

filled shells).

• When an external magnetic field is applied, the orbits of the bound electrons change in accordance with Lenz’s law and set up an orbital

magnetic moment which opposes the external field.

• A second contribution results from free electrons (important in metals) which are forced to move in a circular path with their resulting magnetic

moment counteracting the external field.

𝑩(𝑡) ↑

𝒊 → 𝒊 ↓

𝑬𝑖

𝒎 → 𝒎 ↓

Torque, 𝑑𝐿

𝑑𝑡= 𝜏 = 𝑟𝐹𝐸 − 𝑒𝐸𝑟

𝑑𝐿

𝑑𝑡= −𝑒𝑟

𝜀

2𝜋𝑟= −𝑒𝑟

− Τ𝑑𝜙𝑚 𝑑𝑡

2𝜋𝑟=

𝑒𝑟2

2

𝑑𝐵

𝑑𝑡

Δ𝐿 =𝑒𝑟2

2𝜇0H → Δ𝒎 = −

𝑒

2𝑚Δ𝑳

Δ𝒎 = −𝑒2𝑟2𝜇0

2𝑚𝑯 → ൜

Δ𝒎 ∝ 𝑯opposite direction

Actualy,

𝑟2 →1

3𝑟2 and × 𝑧 (number of electrons

𝑴 = 𝑛Δ𝒎 = −𝑛𝑍𝜇0𝑒

2 𝑟2

6𝑚𝑯

𝜒𝑑𝑖𝑎 =𝑴

𝑯= −

𝑛𝑍𝜇0𝑒2

6𝑚𝑟2


Paramagnetism

In paramagnet materials:

1. 𝑴 is directly proportional to H

2. 𝜒 is small and positive (10−3 ≤ 𝜒 ≤ 10−5)

3. For most of PMs, 𝜒~ Τ1 𝑇 (Curie Law)

4. H = 0: magnetic moments are randomly oriented and 𝑴 = 0

5. H > 0: magnetic moments tend to align with the field direction

counteracted by thermal agitation

• Each atom has a magnetic dipole moment

• They do not interact with each other

• Orbiting electrons are responsible for paramagnetism

• Spin paramagnetism is only slightly temperature-dependent

• The effect is feeble: 𝑘𝐵𝑇 ≫ 𝜇𝑚𝐵

• For itinerant electrons: Pauli paramagnetism ( Τ𝑑𝜒𝑃𝑎𝑢𝑙𝑖 𝑑𝑡 ~0)

Langevin model (localized moments)


Langevin model: localized classical moments

• For 𝐻 = 0, the fraction of moments lying in between 𝜃 and 𝜃 + 𝑑𝜃 is

𝑑𝐴

𝐴=

(2𝜋𝑟 sin) × (𝑟 𝜃𝑑𝜃)

4𝜋𝑟2=

1

2sin 𝜃𝑑𝜃

• The energy is given by 𝐸(𝜃) = −𝒎 ∙ 𝑩 = −𝑚𝐵 cos 𝜃

• The probability of having energy 𝐸 ∝ 𝑒 Τ−𝐸 𝑘𝐵𝑇 = 𝑒 Τ𝑚𝐵 cos 𝜃 𝑘𝐵𝑇

A moment 𝒎 at an angle 𝜃 contributes to the average magnetization along the applied field as:

𝑴𝑧 = 𝑛𝒎 cos 𝜽 = 𝑛𝒎 න𝟎

𝝅

𝑷(𝜽) cos 𝜽𝒅𝜽

𝑴𝑧 = 𝑛𝑚

0

𝜋𝑒 Τ𝑚𝐵 cos 𝜃 𝑘𝐵𝑇 cos 𝜃 sin 𝜃𝑑𝜃

0𝜋

𝑒 Τ𝑚𝐵 cos 𝜃 𝑘𝐵𝑇 sin 𝜃𝑑𝜃

𝒎

The overall probability of magnetic moment point in between 𝜃 and 𝜃 + 𝑑𝜃 is given by:

𝑃 𝜃 =𝑒 Τ𝑚𝐵 cos 𝜃 𝑘𝐵𝑇 sin 𝜃𝑑𝜃

0

𝜋𝑒 Τ𝑚𝐵 cos 𝜃 𝑘𝐵𝑇 sin 𝜃𝑑𝜃


𝑴𝑧 = 𝑛𝑚

0

𝜋𝑒 Τ𝑚𝐵 cos 𝜃 𝑘𝐵𝑇 cos 𝜃 sin 𝜃𝑑𝜃

0𝜋

𝑒 Τ𝑚𝐵 cos 𝜃 𝑘𝐵𝑇 sin 𝜃𝑑𝜃Ƹ𝑧

Here n is the number of magnetic moments per unit volume. Therefore 𝑛𝑚 = 𝑀𝑆 is the saturation magnetization.

𝑴𝑧𝑀𝑆

=𝑒𝑦 + 𝑒−𝑦

𝑒𝑦 − 𝑒−𝑦−

1

𝑦= coth 𝑦 −

1

𝑦= ℒ(𝑦)

Where ℒ(𝑦) is the so called Langevin function.

Langevin model: Langevin function

By replacing 𝑦 = Τ𝑚𝐵 𝑘𝐵𝑇 and 𝑥 = cos 𝜃, we get 𝑑𝑥 = − sin 𝜃 𝑑𝜃 and 0𝜋

→ 1−1

𝑴𝑧 = 𝑛𝑚

−1

1𝑥𝑒𝑥𝑦𝑑𝑥

−1

1𝑒𝑥𝑦𝑑𝑥

Ƹ𝑧

As 𝑥𝑒𝑥𝑦𝑑𝑥 =1

𝑦2𝑒𝑥𝑦(𝑥𝑦 − 1) and 𝑒𝑥𝑦𝑑𝑥 =

1

𝑦𝑒𝑥𝑦, the above equation becomes:


Langevin model: Curie law

Assuming a small external magnetic field (or high temperature), Τ(𝑚𝐵 𝑘𝐵𝑇) ≪ 1

𝑀

𝑀𝑆≅

1

3

𝑚𝐵

𝑘𝐵𝑇+ ℴ

𝑚𝐵

𝑘𝐵𝑇

3

∴ 𝜒 =𝑀

𝐻=

𝑛𝜇0𝑚2

3𝑘𝐵

1

𝑇=

𝐶

𝑇

See “Fundamentals of Magnetism”

(Mathias Getzlaff)

Curie law. C =𝑛𝜇0𝑚

2

3𝑘𝐵

𝑀

𝑀𝑆= coth

𝑚𝐵

𝑘𝐵𝑇−

𝑘𝐵𝑇

𝑚𝐵

• Saturation will occur if y (= Τ𝑚𝐵 𝑘𝐵𝑇) is large enough

• At small y, the magnetization M varies linearly with H

Pierre Curie

(1859-19006)


Paramagnetism: extension to include quantum mechanics

• In the previous calculation we assumed that the magnetic dipole moments can take all possible

orientation with respect to B.

• But due to the spatial quantization it only have discrete orientations, where J is the total

angular moment of the ion

• The effective magnetic dipole moment is given by: 𝑚𝑒𝑓𝑓 = 𝑔𝜇𝐵 𝐽(𝐽 + 1)

• The component along B is given by: 𝑚𝑧 = 𝑔𝜇𝐵𝑀𝐽, where the (2J+1) allowed values of 𝑀𝐽 =

− 𝐽, −𝐽 + 1 … . 𝐽 − 1, 𝐽.

• The value of J may be an integer or half-integer and varies from 𝐽 = Τ1 2 to 𝐽 = ∞.

Textbook statistical mechanics calculation

The partition function: 𝑍 = σ𝑀𝐽=−𝐽𝐽 𝑒

𝑔𝜇𝐵𝑀𝐽𝐵

𝑘𝐵𝑇 =sinh (2𝐽+1) Τ𝑥 2

sinh Τ𝑥 2, where 𝑥 =

𝑔𝜇𝐵𝐵

𝑘𝐵𝑇

The average value of 𝑚𝑧 is: 𝑚𝑧 =σ𝑀𝐽=−𝐽

𝐽𝑀𝐽𝑒

𝑀𝐽𝑥

σ𝑀𝐽=−𝐽𝐽

𝑒𝑀𝐽𝑥

=1

𝑍

𝜕𝑍

𝜕𝑥

The magnetization along z is: 𝑀𝑧 = 𝑛𝑔𝜇𝐵 𝑚𝑧 =𝑛𝑔𝜇𝐵

𝑍

𝜕𝑍

𝜕𝐵

𝜕𝐵

𝜕𝑥= 𝑛𝑘𝐵𝑇

𝜕 ln 𝑍

𝜕𝐵

By using 𝑦 = 𝑥𝐽 =𝑔𝜇𝐵𝐽𝐵

𝑘𝐵𝑇, 𝑀𝑧 = 𝑛𝑔𝜇𝐵𝐽𝐵𝐽 𝑦

Where 𝑀𝑆 = 𝑛𝑔𝜇𝐵𝐽 and 𝐵𝐽 𝑦 =2𝐽+1

2𝐽coth

2𝐽+1

2𝐽𝑦 −

1

2𝐽coth

𝑦

2𝐽

𝐵𝐽 𝑦 is the Brillouin function!

𝑀

𝑀𝑆=

2𝐽 + 1

2𝐽coth

2𝐽 + 1

2𝐽𝑦 −

1

2𝐽coth

𝑦

2𝐽


𝑀

𝑀𝑆= 𝐵𝐽 𝑦 =

2𝐽 + 1

2𝐽coth

2𝐽 + 1

2𝐽𝑦 −

1

2𝐽coth

𝑦

2𝐽𝑦 =

𝑔𝜇𝐵𝐽𝐵

𝑘𝐵𝑇

• 𝐽 = ∞, 𝐵∞ 𝑦 = L(y). The classical situation

• 𝐽 = Τ1 2, 𝐵 Τ1 2 𝑦 = tanh 𝑦. The quantum mechanics situation

By using: 𝑐𝑜𝑡ℎ 𝑥 =1

𝑥+

𝑥

3−

𝑥3

45+ ⋯

For low B, 𝐵𝐽 𝑦 =𝐽+1 𝑦

3𝐽+ ℴ(𝑦3),

Thus 𝑀 = 𝑀𝑆𝑔𝜇𝐵𝐽𝐵

3𝑘𝐵𝑇=

𝑛𝜇0𝐻 𝑔𝜇𝐵 𝐽(𝐽+1)2

3𝑘𝐵𝑇.

And the susceptibility: 𝜒 =𝑀

𝐻=

𝑛𝜇0𝑚𝑒𝑓𝑓2

3𝑘𝐵𝑇

• A measurement of 𝜒 will permit to obtain the

effective magnetic moment 𝑚𝑒𝑓𝑓 = 𝑔𝜇𝐵 𝐽(𝐽 + 1)

• Where the Landé g-factor is given by:

𝑔 =3

2+

𝑆 𝑆 + 1 − 𝐿(𝐿 − 1)

2𝐽(𝐽 + 1)

Paramagnetism: extension to include quantum mechanics


Ground state of free ions: Hund’s rules

We assume that:

1. The total angular moment quantum number J is always a good quantum number

2. If SOI can be ignored → the orbital 𝐿 and spin 𝑆 angular moments are also good quantum numbers

3. The eigenvalues of 𝑳2, 𝑳𝑧, 𝑺2, 𝑺𝑧, 𝑱

2 and 𝑱𝑧 are 𝐿(𝐿 + 1), 𝐿𝑧, S(𝑆 + 1), 𝑆𝑧, 𝐽(𝐽 + 1), 𝐽𝑧, respectively

4. For filled shells, 𝐽 = 𝐿 = 𝑆 = 0. Only partially filled shells contribute to magnetic moment of free ions

5. The L-S coupling lifts some degeneracies associated to electronic state of the free ions, resulting in a

particular arrangement of spins and orbital momenta in the ground state

6. These arrangements are well described by the Hund’s rules

1st rule: First, arrange the electrons to maximize S=∑ms.

(In other words, fill the subshell with spin-up electrons before you add any with spin down.)

2nd rule: As far as possible consistent with the first rule, arrange the electrons to maximize L=∑mℓ.

(In other words, fill the orbitals with maximum mℓ first and those with minimum mℓ last.)

3rd rule: Calculate the total magnetic moment quantum number J according to the following rule:

𝐽 = ൜𝐿 − 𝑆 , if the subshell is less than half full

𝐿 + 𝑆 , if the subshell is more than half full


Hund’s rules (example)

1st: Arrange the electrons to maximize S=∑ms.

(In other words, fill the subshell with spin-up electrons

before you add any with spin down.)

𝑆 = 4 ×1

2= 2

2nd: As far as possible consistent with the first rule,

arrange the electrons to maximize L=∑mℓ.

(In other words, fill the orbitals with maximum mℓ first

and those with minimum mℓ last.)

𝐿 = 2 × 2 + 1 × 1 + 1 × 0 + 1 × −1 + 1 × (−2) = 2

3rd: Calculate the total magnetic moment quantum

number J according to the following rule:

𝐽 = ൜𝐿 − 𝑆 , if the subshell is less than half full𝐿 + 𝑆 , if the subshell is more than half full

𝐽 = 2 + 2 = 4

2𝑆+1𝐿𝐽 =5𝐷4

L = |ΣLz|= 0 1 2 3 4 5 6

= S P D F G H I

6 electrons

left


• Magnetization curve of a paramagnetic salt containing 𝐶𝑟3+ ions

𝐶𝑟3+ ≡ 𝐴𝑟 3𝑑3 ≡−−↑↑↑≡ 4𝐹 Τ3 2 → 𝑆 = Τ3 2 ; 𝐿 = 3; 𝐽 = Τ3 2 (see

Hund´s rules)

• The best fit was obtained by adjusting the Brillouin function to

the data, by using 𝐽 = 𝑆 = Τ3 2 , L = 0. That gives 𝑔 = 2.

• The excellent agreement shows that the magnetic moment of

𝐶𝑟3+ is only due to spin.

• The maximum moment of 𝐶𝑟3+ along the field is 𝑔𝜇𝐵𝑀𝐽 = 2𝜇𝐵3

2=

3𝜇𝐵

• The effective moment is given by 𝑚𝑒𝑓𝑓 = 𝑔𝜇𝐵 𝐽(𝐽 + 1) =

2𝜇𝐵3

2

5

2= 3.87𝜇𝐵

• By using 𝜒 =𝑀

𝐻=


3𝑘𝐵𝑇, we obtain from initial slope 𝑚𝑒𝑓𝑓 =

3.85𝜇𝐵, which is in good agreement

See, Cullity & Graham (Introduction to Magnetic Materials)

Paramagnetism: example


Pauli paramagnetism: free electrons

Using the previous equation of the susceptibility for free electrons (𝑆 = Τ1 2)

𝜒 =𝑀

𝐻=


3𝑘𝐵𝑇=

𝑛𝜇0 𝑔𝜇𝐵 𝑆(𝑆 + 1)2

3𝑘𝐵𝑇=

𝑛𝜇0𝜇𝐵2

𝑘𝐵𝑇

And the expected magnetization

𝑀𝑧 =𝑛𝜇𝐵

2

𝑘𝐵𝑇𝐵

Which has Τ1 𝑇 dependence.

On the other hand, experiments show that in metals:

• 𝜒 does not depend on T.

• 𝜒 has a value, at room temperature, which is 2 orders of magnitude smaller


Pauli paramagnetism: free electrons

• DOS of a free-electron gas

with no external field

• Band energies for spins up &

down are the same

• No magnetic moment

• Remember that 𝝁𝑠 = −𝛾𝑺, and 𝑈 = −𝝁𝑠 ∙ 𝑩

• DOS of a free-electron gas with

an external field ∥ up direction

• Spin split DOS for free

electrons in an external field

• The down (up) spin states are

lowered (raised) in energy, by

an amount 𝜇𝐵𝜇0𝐻

• A spill-over of electrons from

up-spin to down-spin until the

new Fermi levels for up- and

down-spin are equal.

2𝜇0𝜇𝐵𝐻


Pauli paramagnetism (magnetization of a free electron gas)

𝑀 = 𝜇𝐵 𝑛↑ − 𝑛↓ = 𝜇𝐵1

2𝐷 𝐸𝐹 × 2𝜇𝐵𝜇0𝐻 = 𝜇𝐵

2 𝜇0𝐻𝐷 𝐸𝐹

Where the DOS at the Fermi level is 𝐷 𝐸𝐹 =3

2

𝑛

𝐸𝐹

Thus 𝑀 =3𝑛𝜇𝐵

2 𝜇0𝐻

2𝐸𝐹→ 𝜒 =

3𝑛𝜇𝐵2 𝜇0

2𝐸𝐹∴ 𝜒𝑃𝑎𝑢𝑙𝑖 =

3𝑛𝜇𝐵2 𝜇0

2𝑘𝐵𝑇𝐹

Comparing with the Curie susceptibility previously deduced for

𝐽 = 𝑆 = Τ1 2 , 𝐿 = 0:

𝜒𝐶𝑢𝑟𝑖𝑒 =𝑛𝜇0𝜇𝐵

2

𝑘𝐵𝑇∴

𝜒𝑃𝑎𝑢𝑙𝑖𝜒𝐶𝑢𝑟𝑖𝑒

~𝑇

𝑇𝐹

• Pauli susceptibility is independent of temperature

• Pauli susceptibility is reduced by a factor Τ𝑇 𝑇𝐹 ≈ 10−2 (at room temperature)

• Closed shells do not contribute to 𝜒𝑃𝑎𝑢𝑙𝑖. Only s, p and d-electrons of unfilled shells contribute

• Orbital diamagnetism of free electrons contributes with the Landau susceptibility, 𝜒𝐿𝑎𝑛𝑑𝑎𝑢 = −1

3𝜒𝑃𝑎𝑢𝑙𝑖

𝜒𝑓𝑟𝑒𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = 𝜒𝑃𝑎𝑢𝑙𝑖 + 𝜒𝐿𝑎𝑛𝑑𝑎𝑢 =𝑛𝜇𝐵

2 𝜇0𝑘𝐵𝑇𝐹

• Considering the paramagnetic and diamagnetic responses, the free electron susceptibility is given by:


Magnetic susceptibilities of a free electron gas: summary

𝜒𝑡𝑜𝑡 = 𝜒𝐶𝑢𝑟𝑖𝑒 + 𝜒𝑃𝑎𝑢𝑙𝑖 + 𝜒𝐿𝑎𝑛𝑑𝑎𝑢


Magnetic susceptibilities of the first 60 elements


FerromagnetismMean field approach


• In fact many paramagnetic materials do not obey the Curie law

• Instead follow a more general law given by the Curie–Weiss law:

𝜒 =𝐶

𝑇 − 𝜃

• These materials undergo spontaneous ordering and become ferromagnetic below some critical temperature, the Curie

temperature, TC

Curie-Weiss law

• At 𝑇𝐶 the Weiss molecular field is strong enough that it magnetizes the substance even with no external field

Actually, the transition is

gradual due to the presence of

small clusters with aligned

spins even above 𝑇𝐶


Ferromagnetism: mean field theory

• As previously discussed, paramagnets exhibit a net magnetic moment when an external field is applied

• On the other hand ferromagnets exhibit a spontaneous magnetization even with no external applied field

• It means that the atomic dipoles are strongly coupled to align each other

• Why does spontaneous magnetization occurs?

• The first modern theory of ferromagnetism was proposed by Pierre Weiss in

1906, even before the atomic magnetic moment being discovered.

• Weiss postulated that there is an internal molecular field, which is strong

enough to align the localized dipole moments, and is proportional to the local

magnetization:

𝑯𝑊 = 𝜆𝑴

where, 𝜆 is the molecular field constant that depends on the material.

• The total field is given by: 𝑯𝑡𝑜𝑡 = 𝑯 + 𝜆𝑴

• Now the Curie law: 𝜒 =𝑴

𝑯+𝜆𝑴=

𝐶

𝑇→ 𝑴 =

𝐶

𝑇−𝐶𝜆𝑯

The Curie Weiss law: 𝜒 =𝑴

𝑯=

𝐶

𝑇−𝜃


Ferromagnetism: mean field theory

The magnetization is given by the Brillouin function

𝑀

𝑀𝑆= 𝐵𝐽 𝑦 =

2𝐽 + 1

2𝐽coth

2𝐽 + 1

2𝐽𝑦 −

1

2𝐽coth

𝑦

2𝐽

where 𝑦 =𝑔𝜇𝐵𝜇0𝐽 𝜆𝑀+𝐻

𝑘𝐵𝑇.

If the external field is null (𝐻 = 0), 𝑦0 =𝑔𝜇𝐵𝜇0𝐽𝜆𝑀

𝑘𝐵𝑇

𝑀

𝑀𝑆= 𝐵𝐽 𝑦0 (1)

𝑀

𝑀𝑆=

𝑘𝐵𝑇

𝑛 𝑔𝜇𝐵𝐽 2𝜇0𝜆𝑦0 (2)

• Graphical solution of (1) (black curve) and (2) (green line) for 𝐽 = Τ1 2 to find the spontaneous magnetization M when T < TC .

• Equation (2) is also plotted for T = TC (red) and T > TC (blue).

• The effect of an external field is to offset (2), as shown by the dotted line.


1. The critical temperature 𝑇𝐶 can be obtained by matching the slope of the magnetization (at the origin)

from equation 𝑀 =𝑘𝐵𝑇

𝑔𝜇0𝜇𝐵𝐽𝜆𝑦 and equation 𝑀 = 𝑀𝑆𝐵𝐽 𝑦 ≪ = 𝑀𝑆

𝐽+1

3𝐽𝑦 + ℴ(𝑦3), where 𝑀𝑆 = 𝑛𝑔𝜇𝐵𝐽.

𝑘𝐵𝑇𝐶𝑔𝜇0𝜇𝐵𝐽𝜆

= 𝑛𝑔𝜇𝐵𝐽𝐽 + 1

3𝐽𝑦 → 𝑇𝐶 =

𝑛𝜆𝜇0 𝑔𝜇𝐵 𝐽((𝐽 + 1)2

3𝑘𝐵→ 𝑇𝐶 =

𝑛𝜆𝜇0𝑚𝑒𝑓𝑓2

3𝑘𝐵∴ 𝑚𝑒𝑓𝑓

2 = (𝑔𝜇𝐵 𝐽(𝐽 + 1))2

2. To calculate 𝐵𝑊𝑒𝑖𝑠𝑠 we assume that: 𝐽 = Τ1 2, and take typically 𝑇𝐶 ≅ 103 𝐾, as 𝑇𝐶 =

(𝜆𝜇0𝑀𝑆)𝜇𝐵

2𝑘𝐵and as,

𝐵𝑊𝑒𝑖𝑠𝑠 = 𝜇0(𝜆𝑀𝑆), we have:

Estimating the critical temperature and the molecular field value

𝐵𝑊𝑒𝑖𝑠𝑠 = 𝜇0𝜆𝑀𝑆 ≅𝑘𝐵𝑇𝐶

𝜇𝐵~1500 𝑇 ‼!


3d 6 (free atoms. 4𝜇𝐵) 3d 7 (free atoms. 3𝜇𝐵) 3d 8 (free atoms. 2𝜇𝐵)

4 f 7 simple FM. Spins all

lined up in ground state

Saturation magnetization at absolute zero

𝑀𝑆(0) = 𝑛𝑛𝐵𝜇𝐵

𝑛 = no. of unit formulas/vol𝑛𝐵 = no. of 𝜇𝐵 per unit formula

The band model accounts for

the ferromagnetism of the

transition metals Fe, Co, Ni.


Exchange interaction


Let us suppose an atom with two electrons.

The Halmiltonian is given by,

𝐻 = 𝐻1 + 𝐻2 + 𝐻12

where, 𝐻1 = −ℏ2

2𝑚𝑒∆1 −

𝑍𝑒2

4𝜋𝜖0𝑟1

𝐻2 = −ℏ2

2𝑚𝑒∆2 −

𝑧𝑒2

4𝜋𝜖0𝑟2

𝐻12 = −𝑒2

4𝜋𝜖0𝑟12

∆𝑗=𝜕2

𝜕𝑥𝑗2 +

𝜕2

𝜕𝑦𝑗2 +

𝜕2

𝜕𝑧𝑗2.

The wave function for one electron is given by,

𝜓 = 𝜙(𝑟)𝜒

where,

• 𝜙(𝑟) is the spatial wavefunction for one –e with no spin

• 𝜒 spin wavefunction for one electron

Exchange interaction: two electrons atom

The total wavefunction Ψ(1,2) has to be antisymmetric (Pauli exclusion principle): Ψ 1,2 = −Ψ 2,1 .

Ψ𝑠𝑖𝑛𝑔 1,2 = 𝜓𝑆 (𝑟1, 𝑟2)𝜒𝐴𝑆or

Ψ𝑡𝑟𝑖𝑝 1,2 = 𝜓𝐴𝑆 (𝑟1, 𝑟2)𝜒𝑆

𝜓𝑆 𝑟1, 𝑟2 =1

2𝜙1 𝑟1 𝜙2 𝑟2 + 𝜙2 𝑟1 𝜙1 𝑟2

𝜓𝐴𝑆 𝑟1, 𝑟2 =1

2𝜙1 𝑟1 𝜙2 𝑟2 − 𝜙2 𝑟1 𝜙1 𝑟2

Exchange interaction

S = 0

Ψ𝑠𝑖𝑛𝑔 1,2 = 𝜓𝑆 (𝑟1, 𝑟2)𝜒𝐴𝑆 =1

2𝜙1 𝑟1 𝜙2 𝑟2 + 𝜙2 𝑟1 𝜙1 𝑟2 ×

Ψ𝑡𝑟𝑖𝑝 1,2 = 𝜓𝐴𝑆 𝑟1, 𝑟2 𝜒𝑆 =1

2𝜙1 𝑟1 𝜙2 𝑟2 − 𝜙2 𝑟1 𝜙1 𝑟2 ×

S = 1

ms = +1,0, -1

A triplet of symmetric spin

wave functions

A singlet of an anti-

symmetric spin wave function


121221 JKIIE ++=

The energy can be calculated by 𝐸 = ΤΨ 𝐻 Ψ Ψ Ψ , where both spatial and spin wave function are included.

Where,

• 𝐼1 and 𝐼2 are the ionization energies for electrons 1 and 2

• 𝐾12 is the Coulomb interaction between both electrons

• 𝐽12 is the Exchange energy, which has no classical counterpart, and is given by,

• 𝐽12𝑆𝑖𝑛𝑔

= 𝜓𝑆 𝐻12 𝜓𝑆

and

• 𝐽12𝑡𝑟𝑖𝑝

= 𝜓𝐴𝑆 𝐻12 𝜓𝐴𝑆

For S=1, the spatial wave

function is antisymmetric

ۧ|𝝍𝑨𝑺

For S=0, the spatial wave

function is symmetric

ۧ|𝝍𝑺

Charge distribution is different ⇒ electrostatic energy is different

Therefore the singlet and triplet states have different

energies.


Given that Ԧ𝑆 = Ԧ𝑆1 + Ԧ𝑆2 ∴ 𝑆2 = 𝑆1

2 + 𝑆22 + 2 Ԧ𝑆1 ∙ Ԧ𝑆2

2 Ԧ𝑆1 ∙ Ԧ𝑆2 = 𝑆 𝑆 + 1 − 𝑆1 𝑆1 + 1 − 𝑆2(𝑆2 + 1)

Triplet (S=1): 2 Ԧ𝑆1 ∙ Ԧ𝑆2 = 2 − Τ3 4 − Τ3 4 = Τ1 2

→ Τ1 2 + 2 Ԧ𝑆1 ∙ Ԧ𝑆2 = 1

Singlet (S=0): 2 Ԧ𝑆1 ∙ Ԧ𝑆2 = 0 − Τ3 4 − Τ3 4 = − Τ3 2

→ Τ1 2 + 2 Ԧ𝑆1 ∙ Ԧ𝑆2 = −1.

Therefore the expected value for the energy can be written as

𝐸 = 𝐼1 + 𝐼2 + 𝐾12 − Τ1 2 + 2 Ԧ𝑆1 ∙ Ԧ𝑆2 𝐽12

𝐽12 < 0, antiparallel alignment

𝐽12 > 0, parallel alignment

∆𝐸𝑒𝑥𝑐 = ℋ𝑒𝑥 = −2𝐽 Ԧ𝑆1 ∙ Ԧ𝑆2 (Heisenberg interaction)( J is the exchange integral)

To include the spin dependence, we use the following procedure:


Comparing the exchange integral with the Weiss molecular field

Lets us assume two atoms with magnetic moments 𝒎1 and 𝒎2 where 𝒎1 = −𝑔𝜇𝐵𝑺1 and 𝒎2 = −𝑔𝜇𝐵𝑺2.

The Heisenberg Hamiltonian can be written as,

𝑈12 = −2𝐽12𝑺1 ∙ 𝑺2 = −2𝐽12

(𝑔𝜇𝐵)2𝒎1 ∙ 𝒎2.

The effect of spin 𝑺2on the spin 𝑺1 can be represented by an effective field,

𝑩12 =2𝐽12

𝑔𝜇𝐵𝑺2.

As the exchange integral is the same between the near neighbors, the average effective field due to the z1near neighbors atoms acting on 𝑺1 is,

𝐵𝑒𝑓𝑓 =2𝑆𝑧1𝐽1

𝑔𝜇𝐵.

In terms of the magnetization 𝑴𝑆 = 𝑛𝑔𝜇𝐵𝑺𝑗 , the effective field is,

𝑩𝑒𝑓𝑓 =2𝑧𝐽𝑖𝑗𝑺𝑗

𝑔𝜇𝐵=

2𝑧𝐽𝑖𝑗𝑴𝑆

𝑛 𝑔𝜇𝐵2 ≡ 𝑩𝑊𝑒𝑖𝑠𝑠 = 𝜆𝑴𝑆.

As previously shown,

𝜆 =𝑇𝐶𝐶

=3𝑘𝐵𝑇𝐶

𝑛 𝑔𝜇𝐵 2𝑗(𝑗 + 1)≡

2𝑧𝐽𝑖𝑗

𝑛 𝑔𝜇𝐵 2→ 𝐽𝑖𝑗 =

3𝑘𝐵𝑇𝐶2𝑧𝑗(𝑗 + 1)

For Fe (BCC), z = 8, TC = 1043 K, j = 1→ 𝐽𝑖𝑗≅ 0.01 𝑒𝑉𝑈𝑒𝑥𝑐

𝑈𝑑𝑖𝑝−𝑑𝑖𝑝~

10−21𝐽

10−24𝐽~103


Bethe-Slater curve

The exchange interaction between neighboring magnetic

moments is given by,

𝐽 = න 𝜓𝑎∗(𝑟1)𝜓𝑏

∗(2)1

𝑟𝑎𝑏−

1

𝑟𝑎2−

1

𝑟𝑏1+

1

𝑟21𝜓𝑏 (𝑟1)𝜓𝑎 (𝑟2) 𝑑

3𝑥

where 𝑟𝑎𝑏=distance between atom cores; 𝑟21= distancebetween the 2 electrons; 𝑟𝑎2 and 𝑟𝑏1 are the distances between each electron to the respective nuclei.

This integral was evaluated and the result found that:

1. J becomes positive for small r12 and small rab.

2. J becomes positive for large ra2 and rb1.

3. The exchange integral can therefore be plotted against

the ratio rab/rd, where rd is the radius of the 3d orbital.

4. This gives the Bethe-Slater curve, which correctly

separates the FM 3d elements (Ex. Fe, Co abd Ni) from

the AF 3d elements Ex. Cr and Mn).

Although this approach (by Heitler-London,

Heisenberg and Bethe) provides a useful conceptual

framework for discussing the magnetic interactions of

electrons, better description of the values of J has to

be use first principles.


Magnetism of itinerant electron systems


Magnetism of 3d-band materials: Slater-Pauling curve

Introduction

• In magnetism of atoms as well as in oxides, the electrons

remain fairly well localized on the atomic or ionic sites.

• There is little hopping or itinerant character in the valence

electrons.

• In this case the electronic states of the ions are atomic

like and Hund’s rules provide a good starting point for

determining ionic moments.

• However, in most metals, the electrons are itinerant and

the band theory gives better results.

• On the other hand, transition metals that has the

magnetism given by the 3d-band electrons, show some

features of both localization and itinerant electrons.

Slater-Pauling curve shows the average magnetic

moment 𝜇 per transition metal atom. It is 2.2𝜇𝐵 forFe, 1.7𝜇𝐵 for Co, and 0.6𝜇𝐵 for Ni.


Note that 𝑁 ↓ remains approximately 3 electrons/atom on lhs of Slater-Pauling

Note that 𝑁 ↑ remains approximately 5.3 electrons/atom on rhs of Slater-Pauling

Slater-Pauling curve features

Localized

electrons


Energy levels: from atom to solid

Rigid

band


Rigid electron bands

See Inomata lectures.

The s and d bands are rigid in shape as atomic number changes. This simplifies modeling the behavior of different alloys

by simply moving 𝐸𝐹 up or down through the bands according to the number of electrons present.

• This model allows to explain the magnetic behavior of different alloys by only

moving the Fermi energy EF up or down through the majority and minority band

according to the number of electrons being present.

• Magnetic moments arise from unpaired electrons: 𝜇 = (𝑛𝑑↑ − 𝑛𝑑

↓ )𝜇𝐵

• Non-integer magnetic moment per atom is obtained

[𝐴𝑟] 3𝑑104𝑠1

[𝐴𝑟] 3𝑑64𝑠2 [𝐴𝑟] 3𝑑84𝑠2


Rigid electron bands: Fe and Ni

Iron

• Has 8 valence electrons in 3d and 4s bands

• Measurements: 4𝑠0.95, 3𝑑7.05

• # of d-electrons:𝑛𝑑↑ + 𝑛𝑑

↓ = 7.05

• Observed magnetic moment: 𝑛𝑑↑ − 𝑛𝑑

↓ = 2.2

• It means that: 4.62 spin↑ and 2.43 spin↓

• (4.62 − 2.43 ≅ 2.2)

Nickel

• Has 10 valence electrons in 3d and 4s bands

• Measurements: 4𝑠0.6, 3𝑑9.4

• # of d-electrons:𝑛𝑑↑ + 𝑛𝑑

↓ = 9.4

• Observed magnetic moment: 𝑛𝑑↑ − 𝑛𝑑

↓ = 0.6

• It means that: 5 spin↑ and 4.4 spin↓

• The spin-up band is fully occupied

Weak FM

Found at LHS of the S-P curve

Have holes in both spin bands

Ex. Fe

Strong FM

Found at RHS of the S-P curve

Have holes only in the minority band

Ex. Ni


Application of the rigid electron bands: magnetism of the alloy 𝐹𝑒1−𝑥𝑁𝑖𝑥 (naive picture)

The rigid band model can be used to explain the magnetic

properties of alloys such as 𝐴1−𝑥𝐵𝑥, for instance at the rhs of the S-P curve, with slope -1(strong FM)

The average amount of valence electrons of the alloy is

𝑛𝑑𝑎𝑙𝑙𝑜𝑦

= 1 − 𝑥 𝑛𝑑𝐴 + 𝑥𝑛𝑑

𝐵.

For the 𝐹𝑒1−𝑥𝑁𝑖𝑥 alloy, 𝑛𝑑𝐹𝑒 = 8 and 𝑛𝑑

𝑁𝑖 = 10.

Therefore, 𝑛𝑑𝐹𝑒1−𝑥𝑁𝑖𝑥 (𝑥) = 8 1 − 𝑥 + 10𝑥

The magnetic moment is given by 𝜇 = (𝑛𝑑↑ − 𝑛𝑑

↓ )𝜇𝐵 where both

𝑛↑and 𝑛↓ vary with x. However, for strong FMs (rhs) 𝑛𝑑↑ = 5

(spin-up band is fully occupied). Thus the magnetic moment

per atom is

𝜇 = 5 − 𝑛𝑑↓ 𝜇𝐵 = 5 − 𝑛𝑑 + 5 𝜇𝐵 = 10 − 𝑛𝑑 𝜇𝐵

This equation explains the -1 slope for the rhs of the S-P

curve and also explains why the average moment of

Co should be very close to that of Ni50Fe50. Both exhibit the

same valence electron concentration and thus the same

value of nd.

𝑛𝑑𝐹𝑒50𝑁𝑖50 = 8 1 − 0.5 + 10 × 0.5 = 9

See Mathias Getzlaff𝐶𝑜 = [𝐴𝑟] 3𝑑104𝑠1


Domain wall and magnetization curves

𝐻 > 0 → spins of the favorable domains grow at

the expense of unfavorable

domains, until the material

contains one single domain.


Why is the initial magnetization of a FM zero?

Weiss propose the formation of magnetic domains (MD)

(1) Small regions in which all the magnetic dipoles are aligned parallel to each other

(2) With no magnetic field (H = 0) the total magnetization of the sample averages to zero

(3) MD results from the energy minimization

Single domain

Minimum exchange energy

Large magnetostatic energy

(magnetization creates an

external field . The

uncompensated dipoles

create a demagnetizing field

inside the block)

Divide block into

domains

Lower the magnetostatic

energy.

But increased exchange

energy.

With no surface poles

No magnetostatic energy

But increased exchange,

anisotropic and

magnetostriction energies

𝐸𝑚𝑎𝑔 = 𝑉 𝑑𝐸

= − 𝑉𝑩∙𝑑𝒎

𝑑𝑉𝑑𝑉

= −𝜇0 𝑉 𝑯𝑑 ∙ 𝑴𝑑𝑉

𝐸𝑚𝑎𝑔 =𝜇0𝑁𝑀

2

2𝑉

𝐸𝑚𝑎𝑔 decreases

by breaking into

domains

Exchange energy


Transition between domains: domain walls

𝟗𝟎𝒐

𝟏𝟖𝟎𝒐

180o Rotation of the magnetization in a (a) Bloch wall and (b) Néel wall

Domain wall width: competition between exchange and anisotropy energies

𝑈 = −2𝐽𝑺1 ∙ 𝑺2 = −2𝐽𝑆2 cos 𝜑

𝑈 ≈ −2𝐽𝑆2 1 − Τ𝜑2 2

∆𝑈𝑒𝑥 = 𝑈 𝜑 − 𝑈 0 = 2𝐽𝑆2 (1 − cos 𝜑)

For small 𝜑, ∆𝑈𝑒𝑥 ≈ 𝐽𝑆2𝜑2

If the180º spin-flip takes N steps

(N = DW width and 𝜑 = Τ𝜋 𝑁)

∆𝑈𝑒𝑥𝑤𝑎𝑙𝑙

= 𝑁∆𝑈𝑒𝑥 ≈ 𝐽𝑆2𝑁 Τ𝜋 𝑁 2

Larger N decreases the DW energy!

If a is the lattice constant, a Bloch wall the

energy per unit of area is:

∆𝑈𝑒𝑥 ≈ 𝐽𝑆2

𝜋2

𝑁𝑎2

Therefore, the exchange energy favors a

larger DW

On the de other hand, the uniaxial

anisotropy favors a shorter DW


𝑈𝑐𝑟𝑦𝑠(𝑖)

= 𝐾 sin2 𝜑𝑖

𝐾 = uniaxial anisotropy constant [𝑒𝑛𝑒𝑟𝑔𝑦/𝑣𝑜𝑙]

𝑈𝑐𝑟𝑦𝑠(𝑤𝑎𝑙𝑙)

= σ𝑖=1𝑁 𝐾 sin2 𝜑𝑖 ≈

𝑁

𝜋0

𝜋𝐾 sin2 𝜑 𝑑𝜑 =

𝐾𝑁

2

Per unit of area,

𝑈𝑐𝑟𝑦𝑠(𝑤𝑎𝑙𝑙)

=𝐾𝑁𝑎

2

where 𝑎= lattice constant.

Larger N increases the DW energy!

180o domain wall width

The total wall energy is given by,

𝑈𝑤𝑎𝑙𝑙 = 𝐽𝑆2

𝜋2

𝑁𝑎2+

𝐾𝑁𝑎

2= 𝐴𝑒𝑥

𝜋2

𝛿+

𝐾

2𝛿

where, 𝐴𝑒𝑥 = Τ2𝐽𝑆2 𝑎 is the exchange stiffness.

By minimizing 𝑈𝑤𝑎𝑙𝑙,

𝑁 = 𝜋𝑆2𝐽

𝐾𝑎3.

The DW width is given by,

𝛿0 = 𝑁𝑎 = 𝜋2𝐴𝑒𝑥

𝐾

• Larger J makes it thicker

• Larger K makes it smaller.

Typical values: 𝛿0 ≈ 1 𝜇𝑚 in NiFe

𝛿0 ≈ 5 𝑛𝑚 in Nd2Fe14B

Uniaxial crystalline energy

See: “Magnetism in condensed matter”, S. Blundell


𝑈𝑤𝑎𝑙𝑙 = 𝐴𝑒𝑥𝜋2

𝛿+

𝐾

2𝛿

𝛿0 = 𝑁𝑎 = 𝜋2𝐴𝑒𝑥

𝐾


Ferromagnetic hysteresis loop

http://hyperphysics.phy-

astr.gsu.edu/hbase/Solids/hyst.html


Classification FM materials according to the hysteresis loops

http://hyperphysics.phy-

astr.gsu.edu/hbase/Solids/hyst.html

• Hysteresis curve implies loss of

a certain amount of energy

• Energy losses are critical in AC

applications

• Hysteresis loss Phy

𝑃ℎ𝑦 = 𝑓 න𝑙𝑜𝑜𝑝

𝐵𝑑𝑀

• Eddy-current loss Ped

𝑃𝑒𝑑 = Τ𝜋𝑡𝑓𝐵𝑚𝑎𝑥2 6𝜌

• Anomalous losses Pan(due to domain wall motion,

nonuniform magnetization and

sample inhomogeneity)

(See M. Coey)


Ferrimagnetism


Ferrimagnetism: introduction

Temperature dependence Τ1 𝜒 for:(1) A paramagnet with 𝜒 = Τ𝐶 𝑇(2) A ferromagnet with, 𝜒 = Τ𝐶 (𝑇 − 𝜃)(3) An antiferromagnet with 𝜒 = Τ𝐶 (𝑇 + 𝜃)(4) A ferrimagnet

Temperature dependence of the spontaneous

magnetization for a ferrimagnet with two sublattices. The

observed magnetization, M, is the algebraic sum of the

contributions arising from each of the sub-lattices, MA and

MB below a critical temperature Tc. (a) Usually M varies

monotonically as a function of temperature. (b) M shows a

compensation temperature, Tcomp, at which the sign of the

magnetization is reversed.(O. Khan, Nature|Vol 399 |6

May 1999)

• Two sublattices with different magnetizations point in opposite directions

• Show spontaneous magnetization below 𝑇𝐶like a FM, but the temperature dependence can be different

Néel explained the T-dependence of M and 𝜒 using the Weiss theory + assuming antiparallel coupling between magnetic moments of

different magnitude


Ferrimagnetism: mean field theory

Assume that,

• A and B are different ions

• Interactions are ቐ𝐴 − 𝐴𝐵 − 𝐵𝐴 − 𝐵

• 𝑛 #magnetic ions per unit of volume

• 𝛼 = fraction of A’s

• 𝛽 = fraction of B’s (1 − 𝛼)

• 𝑚𝐴 = moment of one A atom

• 𝑚𝐵 = moment of one B atom

Thus,

• Magnetization of A sublattice = 𝑀𝐴 = 𝛼𝑛𝑚𝐴

• Magnetization of B sublattice = 𝑀𝐵 = 𝛽𝑛𝑚𝐵

Total magnetization, 𝑀 = 𝑀𝐴 + 𝑀𝐵 = 𝛼𝑛𝑚𝐴 + 𝛽𝑛𝑚𝐵

• Weiss molecular field on sublattice A is

𝐵𝐴 = 𝐵0 − 𝜆𝐴𝐵𝑀𝐵 + 𝜆𝐴𝐴𝑀𝐴

• Weiss molecular field on sublattice B is

𝐵𝐵 = 𝐵0 − 𝜆𝐴𝐵𝑀𝐴 + 𝜆𝐵𝐵𝑀𝐵

• Assume that,

↑𝐴

↑𝐴

↑𝐵

↑𝐵

↑𝐴

↓𝐵

𝜆𝐴𝐴 = 𝑎𝜆𝐴𝐵 and 𝜆𝐵𝐵 = 𝑏𝜆𝐴𝐵, thus

𝐵𝐴 = 𝐵0 + 𝜆𝐴𝐵 𝑎𝑀𝐴 − 𝑀𝐵

𝐵𝐵 = 𝐵0 + 𝜆𝐴𝐵(𝑏𝑀𝐵 − 𝑀𝐴).

The magnetization of the sublattices are

𝑀𝐴 = 𝑛𝐴𝑔𝐵𝑆𝑔𝜇𝐵𝑆

𝑘𝐵𝑇𝐵𝐴

𝑀𝐵 = 𝑛𝐵𝑔𝐵𝑆𝑔𝜇𝐵𝑆

𝑘𝐵𝑇𝐵𝐵


Ferrimagnetism in the paramagnetic region: mean field theory

For 𝑇 > 𝑇𝐶 (paramagnetic region)

൞𝑀𝐴 =

𝐶𝐴

𝑇𝐵𝐴 ∴ 𝐶𝐴 = Τ𝑛𝐴 𝑔𝜇𝐵

2𝑆(𝑆 + 1) 3𝑘𝐵

𝑀𝐵 =𝐶𝐵

𝑇𝐵𝐵 ∴ 𝐶𝐵 = Τ𝑛𝐵 𝑔𝜇𝐵

2𝑆(𝑆 + 1) 3𝑘𝐵

Substituting 𝐵𝐴 and 𝐵𝐵 in the above equations,

𝑀𝐴 =𝐶𝐴

𝑇𝐵0 + 𝜆𝐴𝐵 𝑎𝑀𝐴 − 𝑀𝐵

𝑀𝐵 =𝐶𝐴

𝑇𝐵0 + 𝜆𝐴𝐵 𝑏𝑀𝐵 − 𝑀𝐴 ,

that can be writen as,

1 −𝐶𝐴𝑇

𝑎𝜆𝐴𝐵𝐶𝐴𝑇

𝜆𝐴𝐵

𝐶𝐵𝑇

𝜆𝐴𝐵 1 −𝐶𝐵𝑇

𝑏𝜆𝐴𝐵

𝑀𝐴

𝑀𝐵

=

𝐶𝐴𝑇

𝐵0

𝐶𝐵𝑇

𝐵0

𝐵0𝑀𝐴 + 𝑀𝐵

=1

𝜒=

𝑇

𝐶+

1

𝜒0−

𝜎

𝑇 − Θ

𝐶 = 𝐶𝐴 + 𝐶𝐵

1

𝜒0=

1

𝐶2𝐶𝐴

2𝜆𝐴𝐴 + 𝐶𝐵2𝜆𝐵𝐵 + 2𝐶𝐴𝐶𝐵𝜆𝐴𝐵

𝜎 =𝐶𝐴𝐶𝐵

𝐶3ሼ

ሽ

𝐶𝐴2 𝜆𝐴𝐴 − 𝜆𝐴𝐵

2 + 𝐶𝐵2 𝜆𝐵𝐵 − 𝜆𝐴𝐵

2 − 2𝐶𝐴𝐶𝐵[

]

𝜆𝐴𝐵2 −

𝜆𝐴𝐴 + 𝜆𝐵𝐵 𝜆𝐴𝐵 + 𝜆𝐴𝐴𝜆𝐵𝐵

Θ = −𝐶𝐴𝐶𝐵

𝐶𝜆𝐴𝐴 + 𝜆𝐵𝐵 − 2𝜆𝐴𝐵

The Néel ferrimagnetic temperature is calculated by making 1

𝜒=0

Solution is


Magnetite ( Fe3O4 = FeO Fe2 O3) where,

• FeO (ferrous oxide) = (Fe2+ : d 6 ) ion: S = 2

• Fe2 O3 (ferric oxide) = (Fe3+ : d 5 ) ion: S = 5/2

a ~ 8A. All J < 0 (antiferromagnetic)|JAB | > |JAA | , |JBB | → AA, BB //, AB anti //.

Spinel MgAl2O4

Mg: 8 tetrahedral sites

Al: 16 octahedral sites

O: 32 vertices

A ferrimagnetic prototype: magnetite (𝐹𝑒3𝑂4)

Normally, ferrites [(MO Fe2 O3), where M = Zn, Cd, Fe, Ni, Co, Mg] are ferrimagnetic materials.

They are mostly poor conductors .

The magnetic moments are distributed as:

cancelled

𝐴

↑ 𝐹𝑒3+𝐵

↓ 𝐹𝑒3+𝐹𝑒2+ ↓


Antiferromagnetism


Antiferromagnetism

At 0 K, two interpenetrating and identical sublattices of

magnetic ions, each spontaneously magnetized to

saturation in zero applied field, but in opposite directions.

M = 0M =saturation

moment 0

M =saturation

moment 0

The evidence of antiferromagnetism is the behavior of

susceptibility above the critical temperature, called the

Néel temperature (TN).

Above TN, the susceptibility obeys the Curie-Weiss law

for paramagnets but with a negative intercept indicating

negative exchange interactions.

TN : Néel temperature

T < TN : AF state

T > TN : paramagnetic

)(

−−=

T

C


Antiferromagnetism: mean field model (𝜒 for T>TN)

𝑀𝐴

𝑀𝐴

Use the same approximation of ferrimagnets

• For T>TN.

• The magnetization of both sublattices are equal

• Curie constants are equal: 𝐶𝐴 = 𝐶𝐵 = 𝐶

• Coupling with nearest neighbors:𝜆𝐴𝐵 ≫ 𝜆, (𝜆 = 𝜆𝐴𝐴= 𝜆𝐵𝐵)

• 𝜎 = 0, 1

𝜒0=

𝜆+𝜆𝐴𝐵

2≅

𝜆𝐴𝐵

2

• Therefore,1

𝜒=

𝑇+𝜆𝐴𝐵𝐶

2𝐶→ 𝜒 =

2𝐶

𝑇+𝜃

Equivalent to the Curie-Weiss law, but 𝑇𝑁 = 𝐶(𝜆𝐴𝐵 ≫ 𝜆)

𝑩𝐴 = −𝜆𝐴𝐴𝑴𝐴 − 𝜆𝐴𝐵𝑴𝐵 + 𝑩𝑎

𝑩𝐵 = −𝜆𝐵𝐵𝑴𝐵 − 𝜆𝐴𝐵𝑴𝐴 + 𝑩𝑎

The energy (𝜆𝐴𝐴= 𝜆𝐵𝐵 = 𝜆)

𝑈 = −𝑴𝐴 ∙ 𝑩𝐴 − 𝑴𝐵 ∙ 𝑩𝐵

𝑈 = 2𝜆 𝑀𝐴2 + 𝑀𝐵

2 + 2𝜆𝐴𝐵𝑴𝐴 ∙ 𝑴𝐵 − 𝑩𝑎 ∙ (𝑴𝐴 + 𝑴𝐵)


For T > TN , 𝜒 is independent of direction of Ba .

For T


Antiferromagnetism: mean field model (𝜒 for T


Antiferromagnetism: mean field model (spin-flop transition)

The SF transition can be understood as competing

effects of the various contributions to the energy of

the antiferrornagnet.

The free energy (Exchange + Zeeman + Anisotropy) is given by

𝐹 = 𝐽𝑴1 ∙ 𝑴2 − 𝑩 ∙ 𝑴1 + 𝑴2 −𝐾1

2cos2 𝜃 + cos2 𝜙

where: 𝑴1 = 𝑴2 = 𝑀, at the SF fase 𝜃 = 𝜙.

The equilibrium state is given by 𝜕𝑖𝐹 = 0.

𝒍 = 𝑴1 − 𝑴2 is the AF vector

The three phases are characterized by:

AF state: 𝜃 = 0 and 𝜙 = 𝜋 → 𝒍 ∥ Ƹ𝑧

SF state: 𝜃 = 𝜙 = 𝜃 → 𝒍 ⊥ Ƹ𝑧

PM state:𝜃 = 𝜙 = 0 → 𝒍 = 𝟎

At the SF transition vector 𝒍 switches: 𝒍 ∥ Ƹ𝑧 to 𝒍 ⊥ Ƹ𝑧, and the energy .

𝐹 = 𝐽𝑀2 cos 2𝜃 − 2𝐵𝑀 cos 𝜃 − 𝐾1 cos2 𝜃

The SF angle is cos 𝜃 =𝑀𝐵

2𝐽𝑀2−𝐾1and the SF energy is

𝐹𝑆𝐹 = −𝐽𝑀2 −

𝑀2𝐵2

2𝐽𝑀2 − 𝐾1

The AF energy (B=0) is 𝐹𝐴𝐹 = −𝐽𝑀2 − 𝐾1, thus the SF field is (𝐹𝑆𝑃 = 𝐹𝐴𝑃)

𝐵𝑆𝐹 = 2𝐽𝐾1 − Τ𝐾1 𝑀 2 ቊIf 𝐾1 ↑∴ 𝐵𝑆𝐹 ↑

If 𝐾1 ↓∴ 𝐵𝑆𝐹 ↓


Magnetic anisotropies


Types of magnetic anisotropies

The magnetization in FM crystals tend to align along certain preferred crystallographic directions.

This directions are called the “easy axes”

It is easiest to magnetize a demagnetized sample to saturation if the field is applied along an easy axis

• Magnetocrystalline anisotropy

Origin: crystal field + spin-orbit coupling

• Shape anisotropy

Origin: dipolar fields

• Surface anisotropy

Origin: broken symmetry at surfaces and interfaces

• Induced anisotropy

Origin: induced by extrinsic manners (applying a field: during

growth or during heat treatment)

• Magnetostrictive anisotropy

Origin: created by stress (thin films grown by tilting the substrate)

𝐻𝐴


Effect of the magnetocrystalline anisotropy on the magnetization curves of single crystals of Fe, Ni and Co

Easy axes

(See M. Coey)


Magnetocrystalline anisotropy

The Heisenberg exchange interaction is isotropic (coherent rotation of a magnet’s spin system does not change the

Heisenberg exchange energy)

𝐻 = −

𝑖


• How does the local magnetic moment (𝒎𝐽) distinguish between different crystallographic directions?

• How is 𝒎𝐽 coupled to the lattice?

Answer:

- By means of the spin orbit coupling

- By the interaction of the atomic orbitals with the crystal field

2D symmetry of the crystalline electric field

2D symmetry of atomic or ionic wavefunctions

𝐿𝑧 = orbital angular moment

Magnetocrystalline anisotropy

• The AO and CEF have less than spherical symmetry for

magnetic anisotropy

• Further, if L-S is appreciable, the spin also will prefer

particular crystallographic directions.

• Solid lines show combinations favorable to strong anisotropy;

dashed lines, weak. See, O’Handley, Modern Magnetic Magnetic Materials


Why rare-earth materials have stronger anisotropy than 3d transition metals and alloys

Answer: competition between the coupling of 𝑳 with the crystal field (ℋ𝑐𝑓) and the 𝑳 − 𝑺 coupling (ℋ𝑠𝑜).

(a) In 3d transition metals, the unpaired electrons occupy the outermost shell, thus ℋ𝑐𝑓 is significant and 𝑚𝐿 is mostly

quenched (because the electric field in which the electrons move is noncentral, resulting in a average zero 𝐿𝑧). The spin magnetic moment 𝑚𝑆 will show a weak anisotropic response.

(b) In 4f rare-earth metals, the unpaired electrons (that occupy 4f shell) are shielded by the filled outer electron shells,

so the effect of ℋ𝑐𝑓 is insignificant and it does destroy 𝐿𝑧. The SOC is strong and 𝑚𝐽 = 𝑚𝐿 + 𝑚𝑆 will respond intensely

to 𝐵𝑒𝑥𝑡, and the pull of low-symmetry crystal field generate high magnetic anisotropies.

In RE ions, the 4f shell is

shielded by the outer

electrons, so the action of

ℋ𝑐𝑓 on the unpaired

electron is small

In 3d ions, the 3d

shell is outermost,

so ℋ𝑐𝑓 acts on the

unpaired electrons

See M. Coey


𝒎 ≡ 𝛼1, 𝛼2, 𝛼3

where,

𝛼1 = cos Τ(𝑚𝑥 𝑚) = sin 𝜃 cos 𝜑𝛼2 = cos Τ(𝑚𝑦 𝑚) = sin 𝜃 sin 𝜑

𝛼3 = cos Τ(𝑚𝑧 𝑚) = cos 𝜃

Magnetocrystalline anisotropy energy

1. The uniaxial Anisotropy

The crystal anisotropy energy density is

often expressed as a power series of the

form

𝐸𝑢𝑎 = 𝑛

𝐾𝑢𝑛 sin2𝑛 𝜃

𝐸𝑢𝑎 = 𝐾𝑢0 + 𝐾𝑢1sin2 𝜃 + 𝐾𝑢2sin

4 𝜃 + ⋯

where

• z is the uniaxial axis

• 𝐾𝑢𝑛 are the uniaxial anisotropy constants (energy/vol)

• 𝐾𝑢0 has no meaning for anisotropic properties because it is independent of the orientation of M.

• 𝐾𝑢1 > 0 implies an easy axis.

• For Co: 𝐾𝑢1 = 4.8 × 105 Τ𝐽 𝑚3 and 𝐾𝑢2 = 1.5 × 10

5 Τ𝐽 𝑚3


Magnetocrystalline anisotropy energy surfaces

• The anisotropy energy can be visualized as a three-

dimensional energy surface.

• Below is the 1st-order anisotropy energy surfaces for Co

• The length of the radius vector to any point on the

surface defines the anisotropy energy density in that

direction, (𝜃, 𝜑)

1. The cubic crystalline anisotropy

For cubic crystals such as Fe and Ni, the anisotropy

energy can be expressed in terms of the direction

cosines 𝛼1, 𝛼2, 𝛼3 of the magnetization vector with respect to the three cube edges.

𝐸𝑐𝑢𝑏 = 𝐸0 +

𝑖

𝑏𝑖𝛼𝑖 +

𝑖𝑗

𝑏𝑖𝑗𝛼𝑖𝛼𝑗 +

𝑖𝑗𝑘

𝑏𝑖𝑗𝑘𝛼𝑖𝛼𝑗𝛼𝑘 + ⋯

By using symmetry operations, the equation can be written as

𝐸𝑐𝑢𝑏 = 𝐾1 𝛼12𝛼2

2 + 𝛼22𝛼3

2 + 𝛼32𝛼1

2 + 𝐾2𝛼12𝛼2

2𝛼32 + 𝐾3 𝛼1

2𝛼22 + 𝛼2

2𝛼32 + 𝛼3

2𝛼12 2 + ⋯

where 𝐾1, 𝐾2, and 𝐾3 are cubic anisotropy constants.

For Fe (20 oC) 𝐾1 = 4.72 × 104 Τ𝐽 𝑚3and 𝐾2 = −0.075 × 10

4 Τ𝐽 𝑚3.

For Ni (23 oC) 𝐾1 = −5.7 × 103 Τ𝐽 𝑚3and 𝐾2 = −2.3 × 10

3 Τ𝐽 𝑚3.

For 100 , 𝛼1 = 1, 𝛼2 = 𝛼3 = 0 → 𝐸𝑐𝑢𝑏[100]

= 0

For 111 , 𝛼1 = 𝛼2 = 𝛼3 = Τ1 3 → 𝐸𝑐𝑢𝑏[111]

=1

3𝐾1 +

1

27𝐾2 +

1

9𝐾3 + ⋯

If 𝐾1 > 0 (as valid for Fe) and neglecting 𝐾2 and 𝐾3 terms, 𝐸𝑐𝑢𝑏[111]

> 𝐸𝑐𝑢𝑏[100]

, Therefore,

100 is the easy axis. By symmetry, 010 and 001 are also easy axes.


Magnetocrystalline anisotropy energy (surface energies for Fe and Ni)

energy surface for K0=1, K1=2 and K2=0

typical for bcc cubic crystals (Fe)energy surface for K0=1, K1=-2 and K2=0

typical for fcc cubic crystals (Ni)


Measuring the magnetocrystalline anisotropy

Torque magnetometer• A easy manner to measure the anisotropy energy is by using a

torque magnetometer

• Consider a disk in which the easy axis (c-axis) makes an angle 𝜑with H

• The torque is given by 𝜏 = − Τ𝜕𝐸 𝜕𝜑

• Suppose a cubic crystal in which the magnetization rotates in the x-

y or (001) plane, where 𝜃 is the angle between M and c-axis.

• As θ = 0, 𝛼1 = cos 𝜑, 𝛼2 = sin 𝜑, 𝛼3 = 0 and 𝐸𝑐𝑢𝑏 = 𝐾1()

𝛼12𝛼2

2 + 𝛼22𝛼3

2 +𝛼3

2𝛼12 + 𝐾2𝛼1

2𝛼22𝛼3

2 + 𝐾3 𝛼12𝛼2

2 + 𝛼22𝛼3

2 + 𝛼32𝛼1

2 2

• 𝐸𝑐𝑢𝑏 =1

8𝐾1 1 − cos 4𝜑 +

1

128𝐾3 3 − 4 cos 4𝜑 + cos 8𝜑 + ⋯

• Considering only the 1st order, 𝐿 = −𝐾1

2sin 4𝜑

Torque curve measured for a

single crystal disk of 4% Si-Fe cut

parallel to (001). Amplitude is 𝐾1

2.


aUniaxial materials are designated with a

superscript u and their values 𝐾𝑢1 and 𝐾𝑢2 are listed under 𝐾1 and 𝐾2, respectively. 𝐾1 > 0 implies an easy axis. Units are erg/cm3; divide

these values by 10 to get J/m3.

bDisordered; 𝐾1 ≈ 0 for ordered phase.cNet moment canted about 30o from [00l] toward [110].

Anisotropy constants for selected materials

Modern magnetic materials : principles and applications /

Robert C. O'Handley.


• If a finite sample has no magnetocrystalline anisotropy (ex. a polycrystalline sample with

grain anisotropy axes randomly oriented), the magnetization tends to orient in order to

minimize the magnetostatic energy

• If the sample is not spherical the magnetostatic energy of the system depends on the

orientation of magnetic moments within the sample.

• This effect is of purely magnetostatic origin and is closely related to demagnetizing fields.

In particular, for a spheroidal

𝐸𝑚𝑎𝑔 =1

2𝜇0𝑀𝑆

2 𝑁𝑥𝛼𝑥2 + 𝑁𝑦𝛼𝑦

2 + 𝑁𝑧𝛼𝑧2 ∴ 𝐸𝑚𝑎𝑔 =

1

2𝜇0𝑀𝑆

2 𝑁𝑥 + (1 − 3𝑁𝑥) cos2 𝜃 , where 𝑁𝑥 = 𝑁𝑦 and 𝑁𝑧 = 1 − 2𝑁𝑥

Shape anisotropy

For a sphere

𝑁𝑥 = 𝑁𝑦 = 𝑁𝑧 = Τ1 3

𝐸𝑚𝑎𝑔 =1

6𝜇0𝑀𝑆

2

(isotropic)

For a long cylinder

𝑁𝑥 = 𝑁𝑦 = Τ1 2 , 𝑁𝑧 = 0

𝐸𝑚𝑎𝑔 =1

4𝜇0𝑀𝑆

2 sin2 𝜃

For a thin film

𝑁𝑥 = 𝑁𝑦 = 0, 𝑁𝑧 = 1

𝐸𝑚𝑎𝑔 =1

2𝜇0𝑀𝑆

2 cos2 𝜃

In particular for thin films

𝐸𝑠ℎ𝑎𝑝𝑒 = 𝐾𝑒𝑓𝑓 cos2 𝜃


Magnetization curve for 50 nm Fe(001)/GaAs(001)

Magnetization curve (effect of the shape anisotropy)

Fe(100) – 50 nm

𝑯

𝐵𝑠𝑎𝑡⊥ = 𝜇0𝐻𝑑𝑒𝑚 = 𝜇0𝑁𝑧𝑀𝑆 = 𝜇0𝑀𝑆 = 2.1 𝑇


Surface and interface anisotropy

The existence of surface interface:

• Translational symmetry breaking affects the orbital motion of electrons

• In ultrathin films and magnetic multilayers the interface contribution can even be dominating

• Perpendicular magnetic anisotropy (PMA) can be strong enough to overcome the shape anisotropy in thin films

• Phenomenologically, the effective anisotropy constant is written as

𝐾𝑒𝑓𝑓 = 𝐾𝑉 +2𝐾𝑆

𝑡

• where, 𝐾𝑉 (energy/vol) is the volume dependent magnetocrystalline anisotropy constant, 𝐾𝑆 (energy/area) is the surface dependent magnetocrystalline anisotropy constant and t is the film thickness. Obs. The factor of 2 takes into

account the existence of two surfaces

Proposed by L. Néel in 1952;

Exp. discovered by Gradmann et al. 1968

Adding the shape anisotropy the surface anisotropy energy for thin films is normally written as

𝐸𝑠𝑢𝑟𝑓 = 𝐾𝑒𝑓𝑓 cos2 𝜃 = 𝐾𝑉 +

2𝐾𝑆𝑡

cos2 𝜃 ∴ ൝For 𝐾𝑒𝑓𝑓 > 0, 𝑀 is in plane

For 𝐾𝑒𝑓𝑓 < 0, 𝑀 is out of plane

𝐸𝑠𝑢𝑟𝑓 =2𝐾𝑆

𝑡cos2 𝜃

It has the same dependence as the shape anisotropy (which is a vol contribution),

𝐸𝑠ℎ𝑎𝑝𝑒 = 𝐾𝑉 cos2 𝜃, (here I used 𝐾𝑉 = 𝐾𝑒𝑓𝑓).


Effect of surface anisotropy

In some papers, people use sin2 𝜃 instead of cos2 𝜃. Therefore the equation for 𝐸𝑠𝑢𝑟𝑓 is written as,

𝐸𝑠𝑢𝑟𝑓 = 𝐾𝑒𝑓𝑓 sin2 𝜃 = 𝐾𝑉 +

2𝐾𝑆𝑡

sin2 𝜃 ∴ ൝For 𝐾𝑒𝑓𝑓 > 0, 𝑀 is out of plane

For 𝐾𝑒𝑓𝑓 < 0, 𝑀 is in plane

𝑡𝐾𝑒𝑓𝑓 = 𝑡𝐾𝑉 + 2𝐾𝑆.

Plotting the dependence of 𝑡𝐾𝑒𝑓𝑓 × 𝑡, both 𝐾𝑉 and 2𝐾𝑆 can be easily determined.

• This plot shows 𝑡𝐾𝑒𝑓𝑓 × 𝑡𝐶𝑜 measured for a multilayer of

𝐶𝑜(𝑡𝐶𝑜)/𝑃𝑑(11 Å)

• For 𝑡𝐶𝑜 ≤ 11Å, 𝐾𝑒𝑓𝑓> 0 which means that 𝑀 ⊥to the plane

• For 𝑡𝐶𝑜 ≥ 11Å, 𝐾𝑒𝑓𝑓> 0 which means that 𝑀 ∥ to the plane

• The y axis intercept equals twice the surface anisotropy 2𝐾𝑆

• Whereas the slope gives the volume contribution 𝐾𝑉

J. Magn. Magn. Mater. 93, 562 (1991)


Induced magnetic anisotropy

Induced magnetic anisotropy can be reached by means of extrinsic manners.

1. Induced during growth (growth induced, stress induced, field induced)

2. Post-deposition magnetic annealing

Oblique deposited films

CoFeB oblique deposited films

Apply field during deposition

Nov. 2015 - Sci.

Rep. 5:17023


Comparing the exchange stiffness of Fe, Ni and Co

A better term for cross material comparison is the exchange stiffness,

𝐴𝑒𝑥 = Τ𝑛𝐽𝑆2 𝑎

n = number of nearest neighbors

J = exchange integral

a = lattice constant.

𝐴𝑒𝑥𝐶𝑜 = 30.2 Τ𝑝𝐽 𝑚

𝐴𝑒𝑥𝐹𝑒 = 20.7 Τ𝑝𝐽 𝑚

𝐴𝑒𝑥𝑁𝑖 = 7.2 Τ𝑝𝐽 𝑚


Types of exchange interaction

Direct exchange• Occurs between moments close enough to have sufficient overlap of their

wavefunctions.

• It gives rise to a strong but short range coupling which decreases rapidly as the ions

are separated.

Super exchange• Important ionic solids with no direct exchange between magnetic ions show magnetic

ordering (MnO, etc.)

• Two magnetic ions with non-overlapping charge distribution interact via their overlap

with a third non-magnetic ion. MnO is a prototype example, in which the p orbitals

from O and d orbitals from Mn can form a direct exchange favoring the singlet state.

• Each Mn2+ ion exhibits 5 electrons in its d shell with all spins being parallel due to

Hund’s rule. The O2− ions possess electrons in p orbitals which are fully occupied

with their spins aligned antiparallel.

• In the excited state, one of the extra electrons of O-- is transferred to the a Mn in such

a way to maximize the spin magnetic moment. The unpaired electron of the p-orbital

of O– will couple with the other Mn ion. Therefore the two Mn ions end up with

antiparallel magnetic moments.

• This superexchange interaction will be strongest along a straight line, because the

strechted p-orbital can maximally overlap the Mn 3d-orbitals in this configuration.

𝑀𝑛 = 1𝑠22 𝑠2𝑝6 3 𝑠2𝑝6𝑑5 4𝑠2 and 𝑂 = 1𝑠22 𝑠2𝑝4


Indirect exchange (RKKY)

• Indirect exchange couples moments over relatively large distances. It is the

dominant exchange interaction in metals where there is little or no direct

overlap between neighboring magnetic electrons, and the conduction

electrons mediate the interaction.

• For instance, in rare-earth metals, whose magnetic electrons in the 4f shell

are shielded by the 5s and 5p electrons, direct exchange is rather weak and

indirect exchange via the conduction electrons gives rise to magnetic order

in these materials.

• This type of exchange is better known as the RKKY interaction named after

Ruderman, Kittel, Kasuya and Yoshida.

• The RKKY exchange coefficient J oscillates from positive to negative as the

separation of the ion changes and has the damped oscillatory nature.

Types of exchange interaction

𝐽(𝑟) Electron gas

Localized

spins

𝐻𝑅𝐾𝐾𝑌 =

𝑖

𝑗

IEEE Mag Society summer school 2018 (USFQ)106

Spin orbit interaction

1) Symmetry-independent (intrinsic):

exists in all types of crystals

stem from SOI in atomic orbitals

2) Symmetry-dependent (extrinsic):

exists only in crystals with break in the spatial inversion symmetry

a) Dresselhaus interaction (bulk): Bulk-Induced-Assymetry

Zincblende (ZnS): GaAs, GaP, InAs, InSb, ZnSe, CdTe …

- z > 0 and z < 0 half-spaces are not equivalent

- ො𝑛 has a specific direction

- Three vectors: ො𝑛, 𝑘, ො𝜎

b) Rashba interaction (surface): Surface-Induced-Asymmetry

There are two types of SOI in solids


Electron behaves as a free particle: 𝐻 =𝑝2

2𝑚∗; ε(k)=

ℏ2

2𝑚∗𝑘2; 𝜓𝑘 𝑟 =

1

Ω𝑒𝑖𝑘∙ Ԧ𝑟

Eigenvalues are degenerated: ε↑ 𝑘 = ε↓ 𝑘

Rashba idea: If the potential is asymmetrical by spatial inversion, then the degeneracy is lifted (even for B = 0)

By expanding the potencial in Taylor series: 𝑉 Ԧ𝑟 = 𝑉0 + 𝑒𝐸 ∙ Ԧ𝑟 + ⋯

The asymmetry is characterized by a electric field 𝐸 ∥ ො𝑛

By surface SOI, originates a magnetic field 𝐵 = Τ−( Ԧ𝑣 × 𝐸 ) 𝑐2

Rashba Hamiltonian: 𝐻𝑅 = − Ԧ𝜇𝑠 ∙ 𝐵 ∝ Ԧ𝑣 × 𝐸 ∙ Ԧ𝜎 =𝛼𝑅

ℏԦ𝜎 × Ԧ𝑝 ∙ Ԧ𝑧

Rashba effect (brief tutorial)

E. I. Rashba and V. I. Sheka: Fiz. Tverd. Tela 3 (1961) 1735; ibid. 1863;

Sov. Phys. Solid State 3 (1961) 1257; ibid. 1357.

Yu. A. Bychkov and E. I. Rashba, JETP Lett. 39, 78 (1984);

Yu. A. Bychkov and E. I. Rashba, J. Phys. C: Solid State Phys. 17 (1984)

6039.

Let us assume a two dimensional electron gas


Hamiltonian (kinetic + Rashba)

𝐻 = 𝐻𝐾 + 𝐻𝑅 =𝑝∕∕

2

2𝑚∗+

𝛼𝑅ℏ

Ԧ𝜎 × Ԧ𝑝∕∕ 𝑧

𝐻 =𝑝∕∕

2

2𝑚∗+

𝛼𝑅ℏ

𝜎𝑥𝑝𝑦 − 𝜎𝑦𝑝𝑥

Ƹ𝑧

𝑘ො𝜎

ො𝑦ො𝑥

Eigenfunctions

𝜓+ 𝑥, 𝑦 = 𝑒𝑖(𝑘𝑥𝑥+𝑘𝑦𝑦)

1

2

01

𝑖𝑒−𝑖𝜃

𝜓− 𝑥, 𝑦 = 𝑒𝑖(𝑘𝑥𝑥+𝑘𝑦𝑦)

1

2

1−𝑖

𝑖𝑒−𝑖𝜃

Eigenvalues

𝜀± 𝑘 =ℏ2

2𝑚∗𝑘 ± 𝑘𝑆𝑂

2 − ∆𝑆𝑂

• Translational symmetry breaking

• Degeneracy is lifted (with no need of B)

• Splitting of the spin sub-bands



Spin state is always ⊥ to 𝑘

𝑘

𝐵𝑒𝑓𝑓

Free electron Efeito Zeeman

𝐵 ≠ 0

With Rashba

interaction

SOI enables a wide variety of fascinating phenomena!

• Rashba SO coupling locks spin to the linear momentum.

• Splitting of the spin sub-bands in energy .

aR ~0.67 × 10-11 eV m

(InAlAs/InGaAs)

aR ~3.7 × 10-10 eV m

(BiAg(111) )



Conclusions

The young generation is very luck because we are in the middle of the “golden age of magnetism”

Thank you very much

Muchas gracias

fundamentals of magnetism - ieee magnetics societyieeemagnetics.org/files/summer_school/1-ieee...

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