Magnetism and Magnetic MaterialsMagnetism and Magnetic Materials

DTU (10313) – 10 ECTSDTU (10313) – 10 ECTSKU – 7.5 ECTSKU – 7.5 ECTS

Sub-atomic – pm-nm

But with some surrounding environment

Module 3

08/02/2001

Crystal fields

Intended Learning Outcomes (ILO)Intended Learning Outcomes (ILO)

(for today’s module)(for today’s module)

1. Explain why paramagnetism is T-dependent whereas diamagnetism is not2. Estimate the value of the Curie constant for a given paramagnetic substance3. Predict the ground state of ions by applying Hund’s rules4. Explain the origin of the spin-orbit interaction, and describe its main effects 5. Compare Hund’s rule predictions with data on 4f and 3d elements6. Describe how crystal fields arise7. Explain phenomena such as crystal field splitting, Jahn-Teller distortions, low/high spin states

FlashbackFlashback

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χ =−Na

Vm

μ 0e2

6me

0 ri2 0

i

∑

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M = ngJμ BJBJ (y) = M S BJ (y)

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BJ (y) =2J +1

2Jcoth

2J +12J

y ⎛ ⎝ ⎜

⎞ ⎠ ⎟−

12J

cothy

2J

⎛ ⎝ ⎜

⎞ ⎠ ⎟

y =gJμ BJB

kBT

Einstein de Haas:-measure g-factor

Diamagnetism:-small-T-independent-Orbital size

Paramagnetism:-small-T-dependent ---> Curie law-Total angular momentum J

Van Vleck paramagnetismVan Vleck paramagnetism

Another contribution to the paramagnetic susceptibility (there’s one more…mobile electrons – Pauli)

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ΔE0 =0 μ B (L + gS) ⋅B n

2

E0 − Enn

∑

If J=0, in principle there is no paramagnetic term.However, if we go second-order, and consider the possibility of excited states (off-diagonal matrix terms) with nonzero J, then we have:

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χ =2μ 0μ B2 N

V

0 (Lz + gSz ) n2

En − E0n

∑ Which is positive (para), and T-independent.

Why is it T-indepenent?? And why was the Langevin term T-dependent instead?

John H. van Vleck, Nobel prize lecture

Lande’ g-value and effective momentLande’ g-value and effective moment

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BJ (y) ≈J +13J

y

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χ = MM S

≈nμ 0μ eff

2

3kBT

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μeff = gJμ B J(J +1)

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gJ =32

+S(S +1) − L(L +1)

2J(J +1)

J=1/2 J=3/2

J=5

Curie law: χ=CC/T

Estimate the Curie constant for a paramagnetic ionic salt with a=0.3 nm, J=S=3/2

Check: where are we?Check: where are we?

All atoms and ions are diamagneticdiamagnetism arises from a perturbation of the ground state

diamagnetism is small and T-independent

Whenever J differs from zero, we observe a paramagnetic responseJ can be either from OAM or from Spin or both

paramagnetism is larger than diamagnetismbut still small at room T

The question now is:What gives angular momentum to an atom?

Why are some atoms “more magnetic” than others?

That’s what we focus on today.

The multi-electron atom and the Hund’s rulesThe multi-electron atom and the Hund’s rules

Find the electronic structure of Fe3+, Ni2+, Nd3+, Dy3+, and determine their spin configuration

With many electrons, it gets messy. How do electrons “choose” which state to occupy?

(1)Arrange the electronic wave function so as to maximize S. In this way, the Coulomb energy is minimized because of the Pauli exclusion principle, which prevents electrons with parallel spins being in the same place, and this reduces Coulomb repulsion.

(2)The next step is to maximize L. This also minimizes the energy and can be understood by imagining that electrons in orbits rotating in the same direction can avoid each other more effectively.

(3)Finally, the value of J is found using J=|L-S| if the shell is less than half-filled, J=L+S is the shell is more than half-filled, J=S (L=0) if the shell is exactly half-filled (obviously). This third rule arises from an attempt to minimize the spin-orbit energy.

€

ˆ H =pi

2

2m−

Ze2

4πε0ri

⎛

⎝ ⎜

⎞

⎠ ⎟

i

∑ +Ze2

4πε0 | ri − ri |i< j

∑

2S+1LJ

Spin-orbit and the fine structureSpin-orbit and the fine structure

This is an opportunity to put QFT in action! Try to re-derive spin-orbit in a fully relativistic framework.

For the atomic Hamiltonian we’ve considered so far, L and S were good quantum numbers. Problem is: they are not…

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ˆ H = ˆ H 0 +μ B (L + gS) ⋅B +e2

8me

(B × ri )2

i

∑ + λS ⋅L

€

λ = Z 4e2h2

2πε0a03n3l(2l +1)(l +1)

€

Bso =μ 0I2r

=μ 0Zev4πr2 ≈

μ 0Zeh4πmer

3

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H so = −12

(gμ BmsS) ⋅Bso = λS ⋅L

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H so = ΛS ⋅LFor multi-electron atoms:

Where the sign of Lambda depends on the shell occupancy.

Spin-orbit in the multi-electron atomSpin-orbit in the multi-electron atom

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H so = ΛS ⋅L = ±λ2S

S ⋅L

For multi-electron atoms:

Where the sign of the energy depends on the shell occupancy (see table).

This justifies Hund’s third rule, whenever spin-orbit is a significant perturbation.

If spin-orbit dominates (large atomic number, as it goes as Z4), the L-S coupling scheme fails. Alternative: j-j coupling.

Composition of angular momentumComposition of angular momentum

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J = L +S

Possibilities:J=L+S, L+S-1…|L-S|

How many?

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(2J +1) = (2L +1)(2S +1)J=|L−S|

L+S

∑

Without spin-orbit, L and S are good quantum numbers (i.e. L and S are conserved), and J is not useful.

With spin-orbit, L and S are not good quantum numbers (i.e. L and S are not conserved, although L2, S2 and J2 are), and J becomes important. States are |L,S,J,MJ>

Summary and exampleSummary and example

Fine structure of the Co2+ ion: 3d7: S=3/2, L=3, J=9/2, gJ=5/3

Data and comparison (4f and 3d)Data and comparison (4f and 3d)

Hund’s rules seem to work well for 4f ions. Not so for many 3d ions. Why?

How do we measure the effective moment?

Origin of crystal fieldsOrigin of crystal fields

When an ion is part of a crystal, the surroundings (the crystal field) play a role in establishing the actual electronic structure (energy levels, degeneracy lifting, orbital “shapes” etc.).

Not good any longer!

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ˆ H = ˆ H 0 + ˆ H so + ˆ H cf + ˆ H Z

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Vcf =1

4πε0

ρ (r)r − r'

∫ d 3r'

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ˆ H cf = ρ 0(r)∫ Vcf (r)d 3r

A new set of orbitalsA new set of orbitals

Octahedral

Tetrahedral

Crystal field splitting; low/high spin statesCrystal field splitting; low/high spin states

The crystal field results in a new set of orbitals where to distribute electrons. Occupancy, as usual, from the lowest to the highest energy. But, crystal field acts in competition with the remaining contributions to the Hamiltonian. This drives occupancy and may result in low-spin or high-spin states.

Orbital quenchingOrbital quenching

Examine again the 3d ions. We notice a peculiar trend: the measured effective moment seems to be S-only. L is “quenched”. This is a consequence of the crystal field and its symmetry.

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Vcf =1

4πε0

ρ (r)r − r'

∫ d 3r'

Is real. No differential (momentum-related) operators. Hence, we need real eigenfunctions. Therefore, we need to combine ml states to yield real functions. This means, combining plus or minus ml, which gives zero net angular momentum.

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px ∝ l =1,ml =1 + l =1,ml = −1

dxy ∝ l = 2,ml = 2 − l = 2,ml = −2

dx2 −y2 ∝ l = 2,ml = 2 + l = 2,ml = −2

dz 2 ∝ l = 2,ml = 0

Examples

Jahn-Teller effectJahn-Teller effect

In some cases, it may be energetically favorable to shuffle things around than to squeeze electrons within degenerate levels.

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δEJT ∝ −Aε + Bε 2

Sneak peekSneak peek

μ1

μ2

Interactions

Ferromagnetism (Weiss)

Wrapping upWrapping up

Next lecture: Friday February 11, 8:15, KU

Interactions (MB)

•Temperature dependencies•Curie Law•Van Vleck paramagnetism•Hund’s rules•Spin-orbit•Crystal field•Orbital quenching•Jahn-Teller distortions