Magnetic susceptibility of different non ferromagnets

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Magnetic susceptibility of different non
ferromagnets
Free spin paramagnetism
?
Van Vleck
Pauli (metal)
T
Diamagnetism (filled shell)
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Diamagnetism of atoms

• ? in CGS for He, Ne, Ar, Kr and Xe are -1.9,
  -7.2,-19.4, -28, -43 times 10-6 cm3/mole.
• ? is negative, this behaviour is called
  diamagnetic.

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Simple theory of the diamagnetism

• Under a magnetic field, there is a change in the
  angular frequency, the change in the centrigual
  force is, m (?0? ?) 2 R-m?02R¼ 2?? ?0 R. This is
  balanced by the force due to the external field,
  e ?0 R B.
• Equating these two forces, we get ? ? e B/2m

B
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Simple diamagnetism

• The current I charge (revolution per unit
  time)(-Ze)(eB/2m)/2?.
• The magnetic moment /atom ?area current -R2
  Ze2 B/4m.
• The magnetic susceptibility is ? - R2 Ze2 /4m

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Quantum treatment

• H(p-eA/c)2/2m.
• EltHgtltp2-2eA? p/ce2 A2/c2gt/2m (p? A0).
• Ar? B/2 Eltp2?eB?L/ice2r?2B2 /4c2 gt/2m.
• For ltLgt0, ltMgt-? E/? B -e2 ltr ? 2 gt B/c2gt/4m
  
• ?ltMgt/B -e2ltr2gt/6mc2 ltr ? 2 gt 2ltr2gt/3

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Homework (1)

• The ground state wavefunction of the hydrogen
  atom is ?e-r/a0(? a03)-1/2 where a00.53 A. What
  is ltr2gt? What is the susceptibility?

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Van Vleck paramagnetism

• This comes from the change of the electronic
  state caused by the external field.
• ? ??j jgtltjg?B J B0gt/? E0j.
• ? ltMgtlt0g?B J? ?gt c.c.

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Homework (2) Van Vleck paramagnetism for Eu3

• Eu3 has 6 f (l3) electrons, from Hunds rule,
  work out the total L, S and J of the ground
  state.
• What is the magnetic moment ltGMGgt of the ground
  state? (M?B(L2S))
• What is the average squared moment ltGM2Ggt?
• Show that ltGM2GgtltGM1gt ? lt1MGgt where 1gt is
  the first excited state.

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Homework

• Assume an energy gap ?255 cm-1, what is the
  Van-Vleck susceptibility for Eu3?

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Hunds rule

• In an atom, because of the Coulomb interaction,
  the electrons repel each other. A simple rule
  that captures this says that the energy of the
  atom is lowered if
• S is maximum
• L is maximum consistent with S
• JL-S for less than half-filled LS for more
  than half filled.

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Illustration Of Hunds rule

• Mn2 has 5 d (l2) electrons, it is possible to
  have all spins up, S5/2. From exclusion
  principle, the orbital wave function has to be
  all different mL-2, -1, 0, 1, 2. This
  completely antisymmetric orbital function
  corresponds to L0, J5/2.
• Ce3 has 1 f electron. S1/2, L3, JL-S5/2.

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Pauli paramagnetism

• For metals, the up and down electrons differ by
  an energy caused by the external field, yet their
  Fermi energies are the same. Some spin up
  electrons are converted into spin down electrons.

EF
up
down
2 ?B B
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Metal paramagnetism

• M?B(? N-? N-) 3N?B2 B/2k TF.
• N? ?0? de f(e-?? B) D(e)/2
• M?(N-N-)? ?0? de f(e-? B)-f(e? B) D(e)/2
  -?2 B ?0? de (? f(e)/? e) D(e)/2 .
• ?M/B?2N(EF).

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Ferromagnetism

• At the Curie Temperature Tc, the magnetism M
  becomes zero.
• Tc is mainly determined by the exchange J.
• As T approaches Tc, M approaches zero in a power
  law manner (critical behaviour).

M
Tc
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Coercive behaviour

• Hc, the coercive field, is mainly determined by
  the anisotropy constant (both intrinsic and
  shape.)

Hc
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Mean field theory of ferromagnetism

• Eexch-J?i,? Si SI?-?iSiHeff,i.
• Heff,i?? JSi?
• ltHeffgt??-JltSgt-zJltSgt.
• P(S)/ exp(-S Heff/kT).
• For continuous spins ltSgts P(S) S dS/s P(S) dS.
• For spin ½, ltSgt?m P(m) m/?m P(m)

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For spin 1/2

• Considet Z?m P(m)2 cosh (x) where xzJltSgt/2kB
  T.
• ltSgtd ln Z/2dx
• ltSgttanh(x)/2.
• This is a nonlinear equation that need to be
  solved numerically in general.

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General graphical solution
tanh(cltSgt/T)
higher T
ltSgt
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Curie Temperature Tc

• ltSgt goes to zero at Tc. Near Tc, xltlt1,
• tanh(x)¼ xx3/3.
• The self-consistent equation becomes
  ltSgtxzJltSgt/4kB Tc.
• Hence TczJ/4kB.

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Critical behaviour near Tc

• tanh(x)¼ xx3/3.
• For T Tc-?, ltSgtyltSgty3ltSgt3/3 where yzJ/4kB T
• 3(1-y)0.5ltSgt
• ltSgt/ ?0.5.
• In general ltSgt/ ??.
• In the mean field approximation, the critical
  exponent ?1/2.

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ltSgt(T)
1/2
ltSgt
T
Tc
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Similar results hold for continuous orientation
of the spins

• Consider the partition function Zs dS
  exp(-HeffS)s-11 d cos(?) expx cos(?) where
  xzJltSgt/kBT.
• We find Z 2 sinh(x)/x.
• ltSgtd ln Z/dx.
• ltSgt2cosh(x)/x-sinh(x)/x2.
• This is a nonlinear equation that needs to be
  solved.

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Coherent rotation model of coercive behaviour

• E-K cos2 (?)MH cos(? -?0).
• ?E/??0 ?2E/?2?0.
• ?E/?? K sin 2(?)-MH sin(? -?0).
• K sin 2?MH sin(? -?0).
• ?2E/?2?2K cos 2(?)-MH cos(? -?0).
• 2K cos 2?MH cos(? -?0).

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Coherent rotation

• K sin 2?MHc sin(? -?0).
• K cos 2?MHc cos(? -?0)/2.
• Hc(?0)(2K/ M)1-(tan?0)2/3 (tan?0)4/3 0.5 /
  (1(tan?0)2/3).

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Special case ?00

• Hc02K/M.
• This is a kind of upper limit to the coercive
  field. In real life, the coercive field can be a
  1/10 of this value because the actual behaviour
  is controlled by the pinning of domain walls.

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Special case ?00, finite T, HltHc

• Hc2K/M.
• In general, at the local energy maximum, cos
  ?mMH/2K.
• Emax -K cos2 ?m MH cos ?m (MH)2/4K.
• E0E(?0)-KMH
• For Hc-H?, UN(Emax-E0)NM2?2/4K.
• Rate of switching, P ? exp(-U/kBT) where ? is
  the attempt frequency

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Special case ?00, H_c(T)

• Hc02K/M.
• For Hc0-H?, UNM2?2/4K.
• Rate of switching, P ? exp(-U/kBT).
• Hc(T) determined by P ? ¼ 1. We get Hc(T)Hc0-4K
  kB T ln(??)/NM20.5
• In general Hc0-Hc(T)/ T?. For ?00, ?1/2 for ?0
  ? 0, ?3/2