Magnetic susceptibility of different
non ferromagnets
Free spin paramagnetism

Van Vleck
Pauli (metal)
T
Diamagnetism (filled shell)
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.
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 Rm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
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
Quantum treatment
•
•
•
•

H=(p-eA/c)2/2m.
E=<H>=<[p2-2eA p/c+e2 A2/c2]>/2m (p A=0).
A=r B/2; E=<[p2+eBL/ic+e2r2B2 /4c2 ]>/2m.
For <L>=0, <M>=- E/ B= -e2 <r  2 > B/c2]>/4m;
=<M>/B= -e2<r2>/6mc2; <r  2 > =2<r2>/3
Homework (1)
• The ground state wavefunction of the
hydrogen atom is =e-r/a0( a03)-1/2 where
a0=0.53 A. What is <r2>? What is the
susceptibility?
Van Vleck paramagnetism
• This comes from the change of the
electronic state caused by the external
field.
•  =j |j><j|gB J¢ B|0>/ E0j.
•  <M>=<0|gB J| > +c.c.
Homework (2): Van Vleck
paramagnetism for Eu3+
• Eu3+ has 6 f (l=3) electrons, from Hund’s
rule, work out the total L, S and J of the
ground state.
• What is the magnetic moment <G|M|G> of
the ground state? (M=B(L+2S))
• What is the average squared moment
<G|M2|G>?
• Show that <G|M2|G>=<G|M|1>  <1|M|G>
where |1> is the first excited state.
Homework
• Assume an energy gap =255 cm-1, what
is the Van-Vleck susceptibility for Eu3+?
Hund’s 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
• J=|L-S| for less than half-filled; L+S for
more than half filled.
Illustration Of Hund’s rule
• Mn2+ has 5 d (l=2) electrons, it is possible
to have all spins up, S=5/2. 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 L=0, J=5/2.
• Ce3+ has 1 f electron. S=1/2, L=3, J=|LS|=5/2.
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
down
up
2 B B
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).
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
Coercive behaviour
• Hc, the coercive
field, is mainly
determined by
the anisotropy
constant (both
intrinsic and
shape.)
Hc
Mean field theory of
ferromagnetism
•
•
•
•
•
Eexch=-Ji, Si¢ SI+=-iSiHeff,i.
Heff,i= –JSi+
<Heff>=-J<S>=-zJ<S>.
P(S)/ exp(-S¢ Heff/kT).
For continuous spins <S>=s P(S) S dS/s
P(S) dS.
• For spin ½, <S>=m P(m) m/m P(m)
For spin 1/2
• Considet Z=m P(m)=2 cosh (x) where
x=zJ<S>/2kB T.
• <S>=d ln Z/2dx
• <S>=tanh(x)/2.
• This is a nonlinear equation that need to
be solved numerically in general.
General graphical solution
tanh(c<S>/T)
higher T
<S>
Curie Temperature Tc
• <S> goes to zero at Tc. Near Tc, x<<1,
• tanh(x)¼ x+x3/3.
• The self-consistent equation becomes
<S>=x=zJ<S>/4kB Tc.
• Hence Tc=zJ/4kB.
Critical behaviour near Tc
• tanh(x)¼ x+x3/3.
• For T = Tc-, <S>=y<S>+y3<S>3/3; where
y=zJ/4kB T
• [3(1-y)]0.5=<S>
• <S>/ 0.5.
• In general <S>/ .
• In the mean field approximation, the
critical exponent =1/2.
<S>(T)
1/2
<S>
Tc
T
Similar results hold for continuous
orientation of the spins
• Consider the partition function Z=s dS
exp(-HeffS)=s-11 d cos() exp[x cos()]
where x=zJ<S>/kBT.
• We find Z= 2 sinh(x)/x.
• <S>=d ln Z/dx.
• <S>=2[cosh(x)/x-sinh(x)/x2].
• This is a nonlinear equation that needs to
be solved.
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).
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).
Special case: 0=0
• Hc0=2K/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.
Special case: 0=0, finite T, H<Hc
• Hc=2K/M.
• In general, at the local energy maximum,
cos m=MH/2K.
• Emax= -K cos2 m +MH cos m= (MH)2/4K.
• E0=E(=0)=-K+MH
• For Hc-H=, U=N(Emax-E0)=NM22/4K.
• Rate of switching, P = exp(-U/kBT) where
 is the attempt frequency
Special case: 0=0, H_c(T)
•
•
•
•
Hc0=2K/M.
For Hc0-H=, U=NM2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()/NM2]0.5
• In general Hc0-Hc(T)/ T. For 0=0, =1/2;
for 0  0, =3/2