1. Magnetic susceptibility of different non ferromagnets  T Free spin paramagnetism Van Vleck Pauli (metal) Diamagnetism (filled shell)

2. Diamagnetism of atoms  in CGS for He, Ne, Ar, Kr and Xe are - 1.9, -7.2,-19.4, -28, -43 times 10 -6 cm 3 /mole.  is negative, this behaviour is called diamagnetic.

3. 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  0 2 R¼ 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

4. Simple diamagnetism The current I= charge £ (revolution per unit time)=(-Ze)(eB/2m)/2 . The magnetic moment /atom  =area £ current =-R 2 Ze 2 B/4m. The magnetic susceptibility is  = - R 2 Ze 2 /4m

5. Quantum treatment H=(p-eA/c) 2 /2m. E= = /2m (p  A=0). A=r  B/2; E= /2m. For =0, =-  E/  B= -e 2 B/c 2 ]>/4m;  = /B= -e 2 /6mc 2 ; =2 /3

6. Homework (1) The ground state wavefunction of the hydrogen atom is  =e -r/a 0 (  a 0 3 ) -1/2 where a 0 =0.53 A. What is ? What is the susceptibility?

7. Van Vleck paramagnetism This comes from the change of the electronic state caused by the external field.   =  j |j> /  E 0j.  = +c.c.

8. Homework (2): Van Vleck paramagnetism for Eu 3+ Eu 3+ 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 of the ground state? (M=  B (L+2S)) What is the average squared moment ? Show that =  where |1> is the first excited state.

9. Homework Assume an energy gap  cm , what is the Van-Vleck susceptibility for Eu 3+ ?

10. 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.

11. Illustration Of Hund’s rule Mn 2+ 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: m L =-2, -1, 0, 1, 2. This completely antisymmetric orbital function corresponds to L=0, J=5/2. Ce 3+ has 1 f electron. S=1/2, L=3, J=|L- S|=5/2.

12. 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    B updown

13. Metal paramagnetism M=  B (  N + -  N - ) =3N  B 2 B/2k T F. 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=  2 N(E F ).

15. 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

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

17. Mean field theory of ferromagnetism E exch =-J  i,  S i ¢ S I+  =-  i S i H eff,i. H eff,i =   –JS i+  =   -J =-zJ. P(S)/ exp(-S¢ H eff /kT). For continuous spins =s P(S) S dS/s P(S) dS. For spin ½, =  m P(m) m/  m P(m)

18. For spin 1/2 Considet Z=  m P(m)=2 cosh (x) where x=zJ /2k B T. =d ln Z/2dx =tanh(x)/2. This is a nonlinear equation that need to be solved numerically in general.

19. General graphical solution tanh(c /T) higher T

20. Curie Temperature T c goes to zero at T c. Near T c, x<<1, tanh(x)¼ x+x 3 /3. The self-consistent equation becomes =x=zJ /4k B T c. Hence T c =zJ/4k B.

21. Critical behaviour near T c tanh(x)¼ x+x 3 /3. For T = T c - , =y +y 3 3 /3; where y=zJ/4k B T [3(1-y)] 0.5 = /  0.5. In general /  . In the mean field approximation, the critical exponent  =1/2.

22. (T) T TcTc 1/2

23. Similar results hold for continuous orientation of the spins Consider the partition function Z=s dS exp(-H eff S)=s -1 1 d cos(  ) exp[x cos(  )] where x=zJ /k B T. We find Z= 2 sinh(x)/x. =d ln Z/dx. =2[cosh(x)/x-sinh(x)/x 2 ]. This is a nonlinear equation that needs to be solved.

25. Coherent rotation model of coercive behaviour E=-K cos 2 (  )+MH cos(  -  0 ).  E/  =0;  2 E/  2  =0.  E/  = K sin 2(  )-MH sin(  -  0 ). K sin 2  =MH sin(  -  0 ).  2 E/  2  =2K cos 2(  )-MH cos(  -  0 ). 2K cos 2  =MH cos(  -  0 ).

26. Coherent rotation K sin 2  =MH c sin(  -  0 ). K cos 2  =MH c cos(  -  0 )/2. H c (  0 )=(2K/ M)[1-(tan  0 ) 2/3 +(tan  0 ) 4/3 ] 0.5 / (1+(tan  0 ) 2/3 ).

27. Special case:  0 =0 H c0 =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.

28. Special case:  0 =0, finite T, H<H c H c =2K/M. In general, at the local energy maximum, cos  m =MH/2K. E max = -K cos 2  m +MH cos  m = (MH) 2 /4K. E 0 =E(  =0)=-K+MH For H c -H= , U=N(E max -E 0 )=NM 2  2 /4K. Rate of switching, P = exp(-U/k B T) where is the attempt frequency

29. Special case:  0 =0, H_c(T) H c0 =2K/M. For H c0 -H= , U=NM 2  2 /4K. Rate of switching, P = exp(-U/k B T). H c (T) determined by P  ¼ 1. We get H c (T)=H c0 -[4K k B T ln(  )/NM 2 ] 0.5 In general H c0 -H c (T)/ T . For  0 =0,  =1/2; for  0  0,  =3/2