1. Magnetism How to describe the physics: (1)Spin model (2)In terms of electrons

2. Spin model: Each site has a spin S i There is one spin at each site. The magnetization is proportional to the sum of all the spins. The total energy is the sum of the exchange energy E exch, the anisotropy energy E aniso, the dipolar energy E dipo and the interaction with the external field E ext.

3. Dipolar interaction The dipolar interaction is the long range magnetostatic interaction between the magnetic moments (spins). E dipo =(1/4  0 )  i,j M ia M jb  ia  jb (1/|R i -R j |). E dipo =(1/4  0 )  i,j M ia M jb [  a,b /R 3 - 3R ij,a R ij,b /R ij 5 ]  0 =4  £ 10 -7 henrys/m For cgs units the first factor is absent.

4. Interaction with the external field E ext =-g  B H S=-HM We have set M=  B S. H is the external field,  B =e~/2mc is the Bohr magneton (9.27£ 10 -21 erg/Gauss). g is the g factor, it depends on the material. 1 A/m=4  times 10 -3 Oe (B is in units of G); units of H 1 Wb/m=(1/4  ) 10 10 G cm 3 ; units of M (emu)

5. Anisotropy energy The anisotropy energy favors the spins pointing in some particular crystallographic direction. The magnitude is usually determined by some anisotropy constant K. Simplest example: uniaxial anisotropy E aniso =-K  i S iz 2

6. Orders of magnitude For Fe, between atomic spins J¼ 522 K K¼ 0.038 K Dipolar interaction =(g  B ) 2 /a 3 ¼ 0.254 K g  B ¼ 1.45£ 10 -4 K/Gauss

7. Last lecture we talk about J a little bit. We discuss the other contribution next First: H ext

8. H ext g factor We give two examples of the calculation of the g factor ， the case of a single atom and the case in semiconductors.

9. Atoms In an atom, the electrons have a orbital angular momentum L, a spin angular momentum S and a total angular momentum J=L+S. The energy in an external field is given by E ext =-g  B by the Wigner-Eckert theorem.

10. Derivation of the orbital contribution: g L =1 E=-H¢ M. The orbital magnetic moment M L = area x current/c; area=  R 2 ; current=e  /(2  ) where  is the angular velocity. Now L=m  R 2 =l~. Thus M L =  emR 2  /(cm2  )= -  0 I  e/(2mc). Recall   =e  /2mc M=  B l. The spin contribution is M S =2  B S Here S does not contain the factor of ~ R

11. Summary E=-M¢ H M=  B ( g L L+g s S) where g L =1, g S =2; the spin g factor comes from Dirac’s equation. We want. One can show that =g for some constant g (W-E theorem). We derive below that g=1+[j(j+1)+s(s+1)-l(l+1)]/[2j(j+1)].

12. Calculation of g M=L+2S=J+S =  j’,m’  = g  j’,m’  = g =g j (j+1). gj(j+1)= = =j(j+1)+. g=1+ /j(j+1).

13. Calculation of g in atoms L=(J-S); L 2 =(J-S) 2 =J 2 +S 2 -2J¢ S. = /2= [j(j+1)+s(s+1)- l(l+1)]/2. Thus g=1+ [j(j+1)+s(s+1)-l(l+1)]/2j(j+1)

14. Another examples: in semiconductors, k¢ p perturbation theory The wave function at a small wave vector k is given by  = exp(ik¢ r)u k (r) where u is a periodic function in space. The Hamiltonian H=-~ 2 r 2 /2m+V(r). The equation for u becomes [-~ 2 r 2 /2m+V-~ k¢ p/2]u=Eu where the k 2 term is neglected.

15. G factor in semiconductors The extra term can be treated as perturbation from the k=0 state, the energy correction is  D ij k i k j =  /[E  -E  ]  In a magnetic field, k is replaced p-eA/c.  The equation for u becomes H’u=Eu;  H’=  D ij (p i -eA i /c)(p j -eA j /c)-  B  ¢ B). Since A=r£ B/2, the D ij term also contains a contribution proportional to B.

16. Calculation of g H’=H 1 +…; H 1 =  e/c)  p  D  A+A  D  p. Since A=r£ B/2, H 1 =  e/2c)  p  D  (r  B)+(r  B)  D  p. A  B  C=A  B  C, for any A, B, C; so H 1 =  e/2c)  (p  D  r  B -B  r  D  p )=g  B  B g= m  (p  D  r - r  D  p)/ . Note p i r j =   ij /im+r j p i g j =  /i  D il  jli +O(p) where  ijk =  1 depending on whether ijk is an even or odd permutation of 123; otherwise it is 0; repeated index means summation.

17. g=D A /i g_z=(D_{xy}-D_{yx})/i, the antisymmetric D. g is inversely proportional to the energy gap. For hole states, g can be large

18. Effect of the dipolar interaction: Shape anisotropy Example: Consider a line of parallel spins along the z axis. The lattice constant is a. The orientation of the spins is described by S=(sin , 0, cos  ). The dipolar enegy /spin is M 0 2  [1/i 3 -3 cos 2  /i 3 ]/4  0 a 3 =A- B cos 2 .  1/i 3 =  (3)¼ 1.2 E=-K eff cos 2 (  ), K eff =1.2 M 0 2 /4  0.

20. Paramagnetism: J=0 Magnetic susceptibility:  =M/B (  0 ) We want to know  at different temperatures T as a function of the magnetic field B for a collection of classical magnetic dipoles. Real life examples are insulating salts with magnetic ions such as Mn 2+, etc, or a gas of atoms.

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

22. Boltzmann distribution Probability P/ exp(-U/k B T) U=-g  B B ¢ J P(m)/ exp(-g  B B m/k B T) =N  B g  m P(m) m/  m P(m) To illustrate, consider the simple case of J=1/2. Then the possible values of m are - 1/2 and 1/2.

23. and  We get =Ng  B [ exp(-x)-exp(x)]/2[exp(- x)+exp(x)] where x=g  B B/(2k B T). Consider the high temperature limit with x ¼ N g  B x/2. We get  =N(g  B ) 2 /2kT At low T, x>>1, =Ng  B /2, as expected.

24. More general J Consider the function Z=  m=-j m=j exp(-mx) For a general geometric series 1+y+y 2 +…y n =(1-y n+1 )/(1-y) We get Z=sinh[(j+1/2)x]/sinh(x/2). =-d ln Z/dx=Ng  B [(j+1/2) coth[(j+1/2)x]-coth(x/2)/2].

25. 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.