1. Micromagnetics 101

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. Exchange energy E exch =-J   S i ¢ S i+  The exchange constant J aligns the spins on neighboring sites . If J>0 (<0), the energy of neighboring spins will be lowered if they are parallel (antiparallel). One has a ferromagnet (antiferromagnet

4. Magnitude of J k B T c /zJ¼ 0.3 Sometimes the exchange term is written as A s d 3 r |r M(r)| 2. A is in units of erg/cm. For example, for permalloy, A= 1.3 £ 10 -6 erg/cm

5. 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)

6. Dipolar interaction The dipolar interaction is the long range magnetostatic interaction between the magnetic moments (spins). E dipo =  i,j M ia M jb [  a,b /R 3 -3R ij,a R ij,b /R ij 5 ] E dipo =  i,j M ia M jb  ia  jb (1/|R i -R j |). E dipo =s r¢ M( R) r¢ M(R’)/|R-R’| If the magnetic charge q M =-r¢ M is small E dipo is small

7. 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

8. Modifies Landau-Gilbert equation  M /  t -  M£H + r¢J m = -  M/   is the thermal noise. Ordinarily the magnetization current J m is zero. H is a sum of contributions from the exchange, H ex ; the dipolar H dipo, the anisotropy and the external field: H=H e + H ex + H dipo +H an ; H ex =Jr  M; H an =K M .

9. Some mathmatical challenges The dipolar field is long range: different scheme has been developed to take care of this. These include using fast Fourier transforms or using the magnetostatic potential. For large systems, the implicit scheme takes a lot of memory. Preconditioner: Just the exchange. (it is sparse.) Physically the exchange energy is usually the largest term.

10. Alternative approach Monte Carlo simulation with the Metroplois algoraithm. This is the same as solving the master equation: dP/dt=TP where T is the transition matrix.

11. Physical understanding

12. Three key ideas at finite temperatures: Nucleation Depinning Spins try to line up parallel to the edge because of the dipolar interaction. The magnetic charge is proportional to, and this is reduced.

13. Approximation Minimize only the exchange and the anisotropy energy with the boundary condition that the spins are parallel to the edge.

14. Two dimension: A spin is characterized by two angles  and . In 2D, they usually lie in the plane in order to minimize the dipolar interaction. Thus it can be characterized by a single variable . The configurations are then obtained as solutions of the imaginary time Sine- Gordon equation r 2  +(K/J) sin  =0 with the “parallel edge” b.c.

15. Edge domain: Simulation vs Analytic approximation.  =tan -1 [sinh(  v(y’- y’ 0 ))/(- v sinh(  (x’- x’ 0 )))], y’=y/l, x’=x/l; the magnetic length l=[J/2K] 0.5;  =1/[1+v 2 ] 0.5 ; v is a parameter.

16. Closure domain: Simulation vs analytic approximation  =tan -1 [A tn(  x', f ) cn(v [1+k g 2 ] 0.5 y', k 1g )/ dn(v [1+k g 2 ] 0.5 y', k 1g )], k g 2 =[A 2  2 (1-A 2 )]/[  2 (1-A 2 ) 2 -1], k 1g 2 =A 2  2 (1-A 2 )/(  2 (1-A 2 )-1),  f 2 =[A 2 +  2 (1-A 2 ) 2 ]/[  2 (1-A 2 )] v 2 =[  2 (1-A 2 ) 2 -1]/[1-A 2 ]. The parameters A and  can be determined by requiring that the component of S normal to the surface boundary be zero

17. For Permalloy For an important class of magnetic material, the intrinsic anisotropy constant is very small. r 2  =0. For this case, conformal mapping ideas are applicable.

18. An example Constraint: M should be parallel to the boundary! For the circle, a simple solution is  =tan -1 y/x. Conformal mapping allows us to get the corresponding solution for the rectangle.

19. Current directions: Current induced torque Magnetic random access memory

20. Nanopillar Technique (Katine, Albert, Emley) -Multilayer film deposited (thermal evaporation, sputtering) on insulating substrate Au (10 nm) Co (3 nm) Cu (6 nm) Co (40 nm) Cu (80 nm) -Current densities of 10 8 A/cm 2 can be sent vertically through pillar -Electron-beam lithography, ion milling form pillar structure (thicker Co layer left as extended film) -Polyimide insulator deposited and Cu top lead connected to pillar Polyimide insulator Cu

21. Magnetic Reversal Induced by a Spin-Polarized Current Large (~10 7 -10 9 A/cm 2 ) spin-polarized currents can controllably reverse the magnetization in small (< 200 nm) magnetic devices Parallel (P) Antiparallel (AP) Ferromagnet 1 Ferromagnet 2 Nonmagnetic Cornell THALES/Orsay NIST Positive Current

22. Modifies Landau-Gilbert equation  M /  t -  M£H + r¢J m = -  M/  The magnetization current J m is nonzero.

23. Charge and magnetization current J e =-  r V -e Dr  n -D M r (  M¢ p 0 ) J=-  M r ( V p 0 ) - D M ' r  M - D' r (  n p 0 ) p 0 =M 0 /|M 0 |; M 0 is the local equilibrium magnetization, V=-Er+W; W(r)=s d 3 r'  n(r')/|r-r’|

24. Two perpendicular wires generate magnetic felds H x and H y Bit is set only if both Hx and Hy are present. For other bits addressed by only one line, either H x or H y is zero. These bits will not be turned on.

25. Coherent rotation Picture The switching boundaries are given by the line AC, for example, a field at X within the triangle ABC can write the bit. If H x =0 or H y =0, the bit will not be turned on. Hx Hy A B C X

26. Bit selectivity problem: Very small (green) “writable” area Different curves are for different bits with different randomness. Cannot write a bit with 100 per cent confidence.

27. Another way recently proposed by the Motorola group: Spin flop switching Electrical current required is too large at the moment

28. Simple picture from the coherent rotation model M1, M2 are the magnetizations of the two bilayers. The external magnetic fields are applied at -135 degree, then 180 degree then 135 degree.

29. Magnetization is not uniform: coherent rotation model is not enough