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## PowerPoint Slideshow about 'Magnetism' - lixue

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### Last lecture we talk about J a little bit. We discuss the other contribution next

Spin model: Each site has a spin Si

- 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 Eexch, the anisotropy energy Eaniso, the dipolar energy Edipo and the interaction with the external field Eext.

Dipolar interaction

- The dipolar interaction is the long range magnetostatic interaction between the magnetic moments (spins).
- Edipo=(1/40)i,j MiaMjbiajb(1/|Ri-Rj|).
- Edipo=(1/40)i,j MiaMjb[a,b/R3-3Rij,aRij,b/Rij5]
- 0=4£ 10-7 henrys/m
- For cgs units the first factor is absent.

Interaction with the external field

- Eext=-gB 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 10-3Oe (B is in units of G); units of H
- 1 Wb/m=(1/4) 1010 G cm3 ; units of M (emu)

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
- Eaniso=-Ki Siz2

Orders of magnitude

- For Fe, between atomic spins
- J¼ 522 K
- K¼ 0.038 K
- Dipolar interaction =(gB)2/a3¼ 0.254 K
- gB¼ 1.45£ 10-4 K/Gauss

First: Hext

H other contribution nextext g factor

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

Atoms other contribution next

- 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 Eext=-gB<Jz> by the Wigner-Eckert theorem.

Derivation of the orbital contribution: g other contribution nextL=1

- E=-H¢ M.
- The orbital magnetic moment ML= area x current/c; area= R2; current=e/(2) where is the angular velocity. Now L=m R2=l~. Thus ML= emR2 /(cm2 )= -0 I e/(2mc). Recall B=e/2mc
- M=B l.
- The spin contribution is MS=2B S
- Here S does not contain the factor of ~

R

Summary other contribution next

- E=-M¢ H
- M=B( gL L+gs S) where gL=1, gS=2; the spin g factor comes from Dirac’s equation.
- We want <j,m|Jz|j,m>.
- One can show that <j,m|M|j,m>=g <j,m|J|j,m> 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)].

Calculation of g other contribution next

- M=L+2S=J+S
- <j,m|J¢M|j,m>=j’,m’<j,m|J|j’m’><j’,m’|M|j,m> = gj’,m’<j,m|J|j’m’> <j’,m’|J|j,m>= g <j,m|J2|j,m> =g j (j+1).
- gj(j+1)=<j,m|J¢ M|j,m>=<j,m|J2+J¢ S|j,m>=j(j+1)+<j,m|J¢ S|j,m>.
- g=1+<j,m|J¢ S|j,m>/j(j+1).

Calculation of g in atoms other contribution next

- L=(J-S); L2=(J-S)2=J2+S2-2J¢ S.
- <J¢ S>=<(J2+S2-L2)>/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)

Another examples: in semiconductors, k other contribution next¢ p perturbation theory

- The wave function at a small wave vector k is given by = exp(ik¢ r)uk(r) where u is a periodic function in space.
- The Hamiltonian H=-~2r2/2m+V(r). The equation for u becomes [-~2r2/2m+V-~ k¢ p/2]u=Eu where the k2 term is neglected.

G factor in semiconductors other contribution next

- The extra term can be treated as perturbation from the k=0 state, the energy correction is
- Dijkikj= <|kipi|><|kjpj|>/[E-E]
- In a magnetic field, k is replaced p-eA/c.
- The equation for u becomes H’u=Eu;
- H’= Dij(pi-eAi/c)(pj-eAj/c)-B¢ B). Since A=r£ B/2, the Dij term also contains a contribution proportional to B.

Calculation of g other contribution next

- H’=H1+…; H1= (e/c)p D A+A D p.
- Since A=r£ B/2, H1= (e/2c)p D (r B)+(r B) D p.
- A B C=A B C, for any A, B, C; so H1= (e/2c)(pDrB -BrDp )=gBB
- g= m(pDr - rDp)/.
- Note pirj=ij/im+rjpi
- gj= /i Diljli+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.

g=D other contribution nextA /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

Effect of the dipolar interaction: Shape anisotropy other contribution next

- 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 M02 [1/i3-3 cos2 /i3]/40 a3=A-B cos2 .
- 1/i3=(3)¼ 1.2
- E=-Keff cos2(), Keff=1.2 M02/40.

Paramagnetism: J=0 other contribution next

- 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 Mn2+, etc, or a gas of atoms.

Magnetic susceptibility of different non ferromagnets other contribution next

Free spin paramagnetism

Van Vleck

Pauli (metal)

T

Diamagnetism (filled shell)

Boltzmann distribution other contribution next

- Probability P/ exp(-U/kB T)
- U=-gB B ¢ J
- P(m)/ exp(-gB B m/kBT)
- <M>=NB gm 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.

<M> and other contribution next

- We get <M>=NgB[ exp(-x)-exp(x)]/2[exp(-x)+exp(x)] where x=gB B/(2kBT).
- Consider the high temperature limit with x<<1, <M>¼ N gB x/2.
- We get =N(gB )2/2kT
- At low T, x>>1, <M>=NgB/2, as expected.

More general J other contribution next

- Consider the function Z= m=-jm=j exp(-mx)
- For a general geometric series 1+y+y2+…yn=(1-yn+1)/(1-y)
- We get Z=sinh[(j+1/2)x]/sinh(x/2).
- <M>=-d ln Z/dx=NgB[(j+1/2) coth[(j+1/2)x]-coth(x/2)/2].

Diamagnetism of atoms other contribution next

- 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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