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Spin Electrons have “spin”. But this is not easy to visualize, as electrons also act like somewhat fuzzy particle distributions with no internal structure… 1 So why do we attribute spin? Stern-Gerlach experiment measures electron magnetic moment Two lines imply only two possible moment values/orientations 2 A ‘classic’ quantum experiment Up and Down spins But what determines ‘up’ and ‘down’? 4 Why ‘spin’ is hard to understand !! Start by filtering up/down spins Further sort them into left/right Classical determinism dictates that one of the branches consists of electrons that are left and up Is it still up? What does the measurement say? 5 Why ‘spin’ is hard to understand !! Instead of purely up electrons on the examined branch, we again get up and down. So ‘Up’ electrons lose their ‘upness’ when their left/right character is probed !!! 6 We thus conjecture… [Sx, Sy] = ASz and circular permutations But what is A? 7 Spin is like angular momentum Recall m can have (2l+1) values between –l and l. For spin, since only 2 streams measured, (2s+1) = 2, meaning s = ½ and ms = ±½ Call these states | > = |½, ½> and | > = |½,-½> 8 By analogy with [Lx,Ly] = iħLz [Sx, Sy] = iħSz and circular permutations 9 Spin algebra in z and z bases |s,sz > Sz |> = (ħ/2)|> Sz |> = -(ħ/2)|> 10 What is the operator representation of Sz? How does Sz behave on an arbitrary state? 11 Define an arbitrary state |A> = a|> + b|> Collect coefficients in up-down basis and write state |A> as a b We thus have, |> = 1 0 |> = 0 1 12 In the z and z basis, Sz is diagonal Sz = |> |> 1 0 |> ħ/2 0 -1 |> |> = Verify ! |> = 1 0 0 1 Sz |> = (ħ/2)|> Sz |> = -(ħ/2)|> 13 What about Sx and Sy ? [Sx, Sy] = iħSz and circular permutations In the z and z basis, Sx,y are off-diagonal 14 Easy to check Sx = ħ/2 0 1 1 0 Sy = ħ/2 0 -i i 0 Sz = ħ/2 1 0 0 -1 Pauli Matrices sx,y,z S = ħs/2 All S eigenvalues are ±ħ/2 15 In the z and z basis, what are x and x ? Expect them to be 50-50 of z andz Eigenstates of Sx and Sy Sx = ħ/2 Sy = ħ/2 0 1 1 0 0 -i i 0 x = 1 /√2 1 x = 1 /√2 -1 y = 1 /√2 i y = 1 /√2 -i 16 Symmetry and Pauli Exclusion •http://dslrs.net/wp-content/uploads/2013/05/coffee-photo-1.jpg Electron spins Just multiply by 2 ? Pauli exclusion (like spins can’t sit atop each other) principle So each orbital state can hold 2 opposite spins Identical particles P(1,2) = P(2,1) |Y(1,2)|2 = |Y(2,1)|2 Y(1,2) = ± Y(2,1) Fermions vs Bosons Spin-Statistics theorem ½ integer spins are fermions Y(1,2) = -Y(2,1) Integer spins are bosons Y(1,2) = Y(2,1) Electrons Y(1,2) = -Y(2,1) Y(1,1) = -Y(1,1) = 0 (Pauli Exclusion) But how to account for this correctly ? Hartree SCF approach Y(1,2) = Fa(r1)Fb(r2) Does not satisfy exchange symmetry Hartree Fock approach Yab(1,2) = [Fa(r1)Fb(r2) - Fa(r2)Fb(r1)]/√2 Slater Determinant Yab(1,2) = 1 √2 Fa(r1) Fb(r2) Fa(r1) Fb(r2) Coulomb Term U = ∫d3r1d3r2|Y(1,2)|2/r12 = ∫d3r1d3r2|Fa(r1)Fb(r2) - Fa(r2)Fb(r1)|2/2r12 = ∫d3r1d3r2na(r1)nb(r2)/r12 –∫d3r1d3r2F*a(r1)F *b(r2)Fa(r2)Fb(r1)/r12 G(r1,r2) = UH – UF A nonlocal correction F = fnlm(r)Ylm(q,j)/r -ħ2/2m.d2f(r)/dr2 + [U(r) + l(l+1)ħ2/2mr2 + UH(r)]f(r) - ∫UF(r,r’)f(r’)dr’ = Ef(r) • Hund’s Rule • Ferromagnetism Summary: Spin is a new variable. It acts like angular momentum Spin rotation imposes symmetry rules which gives Pauli exclusion