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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…. So why do we attribute spin?. Stern-Gerlach experiment measures electron magnetic moment.

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## Spin

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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…**So why do we attribute spin?**Stern-Gerlach experiment measures electron magnetic moment Two lines imply only two possible moment values/orientations**Up and Down spins**But what determines ‘up’ and ‘down’?**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?**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 !!!**We thus conjecture…**[Sx, Sy] = ASz and circular permutations But what is A?**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>**Spin algebra**|s,sz > Sz|½,½> = (ħ/2)|½,½> |> |> Sz|½,-½> = -(ħ/2)|½,-½> |> |>**Other components?**Since Sx upsets Sz measurements, it can take a z-up spin into z-down and a z-down spin into z-up. Same for Sy. But let’s focus on a one-way jump, ie, operators that ONLY take say, down to up. Clearly this must be a mixture of Sx and Sy.**Thus**S+|½,-½> = ħ|½, ½> S-|½, ½> = ħ|½,-½> |> |> |> |> S+|½, ½> = 0 S-|½,-½> = 0 |> |>**Define an arbitrary state**Collect coefficients in up-down basis and write state |A> as |A> = a| > + b| > 0 a 1 0 b 1 <| = [1 0] We thus have, |> = <| = [0 1] |> =**Completeness**| > < | + | > < | = I 0 1 <| = [1 0] 0 1 |> = <| = [0 1] |> =**In this basis, matrix S’s…**Due to completeness of states |, > any operator M = M(| > < | + | > < |). Thus… Sz = ħ/2 0 0 1 0 0 1 0 1 -1 0 0 S+ = 0 + ħ|> <| = ħ S- = 0 + ħ |> <| = ħ 0**In this basis, matrix S’s…**Sx = ħ/2 Pauli Matrices sx,y,z 1 Sy = ħ/2 0 0 -i 0 1 1 i -1 0 0 Sz = ħ/2 0 S = ħs/2 All S eigenvalues are ±ħ/2**Writing ‘left-right’ states**ie, states pointing along x, say In original z-basis, these will correspond to eigenstates of Sx matrix -1 -1 1 1 1 i 1 i 1 1 1 1 |y > = |x > = |y > = |x > = √2 √2 √2 √2**What about arbitrary orientations?**? |q,j> =**What about arbitrary orientations?**Take a projection of S along the direction n Sn = S.n = Sx(sinqcosj) + Sy(sinqsinj) + Sz(cosq) = cosq sinqe-ij sinqeij -cosq Then find its eigenvectors**Arbitrary orientations**n = (sinqcosj, sinqsinj, cosq) cosq/2 -sinq/2 eijcosq/2 eijsinq/2 |n > = |n > = Upto an arbitrary overall global phase factor**Effect on rotation…**Start with initial condition q=0, j=0, with initial states 0 0 -1 -sinq/2 cosq/2 1 But after full rotation, q=2p, j=0, with final states 1 -1 eijcosq/2 0 0 eijsinq/2 |I > = |I > = |n > = |I > = |n > = |I > = Additional phase of p picked up !!!**Only after 2 rotations…**q=4p, j=0, recover initial states Spinors: Pseudo-rotation and Berry phase**Doing a rotation operation…**First try a translation !! Y(x+a) = Y(x) + aY/x + (a2/2)2Y/x2 + … = [exp(a/x)]Y(x) = [exp(iapx/ħ)]Y(x) By analogy: Y(q,j+j0) = [exp(ij0Lz/ħ)]Y(q,j), where Lz = -iħ/j**Thus, rotating a spin:**Y(sn+s0) = eis0sn/ħY(q,j)= eis0s.n/2Y(q,j) = [cos(s0/2) + i(s.n)sin(s0/2)] Y(q,j) Hence the half-angles !!**Symmetry and Pauli Exclusion**http://dslrs.net/wp-content/uploads/2013/05/coffee-photo-1.jpg**Electron spins**Just multiply by 2 ? Pauli exclusion principle(like spins can’t sit atop each other) 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)**Fermions vs Bosons**Spin-Statistics theorem Exchange is a 3600 rotation A moves around B 1800 B moves around A 1800 in the same direction**Rotation operator**Y(q+q0) = eq0d/dqY(q) = eiq0L.n/ħY(q) R(q) = eiq0Ln/ħ**Spin operator**Y(sn+q0) = eiq0s.n/ħY(sn), s = ħs/2 R(q) = eiq0s/ħ Spin-Statistics theorem ½ integer spins are fermions R(2p) = -1 Y(1,2) = -Y(2,1) Integer spins are bosons (including s = 0) R(2p) = 1 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**Fa(r2) Fb(r1)]/ √2 Yab(1,2) = [Fa(r1) Fb(r2) -**Slater Determinant**Fa(r1) Fb(r2) 1 Yab(1,2) = √2 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 = UH– UF G(r1,r2)**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) A nonlocal correction Hund’s Rule Ferromagnetism**Summary: Spin is a new**variable. It acts like angular momentum Spin rotation imposes symmetry rules which gives Pauli exclusion

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