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# Spin

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.

## Spin

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### Presentation Transcript

1. Spin Electrons have “spin”. But this is not easy to visualize, as electrons also act like somewhat fuzzy particle distributions with no internal structure…

2. So why do we attribute spin? Stern-Gerlach experiment measures electron magnetic moment Two lines imply only two possible moment values/orientations

3. A ‘classic’ quantum experiment

4. Up and Down spins But what determines ‘up’ and ‘down’?

5. 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?

6. 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 !!!

7. We thus conjecture… [Sx, Sy] = ASz and circular permutations But what is A?

8. 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>

9. Spin algebra |s,sz > Sz|½,½> = (ħ/2)|½,½> |> |> Sz|½,-½> = -(ħ/2)|½,-½> |> |>

10. 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.

11. Thus S+|½,-½> = ħ|½, ½> S-|½, ½> = ħ|½,-½> |> |> |> |> S+|½, ½> = 0 S-|½,-½> = 0 |> |>

12. 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] |> =

13. Completeness |  > <  | + |  > <  | = I 0 1 <| = [1 0] 0 1 |> = <| = [0 1] |> =

14. 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

15. 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

16. 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

17. What about arbitrary orientations? ? |q,j> =

18. 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

19. Arbitrary orientations n = (sinqcosj, sinqsinj, cosq) cosq/2 -sinq/2 eijcosq/2 eijsinq/2 |n > = |n > = Upto an arbitrary overall global phase factor

20. 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 !!!

21. Only after 2 rotations… q=4p, j=0, recover initial states Spinors: Pseudo-rotation and Berry phase

22. Doing a rotation operation… First try a translation !! Y(x+a) = Y(x) + aY/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

23. 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 !!

24. Symmetry and Pauli Exclusion http://dslrs.net/wp-content/uploads/2013/05/coffee-photo-1.jpg

25. 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

26. Identical particles P(1,2) = P(2,1) |Y(1,2)|2 = |Y(2,1)|2 Y(1,2) = ± Y(2,1)

27. 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)

28. 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

29. Rotation operator Y(q+q0) = eq0d/dqY(q) = eiq0L.n/ħY(q) R(q) = eiq0Ln/ħ

30. 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)

31. Electrons Y(1,2) = -Y(2,1) Y(1,1) = -Y(1,1) = 0 (Pauli Exclusion) But how to account for this correctly ?

32. Hartree SCF approach Y(1,2) = Fa(r1) Fb(r2) Does not satisfy exchange symmetry

33. Hartree Fock approach Fa(r2) Fb(r1)]/ √2 Yab(1,2) = [Fa(r1) Fb(r2) -

34. Slater Determinant Fa(r1) Fb(r2) 1 Yab(1,2) = √2 Fa(r1) Fb(r2)

35. 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)

36. 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

37. Summary: Spin is a new variable. It acts like angular momentum Spin rotation imposes symmetry rules which gives Pauli exclusion

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