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. |s,sz > |> |> |> |> Spin algebra

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>

12. Define an arbitrary state
|A> = a|
> + b|
>
Collect coefficients in up-down basis and write
state |A> as a b 1
We thus have, |> =
<| = [1 0]
|> = 1
<| = [0 1]

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

14. In this basis, matrix S’s…
Due to completeness of states |,  > any operator M
= M(|  > <  | + |  > <  |). Thus… 1 -1
Sz = ħ/2 1
S+ = ħ|> <|
= ħ 1
S- = ħ
|> <|
= ħ

15. In this basis, matrix S’s…
Sx = ħ/2 1 i -i -1
Sy = ħ/2
Sz = ħ/2
Pauli Matrices
sx,y,z
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
|y > = 1 i √2
|x > = 1 √2
|y > = -1 i 1 √2
|x > = -1 1 √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
sinqeij
sinqe-ij
-cosq
Then find its eigenvectors

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

20. Effect on rotation… |n > = |n > = |I > = |I > =
cosq/2
eijsinq/2
|n > =
-sinq/2
eijcosq/2
Start with initial condition q=0, j=0, with initial states
|I > = 1
|I > =
But after full rotation, q=2p, j=0, with final states
|I > = -1
|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

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/dq Y(q) = eiq0L.n/ħ Y(q)
R(q) = eiq0Ln/ħ

30. Spin operator Y(sn+q0) = eiq0s.n/ħ Y(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
Yab(1,2) =
[Fa(r1)
Fb(r2) -
Fa(r2)
Fb(r1)]/ √2

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. 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

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