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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…
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So why do we attribute spin?
Stern-Gerlach experiment measures electron magnetic moment
Two lines imply only two possible moment values/orientations
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A ‘classic’ quantum experiment
Up and Down spins
But what determines ‘up’ and ‘down’?
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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 !!!
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We thus conjecture…
[Sx, Sy] = ASz and circular permutations
But what is A?
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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>
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By analogy with [Lx,Ly] = iħLz
[Sx, Sy] = iħSz and circular permutations
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Spin algebra in z and z bases
|s,sz >
Sz |> = (ħ/2)|>
Sz |> = -(ħ/2)|>
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What is the operator
representation of Sz?
How does Sz behave
on an arbitrary state?
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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
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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)|>
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What about Sx and Sy ?
[Sx, Sy] = iħSz and circular permutations
In the z and z basis, Sx,y are
off-diagonal
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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
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In the z and z basis, what are x and x ?
Expect them to be 50-50 of z andz
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
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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