Lecture notes from the week of 11-17

11-17-08
Stern Gerlach Experiment
Addition of L and S

T=2π/ω   ω=γβ  
Χ(T)= Martix      Upper left=cos(π/2) Upper right= exp(iω/2*2π/ω) -> exp(iπ)  Lower left = sin(α/2)exp(iδ) Lower right same as upper right

So  Χ(T)= -Χ(0)  and Χ(2T)= Χ(0)
l=1 ->Y(lm)(θφ) ~ exp(imφ) with m=±1, 0

Interference 

Χfinal= X1+X2 without the B field
Xfinal=-X1+X2 with B field

SG Expt

See book for diagrams
X(0) = (a, b) (vertical matrix) = aX++bX-

E+ =-γ(B+αZ)1/2 h2
E- =-γ(B+αZ)1/2 h2
When t=T (time evolution)
 (a*exp(-i E+/ h)T   b* exp(+i E+/ h)T)
Ψ -> Ψ(z)= exp(i (something)z) which is plane wave propagating in the z direction

The bottom line: Pz= ±γα h/2
11-19-08
Combining two or more angular momentum

↑↑ ↑↓ ↓↑ ↓↓
1/√2(↑↓ + ↓↑) triplet
1/√2(↑↓ - ↓↑) singlet

H α S(electron)+S(proton)

H equals to a 4x4 (a, b, c, d )matrix times a 1x4 matrix of the arrows. 
H↑↑ where ↑↑ vertical matrix like (1, 0, 0, 0) (except vertical)
So it becomes a↑↑+b↑↓+c↓↑+d↓↓

S(electron) dot S(proton) ↑↑= (h/2)2 ↑↑
 From S(e)S(p)=S(ex)S(px)+S(ey)S(py)+S(ez)S(pz)

S(ez)S(pz) ↑↑= h /2 h /2
S(e+)S(p+) ↑↑=0
S(e-)S(p-) ↑↑=0
S(ez)S(pz) ↑↑= h2↓↓

These lead to  h /4(1 000, 0-1 2 0, 0 2 -1 0, 0 0 0 1)	

S(tot) = (S1+S2)2	The S’s commute so
=S12+S22+2S1 S2
The eigenvalues for S12 and S22 are 3/4 h2
And it becomes S12↑↑=3/4 h2↑↑ and the same for S22

S(tot)2= 2(3/4 h2(Identy Matrix))+ 2S1 S2

Find the determinate -> (1-λ)2((-1-λ)2-4=0 to get the eigenvalues

11-21-08
Take ↑↑ ↑↓ ↓↑ ↓↓ angular spin combinations which is a member of s=1 (s2=1(1+1)h2)

↓↓ ↑↑ have angular momentum of at least one

Start with ↑↑ spin => |S, Sz> = |1, 1> 
S is the total spin 1(1+1) h2) and Sz is the z component of the total

l1 and l2 refer to the magnitude
l1+l2  l1+l2+1 …  | l1-l2|
With the spin ½ particles there is only parallel or antiparallel

Apply S- to the total spin
S-= S(1)-+ S(2)-↑↑
S-|1,1>=(normalization factor)|1,0>
Normalization-> √l(l+1)-m(m-1)|1,0> -> √2|1, 0>

S(1)-+ S(2)- ↑↑= S(1)-↑↑ + S(2)- ↑↑
=↑↓ ↓↑ to be equal to what we got earlier √2|1, 0>

|1,0>=1/√2( ↑↓ + ↓↑)

Summarizing   |1,1>=↑↑
|1,0>=1/√2( ↑↓ + ↓↑)
|1,-1>=↓↓

l=1 S=1/2
lz=1  ↑			S↑ or  ↓
lz=-1  ↓
lz=0   →

Six possible combinations
|l, S> -> |3/2, 3/2> |J, Jz> J is total angular momentum, Jz is z component

Apply the I operator
J-=L-+S-

J-|3/2, 3/2>=(something)|3/2,3/2>
L-= applies to one and S- to the other (from J-)

Chapter 4 typed up in an attempt to clear things up for myself.
Amy Fleischhacker 11/24/08