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classes:2008:fall:phys4101.001:chapter4

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 4×4 (a, b, c, d )matrix times a 1×4 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>

Summarizing |1,1>=↑↑

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

Six possible combinations

 l, S> → 3/2, 3/2>

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 