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--- classes:2008:fall:phys4101.001:chapter3 [2008/11/24 23:43] – x500_flei0029
+++ classes:2008:fall:phys4101.001:chapter3 [2008/12/02 22:38] (current) – yk
@@ Line 2: / Line 2: @@
 Feel free to write anything related to what you have learned in Chapter 3 or wanted to learn but did not learn enough.
-Lecture notes from the week of 11-17
--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
--19-08
-Combining two or more angular momentum
-↑↑ ↑↓ ↓↑ ↓↓
-/√2(↑↓ + ↓↑) triplet
-/√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
--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-)
-Actually Chapter 4 typed up in an attempt to clear things up for myself.
-Amy Fleischhacker 11/24/08

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