Lecture32

# Lecture32 - PHYS 360 Quantum Mechanics Wed Apr 6,...

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Unformatted text preview: 4/6/11 PHYS 360 Quantum Mechanics Wed Apr 6, 2011 Lecture 32: Can you quanEze the spin of a pair of electrons? (Entangled states!) See separate lecture notes based on Problem 4.27 HW #9 due Monday, Apr 11: Ch 4, #29, 34, 49 & Ch 5, # 4 1 2 More than one parEcle: AddiEon of Angular Momentum Consider two spin ­½ parEcles. Here are 4 ways they might be ordered: ↑↑, ↑↓, ↓↑, ↓↓ ↑↑: ↑↓: ↓↑: ↓↓: m =1 m=0 m=0 m = −1 ⎫ ⎪ ⎪ ⎬ but this is wrong! ⎪ ⎪ ⎭ Construct an operator for the total angular momentum: 2 S ≡ S (1) + S ( ) Operate with it on the wave funcEon: S z χ1 χ 2 = S …because there seems to be “extra” state with m=0. (We know m has to change in units of Δm=1. Applying the lowering operator to the top state leads to: ( (1) z +S (2) z ) χ χ = (S χ ) χ 1 2 (1) z 1 2 + χ1 ( S χ 2 ) (2) z = m1χ1 χ 2 + m2 χ1 χ 2 = ( m1 + m2 )χ1 χ 2 This leads to: ↑↑: ↑↓: ↓↑: ↓↓: m =1 m=0 m=0 m = −1 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ⎧ 11 = ↑↑ ⎪ 1 ⎪ ↑↓ + ↓↑ ⎨ 10 = 2 ⎪ ⎪ 1 − 1 = ↓↓ ⎩ ( ) ⎫ ⎪ ⎪ ⎬ s = 1 (triplet ) ⎪ ⎪ ⎭ ⎧ ⎫ 1 ↑↓ − ↓↑ ⎬ s = 0 ( singlet ) ⎨ 00 = ⎪ ⎪ 2 ⎩ ⎭ ( ) 3 (entangled states) 4 Applying operator algebra to the triplet state yields: ⎛ 3 2 3 2 2 ⎞ S2 1 0 = ⎜ + + 2 ⎟ 1 0 = 2 2 1 0 4 4⎠ ⎝4 3 S 2 = s( s + 1) 2 = 1 ( 1 + 1) 2 = 4 2 22 Sz sm = 11 22 S= 3 0.866 2 =↑ +1/2 S Sy Sz This means that s=1, since S2=s(s+1)ħ2=2ħ2. Sx -1/2 S sm = 1 2 (− 1 ) = ↓ 2 triplet S The singlet state yields s=0: ⎛ 3 2 3 2 3 2 ⎞ S2 0 0 = ⎜ + −2 0 0 =0 4 4⎟ ⎝4 ⎠ sm = 11 =↑↑ +1 sm = 10 = 0 1 ↑↓ + ↓↑ 2 ( ) S S 2 = s( s + 1) 2 = 1(1 + 1) 2 = 2 2 S = 2 1.414 5 -1 Sy S sm = 1(−1) =↓↓ 6 Sx 1 ...
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