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Lecture31

# Lecture31 - PHYS 360 Quantum Mechanics Mon Apr 4 2011...

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4/4/11 1 1 PHYS 360 Quantum Mechanics Mon Apr 4, 2011 Lecture 31: How do you build a Stern-­૒Gerlach apparatus? HW #9 due Monday, Apr 11: Ch 4, #28, 29, 34, 49 & Ch 5, # 4 2 4.4.1 Spin ½ The electron has spin one-­૒half, and is the most important example. There are only two eigenstates: sm = 1 2 1 2 = sm = 1 2 ( 1 2 ) = These are the only two basis funcUons in the enUre Hilbert space for spin ½. +1/2 -1/2 S z S y S x S S S 2 = s ( s + 1) 2 = 1 2 ( 1 2 + 1) 2 = 3 4 2 S = 3 2   0.866 sm = 1 2 1 2 = sm = 1 2 ( 1 2 ) = 3 χ = a b = a χ + + b χ χ + = 1 0 χ = 0 1 S 2 χ + = 3 4 2 χ + and S 2 χ = 3 4 2 χ S 2 = 3 4 2 1 0 0 1 4 We write the eigenfuncUons as a 2-­૒element column matrix, called a spinor : The eigenvalue equaUon for S 2 is: Hence we can solve for S 2 : S z χ + = 2 χ + , S z χ = 2 χ S z = 2 1 0 0 1 S + χ = 0 1 0 0 , S χ + = 0 0 1 0 S x = 2 0 1 1 0 , S y = 2 0 i i 0 5 Using the same basis funcUons, consider S z : Solve for S z : By introducing raising & lowering operators… S x and S z can be determined: σ x 0 1 1 0 , σ y 0 i i 0 , σ z 1 0 0 1 6 These can be represented with S = 2 σ …using the Pauli spin matrices:

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4/4/11 2 Eigenvalue equaUon: S z χ ± = ± 2 χ ± The operator: S z = 2 1 0 0 1 The basis funcUons χ + = 1 0 χ = 0 1 The eigenvalues: S z χ + = 2 1 0 0 1 1 0 = + 2 χ + S z χ = 2 1 0 0 1 0 1 = - 2 χ 7 The operator: S x = 2 0 1 1 0
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