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hw2a - PHYS 507 Answers to Homework II(Fall'05 1 cfw_|1 |2...

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PHYS 507 Answers to Homework II (Fall ’05) 1. {| 1 i , | 2 i , | 3 i} is an orthonormal basis for a 3D Hilbert space. | ψ i = 1 2 ( | 1 i + | 2 i ) , | φ i = 1 3 ( | 2 i + (1 + i ) | 3 i ) , A = 2 | ψ i h ψ | + 6 | φ i h φ | . (a) ψ = 1 2 1 1 0 , φ = 1 3 0 1 1 + i . (b) ψ = 1 2 £ 1 1 0 / , φ = 1 3 £ 0 1 1 - i / . (c) M | ψ ih ψ | = ψψ = 1 2 1 1 0 1 1 0 0 0 0 , M | φ ih φ | == φφ = 1 3 0 0 0 0 1 1 - i 0 1 + i 2 , (d) M A = 1 1 0 1 3 2 - 2 i 0 2 + 2 i 4 . (e) A 23 = 2 - 2 i and A 32 = 2 + 2 i (can be read directly from M A ). (f) A | 3 i = 2 | ψ i h ψ | 3 i + 6 | φ i h φ | 3 i = 6 | φ i 1 - i 3 = (2 - 2 i ) | 2 i + 4 | 3 i . Matrix representation is M A 0 0 1 = 0 2 - 2 i 4 . (g) h 3 | A = 2 h 3 | ψ i h ψ | + 6 h 3 | φ i h φ | = 6 1 + i 3 h φ | = (2 + 2 i ) h 2 | + 4 h 3 | . Matrix representation is £ 0 0 1 / M A = £ 0 2 + 2 i 4 / . 1
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2. | A i = 1 2 ( | 1 i + e | 2 i ) , | B i = 1 2 ( | 1 i - e | 2 i ) . (a) Verify h A | A i = h B | B i = 1 and h A | B i = 0. (b) | A i h A | = 1 2 ( | 1 i + e | 2 i ) ( h 1 | + e - h 2 | ) = 1 2 ( | 1 i h 1 | + | 2 i h 2 | + e | 2 i h 1 | + e - | 1 i h 2 | ) | B i h B | = 1 2 ( | 1 i - e | 2 i ) ( h 1 | - e - h 2 | ) = 1 2 ( | 1 i h 1 | + | 2 i h 2 | - e
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