Chapter 3 Problem Solutions

# 0 rehll imhll rehxa bl x b b imhxa bl x b b rehx

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Unformatted text preview: @ImHlLD @HXa + l* X b N H a^ + l b^N F = Â Xa b\ - Â X b a\ + 2 ImHlL X b b\ and require that each partial vanish as 2 ReHXa b\L + 2 Re HlL X b b\ = 0 -2 ImHXa b\L + 2 ImHlL X b b\ = 0 ReHlL = ImHlL = ﬂ ReHXa b\L =- X b b\ ImHXa b\L X b b\ =- ReHX b a\L X b b\ ImHX b a\L X b b\ X b a\ l = - Xb b\ This is the value of l such that a\ + l b\ is the projection of a\ onto the plane orthogonal to b\ X b H a\ + l b\L = X b a\ - X b a\ X b b\ X b b\ = 0 Thus Printed by Mathematica for Students HXa + l* X b L H a\ + l b\L = Xa a\ Xa a\ - X b a\ 2 X b b\ ¥0 X b a\ X b b\ Xa b\ ¥ 0 ﬂ l = - Xb b\ This is the value of l such that a\ + l b\ is the projection of a\ onto the plane orthogonal to b\ X b H a\ + l b\L = X b a\ - X b a\ X b b\ Ch3ProblemSetKey.nb 3 X b b\ = 0 Thus HXa + l* X b L H a\ + l b\L = Xa a\ X b a\ 2 Xa a\ - X b b\ X b a\ X b b\ Xa b\ ¥ 0 ¥0 Therefore Xa a\ X b b\ ¥ X b a\ 2 which is the Schwarz inequality. The inequality is saturated (becomes equality) when Xa H a\ + l b\L = 0 or when the component of a\ remaining after projecting out its b-component is orthogonal to a\ itself, which can only be true if said component is zero ( a\ has no component orthogonal to itself) a\ + l b\ = 0 a\ = -l b\ Therefore, the Schwarz inequality is saturated when a\ and b\ are linearly dependent. Problem 3.17 Given a spin-1 particle in the state y\ 1 2 3Â 1 Ø ` Sz basis 14 ü Part a ` The probabilities that a measurement of Sz will yield the values 8± —, 0&lt; are given by X1, 1 y\ 2 X1, 0 y\ 2 X1, -1 y\ 1 =J 2 2 14 14 = 2 1 14 N= 2 4 14 3Â =J = N= 14 2 9 14 We see that these sum to 1, as they should. Next we take XSz \ = — X1, 1 y\ 2 + 0 X1, 0 y\ 2 - — X1, -1 y\ 2 1 4 9 = — I 14 M + 0 I 14 M - — I 14 M 8 XSz \ = - 14 — Printed by Mathematica for Students 2 XSz \ = — X1, 1 y\ 4 + 0 X1, 0 y\ 2 - — X1, -1 y\ 2 1 4 9 = — I 14 M + 0 I 14 M - — I 14 M 8 Ch3ProblemSetKey.nb XSz \ = - 14 — ü Part b We have ` ` ` S± = S x ± Â S y ` ` 1` Sx = 2 JS+ + S- N Using ` S ± j , m^ = — jH j + 1L - m Hm ± 1L ` Z j , m£ S ± j , m^ = — j, m ± 1_ jH j + 1L - m Hm ± 1L dm£ m ± 1 gives the matrix representations ` ` S+ ØS 0 z 0 0 2 0 ` ` S- ØS and 0 z basis 0 0 2 0 0 0 0 —0 0 basis 2 2 0 — Thus ` ` Sx ØS 010 101 010 — z basis 2 so using matrix mechanics, we calculate the expectation value XSx \ = = = XSx \ = 010 1 2 -3 Â L 1 0 1 H 010 2 H 1 2 -3 Â L 1 + 3 Â 2 — 14 2 — 14 2 — 14 2 @2 + 2 H1 + 3 ÂL - 3 Â H2LD 4— 14 1...
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