1171_Physics ProblemsTechnical Physics

1171_Physics ProblemsTechnical Physics - 512 Quantum...

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512 Quantum Mechanics P41.53 (a) ψ 2 0 1 dx L z = : A x L x L x L x L dx L 22 2 0 16 2 8 2 1 sin sin sin sin ππ π F H G I K J + F H G I K J + F H G I K J F H G I K J L N M O Q P = z A LL x L x L dx A L x L x L dx A x L L L x xL 2 0 0 23 0 2 16 2 8 2 1 17 2 16 17 2 16 3 1 F H G I K J + F H G I K J + F H G I K J F H G I K J L N M M O Q P P = + F H G I K J F H G I K J L N M M O Q P P =+ F H G I K J L N M M O Q P P = z z = = sin sin sin cos sin A L 2 2 17 = , so the normalization constant is A L = 2 17 . (b) 2 1 dx a a z = : A x a B x a AB x a x a dx a a 2 2 2 2 2 2 2 1 cos sin cos sin F H G I K J + F H G I K J + F H G I K J F H G I K J L N M O Q P = z The first two terms are Aa 2 and Ba 2 . The third term is: 2 2 2 4 8 32 0 2 3 x a x a x a dx A B x a x a dx aAB x a a a a a a a cos sin cos cos sin cos F H G I K J F H G I K J F H G I K J L N M O Q P = F H G I K J F H G I K J = F H G I K J = −− zz so that aA B 1 += ej , giving a 1 . *P41.54 (a) xx a ed x ax 0 12 2 0 = F H G I K J = −∞ z , since the integrand is an odd function of x . (b) a xe d x ax 1 3 2 4 0 2 = F H G I K J = −∞ z , since the integrand is an odd function of x . (c) d x x x x x x d x 01 01 2 1 2 1 2 1 2 = + + −∞ −∞ ψψ bg a f a f The first two terms are zero, from (a) and (b). Thus:
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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