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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 2 2 2 0 16 2 8 2 1 sin sin sin sin π π π 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 L L x L x L dx A L x L x L dx A L L x L L L x x L 2 0 2 2 0 2 3 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 π π π π π π A L 2 2 17 = , so the normalization constant is A L = 2 17 . (b) ψ 2 1 dx a a z = : A x a B x a A B x a x a dx a a 2 2 2 2 2 2 2 1 cos sin cos sin π π π 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 The first two terms are A a 2 and B a 2 . The third term is: 2 2 2 2 2 4 2 2 8 3 2 0 2 3 A B x a x a x a dx A B x a x a dx a A B x a a a a a a a cos sin cos cos sin cos π π π π π π π 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 = z z so that a A B 2 2 1 + = e j , giving A B a 2 2 1 + = . *P41.54 (a) x x a e dx ax 0 1 2 2 0 = F H G I K J = −∞ z π , since the integrand is an odd function of x . (b) x x a x e dx ax 1 3 1 2 2 4 0 2 = F H G I K J = −∞ z π , since the integrand is an odd function of x . (c) x x dx x x x x x dx 01 0 1 2 0 1 0 1 1 2 1 2 1 2 = + = + +
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