Introduction to Quantum Mechanics 2nd Edition

Published by Cambridge University Press
ISBN 10: 1107179866
ISBN 13: 978-1-10717-986-8

Chapter 1 - Section 1.4 - Normalization - Problems - Page 14: 1.4

Answer

(a) $A=√(3/b)$ (b) $Ψ(x,0)=(1/(b-a))x+Ab/(b-a)$ (c) Maximum occurs at A (d)$=a/b$

Work Step by Step

(a) $1=∫_(-∞)^∞〖|Ψ(X,0)|^2 dx〗$ $1= ∫_0^b〖|Ψ(X,0)|^2 dx〗$ $ =∫_0^a〖|Ψ(X,0)|^2 dx〗+ ∫_a^b〖|Ψ(X,0)|^2 dx〗$ $ = ∫_0^a|(A(w/a))|^2 dx+ ∫_a^b|(A((b-x)/(b-a)))|^2 dx$ $ =∫_0^a〖(|A|^2 x^2)/a^2 dx+ ∫_a^b(|A|^2 (b-x)^2)/(b-a)^2 〗 dx$ $1=|A|^2/a^2 ∫_0^a〖x^2 dx 〗+|A|^2/(b-a)^2 ∫_a^b〖(b-x)^2 dx〗$ $1=|A|^2 {├ (1/a^2 )(x^3/3)┤|_0^a+ ├ 1/(b-a)^2 [-(b-x)^3/3┤|_a^b }$ $1=|A|^2 {a/3+(b-a)/3}$ $1=|A|^2 (b/3)$ $3=|A|^2 b$ $|A|^2=3/b $ $A=√(3/b)$ (b) $Ψ(x,0)=A(x/a)$ , from $0≤x≤a$ $Ψ(x,0)=A((b-x)/(b-a))$, if $a≤x≤b$ $Ψ(x,0)=(1/(b-a))x+Ab/(b-a)$ See attachment for graph (c) From the graph on b, when x = a, $Ψ(x,0)=A(x/a)=A(a/a)=A$ So, the maximum occurs at x=a (d) $P=∫_0^a〖|Ψ(x,0)|^2 dx$ $=|A|^2/a^2 ∫_0^a〖x^2 dx〗〗$ $=├ |A|^2/a^2 (x^3/3)┤|_0^a$ $=|A|^2 (a/3)$ $=(√(3/b))^2 (a/3)$ $=a/b$ Check cases Suppose b=a $P=a/b=a/a=1$ Suppose b = 2a $P=a/b=a/2a=1/2$ And we see that both are consistent.
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