Q4P At time t = 0 a particle is repr... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 1: Q4P (page 14)

At time t = 0 a particle is represented by the wave function
$蠄 (x, 0) = {\begin{cases} A (x, 0), & 0 鈮� x 鈮� a, \\ A (b 鈭� x) / (b 鈭� a), & a 鈮� x 鈮� b, \\ 0, & otherwise, \end{cases}$ where A, a, and b are (positive) constants.
(a) Normalize $蠄$ (that is, find A, in terms of a and b).
(b) Sketch $蠄 (x, 0)$ , as a function of x.
(c) Where is the particle most likely to be found, at t = 0?.
(d) What is the probability of finding the particle to the left of a? Check your result in the limiting cases b = a and b= 2a.
(e) What is the expectation value of x?

Short Answer

Expert verified

(a) $A = \sqrt{\frac{3}{b}} A = b 3 .$

(d) $P = {鈭�}_{0}^{a} {| A |}^{2} dx = \frac{{| A |}^{2}}{a^{2}} {鈭�}_{0}^{a} x^{2} dx = {| A |}^{2} \frac{a}{3} = \frac{a}{b} P$

(e) $鉄� X 鉄� = \frac{2 a + b}{4} .$

Step by step solution

(a) Normalizing ψ .

$\begin{matrix} 1 = \frac{{| A |}^{2}}{a^{2}} {鈭�}_{0}^{a} x^{2} d x + \frac{{| A |}^{2}}{{(b 鈭� a)}^{2}} {鈭�}_{0}^{b} {(b 鈭� a)}^{2} d x \\ = {| A |}^{2} \{\frac{1}{a^{2}} (\frac{x^{3}}{3}) {|_{0}^{a} + \frac{1}{{(b 鈭� a)}^{2}} (鈭� \frac{{(b 鈭� x)}^{3}}{3})|}_{a}^{b}\} \end{matrix} \begin{matrix} = {| A |}^{2} [\frac{a}{3} + \frac{b 鈭� a}{3}] = {| A |}^{2} \frac{b}{3} \\ A = \sqrt{\frac{3}{b}} A = b 3. \end{matrix}$

(b) Sketching ψ(x,0)

(c) The particle most likely to be found at,

At x = a.

(d) probability of finding the particle.

$\begin{matrix} P = {鈭�}_{0}^{a} {| A |}^{2} d x = \frac{{| A |}^{2}}{a^{2}} {鈭�}_{0}^{a} x^{2} d x = {| A |}^{2} \frac{a}{3} = \frac{a}{b} P \\ = b a, \{\begin{cases} P = 1 & i f b = a \\ P = 1 / 2 & i f b = 2 a \end{cases} . \end{matrix}$

(e) Expectation value of x.

$鉄� x 鉄� = 鈭� x {| 唯 |}^{2} dx = {| A |}^{2} \{\frac{1}{a^{2}} {鈭�}_{0}^{a} x^{3} dx + \frac{1}{{(b 鈭� a)}^{2}} {鈭�}_{a}^{b} x {(b 鈭� x)}^{2} dx\} .$

$= \frac{3}{b} \{\frac{1}{a^{2}} {(\frac{x^{4}}{4})}_{0}^{a} {|+ \frac{1}{{(b 鈭� a)}^{2}} (b^{2} \frac{x^{2}}{2} 鈭� 2 b \frac{x^{3}}{3} + \frac{x^{4}}{4})|}_{a}^{b}\} .$

$= \frac{3}{4 b {(b 鈭� a)}^{2}} [a^{2} {(b 鈭� a)}^{2} + 2 b^{4} 鈭� 8 b^{4} / 3 + b^{4} 鈭� 2 a^{2} b^{2} + 8 a^{3} b / 3 鈭� a^{4}] .$

$= \frac{3}{4 b {(b 鈭� a)}^{2}} (\frac{b^{4}}{3} 鈭� a^{2} b^{2} + \frac{2}{3} a^{3} b) = \frac{1}{4 {(b 鈭� a)}^{2}} (b^{3} 鈭� 3 a^{2} b + 2 a^{3}) = \frac{2 a + b}{4} .$

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Short Answer

Step by step solution

(a) Normalizing ψ .

(b) Sketching ψ(x,0)

(c) The particle most likely to be found at,

(d) probability of finding the particle.

(e) Expectation value of x.

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