Chapter 1: Q4P (page 14)

At time t = 0 a particle is represented by the wave function
$ψ (x, 0) = {\begin{cases} A (x, 0), & 0 \leq x \leq a, \\ A (b - x) / (b - a), & a \leq x \leq b, \\ 0, & otherwise, \end{cases}$ where A, a, and b are (positive) constants.
(a) Normalize $ψ$ (that is, find A, in terms of a and b).
(b) Sketch $ψ (x, 0)$ , 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) $P = \int_{0}^{a} {| A |}^{2} dx = \frac{{| A |}^{2}}{a^{2}} \int_{0}^{a} x^{2} dx = {| A |}^{2} \frac{a}{3} = \frac{a}{b} P$

(e) $⟨ X ⟩ = \frac{2 a + b}{4} .$

Step by step solution

(a) Normalizing ψ .

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

(b) Sketching ψ(x,0)

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

At x = a.

(d) probability of finding the particle.

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

(e) Expectation value of x.

$⟨ x ⟩ = \int x {| Ψ |}^{2} dx = {| A |}^{2} \{\frac{1}{a^{2}} \int_{0}^{a} x^{3} dx + \frac{1}{{(b - a)}^{2}} \int_{a}^{b} x {(b - x)}^{2} dx\} .$

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

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

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

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

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.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Turning Points in Physics

Scientific Method Physics

Dynamics

Physical Quantities And Units

Conservation of Energy and Momentum

Medical Physics

Study anywhere. Anytime. Across all devices.