202104161839 Homework 1 (Q2)

At time $t=0$ a particle is represented by the wave function

$\Psi (x) = \begin{cases} Ax/a, & \textrm{if }0\leq x\leq a \\ A\big( (x-a)^2+1\big), &\textrm{if }a\leq x\leq 2a \\ A(a^2+1)\displaystyle{\frac{3a-x}{a}},&\textrm{if }2a\leq x\leq 3a\\ 0, & \textrm{otherwise} \end{cases}$

where $A$ , $x$ are constants.

(a) Normalize $\Psi$ (that is, find $A$ in terms of $a$ ).
(b) Sketch $\Psi (x,0)$ 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 $a$ .
(e) What is the expectation value of $x$ ?

(a)

$\begin{aligned} 1 & = \int_{-\infty}^{+\infty}|\Psi |^2\,\mathrm{d}x \\ & = \int_{0}^{a}\frac{|A|^2x^2}{a^2}\,\mathrm{d}x + \int_{a}^{2a}|A|^2[(x-a)+1]^2\,\mathrm{d}x+\int_{2a}^{3a}|A|^2(a+1)^2\bigg( \frac{3a-x}{a}\bigg)^2\,\mathrm{d}x \\ & = \frac{|A|^2}{a^2}\bigg[\frac{x^3}{3}\bigg]\bigg|^a_{0} + |A|^2\bigg[\frac{\big( (x-a)+1\big)^3}{3}\bigg]\bigg|_{1}^{a+1} + |A|^2\bigg(\frac{a+1}{a}\bigg)^2\bigg[\frac{(x-3a)^3}{3}\bigg]\bigg|_{-a}^{0} \\ & = \frac{|A|^2a}{3} + |A|^2\bigg(\frac{(2)^3}{3} - \frac{(2-a)^3}{3} \bigg) + |A|^2\bigg( \frac{a+1}{a} \bigg)^2\bigg( \frac{(-3a)^3}{3} - \frac{(-4a)^3}{3} \bigg) \\ \dots \enspace & \textrm{ (to be continued) }\dots \end{aligned}$

Roughwork.

Simplifying the second term,

$\begin{aligned} &\quad |A|^2\bigg(\frac{(2)^3}{3} - \frac{(2-a)^3}{3} \bigg) \\ & = |A|^2\bigg[ \bigg(\frac{8}{3}\bigg) -\bigg( \frac{8-12a+6a^2-a^3}{3}\bigg) \bigg] \\ & = |A|^2\bigg( \frac{4}{3}a - 2a^2 + \frac{a^3}{3}\bigg) \end{aligned}$

Simplifying the third term,

$\begin{aligned} & \quad |A|^2\bigg( \frac{a+1}{a} \bigg)^2\bigg( \frac{(-3a)^3}{3} - \frac{(-4a)^3}{3} \bigg) \\ & = |A|^2\bigg( 1+\frac{2}{a}+\frac{1}{a^2}\bigg) \bigg( \frac{-27a^3}{3} + \frac{64a^3}{3} \bigg)\\ & = |A|^2\bigg( 1+\frac{2}{a}+\frac{1}{a^2}\bigg) \bigg( \frac{37a^3}{3}\bigg) \\ & = |A|^2\bigg( \frac{37}{3}a^3+\frac{74}{3}a^2 + \frac{37}{3}a \bigg) \end{aligned}$

$\begin{aligned} \dots\enspace &\textrm{ (continue) }\dots \\ 1 & = |A|^2 \bigg[ \bigg( \frac{1}{3} + \frac{4}{3} + \frac{37}{3} \bigg) a + \bigg( -2+\frac{74}{3} \bigg) a^2 + \bigg( \frac{1}{3}+\frac{37}{3} \bigg) a^3 \bigg] \\ 1 & = |A|^2 \bigg( 14a + \frac{68}{3}a^2 + \frac{38}{3}a^3 \bigg) \\ A & = \frac{1}{\sqrt{14a+\frac{68}{3}a^2+\frac{38}{3}a^3}} \end{aligned}$

(b)

(c)

at $x=2a$ where $|\Psi |^2$ attains its maximum value.

(d)

$\begin{aligned} P(x\leqslant a) & =\int_{0}^{a}|\Psi |^2\,\mathrm{d}x \\ & = \int_{0}^{a}\bigg|\frac{Ax}{a}\bigg|^2\,\mathrm{d}x \\ & = |A|^2\frac{a^2}{3} \end{aligned}$

(e)

$\begin{aligned} \langle x\rangle & = \int_{0}^{3a}x|\Psi |^2\,\mathrm{d}x \\ & = |A|^2\bigg( \int_{0}^{a}\frac{x^3}{a^2}\,\mathrm{d}x + \int_{a}^{2a}[(x-a)^2+1]^2x\,\mathrm{d}x + \int_{2a}^{3a}(a^2+1)^2\bigg(\frac{3a-x}{a}\bigg)^2 x\,\mathrm{d}x \bigg) \\ \dots\enspace & \textrm{ (to be continued) }\dots \end{aligned}$

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