Question

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

$\mathbf{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\{\begin{array}{ll}A(x,0),& 0\le x\le a,\\ A(b-x)/(b-a),& a\le x\le b,\\ 0,& \mathrm{otherwise},\end{array}$where A, a, and b are (positive) constants.

(a) Normalize $\mathbf{\psi }$(that is, find A, in terms of a and b).

(b) Sketch $\mathbf{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}$, 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

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

(c) x=0

(d)$\mathrm{P}={\int }_0^{\mathrm{a}}{\left|\mathrm{A}\right|}^2\mathrm{dx}=\frac{{\left|\mathrm{A}\right|}^2}{{\mathrm{a}}^2}{\int }_0^{\mathrm{a}}{\mathrm{x}}^2\mathrm{dx}={\left|\mathrm{A}\right|}^2\frac{\mathrm{a}}{3}=\frac{\mathrm{a}}{\mathrm{b}}\mathrm{P}$

(e) $⟨\mathrm{X}⟩=\frac{2\mathrm{a}+\mathrm{b}}{4}.$

Step-by-step solution

(a) Normalizing ψ .

(b) Sketching ψ(x,0)

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

At x = a.

(d) probability of finding the particle.

(e) Expectation value of x.

$⟨\mathrm{x}⟩=\int \mathrm{x}{\left|\mathrm{\Psi }\right|}^2\mathrm{dx}={\left|\mathrm{A}\right|}^2\left\{\frac{1}{{\mathrm{a}}^2}{\int }_0^{\mathrm{a}}{\mathrm{x}}^3\mathrm{dx}+\frac{1}{{(\mathrm{b}-\mathrm{a})}^2}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}{(\mathrm{b}-\mathrm{x})}^2\mathrm{dx}\right\}.$

$=\frac{3}{b}\left\{\frac{1}{a^2}{\left(\frac{x^4}{4}\right)}_0^a{\left|+\frac{1}{{(b-a)}^2}\left(b^2\frac{x^2}{2}-2b\frac{x^3}{3}+\frac{x^4}{4}\right)\right|}_a^b\right\}.$

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

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