Question

A particle is represented (at time t=0) by the wave function

$\mathit{\Psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\left\{A(a^2-x^2)\right\}$; if $\mathbf{-}\mathit{a}\mathbf{\le }\mathit{x}\mathbf{\le }\mathbf{+}\mathit{a}$

$\mathit{\Psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$; Otherwise

a) Determine the normalization constantA?

b) What is the expectation value of p(at time t=0)?

c) What is the expectation value of x(at time t=0)?

d) Find the expectation value of ${\mathit{x}}^{\mathbf{2}}$.

e) Find the expectation value of ${\mathit{p}}^{\mathbf{2}}$.

f) Find the uncertainty inrole="math" localid="1658551318238" $\mathit{x}\mathbf{\left(}{\mathbf{\sigma }}_{\mathbf{x}}\mathbf{\right)}$.

g) Find the uncertainty in $\mathit{p}\mathbf{\left(}{\mathbf{\sigma }}_{\mathbf{x}}\mathbf{\right)}$.

h) Check that your results are consistent with the uncertainty principle.

Short answer

(a) $A=\sqrt{\frac{15}{16a^2}}$,

(b)$\left(x\right)=0$

(c)$\left(p\right)=0$

(d)$X^2=\sqrt{\frac{a^2}{7}}$

(e) ${\mathrm{p}}^2=\sqrt{\frac{5{\mathrm{h}}^2}{2{\mathrm{a}}^2}}$

(f) ${\sigma }_x=\sqrt{\frac{a^2}{7}}$

(g) role="math" localid="1658551495957" ${\mathrm{\sigma }}_{\mathrm{y}}=\sqrt{\frac{5{\mathrm{h}}^2}{2{\mathrm{a}}^2}}$

(h) It’s Consistent

Step-by-step solution

Step 1: Define the Schrödinger equation

A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

Time-dependent Schrödinger equation is represented as

$\mathbf{ih}\mathbf{}\frac{\mathbf{d}}{\mathbf{dt}\mathbf{}}\mathbf{\left|}\mathbf{\psi }\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right|}\mathbf{\equiv }\stackrel{\mathbf{⏜}}{\mathbf{H}}\mathbf{\left|}\mathbf{\psi }\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right|}$

Calculation

$$\begin{gathered} \left(a\right)1={\int }_{-\infty }^{\infty }{\left[\psi \left(x,0\right)\right]}^{2}dx={\int }_{-a}^a{A}^2{\left(a^2-x^2\right)}^{2}dx \\ 1=A^2{\int }_{-a}^a\left(a^4-2a^2{x}^4\right)=A^2{\left[a^{4}x-\frac{2}{3}{a}^2{x}^3+\frac{x^5}{5}\right]}_{-a}^a \\ 1=A^2\frac{16}{15}{a}^2 \\ A=\sqrt{\frac{15}{16a^2}} \end{gathered}$$

(b)

$$\begin{gathered} \left(x\right)={\int }_{-\infty }^{\infty }{\psi }^{*}x\psi dx \\ ={\int }_{-a}^{a}A\left(a^2-x^2\right)xA\left(a^2-x^2\right)dx \\ =0 \end{gathered}$$

(c)

$$\begin{gathered} \left(p\right)={\int }_{-\infty }^{\infty }{\psi }^{*}p\psi dx \\ =-ih{A}^2{\int }_{-a}^a\left(a^2-x^2\right)\frac{\partial }{\partial x}\left(a^2-x^2\right)dx \\ =-ih{A}^2{\int }_{-a}^a\left(2a^{2}x-2x^3\right) \\ =0 \end{gathered}$$

(d)

$$\begin{gathered} X^2={\int }_{-\infty }^{\infty }{\psi }^{*}{x}^2\psi dx \\ ={\int }_{-a}^{a}A\left(a^2-x^2\right)x^{2}A\left(a^2-x^2\right)dx \\ =A^2{\int }_{-a}^a\left(x^2{a}^4-2a^2{x}^4+x^6\right)dx \\ =\frac{a^2}{7} \end{gathered}$$

(e)

$$\begin{gathered} p^2={\int }_{-\infty }^{\infty }{\psi }^{*}{p}^2\psi dx \\ =-4h^2{A}^2{\int }_0^a\left(a^2-x^2\right)dx \\ =-4h^2{A}^2{\left[a^{2}x-\frac{x^3}{3}\right]}_0^a \\ =\frac{5h^2}{2a^2} \end{gathered}$$

(f)

$$\begin{gathered} {\sigma }_x=\sqrt{x^2-x} \\ =\sqrt{\frac{a^2}{7}} \end{gathered}$$

(g)

$$\begin{gathered} {\sigma }_p=\sqrt{p^2-p} \\ =\sqrt{\frac{5h^2}{2a^2}} \end{gathered}$$

(h)

$$\begin{gathered} {\sigma }_x{\sigma }_p=\sqrt{\frac{a^2}{7}}\sqrt{\frac{5h^2}{2a^2}} \\ =\sqrt{\frac{10}{7}}\frac{h}{2}>\frac{h}{2} \end{gathered}$$