Question

a) Construct ${\mathit{\psi }}_{\mathbf{2}}\left(x\right)$

b) Sketch ${\mathit{\psi }}_{\mathbf{0}}\mathbf{,}{\mathit{\psi }}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathit{\psi }}_{\mathbf{2}}$

c) Check the orthogonality of${\mathit{\psi }}_{\mathbf{0}}{\mathit{\psi }}_{\mathbf{1}}{\mathit{\psi }}_{\mathbf{2}}$ by explicit integration.

Hint:If you exploit the even-ness and odd-ness of the functions, there is really only one integral left to do.

Short answer

(a)The value of is${\psi }_{2}is\frac{1}{\sqrt{2}}{\left(\frac{m\omega }{\mathrm{\pi ħ}}\right)}^{1/4}\left(\frac{2m\omega }{ħ}x2-1\right)e^{-\frac{m\omega }{2ħ}{x}^2}$

(b)The diagrams are in step 2 .

(c)The${\psi }_0{\psi }_1{\psi }_2$ is orthogonal

Step-by-step solution

Step 1:Definition of orthogonal function

Two orthogonal wave functions $\mathit{\phi }\left(x\right)$and$\mathit{\psi }\left(x\right)$ represent mutually exclusive physical states: if one is true, in the sense that it is a valid description of the quantum system, the other is false, in the sense that it is an incorrect description of the quantum system.

Step 2:Calculation of the value of ψ2

(a)

For construction the value of ${\psi }_2$

localid="1656345370561" $$\begin{gathered} a_{+}{\psi }_0=\frac{1}{\sqrt{2ħm\omega }}\left(-ħ\frac{d}{dx}+m\omega x\right){\left(\frac{m\omega }{\mathrm{\pi ħ}}\right)}^{1/4}{e}^{-\frac{m\omega }{2ħ}} \\ =\frac{1}{\sqrt{2ħm\omega }}{\left(\frac{m\omega }{\mathrm{\pi ħ}}\right)}^{1/4}\left[-ħ\left(\frac{m\omega }{2ħ}\right)2x+m\omega x\right]e^{-\frac{m\omega }{2ħ}} \\ =\frac{1}{\sqrt{2ħm\omega }}{\left(\frac{m\omega }{\mathrm{\pi ħ}}\right)}^{1/4}2m\omega x{e}^{-\frac{m\omega }{2ħ}} \\ {\left(a_{+}\right)}^2{\psi }_0=\frac{1}{\sqrt{2ħm\omega }}{\left(\frac{m\omega }{\mathrm{\pi ħ}}\right)}^{1/4}2m\omega \left(-ħ\frac{d}{dx}+m\omega x\right)+e^{-\frac{m\omega }{2ħ}{x}^2} \\ =\frac{1}{ħ}{\left(\frac{m\omega }{\mathrm{\pi ħ}}\right)}^{1/4}2m\omega \left[-ħ\left(1-x\frac{m\omega }{2ħ}2x\right)+m\omega x^2\right]e^{-\frac{m\omega }{2ħ}{x}^2} \\ ={\left(\frac{m\omega }{\mathrm{\pi ħ}}\right)}^{1/4}\left(\frac{2m\omega }{ħ}{x}^2-1\right)e^{-\frac{m\omega }{2ħ}{x}^2} \\ {\psi }_2=\frac{1}{\sqrt{2}}{\left(a_x\right)}^2{\psi }_0 \\ =\frac{1}{\sqrt{2}}{\left(\frac{m\omega }{\mathrm{\pi ħ}}\right)}^{1/4}\left(\frac{2m\omega }{ħ}{x}^2-1\right)e^{-\frac{m\omega }{2ħ}{x}^2} \\ \mathrm{The}\mathrm{value}\mathrm{of}{\mathrm{\psi }}_2\mathrm{is}\frac{1}{\sqrt{2}}{\left(\frac{\mathrm{m\omega }}{\mathrm{\pi ħ}}\right)}^{1/4}\left(\frac{2\mathrm{m\omega }}{\mathrm{ħ}}{\mathrm{x}}^2-1\right){\mathrm{e}}^{-\frac{\mathrm{m\omega }}{2\mathrm{ħ}}{\mathrm{x}}^2} \end{gathered}$$

The diagram of ψ0,ψ1 and ψ2

(b)

The above diagrams are of three functions.

The orthogonality of ψ0ψ1ψ2

(c)

Since ${\psi }_0$and ${\psi }_2$are even, whereas ${\psi }_1$is odd. $\int {{\psi }_0}^{*}{\psi }_{1}dxand\int {{\psi }_2}^{*}{\psi }_{1}dx$vanish automatically. The only one we need to check is $\int {{\psi }_2}^{*}{\psi }_{1}dx$:

localid="1656345463405" $$\begin{gathered} \int {{\psi }_2}^{*}{\psi }_{1}dx=\frac{1}{\sqrt{2}}\sqrt{\frac{m\omega }{\mathrm{\pi ħ}}}{\int }_{-\infty }^{\infty }(\frac{2m\omega }{\mathrm{ħ}}{x}^2-1e^{-\frac{m\omega }{2\mathrm{ħ}}} \\ =\sqrt{\frac{m\omega }{2\mathrm{\pi ħ}}}\left({\int }_{-\infty }^{\infty }{e}^{-\frac{m\omega }{2\mathrm{ħ}}}dx-\frac{2m\omega }{\mathrm{ħ}}{\int }_{-\infty }^{\infty }{x}^2{e}^{-\frac{m\omega }{2\mathrm{ħ}}}dx\right) \\ =\sqrt{\frac{m\omega }{2\mathrm{\pi ħ}}}\left(\sqrt{\frac{\mathrm{\pi ħ}}{m\omega }-}\frac{2m\omega }{\mathrm{ħ}}\frac{\mathrm{ħ}}{2m\omega }\sqrt{\frac{\mathrm{\pi ħ}}{m\omega }-}\right) \\ =0 \\ so,{\psi }_0{\psi }_1{\psi }_2\mathrm{is}\mathrm{orthogonal} \end{gathered}$$