Repeat the exercise 60-62 for the first excited state of harmonic oscillator.

The given data from exercise 60-62 can be listed below,

• The uncertainty in the position is $\Delta x=\sqrt{\frac{1}{\sqrt{2}}}{\left(\frac{{\hslash }^2}{m\kappa }\right)}^{1/4}$,
• The uncertainty in the position is,$\Delta p=\sqrt{\frac{\hslash }{2}}{\left(m\kappa \right)}^{1/4}$

A harmonic oscillator is essentially a system in which, if an item is moved by a distance X, a restoring force F will be applied to it in the direction that is opposite to the direction of the movement.

The expectation value of position is given by,

$\begin{array}{c}\overline{x}={\int }_{-\infty }^{+\infty }x{\psi }^{2}dx\\ ={\int }_{-\infty }^{+\infty }x{\left({\left(\frac{b}{2\sqrt{\pi }}\right)}^{1/2}\left(2bx\right)e^{-(0/2)b^2{x}^2}\right)}^{2}dx\\ =\frac{b}{2\sqrt{\pi }}4b^2{\int }_{-\infty }^{+\infty }{x}^3{e}^{-b^2{x}^2}dx=0\text{(odd)}\end{array}$

The expectation value of $x^2$is given by

$\begin{array}{c}{\overline{x}}^2={\int }_{-\infty }^{+\infty }{x}^2{\psi }^{2}dx\\ ={\int }_{-\infty }^{+\infty }{x}^2{\left({\left(\frac{b}{2\sqrt{\pi }}\right)}^{1/2}\left(2bx\right)e^{-(1/2)b^2{x}^2}\right)}^{2}dx\\ =\frac{b}{2\sqrt{\pi }}4b^2{\int }_{-\infty }^{+\infty }{x}^4{e}^{-b^2{x}^2}dx\end{array}$

$\begin{array}{c}\overline{x^2}=\frac{b}{2\sqrt{\pi }}4b^2\frac{3}{4}\frac{\sqrt{\pi }}{b^5}\\ =\frac{3}{2}\frac{1}{b^2}\cdot \text{Δ}x\\ =\sqrt{\frac{3}{2b^2}-0}\\ =\sqrt{3/2}{({ħ}^2/m\kappa )}^{1/4}\end{array}$

The expectation value of momentum for first excited state of harmonic oscillator is given by,

$\begin{array}{c}\overline{p}={\int }_{-\infty }^{+\infty }\psi \left(x\right)(-i\hslash \frac{\partial }{\partial x})\psi \left(x\right)dx\\ ={\int }_{-\infty }^{+\infty }\left({\left(\frac{b}{2\sqrt{\pi }}\right)}^{1/2}\left(2bx\right)e^{-(1/2)b^2{a}^2}\right)(-i\hslash \frac{\partial }{\partial x})\left({\left(\frac{b}{2\sqrt{\pi }}\right)}^{1/2}\left(2bx\right)e^{-(1/2)e^2{x}^2}\right)dx\\ ={\int }_{-\infty }^{+\infty }{\left(\frac{b}{2\sqrt{\pi }}\right)}^{1/2}\left(2bx\right)e^{-(1/2)b^2{x}^2}{\left(\frac{b}{2\sqrt{\pi }}\right)}^{1/2}(-i\hslash )\left(\left(2b\right)e^{-(1/2)b^2{x}^2}-b^{2}x\left(2bx\right)e^{-(1/2)e^2{x}^2}\right)dx\\ =-i\hslash \left(\frac{b}{2\sqrt{\pi }}\right){\int }_{-}^{+\infty }{e}^{-b^2{x}^2}(\left(4b^{2}x\right)-\left(4b^4{x}^3\right))dx\\ =0\end{array}$

Both the terms are odd so the expectation value of momentum is zero.

The expectation value of momentum square is given by,

$\begin{array}{c}\overline{p^2}={\int }_{-\infty }^{+\infty }\psi \left(x\right){(-i\hslash \frac{\partial }{\partial x})}^2\psi \left(x\right)dx\\ =-{\hslash }^2\left(\frac{b}{2\sqrt{\pi }}\right){\int }_{-\infty }^{+\infty }\left(2bx\right)e^{-(1/2)b^2{x}^2}\left[\left(2b\right)(-b^{2}x)e^{-(1/2)b^2{x}^2}-\left(4b^{3}x\right)e^{-(1/2)b^2{x}^2}-\left(2b^3{x}^2\right)(-b^{2}x)e^{-(1/2)b^2{x}^2}\right]dx\\ =-{\hslash }^2\left(\frac{b}{2\sqrt{\pi }}\right){\int }_{-\infty }^{+\infty }(-12b^4{x}^2+4b^6{x}^4)e^{-b^2{x}^2}dx\end{array}$

Substitute all the values in the above,

$\begin{array}{c}\overline{p^2}=-{\hslash }^2\left(\frac{b}{2\sqrt{\pi }}\right)(-12b^4{\int }_{-\infty }^{+\infty }{x}^2{e}^{-b^2{x}^2}dx+4b^6{\int }_{-}^{+}{x}^4{e}^{-b^2{x}^2}dx)\\ =-{\hslash }^2\left(\frac{b}{2\sqrt{\pi }}\right)(-3b\sqrt{\pi })\\ =\frac{3}{2}{\hslash }^2{\text{b}}^2\\ =\frac{3}{2}{\hslash }^2\frac{\sqrt{m\kappa }}{\hslash }\end{array}$

The uncertainty of the position and momentum is given by,

$\begin{array}{c}\Delta \text{x}\Delta \text{p}=\sqrt{\frac{3}{2}}{({\hslash }^2/m\kappa )}^{1/4}\times \sqrt{\frac{3\hslash }{2}}{\left(m\kappa \right)}^{1/4}\\ =\frac{3}{2}\hslash \end{array}$

Thus, the uncertainty found is $\frac{3\hslash }{2}$ which is more than $\frac{\hslash }{2}$. So, the wave function is Gaussian but it is not only a simple Gaussian function.