Question

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

Short answer

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

Step-by-step solution

Identification of given data

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,$\Delta p=\sqrt{\frac{\hslash }{2}}{\left(m\kappa \right)}^{1/4}$

Concept/Significance of harmonic oscillator

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.

Determination of the first excited state of harmonic oscillator.

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.