91影视

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

• The uncertainty in the position is $螖x=\sqrt{\frac{1}{\sqrt{2}}}{\left(\frac{{\mathrm{鈩�}}^2}{m魏}\right)}^{1/4}$,
• The uncertainty in the position is,$螖p=\sqrt{\frac{\mathrm{鈩�}}{2}}{\left(m魏\right)}^{1/4}$

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,

$\begin{array}{c}\stackrel{炉}{x}={鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}x{蠄}^{2}dx\\ ={鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}x{\left({\left(\frac{b}{2\sqrt{蟺}}\right)}^{1/2}\left(2bx\right)e^{鈭�(0/2)b^2{x}^2}\right)}^{2}dx\\ =\frac{b}{2\sqrt{蟺}}4b^2{鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}{x}^3{e}^{鈭�b^2{x}^2}dx=0\text{(odd)}\end{array}$

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

$\begin{array}{c}{\stackrel{炉}{x}}^2={鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}{x}^2{蠄}^{2}dx\\ ={鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}{x}^2{\left({\left(\frac{b}{2\sqrt{蟺}}\right)}^{1/2}\left(2bx\right)e^{鈭�(1/2)b^2{x}^2}\right)}^{2}dx\\ =\frac{b}{2\sqrt{蟺}}4b^2{鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}{x}^4{e}^{鈭�b^2{x}^2}dx\end{array}$

$\begin{array}{c}\stackrel{炉}{x^2}=\frac{b}{2\sqrt{蟺}}4b^2\frac{3}{4}\frac{\sqrt{蟺}}{b^5}\\ =\frac{3}{2}\frac{1}{b^2}鈰�\text{螖}x\\ =\sqrt{\frac{3}{2b^2}鈭�0}\\ =\sqrt{3/2}{({魔}^2/m魏)}^{1/4}\end{array}$

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

$\begin{array}{c}\stackrel{炉}{p}={鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}蠄\left(x\right)(鈭�i\mathrm{鈩�}\frac{鈭�}{鈭�x})蠄\left(x\right)dx\\ ={鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}\left({\left(\frac{b}{2\sqrt{蟺}}\right)}^{1/2}\left(2bx\right)e^{鈭�(1/2)b^2{a}^2}\right)(鈭�i\mathrm{鈩�}\frac{鈭�}{鈭�x})\left({\left(\frac{b}{2\sqrt{蟺}}\right)}^{1/2}\left(2bx\right)e^{鈭�(1/2)e^2{x}^2}\right)dx\\ ={鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}{\left(\frac{b}{2\sqrt{蟺}}\right)}^{1/2}\left(2bx\right)e^{鈭�(1/2)b^2{x}^2}{\left(\frac{b}{2\sqrt{蟺}}\right)}^{1/2}(鈭�i\mathrm{鈩�})\left(\left(2b\right)e^{鈭�(1/2)b^2{x}^2}鈭�b^{2}x\left(2bx\right)e^{鈭�(1/2)e^2{x}^2}\right)dx\\ =鈭�i\mathrm{鈩�}\left(\frac{b}{2\sqrt{蟺}}\right){鈭�}_{鈭�}^{+\mathrm{鈭�}}{e}^{鈭�b^2{x}^2}(\left(4b^{2}x\right)鈭�\left(4b^4{x}^3\right))dx\\ =0\end{array}$

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

The expectation value of momentum square is given by,

$\begin{array}{c}\stackrel{炉}{p^2}={鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}蠄\left(x\right){(鈭�i\mathrm{鈩�}\frac{鈭�}{鈭�x})}^2蠄\left(x\right)dx\\ =鈭�{\mathrm{鈩�}}^2\left(\frac{b}{2\sqrt{蟺}}\right){鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}\left(2bx\right)e^{鈭�(1/2)b^2{x}^2}\left[\left(2b\right)(鈭�b^{2}x)e^{鈭�(1/2)b^2{x}^2}鈭�\left(4b^{3}x\right)e^{鈭�(1/2)b^2{x}^2}鈭�\left(2b^3{x}^2\right)(鈭�b^{2}x)e^{鈭�(1/2)b^2{x}^2}\right]dx\\ =鈭�{\mathrm{鈩�}}^2\left(\frac{b}{2\sqrt{蟺}}\right){鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}(鈭�12b^4{x}^2+4b^6{x}^4)e^{鈭�b^2{x}^2}dx\end{array}$

Substitute all the values in the above,

$\begin{array}{c}\stackrel{炉}{p^2}=鈭�{\mathrm{鈩�}}^2\left(\frac{b}{2\sqrt{蟺}}\right)(鈭�12b^4{鈭�}_{鈭�\mathrm{鈭�}}^{+\mathrm{鈭�}}{x}^2{e}^{鈭�b^2{x}^2}dx+4b^6{鈭�}_{鈭�}^{+}{x}^4{e}^{鈭�b^2{x}^2}dx)\\ =鈭�{\mathrm{鈩�}}^2\left(\frac{b}{2\sqrt{蟺}}\right)(鈭�3b\sqrt{蟺})\\ =\frac{3}{2}{\mathrm{鈩�}}^2{\text{b}}^2\\ =\frac{3}{2}{\mathrm{鈩�}}^2\frac{\sqrt{m魏}}{\mathrm{鈩�}}\end{array}$

The uncertainty of the position and momentum is given by,

$\begin{array}{c}螖\text{x}螖\text{p}=\sqrt{\frac{3}{2}}{({\mathrm{鈩�}}^2/m魏)}^{1/4}脳\sqrt{\frac{3\mathrm{鈩�}}{2}}{\left(m魏\right)}^{1/4}\\ =\frac{3}{2}\mathrm{鈩�}\end{array}$

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