91影视

Calculating for A by normalizing the wave function.

$\begin{array}{c}1=\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{|{\mathrm{蠄}}_{(\mathrm{x},\mathrm{t})}|}^2\mathrm{dx}\\ 1=\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}\mathrm{蠄}{*}_{(\mathrm{x},\mathrm{t})}{\mathrm{蠄}}_{(\mathrm{x},\mathrm{t})}\mathrm{dx}\\ 1=\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}\left({\mathrm{Ae}}^{鈭�\mathrm{a}\left[\right({\mathrm{mx}}^2/\sout{\mathrm{h}})+\mathrm{it}]}\right)\left({\mathrm{Ae}}^{鈭�\mathrm{a}\left[\right({\mathrm{mx}}^2/\sout{\mathrm{h}})鈭�\mathrm{it}]}\right)\mathrm{dx}\\ 1={\mathrm{A}}^2\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{e}}^{鈭�2{\mathrm{amx}}^2/\sout{\mathrm{h}}}\mathrm{dx}\end{array}$

Further solving above equation,

Thus, the value of A is ${\left(\frac{2\mathrm{am}}{\mathrm{蟺}\sout{\mathrm{h}}}\right)}^{1/4}$.

Schrodinger equation is given by,

$\frac{鈭�\mathrm{蠄}}{鈭�\mathrm{t}}=\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\frac{{鈭�}^2\mathrm{蠄}}{鈭�{\mathrm{x}}^2}鈭�\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{V}}_{(\mathrm{x},\mathrm{t})}{\mathrm{蠄}}_{(\mathrm{x},\mathrm{t})}$

Solving for ${\mathrm{V}}_{(\mathrm{x},\mathrm{t})}$,

Further solving above equation,

Thus, the potential energy function is${\mathrm{V}}_{(\mathrm{x},\mathrm{t})}=2{\mathrm{ma}}^2{\mathrm{x}}^2$ .

Solving for the expectation values.

Since, integral of an odd function over a symmetric interval is zero.

Therefore, the expectation value is zero $鉄�\mathrm{x}鉄�=0$.

Now, for $鉄�{\mathrm{x}}^2鉄�$

$\begin{array}{c}鉄�{\mathrm{x}}^2鉄�=\frac{{\displaystyle \underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{x}}^2|{{\mathrm{蠄}}_{(\mathrm{x},\mathrm{t})}|}^2\mathrm{dx}}}{{\displaystyle \underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}|{{\mathrm{蠄}}_{(\mathrm{x},\mathrm{t})}|}^2\mathrm{dx}}}\\ 鉄�{\mathrm{x}}^2鉄�=\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{x}}^2{\mathrm{蠄}}_{(\mathrm{x},\mathrm{t})}\mathrm{蠄}{*}_{(\mathrm{x},\mathrm{t})}\mathrm{dx}\\ 鉄�{\mathrm{x}}^2鉄�=\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{x}}^2\left[{\left(\frac{2\mathrm{am}}{\mathrm{蟺}\sout{\mathrm{h}}}\right)}^{1/4}{\mathrm{e}}^{鈭�\mathrm{a}\left[\right({\mathrm{mx}}^2/\sout{\mathrm{h}})+\mathrm{it}]}\right]\left[{\left(\frac{2\mathrm{am}}{\mathrm{蟺}\sout{\mathrm{h}}}\right)}^{1/4}{\mathrm{e}}^{鈭�\mathrm{a}\left[\right({\mathrm{mx}}^2/\sout{\mathrm{h}})鈭�\mathrm{it}]}\right]\mathrm{dx}\\ 鉄�{\mathrm{x}}^2鉄�=\sqrt{\frac{2\mathrm{am}}{\mathrm{蟺}\sout{\mathrm{h}}}}\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{x}}^2{\mathrm{e}}^{鈭�2{\max }^2/\sout{\mathrm{h}}}\mathrm{dx}\end{array}$

$\begin{array}{l}鉄�{\mathrm{x}}^2鉄�=2\sqrt{\frac{2\mathrm{am}}{\mathrm{蟺}\sout{\mathrm{h}}}}\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{x}}^2{\mathrm{e}}^{鈭�{\mathrm{x}}^2/{\left[\sqrt{\sout{\mathrm{h}}/\left(2\mathrm{am}\right)}\right]}^2}\mathrm{dx}\\ 鉄�{\mathrm{x}}^2鉄�=2\sqrt{\frac{2\mathrm{am}}{\mathrm{蟺}\sout{\mathrm{h}}}}.\sqrt{\mathrm{蟺}}\frac{2!}{1}{\left(\frac{\sqrt{\sout{\mathrm{h}}/2\mathrm{am}}}{2}\right)}^3\\ 鉄�{\mathrm{x}}^2鉄�=\frac{\sout{\mathrm{h}}}{4\mathrm{am}}\end{array}$

Therefore, the value of $鉄�{\mathrm{x}}^2鉄�$ is $\frac{\sout{\mathrm{h}}}{4\mathrm{am}}$.

Using the Ehrenfest鈥檚 theorem,

$\begin{array}{l}鉄�\mathrm{p}鉄�=\mathrm{m}鉄�\mathrm{v}鉄�\\ 鉄�\mathrm{p}鉄�=\mathrm{m}\frac{\mathrm{d}鉄�\mathrm{x}鉄�}{\mathrm{dt}}\end{array}$

Since $鉄�\mathrm{x}鉄�=0$.

Therefore, the value of $鉄�\mathrm{p}鉄�$ is zero.

Now, solving for $鉄�{\mathrm{p}}^2鉄�$.

Further solving above equation,

Further solving above equation,

$\begin{array}{l}鉄�{\mathrm{p}}^2鉄�={\sout{\mathrm{h}}}^2\left(\frac{4\mathrm{am}}{\sout{\mathrm{h}}}\right)\left(\frac{1}{2}鈭�\frac{1}{4}\right)\\ 鉄�{\mathrm{p}}^2鉄�=\sout{\mathrm{h}}\mathrm{am}\end{array}$

Therefore, the value of $鉄�{\mathrm{p}}^2鉄�$ is $\sout{\mathrm{h}}\mathrm{am}$.

The standard variation of an observable gives the uncertainty, hence,

$\begin{array}{l}{\mathrm{蟽}}_{\mathrm{x}}=\mathrm{螖虫}\\ \mathrm{螖虫}=\sqrt{鉄�{\mathrm{x}}^2鉄�鈭�{鉄�\mathrm{x}鉄�}^2}\\ \mathrm{螖虫}=\sqrt{\frac{\sout{\mathrm{h}}}{4\mathrm{am}}鈭�0}\\ \mathrm{螖虫}=\frac{1}{2}\sqrt{\frac{\sout{\mathrm{h}}}{\mathrm{am}}}\end{array}$

The calculation of standard deviation ${\mathrm{蟽}}_{\mathrm{p}}$ is,

$\begin{array}{c}{\mathrm{蟽}}_{\mathrm{p}}=\mathrm{螖辫}\\ \mathrm{螖辫}=\sqrt{鉄�{\mathrm{p}}^2鉄�鈭�{鉄�\mathrm{p}鉄�}^2}\\ \mathrm{螖辫}=\sqrt{\sout{\mathrm{h}}\mathrm{am}鈭�0}\\ \mathrm{螖辫}=\sqrt{\sout{\mathrm{h}}\mathrm{am}}\end{array}$

The uncertainty can be written as,

$\begin{array}{l}\mathrm{螖虫螖辫}=\frac{1}{2}\sqrt{\frac{\mathrm{h}}{\mathrm{am}}}\sqrt{\mathrm{ham}}\\ \mathrm{螖虫螖辫}=\frac{\mathrm{h}}{2}\end{array}$

And Heisenberg Uncertainty Principle states that

$\mathrm{螖虫螖辫}鈮�\frac{\mathrm{h}}{2}$

Therefore, it is consistent with the Heisenberg Uncertainty Principle.