91影视

In this method, the accelerations and velocities at a specific time point are taken to be constant during a time interval and are utilised to find the location of the following time point.

(a)

The function ${\mathrm{蠄}}_0$ is even, and ${蠄}_1$is odd. In either case ${\left|\mathrm{蠄}\right|}^2$is even, so the value of $鉄�x鉄�=鈭�x{\left|蠄\right|}^{2}dx=0$. Therefore $鉄�\mathrm{p}鉄�=\frac{\mathrm{md}鉄�\mathrm{x}鉄�}{\mathrm{dt}}=0$

Thus, the value of $鉄�x鉄�$ is $鈭�\mathrm{x}{\left|\mathrm{蠄}\right|}^2\mathrm{dx}=0$ and the value of $鉄�p鉄�$is$\frac{\mathrm{md}鉄�\mathrm{x}鉄�}{\mathrm{dt}}=0$.

Now, evaluate the value of $鉄�{\mathrm{x}}^2鉄�$and $鉄�{\mathrm{p}}^2鉄�$.

The below values are known that

$\begin{array}{l}{\mathbf{蠄}}_{\mathbf{0}}\mathbf{=}{\mathbf{伪别}}^{\raisebox{1ex}{$\mathbf{鈭�}{\mathbf{尉}}^{\mathbf{2}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\\ {\mathbf{蠄}}_{\mathbf{1}}\mathbf{=}\sqrt{\mathbf{2}}{\mathbf{伪尉别}}^{\raisebox{1ex}{$\mathbf{鈭�}{\mathbf{尉}}^{\mathbf{2}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\end{array}$

For $\mathrm{n}=0$, the value of $鉄�{\mathrm{x}}^2鉄�$ will be

$鉄�{\mathrm{x}}^2鉄�={\mathrm{伪}}^2{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\mathrm{x}}^2{\mathrm{e}}^{鈭�{\mathrm{尉}}^2/2}\mathrm{dx}$ ...(1)

For given equation-

$\begin{array}{c}\mathrm{尉}鈮�\sqrt{\frac{\mathrm{尘蝇}}{\mathrm{鈩�}}}\mathrm{x}\\ \mathrm{x}=\mathrm{尉}\sqrt{\frac{\mathrm{鈩�}}{\mathrm{尘蝇}}}\\ {\mathrm{x}}^2={\mathrm{尉}}^2\left(\frac{\mathrm{鈩�}}{\mathrm{尘蝇}}\right)\end{array}$

And

$\begin{array}{c}\mathrm{x}=\mathrm{尉}\sqrt{\frac{\mathrm{鈩�}}{\mathrm{尘蝇}}}\\ \mathrm{dx}=\mathrm{诲尉}脳\sqrt{\frac{\mathrm{鈩�}}{\mathrm{尘蝇}}}\end{array}$

Substitute the value of $x^2$ and $dx$ in equation (1).

$\begin{array}{c}鉄�{\mathrm{x}}^2鉄�={\mathrm{伪}}^2{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}\left({\mathrm{尉}}^2\left(\frac{\mathrm{鈩�}}{\mathrm{尘蝇}}\right)\right)脳{\mathrm{e}}^{鈭�{\mathrm{尉}}^2/2}脳\mathrm{诲尉}脳\sqrt{\frac{\mathrm{鈩�}}{\mathrm{尘蝇}}}\\ 鉄�{\mathrm{x}}^2鉄�={\mathrm{伪}}^2{\left(\frac{\mathrm{鈩�}}{\mathrm{尘蝇}}\right)}^{3/2}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\mathrm{尉}}^2{\mathrm{e}}^{鈭�{\mathrm{尉}}^2}\mathrm{诲尉}\\ =\frac{1}{\sqrt{\mathrm{蟺}}}\left(\frac{\mathrm{鈩�}}{\mathrm{尘蝇}}\right)\frac{\sqrt{\mathrm{蟺}}}{2}\\ =\frac{\mathrm{鈩�}}{2\mathrm{尘蝇}}\end{array}$

And the value of $鉄�P^2鉄�$ is

For $n=1$, the value of ${鉄�x鉄�}^2$will be

Substitute the value of $x^2$ and $dx$ in the above equation. So,

And the value of $鉄�{\mathrm{P}}^2鉄�$ is

Therefore, the value of $鉄�x^2鉄�$is $\frac{3\mathrm{鈩�}}{2m蝇}$ and is$\frac{3m\mathrm{鈩�}蝇}{2}$ .

(b)

The value of ${蟽}_x$ will be for $n=0:$

$\begin{array}{c}{\mathrm{蟽}}_{\mathrm{x}}=\sqrt{鉄�{\mathrm{x}}^2)鈭�{鉄�\mathrm{x}鉄�}^2}\\ =\sqrt{\frac{\mathrm{鈩�}}{2\mathrm{尘蝇}}}\\ {\mathrm{蟽}}_{\mathrm{p}}=\sqrt{鉄�{\mathrm{p}}^2鉄�鈭�{鉄�\mathrm{p}鉄�}^2}\\ =\sqrt{\frac{\mathrm{尘蝇}\mathrm{鈩�}}{2}}\\ {\mathrm{蟽}}_{\mathrm{x}}{\mathrm{蟽}}_{\mathrm{p}}=\sqrt{\frac{\mathrm{鈩�}}{2\mathrm{尘蝇}}}\sqrt{\frac{\mathrm{尘蝇}\mathrm{鈩�}}{2}}\\ =\frac{\mathrm{鈩�}}{2}\text{(Rightattheuncertaintymoment)}\end{array}$

For $n=1$, the value will be :

$\begin{array}{c}{\mathrm{蟽}}_{\mathrm{x}}=\sqrt{\frac{3\mathrm{鈩�}}{2\mathrm{尘蝇}}}\\ {\mathrm{蟽}}_{\mathrm{p}}=\sqrt{\frac{3\mathrm{尘蝇}\mathrm{鈩�}}{2}}\\ {\mathrm{蟽}}_{\mathrm{x}}{\mathrm{蟽}}_{\mathrm{p}}=3\frac{\mathrm{鈩�}}{2}>\frac{\mathrm{鈩�}}{2}\end{array}$

The uncertainty principal for these states is greater than $\frac{\mathrm{鈩�}}{2}$.

c)

The average kinetic energy is:

$\begin{array}{c}鉄�T鉄�=\frac{1}{2m}鉄�P^2鉄�\\ =\left\{{}_{\frac{3}{4}蝇\mathrm{鈩�}(n=1)}^{\frac{1}{4}蝇\mathrm{鈩�}(n=0)}\right\}\end{array}$

The average potential energy is

$\begin{array}{c}鉄�V鉄�=\frac{1}{2}m{蝇}^2鉄�x^2鉄�\\ =\left\{{}_{\frac{3}{4}蝇\mathrm{鈩�}(n=1)}^{\frac{1}{4}蝇\mathrm{鈩�}(n=0)}\right\}\end{array}$

$鉄�T鉄�+鉄�V鉄�=鉄�H鉄�=\left\{{}_{\frac{3}{2}蝇\mathrm{鈩�}(n=1)=E_1}^{\frac{1}{2}蝇\mathrm{鈩�}(n=0)=E_0}\right\}$ , as expected.

So, the value of the average kinetic energy is $\left\{{}_{\frac{3}{4}蝇\mathrm{鈩�}(n=1)}^{\frac{1}{4}蝇\mathrm{鈩�}(n=0)}\right\}$ and the average potential energy is $\left\{{}_{\frac{3}{4}蝇\mathrm{鈩�}(n=1)}^{\frac{1}{4}蝇\mathrm{鈩�}(n=0)}\right\}$.The sum of these two is same as expected value.