91影视

The wave function of the particle is$\text{蠄}\left(\text{x}\right)$.

$\text{蠄(虫)=}\{\begin{array}{ll}\text{2}\sqrt{{\text{a}}^{\text{3}}}{\text{xe}}^{鈭�\text{ax}}& \text{x>0}\\ \text{0}& \text{x<0}\end{array}$

The value of$螖x$is, $0.866/a$whereis the atomic radius and the value of$螖p$is$a\mathrm{鈩�}$.

To find the expectation value of particle鈥檚 momentum.

To calculate the uncertainty in the momentum,${蟽}_p$one must first calculate the expectation value of the momentum and the expectation value of the square of the momentum. The momentum's expectation value is twave function's integral squared${\mathit{蠄}}^{\mathbf{*}}\mathit{蠄}$multiplied by the momentum operator$\widehat{p}$.

Given:

Formula used:

Write the expression for the Uncertainty principle.

$螖x螖p鈮�\frac{\mathrm{鈩�}}{2}鈥�\mathrm{..}\text{(1)}$

Here,$螖x$is the position uncertainty,$螖p$is the momentum uncertainty and$h$is Planck's constant.

Calculation:

Substitute$0.866/a$for$螖x$and $a h$ for$螖p$in equation (1).

$$螖x螖p=\left(\frac{0.866}{a}\right)=0.866hn$$

The minimum theoretical value of the product is$\mathrm{鈩�}/2$.

Conclusion:

Thus, the value of the product is$0.866\mathrm{鈩�}$and it is greater than the theoretical minimum value.

The momentum's expectation value is twave function's integral squared${蠄}^{*}蠄$multiplied by the momentum operator$\widehat{p}$

$\begin{array}{c}谩p帽=4a^3{鈭�}_0^{楼}\left(x{e}^{-ax}\right)\left(\widehat{p}\right)\left(x{e}^{-ax}\right)dx\\ =4a^3{鈭�}_0^{楼}\left(x{e}^{-ax}\right)(-ih)\frac{d}{dx}\left(x{e}^{-ax}\right)dx\\ =-4ih{a}^3{鈭�}_0^{楼}\left(x{e}^{-ax}\right)(e^{-ax}-ax{e}^{-ax})dx\\ =-4ih{a}^3{鈭�}_0^{楼}\left(x{e}^{-2ax}\right)(1-ax)dx\end{array}$

Use integration by parts to evaluate the integral

$\begin{array}{c}鉄�p鉄�=鈭�4i\mathrm{鈩�}{a}^3{鈭�}_0^{\mathrm{鈭�}}\left(e^{鈭�2ax}\right)(x鈭�a{x}^2)dx\\ =鈭�4i\mathrm{鈩�}{a}^3{\left[\left(e^{鈭�2ax}\right)(鈭�\frac{x鈭�a{x}^2}{2a}鈭�\frac{1鈭�2ax}{4a^2}+\frac{2a}{8a^3})\right]}_0^{\mathrm{鈭�}}\\ =鈭�4i\mathrm{鈩�}{a}^3{\left[\left(e^{鈭�2ax}\right)(鈭�\frac{4a^{2}x}{8a^3}+\frac{4a^3{x}^2}{8a^3}鈭�\frac{2a}{8a^2}+\frac{4a^{3}x}{8a^3}+\frac{2a}{8a^3})\right]}_0^{\mathrm{鈭�}}\\ =0\end{array}$

Now calculate the expectation value of the square of the momentum${}^2>$

$\begin{array}{c}鉄�p^2鉄�=4a^3{鈭�}_0^{\mathrm{鈭�}}\left(x{e}^{鈭�ax}\right)\left({\widehat{p}}^2\right)\left(x{e}^{鈭�ax}\right)dx\\ =4a^3{鈭�}_0^{\mathrm{鈭�}}\left(x{e}^{鈭�ax}\right)(鈭�{\mathrm{鈩�}}^2)\frac{d^2}{d{x}^2}\left(x{e}^{鈭�ax}\right)dx\\ =鈭�4{\mathrm{鈩�}}^2{a}^3{鈭�}_0^{\mathrm{鈭�}}\left(x{e}^{鈭�ax}\right)\frac{d}{dx}(e^{鈭�ax}鈭�ax{e}^{鈭�ax})dx\\ =鈭�4{\mathrm{鈩�}}^2{a}^3{鈭�}_0^{\mathrm{鈭�}}\left(x{e}^{鈭�2ax}\right)(a^{2}x鈭�2a)dx\end{array}$

Use integration by parts to evaluate the integral

$\begin{array}{c}鉄�p^2鉄�=鈭�4{\mathrm{鈩�}}^2{a}^3{鈭�}_0^{\mathrm{鈭�}}\left(e^{鈭�2ax}\right)(a^2{x}^2鈭�2ax)dx\\ =鈭�4{\mathrm{鈩�}}^2{a}^3{\left[e^{鈭�2ax}(鈭�\frac{2a^{2}x鈭�2a}{2a}鈭�\frac{2a^2}{4a^2})\right]}_0^{\mathrm{鈭�}}\\ =鈭�4{\mathrm{鈩�}}^2{a}^3\left(\frac{鈭�1}{4a}\right)\\ ={\mathrm{鈩�}}^2{a}^2\end{array}$

The uncertainty in the momentum ${蟽}_p$is the square root of the difference between tile expectation value of the square of the momentum and the expectation value of the momentum squared${鉄�p鉄�}^2$, .

$\begin{array}{c}{蟽}_p=\sqrt{{\mathrm{鈩�}}^2{a}^2鈭�0^2}\\ =\mathrm{鈩�}a\end{array}$

The uncertainty in the particle's momentum is${蟽}_p=ha$