Question

What is the product of$\mathit{\Delta }\mathit{x}$and$\mathit{\Delta }\mathit{p}$(obtained in Exercise 83 and 85)? How does it compare with the minimum theoretically possible? Explain.

Short answer

The value of the product is $0.866h$and it is greater than the theoretical minimum value.

Step-by-step solution

Given information

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

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

The value of$\Delta x$is, $0.866/a$whereis the atomic radius and the value of$\Delta p$is$a\hslash$.

To find the expectation value of particle’s momentum.

Step 2:Understandingthe concept of uncertainty

To calculate the uncertainty in the momentum,${\sigma }_p$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${\mathit{\psi }}^{\mathbf{*}}\mathit{\psi }$multiplied by the momentum operator$\widehat{p}$.

Use formula to calculate the expectation values of particles momentum 

Given:

Formula used:

Write the expression for the Uncertainty principle.

$\Delta x\Delta p\ge \frac{\hslash }{2}\dots \mathrm{..}\text{(1)}$

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

Calculation:

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

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

The minimum theoretical value of the product is$\hslash /2$.

Conclusion:

Thus, the value of the product is$0.866\hslash$and it is greater than the theoretical minimum value.

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

$\begin{array}{c}ápñ=4a^3{\int }_0^{\yen }\left(x{e}^{-ax}\right)\left(\widehat{p}\right)\left(x{e}^{-ax}\right)dx\\ =4a^3{\int }_0^{\yen }\left(x{e}^{-ax}\right)(-ih)\frac{d}{dx}\left(x{e}^{-ax}\right)dx\\ =-4ih{a}^3{\int }_0^{\yen }\left(x{e}^{-ax}\right)(e^{-ax}-ax{e}^{-ax})dx\\ =-4ih{a}^3{\int }_0^{\yen }\left(x{e}^{-2ax}\right)(1-ax)dx\end{array}$

Use integration by parts to evaluate the integral

$\begin{array}{c}⟨p⟩=-4i\hslash a^3{\int }_0^{\infty }\left(e^{-2ax}\right)(x-a{x}^2)dx\\ =-4i\hslash a^3{\left[\left(e^{-2ax}\right)(-\frac{x-a{x}^2}{2a}-\frac{1-2ax}{4a^2}+\frac{2a}{8a^3})\right]}_0^{\infty }\\ =-4i\hslash a^3{\left[\left(e^{-2ax}\right)(-\frac{4a^{2}x}{8a^3}+\frac{4a^3{x}^2}{8a^3}-\frac{2a}{8a^2}+\frac{4a^{3}x}{8a^3}+\frac{2a}{8a^3})\right]}_0^{\infty }\\ =0\end{array}$

Use integration by parts to evaluate the integral.

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

$\begin{array}{c}⟨p^2⟩=4a^3{\int }_0^{\infty }\left(x{e}^{-ax}\right)\left({\widehat{p}}^2\right)\left(x{e}^{-ax}\right)dx\\ =4a^3{\int }_0^{\infty }\left(x{e}^{-ax}\right)(-{\hslash }^2)\frac{d^2}{d{x}^2}\left(x{e}^{-ax}\right)dx\\ =-4{\hslash }^2{a}^3{\int }_0^{\infty }\left(x{e}^{-ax}\right)\frac{d}{dx}(e^{-ax}-ax{e}^{-ax})dx\\ =-4{\hslash }^2{a}^3{\int }_0^{\infty }\left(x{e}^{-2ax}\right)(a^{2}x-2a)dx\end{array}$

Use integration by parts to evaluate the integral

$\begin{array}{c}⟨p^2⟩=-4{\hslash }^2{a}^3{\int }_0^{\infty }\left(e^{-2ax}\right)(a^2{x}^2-2ax)dx\\ =-4{\hslash }^2{a}^3{\left[e^{-2ax}(-\frac{2a^{2}x-2a}{2a}-\frac{2a^2}{4a^2})\right]}_0^{\infty }\\ =-4{\hslash }^2{a}^3\left(\frac{-1}{4a}\right)\\ ={\hslash }^2{a}^2\end{array}$

The uncertainty in the momentum ${\sigma }_p$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${⟨p⟩}^2$, .

$\begin{array}{c}{\sigma }_p=\sqrt{{\hslash }^2{a}^2-0^2}\\ =\hslash a\end{array}$

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