Show that$\mathit{\Delta }\mathit{p}\mathbf{=}\mathbf{0}\mathbf{\Rightarrow }\widehat{\mathbf{p}}\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\overline{\mathbf{p}}\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ that is, verify that unless the wave function is an Eigen function of the momentum operator, there will be a nonzero uncertainty in the momentumstarts with showing that the quantity

$\underset{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{s}\mathbf{p}\mathbf{a}\mathbf{c}\mathbf{e}}{\mathbf{\int }}{\mathit{\psi }}^{\mathbf{\ast }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{(}\widehat{\mathbf{p}}\mathbf{-}\overline{\mathbf{p}}\mathbf{)}}^{\mathbf{2}}\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{d}\mathit{x}$

Is ${\mathbf{\left(}\mathbf{\Delta }\mathbf{p}\mathbf{\right)}}^{\mathbf{2}}$. Then using the differential operator form of$\widehat{\mathbf{p}}$and integration by parts, show that it is also,

$\underset{\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{s}\mathbf{p}\mathbf{a}\mathbf{c}\mathbf{e}}{\mathbf{\int }}{\mathbf{\left\{}\mathbf{(}\widehat{\mathbf{p}}\mathbf{-}\overline{\mathbf{p}}\mathbf{)}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right\}}}^{\mathbf{\ast }}\mathbf{\left\{}\mathbf{(}\widehat{\mathbf{p}}\mathbf{-}\overline{\mathbf{p}}\mathbf{)}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right\}}\mathit{d}\mathit{x}$

Together these show that if$\mathit{\Delta }\mathit{p}$is. 0. then the preceding quantity must be 0. However, the Integral of the complex square of a function(the quantity in the brackets) can only be 0 if the function is identically 0, so the assertion is proved.

The given data can be listed below,

The difference in momentum is$\Delta p=0$,

If the translation operator is unitary and the momentum operator is the infinitesimal generator of translations, then the momentum operator is Hermitian.

The$\text{Δp}$ is given by,

$I={\int }_{allspace}{\psi }^{\ast }\left(x\right){(\widehat{p}-\overline{p})}^2\psi \left(x\right)dx$

Solve the above equation.

$I=\underset{allspace}{\int }{\psi }^{\ast }\left(x\right){\widehat{p}}^2\psi \left(x\right)dx-2\overline{p}\underset{allspace}{\int }{\psi }^{\ast }\left(x\right)\widehat{p}\psi \left(x\right)dx+{\overline{p}}^2\underset{allspace}{\int }{\psi }^{\ast }\left(x\right)\psi \left(x\right)dx$

The value of the third integral is 1.

The solution for the integration is given by,

$\begin{array}{c}{\left(\Delta p\right)}^2=\underset{allspace}{\int }{\psi }^{\ast }\left(x\right)(-i\hslash \frac{\partial }{\partial x})(-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)dx-\\ \underset{allspace}{\int }{\psi }^{\ast }\left(x\right)\overline{p}(-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)dx\\ =-i\hslash {\psi }^{\ast }\left(x\right)(-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)-\underset{allspace}{\int }(-i\hslash \frac{\partial {\psi }^{\ast }\left(x\right)}{\partial x})(-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)dx\\ -\underset{allspace}{\int }{\psi }^{\ast }\left(x\right)\overline{p}(-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)dx\\ =0-\underset{allspace}{\int }(-i\hslash \frac{\partial {\psi }^{\ast }\left(x\right)}{\partial x})(-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)dx-\\ \underset{allspace}{\int }{\psi }^{\ast }\left(x\right)\overline{p}(-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)dx\end{array}$

Rearrange the equation for${\left(\text{Δp}\right)}^{\text{2}}$ noting that$\overline{p}$ is real and that the complex conjugate of the sum of complex conjugates is itself a complex conjugate

${\left(\Delta p\right)}^2=\underset{allspace}{\int }{\left((-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)\right)}^{\ast }\left((-i\hslash \frac{\partial }{\partial x}-\overline{p})\psi \left(x\right)\right)dx$

From, the above equation it is clear that

$\widehat{p}\psi \left(x\right)=\overline{p}\psi \left(x\right)$

Thus, the $\Delta p=0$is implies that $\widehat{p}\psi \left(x\right)=\overline{p}\psi \left(x\right)$is proved.