Question

Show that

$<X>\mathbf{=}\mathbf{\int }{\mathbf{\Phi }}^{\mathbf{*}}\left(-\frac{\sout{h}}{i}\frac{\partial }{\partial p}\right)\mathbf{\Phi dp}\mathbf{.}$

Hint: Notice that$\mathbf{x}\mathbf{}\mathbf{exp}\mathbf{(}\mathbf{ipx}\mathbf{/}\sout{\mathbf{h}}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{i}\sout{\mathbf{h}}\left(d/\mathrm{dp}\right)\mathbf{exp}\left(\mathrm{ip}/\sout{h}\right)\mathbf{.}$

In momentum space, then, the position operator is $\mathbf{i}\sout{\mathbf{h}}\mathbf{\partial }\mathbf{/}\mathbf{\partial }\mathbf{p}$ . More generally,

$\mathbf{<}\mathbf{Q}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{p}\mathbf{)}\mathbf{=}\mathbf{\{}\begin{array}{c}\mathbf{\int }{\mathbf{\psi }}^{\mathbf{*}}\hat{\mathbf{Q}}\mathbf{\}}\mathbf{(}\mathbf{x}\mathbf{,}\frac{\sout{\mathbf{h}}}{\mathbf{i}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\mathbf{)}\mathbf{\psi dx}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{in}\mathbf{}\mathbf{position}\mathbf{}\mathbf{space}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{\int }{\mathbf{\Phi }}^{\mathbf{*}}\hat{\mathbf{Q}}\mathbf{\}}\mathbf{(}\mathbf{-}\frac{\sout{\mathbf{h}}}{\mathbf{i}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{p}}\mathbf{,}\mathbf{p}\mathbf{)}\mathbf{\Phi dp}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{in}\mathbf{}\mathbf{momentum}\mathbf{}\mathbf{space}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$In principle you can do all calculations in momentum space just as well (though not always as easily) as in position space.

Short answer

The prove

Step-by-step solution

Concept used

The coordination representation:

$\mathrm{\psi }\left(\mathrm{x}\right)=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\int }_{-\propto }^{\infty }{\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\mathrm{\Phi }\left(\mathrm{p}\right)\mathrm{dp}$

Given information from question

The transformation laws between coordinate space and momentum space. Because a Fourier transform connects the two bases, the coordinate representation is as follows:

$\mathrm{\psi }\left(\mathrm{x}\right)=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\int }_{-\propto }^{\infty }{\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\mathrm{\Phi }\left(\mathrm{p}\right)\mathrm{dp}$ ……. (1)

Now we may try to describe the anticipated value of the position in momentum space for the provided wave function. We start by a definition of the expectation value

$$\begin{gathered} <\mathrm{\psi }\left|\hat{\mathrm{x}}\right|\mathrm{\psi }>={\int }_{\propto }^{\infty }\mathrm{\psi }*\left(\mathrm{x}\right)\mathrm{x\psi }\left(\mathrm{x}\right)\mathrm{dx} \\ ={\int }_{\propto }^{\infty }\left(\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}\right)\left({\int }_{-\propto }^{\infty }{\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\mathrm{\Phi }\left(\mathrm{p}\right)\mathrm{dp}\right)\mathrm{dx} \end{gathered}$$ ……. (2)

We can notice that we can express ${\mathrm{xe}}^{-\mathrm{ipx}/\sout{\mathrm{h}}}$as following:

$\mathrm{x}=-\mathrm{i}\sout{\mathrm{h}}\frac{\mathrm{d}}{\mathrm{dp}}\left({\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\right)$

As a result, the part of the following integral in the previous statement can be simplified as:

localid="1656314346606" $$\begin{gathered} {\int }_{\propto }^{\infty }{\mathrm{xe}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\mathrm{\Phi }\left(\mathrm{p}\right)\mathrm{dp}=-{\int }_{-\infty }^{\infty }\mathrm{i}\sout{\mathrm{h}}\frac{\mathrm{d}}{\mathrm{dp}}\left({\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\right)\mathrm{\Phi }\left(\mathrm{p}\right)\mathrm{dp} \\ \mathrm{Integrating}\mathrm{by}\mathrm{parts},\mathrm{we}\mathrm{get}, \\ -{\int }_{-\infty }^{\infty }\mathrm{i}\sout{\mathrm{h}}\frac{\mathrm{d}}{\mathrm{dp}}\left({\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\right)\mathrm{\Phi }\left(\mathrm{p}\right)\mathrm{dp}=\mathrm{i}\sout{\mathrm{h}}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\frac{\mathrm{d}}{\mathrm{dp}}\mathrm{\Phi }\left(\mathrm{p}\right)\mathrm{dp} \\ \mathrm{Insert}\mathrm{the}\mathrm{values}\mathrm{into}\mathrm{expression}\left(2\right),\mathrm{we}\mathrm{get}, \\ =\mathrm{i}\sout{\mathrm{h}}\frac{1}{2\mathrm{\pi }\sout{\mathrm{h}}}\iiint {\mathrm{e}}^{-\mathrm{ipx}/\sout{\mathrm{h}}}\frac{\mathrm{d}}{\mathrm{dp}}\mathrm{\Phi }\left(\mathrm{p}\right){\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\mathrm{\Phi }*\left(\mathrm{p}\right)\mathrm{dpdp}\mathrm{dx} \\ =\frac{\mathrm{i}}{2\mathrm{\pi }}\iiint {\mathrm{e}}^{\mathrm{ix}\left(\mathrm{p}-\mathrm{p}\right)\mathrm{I}\sout{\mathrm{h}}}\frac{\mathrm{d}}{\mathrm{dp}}{\mathrm{e}}^{\mathrm{ipx}/\sout{\mathrm{h}}}\mathrm{\Phi }*\left(\mathrm{p}\right)\mathrm{dpdp}\mathrm{dx} \end{gathered}$$

We can see that the previous form reminds us of the delta function, however to obtain the right form we need to use a substitution $\mathrm{z}=\mathrm{x}/\sout{\mathrm{h}}\to \mathrm{dz}=\mathrm{dx}/\sout{\mathrm{h}}$Then, the expression becomes:

We proved that the position operator in momentum representation is given as stated in the problem. It's noteworthy that the momentum operator in position space is identical to the momentum operator. The reason for this is that position and momentum are both conjugate variables.