Question

The (time-independent) momentum space wave function in three dimensions is defined by the natural generalization of Equation 3.54:

$$\begin{gathered} \mathbf{\Phi }\mathbf{(}\mathbf{p}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi }\sout{\mathbf{h}}}}{\mathbf{\int }}_{\mathbf{\infty }}^{\mathbf{\infty }}{\mathbf{e}}^{\mathbf{-}\mathbf{ipx}\mathbf{/}\sout{\mathbf{h}}}\mathbf{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{dx} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{54}\mathbf{)}\mathbf{.} \\ \mathbf{\varphi }\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{\equiv }\frac{\mathbf{1}}{{\mathbf{\left(}\mathbf{2}\mathbf{\pi }\sout{\mathbf{h}}\mathbf{\right)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}\mathbf{\int }{\mathbf{e}}^{\mathbf{-}\mathbf{i}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{r}\mathbf{)}\mathbf{I}\sout{\mathbf{h}}}\mathbf{\psi }\mathbf{\left(}\mathbf{r}\mathbf{\right)}{\mathbf{d}}^{\mathbf{3}}\mathbf{r}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{223}\mathbf{)}\mathbf{.} \end{gathered}$$

(a)Find the momentum space wave function for the ground state of hydrogen (Equation 4.80). Hint: Use spherical coordinates, setting the polar axis along the direction of p. Do the θ integral first. Answer:

$$\begin{gathered} {\mathbf{\psi }}_{\mathbf{100}}\left(r,\theta ,\varphi \right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{\pi a}}^{\mathbf{3}}}}{\mathbf{e}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{a}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(4.80\right)\mathbf{.} \\ \mathbf{\varphi }\left(p\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\left(\frac{2a}{\sout{h}}\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{1}}{{\left[1+{\left(\mathrm{ap}/\sout{h}\right)}^2\right]}^{\mathbf{2}\mathbf{.}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(4.224\right)\mathbf{.} \end{gathered}$$

(b) Check that $\mathbf{\Phi }\left(p\right)$is normalized.

(c) Use $\mathbf{\Phi }\left(p\right)$to calculate $<p^2>$, in the ground state of hydrogen.

(d) What is the expectation value of the kinetic energy in this state? Express your answer as a multiple of ${\mathbf{E}}_{\mathbf{1}}$, and check that it is consistent with the virial theorem (Equation 4.218).

$<T>\mathbf{=}\mathbf{-}{\mathbf{E}}_{\mathbf{n}}\mathbf{;}\mathbf{}<V>\mathbf{=}\mathbf{2}{\mathbf{E}}_{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(4.218\right)\mathbf{.}$

Short answer

Step-by-step solution

(a) Finding the momentum of space wave function for the ground state of hydrogen.

$$\begin{gathered} \mathrm{\psi }=\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}\Rightarrow \mathrm{\varphi }\left(\mathrm{p}\right)=\frac{1}{{\left(2\mathrm{\pi }\sout{\mathrm{h}}\right)}^{3/2}}\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}\int {\mathrm{e}}^{-\mathrm{ip}.\mathrm{rI}\sout{\mathrm{h}}}{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}{\mathrm{r}}^2\mathrm{sin\theta drd\theta d\varphi }. \\ \mathrm{With}\mathrm{axes}\mathrm{as}\mathrm{suggested},\mathrm{p}.\mathrm{r}=\mathrm{pr}\mathrm{cos\theta }.\mathrm{Doing}\mathrm{the}\left(\mathrm{trivial}\right)\mathrm{\Phi }\mathrm{integral}: \\ \mathrm{\varphi }\left(\mathrm{p}\right)=\frac{2\mathrm{\pi }}{{\left(2\mathrm{\pi a}\sout{\mathrm{h}}\right)}^{3/2}}\frac{1}{\sqrt{\mathrm{\pi }}}{\int }_0^{\infty }{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}\left[{\int }_0^{\mathrm{\pi }}{\mathrm{e}}^{-\mathrm{ipr}\mathrm{cos\theta }/\sout{\mathrm{h}}}\sin \mathrm{\theta d\theta }\right]\mathrm{dr}. \\ {\int }_0^{\mathrm{\pi }}{\mathrm{e}}^{-\mathrm{ipr}\mathrm{cos\theta }/\sout{\mathrm{h}}}\mathrm{sin\theta d\theta }=\frac{\sout{\mathrm{h}}}{\mathrm{ipr}}{\overline{){\mathrm{e}}^{-\mathrm{ipr}\mathrm{cos\theta }/\sout{\mathrm{h}}}}}_0^{\mathrm{\pi }}=\frac{\sout{\mathrm{h}}}{\mathrm{ipr}}\left|\right({\mathrm{e}}^{\mathrm{ipr}/\sout{\mathrm{h}}}-{\mathrm{e}}^{{}^{\mathrm{ipr}/\sout{\mathrm{h}}}})=\frac{2\sout{\mathrm{h}}}{\mathrm{pr}}\sin \left(\frac{\mathrm{pr}}{\sout{\mathrm{h}}}\right). \\ \mathrm{\varphi }\left(\mathrm{p}\right)=\frac{1}{\mathrm{\pi }\sqrt{2}}\frac{1}{{\left(\mathrm{a}\sout{\mathrm{h}}\right)}^{3/2}}\frac{2\sout{\mathrm{h}}}{\mathrm{p}}{\int }_0^{\infty }{\mathrm{re}}^{-\mathrm{r}/\mathrm{a}}\sin \left(\frac{\mathrm{pr}}{\sout{\mathrm{h}}}\right)\mathrm{dr}. \\ {\int }_0^{\infty }{\mathrm{re}}^{-\mathrm{r}/\mathrm{a}}\sin \left(\frac{\mathrm{pr}}{\sout{\mathrm{h}}}\right)\mathrm{dr}=\frac{1}{2\mathrm{i}}\left[{\int }_0^{\infty }{\mathrm{re}}^{-\mathrm{r}/\mathrm{a}}{\mathrm{e}}^{\mathrm{ipr}/\sout{\mathrm{h}}}\mathrm{dr}-{\int }_0^{\infty }{\mathrm{re}}^{-\mathrm{r}/\mathrm{a}}{\mathrm{e}}^{\mathrm{ipr}/\sout{\mathrm{h}}}\mathrm{dr}\right]. \\ =\frac{1}{2\mathrm{i}}\left[\frac{1}{{\left(1/\mathrm{a}-\mathrm{ip}/\sout{\mathrm{h}}\right)}^2}-\frac{1}{{\left(1/\mathrm{a}+\mathrm{ip}/\sout{\mathrm{h}}\right)}^2}\right]=\frac{1}{2\mathrm{i}}\frac{\left(2\mathrm{ip}/\mathrm{a}\sout{\mathrm{h}}\right)2}{{\left[{\left(1/\mathrm{a}\right)}^2+{\left(\mathrm{p}/\sout{\mathrm{h}}\right)}^2\right]}^{2.}} \\ =\frac{\left(2\mathrm{p}/\sout{\mathrm{h}}\right){\mathrm{a}}^3}{{\left[1+{\left(\mathrm{ap}/\sout{\mathrm{h}}\right)}^2\right]}^{2.}} \\ \mathrm{\varphi }\left(\mathrm{p}\right)=\sqrt{\frac{2}{\sout{\mathrm{h}}}}\frac{1}{{\mathrm{a}}^{3/2}}\frac{1}{\mathrm{\pi p}}\frac{2{\mathrm{pa}}^3}{\sout{\mathrm{h}}}\frac{1}{{\left[1+{\left(\mathrm{ap}/\sout{\mathrm{h}}\right)}^2\right]}^2}=\frac{1}{\mathrm{\pi }}{\left(\frac{2\mathrm{a}}{\sout{\mathrm{h}}}\right)}^{3/2}\frac{1}{{\left[1+{\left(\mathrm{ap}/\sout{\mathrm{h}}\right)}^2\right]}^{2.}} \end{gathered}$$

 Step2: (b) Checking that Φ(p) is normalized.

$$\begin{gathered} \int {\left|\mathrm{\varphi }\right|}^2{\mathrm{d}}^3\mathrm{p}=4\mathrm{\pi }{\int }_0^{\infty }{\mathrm{p}}^2{\left|\mathrm{\varphi }\right|}^2\mathrm{dp}=4\mathrm{\pi }\frac{1}{{\mathrm{\pi }}^2}{\left(\frac{2\mathrm{a}}{\sout{\mathrm{h}}}\right)}^3{\int }_0^{\infty }\frac{{\mathrm{p}}^2}{{\left[1+{\left(\mathrm{ap}/\sout{\mathrm{h}}\right)}^2\right]}^{4.}} \\ \mathrm{From}\mathrm{math}\mathrm{tables}:{\int }_0^{\infty }\frac{{\mathrm{x}}^2}{{\left(\mathrm{m}+{\mathrm{x}}^2\right)}^4}\mathrm{dx}=\frac{\mathrm{\pi }}{32}{\mathrm{m}}^{-5/2},\mathrm{so} \\ {\int }_0^{\infty }\frac{{\mathrm{p}}^2}{{\left[1+{\left(\mathrm{ap}/\sout{\mathrm{h}}\right)}^2\right]}^4}\mathrm{dp}={\left(\frac{\sout{\mathrm{h}}}{\mathrm{a}}\right)}^8\frac{\mathrm{\pi }}{32}{\left(\frac{\sout{\mathrm{h}}}{\mathrm{a}}\right)}^{-5}=\frac{\mathrm{\pi }}{32}\left(\frac{\sout{\mathrm{h}}}{\mathrm{a}}\right);\int {\left|\mathrm{\varphi }\right|}^2{\mathrm{d}}^3\mathrm{p}=\frac{32}{\mathrm{\pi }}{\left(\frac{\mathrm{a}}{\sout{\mathrm{h}}}\right)}^3\frac{\mathrm{\pi }}{32}{\left(\frac{\sout{\mathrm{h}}}{\mathrm{a}}\right)}^3. \end{gathered}$$

(c) Calculating <p2>

 Step4: (d) Expressing the answer as a multiple of  E1