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

$$\begin{gathered} \mathbf{桅}\mathbf{(}\mathbf{p}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{蟺}\sout{\mathbf{h}}}}{\mathbf{鈭�}}_{\mathbf{鈭�}}^{\mathbf{鈭�}}{\mathbf{e}}^{\mathbf{-}\mathbf{ipx}\mathbf{/}\sout{\mathbf{h}}}\mathbf{蠄}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{dx} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{54}\mathbf{)}\mathbf{.} \\ \mathbf{蠒}\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{鈮�}\frac{\mathbf{1}}{{\mathbf{\left(}\mathbf{2}\mathbf{蟺}\sout{\mathbf{h}}\mathbf{\right)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}\mathbf{鈭�}{\mathbf{e}}^{\mathbf{-}\mathbf{i}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{r}\mathbf{)}\mathbf{I}\sout{\mathbf{h}}}\mathbf{蠄}\mathbf{\left(}\mathbf{r}\mathbf{\right)}{\mathbf{d}}^{\mathbf{3}}\mathbf{r}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{223}\mathbf{)}\mathbf{.} \end{gathered}$$

(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 胃 integral first. Answer:

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

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

(c) Use $\mathbf{桅}\left(p\right)$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).

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

Step by step solution

01

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

$$\begin{gathered} \mathrm{蠄}=\frac{1}{\sqrt{{\mathrm{蟺补}}^3}}{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}鈬�\mathrm{蠒}\left(\mathrm{p}\right)=\frac{1}{{\left(2\mathrm{蟺}\sout{\mathrm{h}}\right)}^{3/2}}\frac{1}{\sqrt{{\mathrm{蟺补}}^3}}鈭�{\mathrm{e}}^{-\mathrm{ip}.\mathrm{rI}\sout{\mathrm{h}}}{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}{\mathrm{r}}^2\mathrm{蝉颈苍胃诲谤诲胃诲蠒}. \\ \mathrm{With}\mathrm{axes}\mathrm{as}\mathrm{suggested},\mathrm{p}.\mathrm{r}=\mathrm{pr}\mathrm{肠辞蝉胃}.\mathrm{Doing}\mathrm{the}\left(\mathrm{trivial}\right)\mathrm{桅}\mathrm{integral}: \\ \mathrm{蠒}\left(\mathrm{p}\right)=\frac{2\mathrm{蟺}}{{\left(2\mathrm{蟺补}\sout{\mathrm{h}}\right)}^{3/2}}\frac{1}{\sqrt{\mathrm{蟺}}}{鈭�}_0^{鈭�}{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}\left[{鈭�}_0^{\mathrm{蟺}}{\mathrm{e}}^{-\mathrm{ipr}\mathrm{肠辞蝉胃}/\sout{\mathrm{h}}}\sin \mathrm{胃诲胃}\right]\mathrm{dr}. \\ {鈭�}_0^{\mathrm{蟺}}{\mathrm{e}}^{-\mathrm{ipr}\mathrm{肠辞蝉胃}/\sout{\mathrm{h}}}\mathrm{sin胃诲胃}=\frac{\sout{\mathrm{h}}}{\mathrm{ipr}}{\overline{){\mathrm{e}}^{-\mathrm{ipr}\mathrm{肠辞蝉胃}/\sout{\mathrm{h}}}}}_0^{\mathrm{蟺}}=\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{蠒}\left(\mathrm{p}\right)=\frac{1}{\mathrm{蟺}\sqrt{2}}\frac{1}{{\left(\mathrm{a}\sout{\mathrm{h}}\right)}^{3/2}}\frac{2\sout{\mathrm{h}}}{\mathrm{p}}{鈭�}_0^{鈭�}{\mathrm{re}}^{-\mathrm{r}/\mathrm{a}}\sin \left(\frac{\mathrm{pr}}{\sout{\mathrm{h}}}\right)\mathrm{dr}. \\ {鈭�}_0^{鈭�}{\mathrm{re}}^{-\mathrm{r}/\mathrm{a}}\sin \left(\frac{\mathrm{pr}}{\sout{\mathrm{h}}}\right)\mathrm{dr}=\frac{1}{2\mathrm{i}}\left[{鈭�}_0^{鈭�}{\mathrm{re}}^{-\mathrm{r}/\mathrm{a}}{\mathrm{e}}^{\mathrm{ipr}/\sout{\mathrm{h}}}\mathrm{dr}-{鈭�}_0^{鈭�}{\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{蠒}\left(\mathrm{p}\right)=\sqrt{\frac{2}{\sout{\mathrm{h}}}}\frac{1}{{\mathrm{a}}^{3/2}}\frac{1}{\mathrm{蟺辫}}\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{蟺}}{\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}$$

02

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

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

03

(c) Calculating <p2>

04

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

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