Question

Show that${\mathit{P}}^{\mathbf{2}}$is Hermitian, but${\mathit{P}}^{\mathbf{4}}$is not, for hydrogen states with$\mathit{l}\mathbf{=}\mathbf{0}$. Hint: For such states$\mathit{\psi }$is independent of$\mathit{\theta }$and$\mathit{\varphi }$, so

localid="1656070791118" ${\mathit{p}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\frac{{\mathbf{\hslash }}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{r}}\mathbf{\left(}{\mathbf{r}}^{\mathbf{2}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{r}}\mathbf{\right)}$

(Equation 4.13). Using integration by parts, show that

localid="1656069411605" $\mathbf{<}\mathit{f}\mathbf{\mid }{\mathit{p}}^{\mathbf{2}}\mathit{g}\mathbf{>}\mathbf{=}\mathbf{-}\mathbf{├}\mathbf{4}\mathit{\pi }{\mathit{h}}^{\mathbf{2}}\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{f}\frac{\mathbf{d}\mathbf{g}}{\mathbf{d}\mathbf{r}}\mathbf{-}{\mathbf{r}}^{\mathbf{2}}\mathbf{g}\frac{\mathbf{d}\mathbf{f}}{\mathbf{d}\mathbf{r}}\mathbf{)}\mathbf{}\overline{)\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{}}}\mathbf{}}\mathbf{+}\mathbf{<}{\mathbf{p}}^{\mathbf{2}}\mathbf{f}\mathbf{\mid }\mathbf{g}\mathbf{>}$

Check that the boundary term vanishes for${\mathit{\psi }}_{\mathbf{n}\mathbf{00}}$, which goes like

${\mathit{\psi }}_{\mathbf{n}\mathbf{00}}\mathbf{~}\frac{\mathbf{1}}{\sqrt{\mathbf{\pi }\mathbf{(}\mathbf{n}\mathbf{a}{\mathbf{)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}}\mathit{e}\mathit{x}\mathit{p}\mathbf{(}\mathbf{-}\mathit{r}\mathbf{/}\mathit{n}\mathit{a}\mathbf{)}$

near the origin. Now do the same for${\mathit{p}}^{\mathbf{4}}$, and show that the boundary terms do not vanish. In fact:

$<{\psi }_{n00}\mid p^4{\psi }_{m00}>\mathbf{=}\frac{\mathbf{8}{\mathbf{\hslash }}^{\mathbf{4}}}{{\mathbf{a}}^{\mathbf{4}}}\frac{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{m}\mathbf{)}}{\mathbf{(}\mathbf{n}\mathbf{m}{\mathbf{)}}^{\mathbf{5}\mathbf{/}\mathbf{2}}}\mathbf{+}<p^4{\psi }_{n00}\mid {\psi }_{m00}>$

Short answer

Thus,$p^2$ is Hermitian, but$p^4$ is not

Step-by-step solution

Hermitian for  P2

${\mathit{P}}^{\mathbf{2}}$is Hermitian, but ${\mathit{P}}^{\mathbf{4}}$is not,

${\mathit{P}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{-}{\mathbf{\hslash }}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{r}}\mathbf{\left(}{\mathbf{r}}^{\mathbf{2}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{r}}\mathbf{\right)}$

The ${\mathit{p}}^{\mathbf{2}}$is Hermitian if,

$\mathbf{⟨}\mathit{f}\mathbf{\mid }{\mathit{P}}^{\mathbf{2}}\mathit{g}\mathbf{⟩}\mathbf{=}\mathbf{⟨}{\mathit{P}}^{\mathbf{2}}\mathit{f}\mathbf{\mid }\mathit{g}\mathbf{⟩}$

so, to show that, we use:

$$\begin{gathered} ⟨\mathrm{f}\mid {\mathrm{P}}^2\mathrm{g}⟩=-{\hslash }^2{\int }_0^{\infty }\mathrm{f}\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)4{\mathrm{\pi r}}^2\mathrm{dr} \\ =-4\mathrm{\pi }{\hslash }^2{\int }_0^{\infty }\mathrm{f}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\mathrm{dr} \end{gathered}$$

use integration by parts to solve for Hermitian of P2  .

To evaluate this integral, we use the integration by parts technique:

$$\begin{gathered} \mathrm{u}=\mathrm{f},\hspace{1em}\mathrm{dv}=\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\mathrm{dr} \\ \mathrm{du}=\frac{\mathrm{df}}{\mathrm{dr}}\mathrm{dr},\hspace{1em}\mathrm{v}={\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}} \\ \mathrm{l}={\mathrm{r}}^2\mathrm{f}\frac{\mathrm{dg}}{\mathrm{dr}}\underset{0}{\overset{\infty }{\overline{)}}}-{\int }_0^{\infty }{\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\frac{\mathrm{dg}}{\mathrm{dr}}\mathrm{dr} \end{gathered}$$

 again, use integration by parts

By using the integration by parts once more,

$$\begin{gathered} \mathrm{u}={\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}},\hspace{1em} \\ \mathrm{dv}=\frac{\mathrm{dg}}{\mathrm{dr}}\mathrm{dr} \\ \mathrm{du}=\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\mathrm{dr},\hspace{1em} \\ \mathrm{v}={\mathrm{gI}}^{\text{'}} \\ ={\mathrm{r}}^2\mathrm{g}\frac{\mathrm{df}}{\mathrm{dr}}{|}_0^{\infty }-{\int }_0^{\infty }\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\mathrm{gdr} \end{gathered}$$

then,

$<\mathrm{f}\mid {\mathrm{p}}^2\mathrm{g}>=-4{\mathrm{\pi h}}^2({\mathrm{r}}^2\mathrm{f}\frac{\mathrm{dg}}{\mathrm{dr}}-{\mathrm{r}}^2\mathrm{g}\frac{\mathrm{df}}{\mathrm{dr}})\overline{)\underset{0}{\overset{\infty }{}}}+{\int }_0^{\infty }\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\mathrm{gdr}$

check at boundary

- If

$$\begin{gathered} \mathrm{r}\to 0\Rightarrow =0\mathrm{r}\to \infty \Rightarrow \mathrm{g} \\ \mathrm{f}\to 0\left({\mathrm{r}}^2\mathrm{f}\frac{\mathrm{dg}}{\mathrm{dr}}-{\mathrm{r}}^2\mathrm{g}\frac{\mathrm{df}}{\mathrm{dr}}\right){|}_0^{\infty } \end{gathered}$$

vanishes at the boundaries.

Multiplying the R.H.S. by$\frac{r^2}{r^2}$ ,

solve further

solve further

$$\begin{gathered} ⟨\mathrm{f}\mid {\mathrm{P}}^2\mathrm{g}⟩=-{\hslash }^2{\int }_0^{\infty }\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{df}{\mathrm{dr}}\right)g.4{\mathrm{\pi r}}^2\mathrm{dr} \\ \Rightarrow ⟨\mathrm{f}\mid {\mathrm{P}}^2\mathrm{g}⟩=⟨{\mathrm{P}}^2\mathrm{f}\lg ⟩ \end{gathered}$$

Then,$P^2$ is Hermitian.

solve for P4

Now, for $P^4$:

${\mathrm{P}}^4=\frac{{\hslash }^4}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left[{\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\right)\right)\right]$

By applying the same method,

$$\begin{gathered} ⟨\mathrm{f}\mid {\mathrm{P}}^4\mathrm{g}⟩={\hslash }^4{\int }_0^{\infty }\frac{1}{{\mathrm{r}}^2}\mathrm{f}\frac{\mathrm{d}}{\mathrm{dr}}\left[{\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\right]4{\mathrm{\pi r}}^2\mathrm{dr} \\ =4\mathrm{\pi }{\hslash }^4{\int }_0^{\infty }\mathrm{f}\frac{\mathrm{d}}{\mathrm{dr}}\left[{\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\right)\right]\mathrm{dr} \end{gathered}$$

use integration by parts

By using integration by parts:

$$\begin{gathered} \mathrm{u}=\mathrm{f},\hspace{1em}\mathrm{dv}=\frac{\mathrm{d}}{\mathrm{dr}}\left[{\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\right)\right]\mathrm{dr} \\ \mathrm{du}=\frac{\mathrm{df}}{\mathrm{dr}}\mathrm{dr},\hspace{1em}\mathrm{v}={\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\right) \end{gathered}$$

Solve further

$$\begin{gathered} {\mathrm{I}}_1={\mathrm{r}}^2\mathrm{f}\frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\right]{|}_0^{\infty }-{\int }_0^{\infty }{\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\right]\mathrm{dr} \\ \mathrm{u}={\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}},\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{dv}=\frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\right]\mathrm{dr} \\ \mathrm{du}=\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\mathrm{dr} \\ \mathrm{v}=\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right) \end{gathered}$$

solve further

localid="1656073604117"

localid="1656131027955" role="math" $$\begin{gathered} \mathrm{u}=\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right) \\ {\mathrm{l}}_2=\frac{\mathrm{df}}{\mathrm{dr}}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right){\overline{)}}_0^{\infty }-{\int }_0^{\infty }\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)\mathrm{drdr} \\ {\mathrm{I}}_3=\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\frac{\mathrm{dg}}{\mathrm{dr}}{|}_0^{\infty }-{\int }_0^{\infty }{\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\right]\frac{\mathrm{dg}}{\mathrm{dr}}\mathrm{dr} \\ \mathrm{u}={\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}/\mathrm{dr}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\right] \end{gathered}$$

check the boundary terms

To check the boundary terms:

Take,

$$\begin{gathered} \mathrm{f}\left(\mathrm{r}\right)={\mathrm{e}}^{{}^{\frac{-\mathrm{r}}{\mathrm{na}}}}\hspace{1em} \\ \mathrm{g}\left(\mathrm{r}\right)={\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}} \end{gathered}$$

By substituting into the last equation.

$$\begin{gathered} \frac{\mathrm{dg}}{\mathrm{dr}}=\frac{-1}{\mathrm{ma}}{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}} \\ \frac{\mathrm{df}}{\mathrm{dr}}=\frac{-1}{\mathrm{na}}{\mathrm{e}}^{\frac{-\mathrm{r}}{{\mathrm{nar}}^2}} \\ \frac{\mathrm{dg}}{\mathrm{dr}}=\left(\frac{-{\mathrm{r}}^2}{\mathrm{ma}}\right){\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}} \end{gathered}$$

Solve further

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dr}}\left[{\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right]=-\frac{2\mathrm{r}}{\mathrm{ma}}{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}}+\frac{{\mathrm{r}}^2}{{\mathrm{m}}^2{\mathrm{a}}^2}{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}} \\ \frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left[{\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right]=\left[\frac{-2}{\mathrm{mar}}+\frac{1}{{\mathrm{m}}^2{\mathrm{a}}^2}\right]{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}} \\ \frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left[{\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right]\right]=\frac{-1}{\mathrm{ma}}\left[\frac{-2}{\mathrm{mar}}+\frac{1}{{\mathrm{m}}^2{\mathrm{a}}^2}\right]{\mathrm{e}}^{{}^{\frac{-\mathrm{r}}{\mathrm{ma}}}}+\left(\frac{2}{{\mathrm{mar}}^2}\right){\mathrm{e}}^{{}^{\frac{-\mathrm{r}}{\mathrm{ma}}}} \end{gathered}$$

solve further

$$\begin{gathered} {\mathrm{r}}^2\mathrm{f}\frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left[{\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right]\right]=\left(\frac{2\mathrm{r}}{{\mathrm{m}}^2{\mathrm{a}}^2}-\frac{{\mathrm{r}}^2}{{\mathrm{m}}^3{\mathrm{a}}^3}+\frac{2}{\mathrm{ma}}\right){\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}}{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{na}}} \\ {\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}=\frac{-{\mathrm{r}}^2}{\mathrm{ma}}{\mathrm{e}}^{{}^{\frac{-\mathrm{r}}{\mathrm{ma}}}} \\ \frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)=\left(\frac{-2\mathrm{r}}{\mathrm{ma}}+\frac{{\mathrm{r}}^2}{{\mathrm{m}}^2{\mathrm{a}}^2}\right){\mathrm{e}}^{{}^{\frac{-\mathrm{r}}{\mathrm{ma}}}} \end{gathered}$$

Solve further

$$\begin{gathered} \frac{\mathrm{df}}{\mathrm{dr}}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{dg}}{\mathrm{dr}}\right)=\frac{-1}{\mathrm{na}}\left(\frac{-2\mathrm{r}}{\mathrm{ma}}+\frac{{\mathrm{r}}^2}{{\mathrm{m}}^2{\mathrm{a}}^2}\right){\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}}{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{na}}}{\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}} \\ =\frac{-{\mathrm{r}}^2}{\mathrm{na}}{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{na}}}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right) \\ =\left(\frac{-2\mathrm{r}}{\mathrm{na}}+\frac{{\mathrm{r}}^2}{{\mathrm{n}}^2{\mathrm{a}}^2}\right){\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{na}}}\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right) \\ \left(\frac{-2}{\mathrm{nar}}+\frac{1}{{\mathrm{n}}^2{\mathrm{a}}^2}\right){\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{na}}}\frac{\mathrm{dr}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\frac{\mathrm{dg}}{\mathrm{dr}}=\frac{-1}{\mathrm{ma}}\left(\frac{-2\mathrm{r}}{\mathrm{na}}+\frac{{\mathrm{r}}^2}{{\mathrm{n}}^2{\mathrm{a}}^2}\right){\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}}{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{na}}}\frac{\mathrm{df}}{\mathrm{dr}} \\ =\left(\frac{-2\mathrm{r}}{\mathrm{na}}+\frac{{\mathrm{r}}^2}{{\mathrm{n}}^2{\mathrm{a}}^2}\right){\mathrm{e}}^{{}^{\frac{-\mathrm{r}}{\mathrm{na}}}}\mathrm{dx} \\ \frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\right]=\left[\frac{-1}{\mathrm{na}}\left(\frac{-2}{\mathrm{nar}}+\frac{1}{{\mathrm{n}}^2{\mathrm{a}}^2}\right)+\frac{2}{{\mathrm{nar}}^2}\right]{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{na}}} \\ {\mathrm{r}}^2\mathrm{g}\frac{\mathrm{d}}{\mathrm{dr}}\left[\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{df}}{\mathrm{dr}}\right)\right]=\left[\frac{2\mathrm{r}}{{\mathrm{n}}^2{\mathrm{a}}^2}-\frac{{\mathrm{r}}^2}{{\mathrm{n}}^3{\mathrm{a}}^3}+\frac{2}{\mathrm{na}}\right]{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{na}}}{\mathrm{e}}^{\frac{-\mathrm{r}}{\mathrm{ma}}} \end{gathered}$$

reduce the equation

When , $\mathrm{r}\to 0$the terms (2) and (3) vanish, so the equation can be reduced to;

role="math" localid="1656136148574"

non-Hermitian of P4

Then, $P^4$is non-Hermitian.