Q15P Show that聽聽P2is Hermitian, but... [FREE SOLUTION]

Chapter 6: Q15P (page 270)

Show that $P^{2}$ is Hermitian, but $P^{4}$ is not, for hydrogen states with $l = 0$ . Hint: For such states $蠄$ is independent of $胃$ and $蠒$ , so
localid="1656070791118" $p^{2} = - \frac{{鈩�}^{2}}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r})$
(Equation 4.13). Using integration by parts, show that
localid="1656069411605" $< f 鈭� p^{2} g > = - 鈹� 4 蟺 h^{2} (r^{2} f \frac{d g}{d r} - r^{2} g \frac{d f}{d r})_{0}^{鈭�} + < p^{2} f 鈭� g >$
Check that the boundary term vanishes for $蠄_{n 00}$ , which goes like
$蠄_{n 00} ~ \frac{1}{\sqrt{蟺 (n a)^{3 / 2}}} e x p (- r / n a)$
near the origin. Now do the same for $p^{4}$ , and show that the boundary terms do not vanish. In fact:
$< 蠄_{n 00} 鈭� p^{4} 蠄_{m 00} > = \frac{8 {鈩�}^{4}}{a^{4}} \frac{(n - m)}{(n m)^{5 / 2}} + < p^{4} 蠄_{n 00} 鈭� 蠄_{m 00} >$

Short Answer

Expert verified

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

Step by step solution

Hermitian for P2

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

$P^{2} = \frac{- {鈩�}^{2}}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r})$

The $p^{2}$ is Hermitian if,

$鉄� f 鈭� P^{2} g 鉄� = 鉄� P^{2} f 鈭� g 鉄�$

so, to show that, we use:

$鉄� f 鈭� P^{2} g 鉄� = - {鈩�}^{2} {鈭�}_{0}^{鈭�} f \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr}) 4 {蟺谤}^{2} dr = - 4 蟺 {鈩�}^{2} {鈭�}_{0}^{鈭�} f \frac{d}{dr} (r^{2} \frac{dg}{dr}) dr$

use integration by parts to solve for Hermitian of P2 .

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

$u = f, 鈥� dv = \frac{d}{dr} (r^{2} \frac{dg}{dr}) dr du = \frac{df}{dr} dr, 鈥� v = r^{2} \frac{dg}{dr} l = r^{2} f \frac{dg}{dr}_{0}^{鈭�} - {鈭�}_{0}^{鈭�} r^{2} \frac{df}{dr} \frac{dg}{dr} dr$

again, use integration by parts

By using the integration by parts once more,

$u = r^{2} \frac{df}{dr}, 鈥� dv = \frac{dg}{dr} dr du = \frac{d}{dr} (r^{2} \frac{df}{dr}) dr, 鈥� v = {gI}^{'} = r^{2} g \frac{df}{dr} |_{0}^{鈭�} - {鈭�}_{0}^{鈭�} \frac{d}{dr} (r^{2} \frac{df}{dr}) gdr$

then,

$< f 鈭� p^{2} g > = - 4 {蟺丑}^{2} (r^{2} f \frac{dg}{dr} - r^{2} g \frac{df}{dr})_{0}^{鈭�} + {鈭�}_{0}^{鈭�} \frac{d}{dr} (r^{2} \frac{df}{dr}) gdr$

check at boundary

- If

$r 鈫� 0 鈬� = 0 r 鈫� 鈭� 鈬� g f 鈫� 0 (r^{2} f \frac{dg}{dr} - r^{2} g \frac{df}{dr}) |_{0}^{鈭�}$

vanishes at the boundaries.

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

solve further

$鉄� f 鈭� P^{2} g 鉄� = - {鈩�}^{2} {鈭�}_{0}^{鈭�} \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{d f}{dr}) g . 4 {蟺谤}^{2} dr 鈬� 鉄� f 鈭� P^{2} g 鉄� = 鉄� P^{2} f \lg 鉄�$

Then, $P^{2}$ is Hermitian.

solve for P4

Now, for $P^{4}$ :

$P^{4} = \frac{{鈩�}^{4}}{r^{2}} \frac{d}{dr} [r^{2} \frac{d}{dr} (\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{d}{dr}))]$

By applying the same method,

$鉄� f 鈭� P^{4} g 鉄� = {鈩�}^{4} {鈭�}_{0}^{鈭�} \frac{1}{r^{2}} f \frac{d}{dr} [r^{2} \frac{d}{dr} (r^{2} \frac{dg}{dr})] 4 {蟺谤}^{2} dr = 4 蟺 {鈩�}^{4} {鈭�}_{0}^{鈭�} f \frac{d}{dr} [r^{2} \frac{d}{dr} (\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr}))] dr$

use integration by parts

By using integration by parts:

$u = f, 鈥� dv = \frac{d}{dr} [r^{2} \frac{d}{dr} (\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr}))] dr du = \frac{df}{dr} dr, 鈥� v = r^{2} \frac{d}{dr} (\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr}))$

Solve further

$I_{1} = r^{2} f \frac{d}{dr} [\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr})] |_{0}^{鈭�} - {鈭�}_{0}^{鈭�} r^{2} \frac{df}{dr} \frac{d}{dr} [\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr})] dr u = r^{2} \frac{df}{dr}, 鈥� 鈥� 鈥� dv = \frac{d}{dr} [\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr})] dr du = \frac{d}{dr} (r^{2} \frac{df}{dr}) dr v = \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr})$

solve further

localid="1656073604117" $<f 鈭� P^{4} g> = 4 蟺 {鈩�}^{4} [\begin{matrix} (r^{2} f \frac{d}{dr} [\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dg}{dr})]) - \frac{df}{dr} \frac{d}{dr} (r^{2} \frac{dg}{dr}) + \frac{d}{dr} (r^{2} \frac{df}{dr}) \frac{dg}{dr} \\ - r^{2} g \frac{d}{dr} [\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr})] \end{matrix}] |_{0}^{鈭�} + <P^{4} f 鈭� g>$

localid="1656131027955" role="math" $u = \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr}) l_{2} = \frac{df}{dr} \frac{d}{dr} (r^{2} \frac{dg}{dr})_{0}^{鈭�} - {鈭�}_{0}^{鈭�} \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr}) \frac{d}{dr} (r^{2} \frac{dg}{dr}) drdr I_{3} = \frac{d}{dr} (r^{2} \frac{df}{dr}) \frac{dg}{dr} |_{0}^{鈭�} - {鈭�}_{0}^{鈭�} r^{2} \frac{d}{dr} [\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr})] \frac{dg}{dr} dr u = r^{2} \frac{d}{dr} / dr [\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr})]$

$dv = \frac{dg}{dr} dr du = \frac{d}{dr} [r^{2} \frac{d}{dr} ((\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr}))] dr, 鈥� 鈥� 鈥� v = g, I_{4} = r^{2} g \frac{d}{dr} [\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr})] |_{0}^{鈭�} - {鈭�}_{0}^{鈭�} \frac{d}{dr} [r^{2} \frac{d}{dr} (\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr}))] gdr <鈭� g> = {鈭�}_{0}^{鈭�} \frac{d}{dr} [r^{2} \frac{d}{dr} (\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{df}{dr}))] gdr$

check the boundary terms

To check the boundary terms:

Take,

$f (r) = e^{^{\frac{- r}{na}}} 鈥� g (r) = e^{\frac{- r}{ma}}$

By substituting into the last equation.

$\frac{dg}{dr} = \frac{- 1}{ma} e^{\frac{- r}{ma}} \frac{df}{dr} = \frac{- 1}{na} e^{\frac{- r}{{nar}^{2}}} \frac{dg}{dr} = (\frac{- r^{2}}{ma}) e^{\frac{- r}{ma}}$

Solve further

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

solve further

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

Solve further

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

reduce the equation

When , $r 鈫� 0$ the terms (2) and (3) vanish, so the equation can be reduced to;

role="math" localid="1656136148574" $<f 鈭� P^{4} g> = 4 蟺 {鈩�}^{4} [\frac{2}{ma} - \frac{2}{na}] + <P^{4} f 鈭� g> = \frac{8 蟺 {鈩�}^{4}}{a} (\frac{1}{m} - \frac{1}{n}) + <P^{4} f 鈭� g>$

non-Hermitian of P4

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

$蠄_{n 00} 鈮� {\frac{1}{\sqrt{蟺} {(na)}^{\frac{3}{2}}}}^{e^{\frac{- r}{a}}} 蠄_{m 00} 鈮� \frac{1}{\sqrt{蟺} {(ma)}^{\frac{3}{2}}} e^{\frac{- r}{ma}} <蠄_{n 00} 鈭� P^{4} 蠄_{m 00}> = \frac{8 蟺 {鈩�}^{4}}{a} (\frac{n - m}{nm}) \frac{1}{{蟺补}^{3} {(nm)}^{\frac{3}{2}}} + <P^{4} 蠄_{n 00} 鈭� 蠄_{m 00}> = \frac{8 {鈩�}^{4}}{a^{4}} \frac{(n - m)}{{(nm)}^{\frac{5}{2}}} + <P^{4} 蠄_{n 00} 鈭� 蠄_{m 00}> 鉄� P^{4} 蠄_{n 00} 鈭� 蠄_{m 00} 鉄� = \frac{8 {鈩�}^{4}}{a^{4}} \frac{(n - m)}{{(nm)}^{\frac{5}{2}}} + <P^{4} 蠄_{n 00} 鈭� 蠄_{m 00}>$

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Short Answer

Step by step solution

Hermitian for P2

use integration by parts to solve for Hermitian of P2 .

again, use integration by parts

check at boundary

solve further

solve for P4

use integration by parts

check the boundary terms

solve further

reduce the equation

non-Hermitian of P4

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