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### Q: [1] pr 5.16

For a particle in a spherically symmetric potential \( V(r) \) show that

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:20}

\Abs{\psi(0)}^2 = \frac{m}{2 \pi \Hbar^2} \expectation{ \frac{dV}{dr} },

\end{equation}

for all s-states, ground or excited.

Then show this is the case for the 3D SHO and hydrogen wave functions.

### A:

The text works a problem that looks similar to this by considering the commutator of an operator \( A \), later set to \( A = p_r = -i \Hbar \PDi{r}{} \) the radial momentum operator. First it is noted that

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:40}

0 = \bra{nlm} \antisymmetric{H}{A} \ket{nlm},

\end{equation}

since \( H \) operating to either the right or the left is the energy eigenvalue \( E_n \). Next it appears the author uses an angular momentum factoring of the squared momentum operator. Looking earlier in the text that factoring is found to be

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:60}

\frac{\Bp^2}{2m}

= \inv{2 m r^2} \BL^2 – \frac{\Hbar^2}{2m} \lr{ \PDSq{r}{} + \frac{2}{r} \PD{r}{} }.

\end{equation}

With

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:80}

R = – \frac{\Hbar^2}{2m} \lr{ \PDSq{r}{} + \frac{2}{r} \PD{r}{} }.

\end{equation}

we have

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:100}

\begin{aligned}

0

&= \bra{nlm} \antisymmetric{H}{p_r} \ket{nlm} \\

&= \bra{nlm} \antisymmetric{\frac{\Bp^2}{2m} + V(r)}{p_r} \ket{nlm} \\

&= \bra{nlm} \antisymmetric{\inv{2 m r^2} \BL^2 + R + V(r)}{p_r} \ket{nlm} \\

&= \bra{nlm} \antisymmetric{\frac{-\Hbar^2 l (l+1)}{2 m r^2} + R + V(r)}{p_r} \ket{nlm}.

\end{aligned}

\end{equation}

Let’s consider the commutator of each term separately. First

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:120}

\begin{aligned}

\antisymmetric{V}{p_r} \psi

&=

V p_r \psi

–

p_r V \psi \\

&=

V p_r \psi

–

(p_r V) \psi

–

V p_r \psi \\

&=

–

(p_r V) \psi \\

&=

i \Hbar \PD{r}{V} \psi.

\end{aligned}

\end{equation}

Setting \( V(r) = 1/r^2 \), we also have

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:160}

\antisymmetric{\inv{r^2}}{p_r} \psi

=

-\frac{2 i \Hbar}{r^3} \psi.

\end{equation}

Finally

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:180}

\begin{aligned}

\antisymmetric{\PDSq{r}{} + \frac{2}{r} \PD{r}{} }{ \PD{r}{}}

&=

\lr{ \partial_{rr} + \frac{2}{r} \partial_r } \partial_r

–

\partial_r \lr{ \partial_{rr} + \frac{2}{r} \partial_r } \\

&=

\partial_{rrr} + \frac{2}{r} \partial_{rr}

–

\lr{

\partial_{rrr} -\frac{2}{r^2} \partial_r + \frac{2}{r} \partial_{rr}

} \\

&=

-\frac{2}{r^2} \partial_r,

\end{aligned}

\end{equation}

so

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:200}

\antisymmetric{R}{p_r}

=-\frac{2}{r^2} \frac{-\Hbar^2}{2m} p_r

=\frac{\Hbar^2}{m r^2} p_r.

\end{equation}

Putting all the pieces back together, we’ve got

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:220}

\begin{aligned}

0

&= \bra{nlm} \antisymmetric{\frac{-\Hbar^2 l (l+1)}{2 m r^2} + R + V(r)}{p_r} \ket{nlm} \\

&=

i \Hbar

\bra{nlm} \lr{

\frac{\Hbar^2 l (l+1)}{m r^3} – \frac{i\Hbar}{m r^2} p_r +

\PD{r}{V}

}

\ket{nlm}.

\end{aligned}

\end{equation}

Since s-states are those for which \( l = 0 \), this means

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:240}

\begin{aligned}

\expectation{\PD{r}{V}}

&= \frac{i\Hbar}{m } \expectation{ \inv{r^2} p_r } \\

&= \frac{\Hbar^2}{m } \expectation{ \inv{r^2} \PD{r}{} } \\

&= \frac{\Hbar^2}{m } \int_0^\infty dr \int_0^\pi d\theta \int_0^{2 \pi} d\phi r^2 \sin\theta \psi^\conj(r,\theta, \phi) \inv{r^2} \PD{r}{\psi(r,\theta,\phi)}.

\end{aligned}

\end{equation}

Since s-states are spherically symmetric, this is

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:260}

\expectation{\PD{r}{V}}

= \frac{4 \pi \Hbar^2}{m } \int_0^\infty dr \psi^\conj \PD{r}{\psi}.

\end{equation}

That integral is

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:280}

\int_0^\infty dr \psi^\conj \PD{r}{\psi}

=

\evalrange{\Abs{\psi}^2}{0}{\infty} – \int_0^\infty dr \PD{r}{\psi^\conj} \psi.

\end{equation}

With the hydrogen atom, our radial wave functions are real valued. It’s reasonable to assume that we can do the same for other real-valued spherical potentials. If that is the case, we have

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:300}

2 \int_0^\infty dr \psi^\conj \PD{r}{\psi}

=

\Abs{\psi(0)}^2,

\end{equation}

and

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:320}

\boxed{

\expectation{\PD{r}{V}}

= \frac{2 \pi \Hbar^2}{m } \Abs{\psi(0)}^2,

}

\end{equation}

which completes this part of the problem.

### A: show this is the case for the 3D SHO and hydrogen wave functions

For a hydrogen like atom, in atomic units, we have

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:360}

\begin{aligned}

\expectation{

\PD{r}{V}

}

&=

\expectation{

\PD{r}{} \lr{ -\frac{Z e^2}{r} }

} \\

&=

Z e^2

\expectation

{

\inv{r^2}

} \\

&=

Z e^2 \frac{Z^2}{n^3 a_0^2 \lr{ l + 1/2 }} \\

&=

\frac{\Hbar^2}{m a_0} \frac{2 Z^3}{n^3 a_0^2} \\

&=

\frac{2 \Hbar^2 Z^3}{m n^3 a_0^3}.

\end{aligned}

\end{equation}

On the other hand for \( n = 1 \), we have

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:380}

\begin{aligned}

\frac{2 \pi \Hbar^2}{m} \Abs{R_{10}(0)}^2 \Abs{Y_{00}}^2

&=

\frac{2 \pi \Hbar^2}{m} \frac{Z^3}{a_0^3} 4 \inv{4 \pi} \\

&=

\frac{2 \Hbar^2 Z^3}{m a_0^3},

\end{aligned}

\end{equation}

and for \( n = 2 \), we have

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:400}

\begin{aligned}

\frac{2 \pi \Hbar^2}{m} \Abs{R_{20}(0)}^2 \Abs{Y_{00}}^2

&=

\frac{2 \pi \Hbar^2}{m} \frac{Z^3}{8 a_0^3} 4 \inv{4 \pi} \\

&=

\frac{\Hbar^2 Z^3}{4 m a_0^3}.

\end{aligned}

\end{equation}

These both match the potential derivative expectation when evaluated for the s-orbital (\( l = 0 \)).

For the 3D SHO I verified the ground state case in the Mathematica notebook sakuraiProblem5.16bSHO.nb

There it was found that

\begin{equation}\label{eqn:symmetricPotentialDerivativeExpectation:420}

\expectation{\PD{r}{V}}

= \frac{2 \pi \Hbar^2}{m } \Abs{\psi(0)}^2

= 2 \sqrt{\frac{m \omega ^3 \Hbar}{ \pi }}

\end{equation}

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.