Moments of hydrogen orbitals/Transition dipole moments

Question

I'm interested in calculating transition dipole elements for atomic transitions. This means I would like to calculate things like

$$ r_{nlm,q}^{n'l'm'} = \langle\psi_{nlm}|r_q|\psi_{n'l'm'}\rangle $$

Where $|\psi_{nlm}\rangle$ is a hydrogen wavefunction and \begin{align} r_{\pm 1} =& \frac{1}{\sqrt{2}}\left(\mp\hat{x}-i\hat{y}\right)\\ r_0 = \hat{z} \end{align} are the spherical basis vectors. I am curious for a reference, or someone demonstrating a calculation in the answer, for how to calculate $r_{nlm,q}^{n'l'm'}$ in general. This can be calculated numerically and tabulated pretty easily using the formula for $|\psi_{nlm}\rangle$, but I wonder if there is a closed form formula for it?

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More work:

From the Wigner-Eckart theorem (using the convention in the Steck references) we get

$$ r_{nlm,q}^{n'l'm'} = \langle nl||r||n'l'\rangle \langle lm|l'm',1q\rangle $$ Where the first term on the right is the $m$-, $m'$-independent reduced matrix element and the second term on the right is a Clebsch-Gordon coefficient.

My question then reduces to asking for a closed form for $\langle nl||r||n'l'\rangle$.

Again, using the Steck reference, we can calculate

$$ \langle nl||r||n'l'\rangle = \sum_{m'q}\langle \psi_{nlm}|r_q|\psi_{n'l'm'}\rangle\langle{lm|l'm',1q}\rangle $$

So I want to emphasize that use of the Wigner-Eckart theorem has not absolved us from having to calculate $\langle \psi_{nlm}|r_q|\psi_{n'l'm'}\rangle$, it has just saved us from having to tabulate separately values for each $m, m'$ pair.

Answer 1

We can break this down.

\begin{align} \psi_{nlm}(r, \theta, \phi) =& R_{nl}(r)Y_l^m(\theta, \phi)\\ R_{nl}(r) =& N_{nl} e^{-\frac{r}{na_0}} \left(\frac{2r}{na_0}\right)^l L_{n-l-1}^{2l+1}\left(\frac{2r}{na_0}\right)\\ N_{nl} =& \sqrt{\left(\frac{2}{na_0}\right)^3 \frac{(n-l-1)!}{2n(n+l)!}} \end{align} Where $Y_l^m$ is a spherical harmonic and $L_{n-l-1}^{2l+1}$ is a generalized Laguerre polynomial.

We seek a formula for $$ r_{nlm}^{n'l'm',q} = \langle nlm |\hat{r}_q|n'l'm'\rangle $$ which we expand as $$ r_{nlm}^{n'l'm',q} = \sqrt{\frac{4\pi}{3}}\int R_{nl}(r)R_{n'l'}(r) r^3 dr \int\int Y_l^{m*}(\theta, \phi)Y_{l'}^{m'}(\theta, \phi)Y_1^q(\theta, \phi)\sin(\theta) d\theta d\phi $$ Where we've used $r_q = \sqrt{\frac{4\pi}{3}} r Y_1^q(\theta, \phi)$ and $dV = r^2\sin(\theta)drd\theta d\phi$. The term on the right can be calculated as $$ Y_{ll'1}^{mm'q} = \int\int Y_l^{m*}(\theta, \phi)Y_{l'}^{m'}(\theta, \phi)Y_1^q(\theta, \phi)\sin(\theta) d\theta d\phi = (-1)^m\sqrt{\frac{(2l+1)(2l'+1)(2\times 1 + 1)}{4\pi}}\begin{pmatrix}l && l' && 1\\0 && 0 && 0\end{pmatrix}\begin{pmatrix}l && l' && 1\\-m && m' && q\end{pmatrix} $$ Where the expressions in parentheses are Wigner 3j symbols.

We now turn to the calculation $$ R_{nl}^{n'l'} = \int R_{nl}(r)R_{n'l'}(r) r^3 dr $$ Exactly this expression is calculated in this answer for $k=1$ so we'll consider $R_{nl}^{n'l'}$ to be known.

So we have $$ r_{nlm}^{n'l'm',q} = \sqrt{\frac{4\pi}{3}}R_{nl}^{n'l'}Y_{ll'1}^{mm'q} $$ all of which are known. This answers the problem in the question.

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There is a related question about determining the reduced matrix element $\langle nl|\hat{r}|n'l'\rangle$ that appears in the Wigner-Eckart theorem for this problem: $$ r_{nlm}^{n'l'm',q} = \langle nlm|\hat{r}_q|n'l'm'\rangle = \langle nl||\hat{r}||n'l'\rangle \langle lm|l'm', 1q\rangle $$ There are two ways we can evaluate the reduced matrix element: \begin{align} \langle nl||\hat{r}||n'l'\rangle =& \frac{\langle nlm|\hat{r}_q|n'l'm'\rangle}{\langle{lm|l'm',1q\rangle}}\\ \langle nl||\hat{r}||n'l'\rangle =& \sum_{m'q} \langle nlm|\hat{r}_q|n'l'm'\rangle \langle lm|l'm', 1q\rangle \end{align}

We'll work on the latter expression. We note we must arbitrarily choose a value for $m$ with $-l\le m \le l$ which will eventually collapse the $m'$ sum to $\delta_{m', m-q}$. We expand

$$ \langle nl||\hat{r}||n'l'\rangle = R_{nl}^{n'l'}\sqrt{\frac{4\pi}{3}}\sum_{m'q} Y_{ll'1}^{mm'q} \langle lm|l'm', 1q\rangle $$ So we see that $\langle nl||\hat{r}||n'l'\rangle$ is directly equal to $R_{nl}^{n'l'}$ multiplied by a complicated, but easy to numerically evaluate, sum over products of Wigner 3j/Clebsch Gordon coefficients.