Long-range approximations of the Uehling interaction

Question

A common approximation to the \begin{equation} U(\vec{r})=-m\frac{\alpha(Z\alpha)}{\pi} \int_1^\infty\mathrm{d}u\frac{\sqrt{u^2-1}\left(2u^2+1\right)}{3u^4}\frac{\exp(-2mur)}{mr} \tag{$\star$} \end{equation} Uehling interaction energy is a delta function approximation: \begin{equation} U_\delta(\vec{r})=-\frac{4}{15}\frac{\alpha(Z\alpha)}{m^2}\delta(\vec{r}) \ . \end{equation} This is usually not derived directly from $(\star)$, but by neglecting the higher momentum contributions of the one-loop vacuum polarization function $\Pi(-q^2)$ in momentum space (with $q=|\vec{q}|$): \begin{equation} U(\vec{q})=-Z\alpha\frac{4\pi}{q^2}\Pi(-q^2)=-Z\alpha\frac{4\pi}{q^2}\left(\frac{\alpha}{15\pi}\frac{q^2}{m^2}+{\cal{O}}(q^4/m^4)\right)\approx-\frac{4}{15}\frac{\alpha(Z\alpha)}{m^2} \ , \end{equation} from which $U_\delta$ is found by a Fourier transform. Clearly, this works under the assumption $q\ll m$, which in coordinate space means $r\gg1/m$: this is a long-range approximation for distances much larger than the (reduced) Compton wavelength.

However, in the literature, something different is meant by the $r\gg1/m$ approximation: \begin{equation} U_\text{h}(\vec{r})=-m\frac{\alpha(Z\alpha)}{4\sqrt{\pi}(mr)^{5/2}}\exp(-2mr) \ , \end{equation} which is found from a direct asymptotic expansion of $(\star)$. I often hear the claim (e.g. in Sec. 10.4.4 of Jentschura & Adkins) that $U_\text{h}$ should be thought of as an improvement over $U_\delta$, but I do not see how the two are related.

In order to compare the two approximations, I tried to approximately turn $U_\text{h}$ into a delta function form by using the following representation of the nascent delta function: \begin{equation} \delta(\vec{r};\epsilon)=\frac{1}{\epsilon^3}f(\vec{r}/\epsilon) \ \ , \ \ f(\vec{r})=\sqrt{\frac{2}{\pi}}\frac{\exp(-2r)}{4\pi r^{5/2}} \ . \end{equation} This might seem weird, but all requirements for the nascent delta function are satisfied: the function is absolutely integrable on $\mathbb{R}^3$, and we have \begin{align} \lim_{\epsilon\rightarrow0}\delta(\vec{r};\epsilon)= \begin{cases} 0 \ \ \ &\text{if $\vec{r}\neq\vec{\emptyset}$} \\ \infty \ \ \ &\text{if $\vec{r}=\vec{\emptyset}$} \end{cases} \ \ \ \ , \ \ \ \ \int\mathrm{d}^3r\delta(\vec{r};\epsilon)=1 \ . \end{align} Then we can write \begin{equation} U_\text{h}(\vec{r})=-\frac{\pi}{\sqrt{2}}\frac{\alpha(Z\alpha)}{m^2}\delta\left(\vec{r};\frac{1}{m}\right)\approx-\frac{\pi}{\sqrt{2}}\frac{\alpha(Z\alpha)}{m^2}\delta(\vec{r}) \ , \end{equation} where we used $1/m\ll r$ again in the last step. We ended up with a delta function, but with a wrong prefactor, so the claim that $U_\delta$ is just a cruder form of $U_\text{h}$ does not seem to be true. Interestingly, the same trick does lead to $U_\delta$ when applied directly on $(\star)$! Using \begin{equation} \delta(\vec{r};\epsilon)=\frac{1}{\epsilon^3}f(\vec{r}/\epsilon) \ \ , \ \ f(\vec{r})=\frac{\exp(-r)}{4\pi r} \ , \end{equation} we find \begin{align} U(\vec{r}) &= -m\frac{\alpha(Z\alpha)}{\pi} \int_1^\infty\mathrm{d}u\frac{\sqrt{u^2-1}\left(2u^2+1\right)}{3u^4}\frac{\pi}{m^3u^2}\delta\left(\vec{r};\frac{1}{2mu}\right) \\ &= -\frac{\alpha(Z\alpha)}{m^2} \int_1^\infty\mathrm{d}u\frac{\sqrt{u^2-1}\left(2u^2+1\right)}{3u^6}\delta\left(\vec{r};\frac{1}{2mu}\right) \\ &\approx-\frac{\alpha(Z\alpha)}{m^2}\delta(\vec{r})\int_1^\infty\mathrm{d}u\frac{\sqrt{u^2-1}\left(2u^2+1\right)}{3u^6} \\ &= -\frac{4}{15}\frac{\alpha(Z\alpha)}{m^2}\delta(\vec{r}) \ . \end{align} So, my main question is: how to properly understand $U_\delta$? Is it really "just" a long-range approximation of the Uehling interaction, or are there some other conditions/properties that I do not see? It cannot be obtained by further approximating the "usual" long-range approximation $U_\text{h}$, but then what is the actual relation between these two?

While $U_\delta$ is used routinely to compute the leading-order perturbative QED correction for atoms, calculating $\langle U_\text{h}\rangle$ (with or without the above delta function approximation) for e.g. the $1S$ ground state of hydrogen would give an incorrect contribution at ${\cal{O}}(m\alpha(Z\alpha)^4)$; higher order terms in the asymptotic expansion would be even divergent for $S$ states (the next term scales as $\sim\exp(-2mr)/r^{7/2}$). On one hand, this problematic behaviour is understandable, since the ground state wave function has an important part for $r<1/m$ too. But if both $U_\delta$ and $U_\text{h}$ are "just" long-range approximations, then why does one of them work, while the other does not?

Edit

Just to make my point more clear, let us compare the expectation values of $U_\delta$ and $U_\text{h}$ for the ground state of hydrogen (in the infinite nuclear mass limit): \begin{equation} \psi(\vec{r})=\sqrt{\frac{(m Z\alpha)^3}{\pi}}\exp(-m Z\alpha r) \ . \end{equation} We find \begin{equation} \langle\psi|U_\delta|\psi\rangle=-\frac{4}{15\pi}m\alpha(Z\alpha)^4 \ , \end{equation} but \begin{equation} \langle\psi|U_\text{h}|\psi\rangle=-\frac{1}{\sqrt{2(1+Z\alpha)}}m\alpha(Z\alpha)^4\approx-\frac{1}{\sqrt{2}}m\alpha(Z\alpha)^4 \ . \end{equation} We would get the same latter result with the delta function approximation of $U_\text{h}$. We know from the precision spectroscopy of hydrogen that $\langle U_\delta\rangle$ gives the correct ${\cal{O}}(m\alpha(Z\alpha)^4)$ vacuum polarization contribution to the energy shift (see e.g. p. 23 and 32-33 of the review by Eides, Grotch and Shelyuto). But then what is wrong with $\langle U_\text{h}\rangle$?

Answer 1

Let me just note that the long-range behavior of the potential is power-type (it is defined by higher terms in $\alpha$), not exponential (E. H. Wichmann, N. M. KrollL, Vacuum Polarization in a Strong Coulomb Field, Phys. Rev. (1956), v. 101, p. 843, eq. 60 and Appendix III, or A.M. Frolov, D.M. Wardlaw, Analytical formula for the Uehling potential, Eur. Phys. J. B (2012) 85: 348, eq. 10).

Answer 2

I found out the answer; it might be useful to someone in the future.

The two approximate interactions, $$ U_\delta=-\frac{4}{15}\frac{\alpha(Z\alpha)}{m^2}\delta(\vec{r}) \ \ \ , \ \ \ U_\text{h}=-m\frac{\alpha(Z\alpha)}{4\sqrt{\pi}(mr)^{5/2}}\exp(-2mr) \ , $$ are the leading terms of two different asymptotic expansions of the Uehling interaction, both being valid for $r\gg1/m$. However, as we will see, only $U_\delta$ is useful for extracting the leading vacuum polarization correction to atomic energy levels.

It is not hard to guess the form of the higher order terms in the delta function approximation: since $\Pi(-q^2)$ only depends on the square of $\vec{q}$, we must have for $q\ll m$ $$ U(\vec{q})=-Z\alpha\frac{4\pi}{q^2}\Pi(-q^2)=-\frac{4}{15}\frac{\alpha(Z\alpha)}{m^2}\left[1+c_1\left(-\frac{q^2}{m^2}\right)+c_2\left(-\frac{q^2}{m^2}\right)^2+...\right] \ , $$ which, after using $$ \int\frac{\mathrm{d}^3q}{(2\pi)^3}q^2\exp(i\vec{q}\cdot\vec{r})=-\nabla^2\int\frac{\mathrm{d}^3q}{(2\pi)^3}\exp(i\vec{q}\cdot\vec{r})=-\nabla^2\delta(\vec{r}) \ , $$ can be written in coordinate space as $$ U(\vec{r})=-\frac{4}{15}\frac{\alpha(Z\alpha)}{m^2}\left[1+c_1\left(\frac{\nabla^2}{m^2}\right)+c_2\left(\frac{\nabla^2}{m^2}\right)^2+...\right]\delta(\vec{r}) \ . \tag{$\ast$} $$ This expansion of course should be understood in a distributional sense. The actual value of the expansion coefficients is not interesting; it is sufficient to know that they are dimensionless, and do not depend on $\alpha$ or $Z\alpha$.

One could also continue the direct asymptotic analysis of $U(\vec{r})$ (see $(\star)$ in the question) to higher orders to find $$ U(\vec{r})=-m\frac{\alpha(Z\alpha)}{4\sqrt{\pi}(mr)^{5/2}}\exp(-2mr) \left[1+\frac{d_1}{mr}+\frac{d_2}{(mr)^2}+...\right] \ . \tag{$\ast\ast$} $$ The same remarks hold for these expansion coefficients as before; see Eq. (10.246) of Quantum Electrodynamics by Jentschura & Adkins for the numerical values of the first few of them.

Neither of these expansions seems useful for calculating the energy shift $\langle\psi_k|U|\psi_k\rangle$ for hydrogenic states, because the separate terms are all divergent starting from a sufficiently high order: for example, already the second term is divergent for $S$ states in both expansions!

Let us try to extract only the leading, ${\cal{O}}(m\alpha(Z\alpha)^4)$ energy contribution from $\langle\psi_k|U|\psi_k\rangle$ (with $k=\{n,l,j,m_j\}$). In the delta function expansion of $(\ast)$, the $p$-th term (after integrating by parts $2p$ times) gives $$ \begin{aligned} \frac{\alpha(Z\alpha)}{m^2}\int\mathrm{d}^3r|\psi_k(\vec{r})|^2\left(\frac{\nabla^2}{m^2}\right)^p\delta(\vec{r}) &= \frac{\alpha(Z\alpha)}{m^2}\int\mathrm{d}^3r\delta(\vec{r})\left(\frac{\nabla^2}{m^2}\right)^p|\psi_k(\vec{r})|^2 \\ &= \frac{\alpha(Z\alpha)}{m^2}\left[\left(\frac{\nabla^2}{m^2}\right)^p|\psi_k(\vec{r})|^2\right]_{\vec{r}=\vec{\emptyset}} \\ &= m \alpha(Z\alpha)^{4+2p}\left[\left(\nabla_{\vec{u}}^2\right)^p|f_k(\vec{u})|^2\right]_{\vec{u}=\vec{\emptyset}} \\ &\sim m\alpha(Z\alpha)^{4+2p} \ , \end{aligned} $$ where we used the $$ \psi_k(\vec{r})=(mZ\alpha)^{3/2}f_k(mZ\alpha\vec{r}) $$ form of the hydrogenic functions to scale out the $Z\alpha$-dependence. The above result shows that the complete ${\cal{O}}(m\alpha(Z\alpha)^4)$ contribution of the vacuum polarization to QED energy shifts can be obtained from the first ($p=0$) term: $$ \Delta E_{\text{VP}}=-\frac{4}{15}\frac{\alpha(Z\alpha)}{m^2}|\psi_{k}(\vec{\emptyset})|^2=-\frac{4}{15\pi n^3}m\alpha(Z\alpha)^4\delta_{l0} \ . $$ The other terms only count in higher $Z\alpha$ orders (their naive divergences regularized by other similarly divergent QED contributions of the same $Z\alpha$ order).

Now let us look at the $p$-th term of the $(\ast\ast)$ expansion. To keep things simple, let us consider the ground state of hydrogen: $$ \begin{aligned} m\alpha(Z\alpha)\int\mathrm{d}^3r|\psi_0(\vec{r})|^2\frac{\exp(-2mr)}{(mr)^{p+5/2}} &= 4m^2\alpha(Z\alpha)^4 \int_0^\infty\mathrm{d}r\frac{\exp(-2m(1+Z\alpha)r)}{(mr)^{p+1/2}} \\ &= 4m\alpha(Z\alpha)^4(1+Z\alpha)^{p-1/2} \int_0^\infty\mathrm{d}u\frac{\exp(-2u)}{u^{p+1/2}} \\ &={\cal{O}}(m\alpha(Z\alpha)^4) \ , \end{aligned} $$ which shows that all terms of this expansion contribute to ${\cal{O}}(m\alpha(Z\alpha)^4)$ (more generally, to ${\cal{O}}(m\alpha(Z\alpha)^{4+2l})$ for a state of angular momentum $l$). If we wanted to extract the leading QED contribution from this expansion, we should somehow tame the divergences of the separate terms, and resum the series, which would be an extremely and unnecessarily complicated way to find the same simple result as before.

This is how we know that $U_\delta$ yields the correct result to ${\cal{O}}(m\alpha(Z\alpha)^4)$ accuracy, and why $U_\text{h}$ could not reproduce the same result in my question.

Overall, I think the claim often found in textbooks (e.g. in Peskin & Schroeder, Greiner & Reinhardt, Itzykson & Zuber or Jentschura & Adkins) about $U_\delta$ being an approximation to $U_\text{h}$ should be taken with some reservation; the two are not directly related, and while $U_\text{h}$ gives a deeper insight into the physics of vacuum polarization, $U_\delta$ is superior when it comes to bound state QED calculations.

Of course, the $Z\alpha$ expansion becomes meaningless for high $Z$ (as in highly charged one-electron ions), in which case a fully relativistic calculation is required with the complete vacuum polarization potential.