May 5, 2015

Atomic Form Factors for Hydrogen

The form factor of an atom is defined, in the non-relativistic quantum mechanics, as the Fourier transformation of the probability density: $F(\vec q) = \int d^3 r e^{i\vec q\cdot \vec r} \psi^*(\vec r) \psi(\vec r),$ where $\psi$ is the wavefunction. The form factor characterizes the charge distribution of the atoms, as is probed by an external source (e.g., X Ray, E & M field etc). For example, $F(q \to 0) = Z$ gives the total charge number.

At small momentum transfer, the atomic form factor is approximate the same as the form factor $F_1(Q^2)$ computed from relativistic quantum mechanics, esp. QFT. These are non-perturbative objects. Therefore, the analytical expression of the atomic form factor provides a useful benchmark for non-perturbative solution of bound states in the relativistic quantum mechanics. In QM, the readily known closed-form bound states include the Coulomb potential bound states (e.g. hydrogen atom) and the harmonic oscillator potential. The Coulomb potential can be used to compare with the heavy quarkonium.

Form Factor for the Two-Body Coulomb Bound States

Here in this post, I evaluate the integral for the Coulomb potential bound states, i.e., the hydrogen like atoms. These results are well known in atomic and nuclear physics. For example, the form factor for the S-wave ground state is well known, as $F(q^2) = (1 + 4 a^2 q^2 )^{-2}$ where $a = \hbar / (\alpha m_e c)$  is the Bohr radius. In this post, we try to discuss some high excited states.

The wavefunction of the hydrogen atom in the spherical coordinate is $\psi_{n\ell{}m}(r, \theta, \phi) = R_{n\ell}(r) Y_\ell^m(\theta, \phi)$ where $R(r)$ is the radial wavefunction and $Y_\ell^m$ is the spherical harmonics. This is the general structure of the wavefunction for central forces. In the particular example of the Coulomb potential, $R_{n\ell}(r) = (2n a)^{-\frac{3}{2}} \sqrt{\frac{(n-\ell-1)!}{2n(n+\ell)!}} e^{-\rho/2} \rho^\ell L_{n-\ell-1}^{2\ell+1}(\rho)$ where $\rho = r/(2na)$.

There are two vectors in the picture: the polarization vector, chosen as the $\hat z$ direction; the vector $\hat q$. The form factor depends on $q^2 = \vec q^2$ and $\hat q\cdot \hat z = \cos\vartheta$. Let's first expand the wave factor in the spherical harmonics: $e^{i \vec q \cdot \vec r} = 4\pi \sum_{L=0}^\infty \sum_{M=-L}^L i^L j_L(qr) Y_L^{M*}(\vartheta,\varphi) Y_L^M(\theta, \phi)$ Here $j_L(z)$ is the spherical Bessel function and $\vec q = (q, \vartheta, \varphi)$. Then, the form factor becomes, $F(\vec q) = \int \rho^2 d\rho R_{n\ell}(\rho) \sum_{L=0}^\infty \sum_{M=-L}^L i^L j_L(\Delta \rho) Y_L^{M*}(\vartheta, \varphi) \int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\phi Y_L^M(\theta, \phi) Y_\ell^m(\theta, \phi) Y_\ell^{-m}(\theta, \phi)$ There exists a standard formula for the angular integral using the Wigner 3j-symbols: $\int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\phi Y_L^M(\theta, \phi) Y_\ell^m(\theta, \phi) Y_\ell^{-m}(\theta, \phi) \\ = (2\ell+1) \sqrt{\frac{2L+1}{4\pi}} \begin{pmatrix} \ell & \ell & L \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} \ell & \ell & L \\ -m & m & M \end{pmatrix}.$ Apply this formula, the form factor becomes, $F(\vec q) = (2\ell+1) \sum_{k=0}^{\ell} (-1)^k (4k+1) \begin{pmatrix} \ell & \ell & 2k \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} \ell & \ell & 2k \\ -m & m & 0 \end{pmatrix} P_{2k}(\cos\vartheta) \\ \times \int\limits_0^\infty d\rho \, R_{n\ell}^2(\rho) j_{2k}(\Delta \rho).$ We have taken into account the properties of the 3j-symbols. I didn't find a general formula for Coulomb potential or the harmonic oscillator potential, though closed-form expression exists for some special cases. Nevertheless, the radial integral is not hard to do for given $n,\ell,m$ (one can for example, expand the Laguerre polynomial. But the result is not particularly illuminating.).

Let's make some observations. First of all, the result does not depend on the azimuthal angle $\varphi$, as we expect. Second, for S-states, it only depend on $q^2 = \vec q^2$. For non-vanishing orbital angular momentum $\ell$, the result also depends on the dot product $\cos\vartheta = \hat z \cdot \hat q$. Last but not least, the atomic form factor contains all kinds of multipole contributions: the monopole $k=0$, the quadrupole $P_2(\cos\vartheta)$ etc, up to $P_{2\ell}(\cos\vartheta)$.

Finally, let's look at the rms radius, which is defined as $R = \sqrt{\langle r^2 \rangle}$, and $\langle \vec r^2 \rangle = \int d^3 r \, \vec r^2 \psi^*(\vec r) \psi(\vec r)$ For S-states, the rms radius is related to the origin of the form factor by the famous formula, $\langle r^2 \rangle = -6 \frac{\partial}{\partial q^2} F(q^2)\Big|_{q \to 0}$ More generally, it can be shown that $\langle \vec r^2 \rangle = -\nabla^2_{\vec q} F(\vec q)\Big|_{q\to 0} = 6 \bigg[ - \frac{\partial }{ \partial q^2} + \frac{P_2(\cos\vartheta)}{q^2} \frac{\partial }{\partial P_2(\cos\vartheta)} \bigg] F(q, \cos\vartheta) \Big|_{q\to 0}.$ Here $P_2(\cos\vartheta) = 3\cos^2\vartheta - 1$ is the 2nd Legendre polynomial.
In relativistic quantum mechanics, the form factors are defined through the matrix element of the current operator: $\langle p', \sigma' | J^\mu | p, \sigma \rangle.$