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Derivation of the photoionization cross section

  1. Oct 12, 2011 #1
    Hallo Physics Forum,

    this is my first post, so please be gentle to me :). I am currently studying the relativistic photoionization process, and especially its differential cross section. Let us assume that we have a highly energetic, unpolarized photon beam traveling on the z-axis and colliding with unpolarized atoms. As a result these atoms get ionized by the absorption of photons. Here we do not observe the spin of these electrons. In Physics Reports 439 (2007) 1-99 it is stated on page 36 that the differential cross section for this process can be written as
    \begin{align}
    \frac{d \sigma^{ph}(\theta)}{d\theta} = \frac{\alpha \ m_e \ c^2}{4 \ \hbar \ \omega} \frac{\lambda_c^2}{2 (2 j_n+1)} \sum_{\mu_n} \ \sum_{m_s=\pm 1/2} \ \sum_{\lambda=\pm 1} |M_{p,n}(m_s,\lambda,\mu_n)|^2,
    \end{align}
    where [itex]\alpha[/itex] is the finestructure constant and [itex]\lambda_c = \frac{\hbar}{m_e c}[/itex] the Compton wavelength.
    The matrix element is written as
    \begin{align}
    M_{p,n}(m_s,\lambda,\mu_n) = \int \psi_{p,m_s}^\dagger(\mathbf{r}) \ \alpha \ u_\lambda \ \exp(i \mathbf{k r}) \psi_{j_n, \mu_n}(\mathbf{r}) \mathrm{d}^3 r,
    \end{align}
    where [itex]u_\lambda[/itex] is the polarization vector of the electromagnetic potential, [itex]\alpha[/itex] a vector of Dirac matrices, [itex]\psi_{j_n, \mu_n}(\mathbf{r})[/itex] the initial atomic wavefunction and [itex]\psi_{p,m_s}^\dagger(\mathbf{r})[/itex] the final continuous electron wavefunction. In the mentioned reference they state that this result has been obtained from a review article of Pratt (R.H. Pratt, A. Ron, H.K. Tseng, Rev. Mod. Phys. 45 (1973) 273). However, the given formula is not included in the article so I tried to understand it by myself.

    The starting point for me is Fermis Golden Rule under the condition that we average over the photon polarization (photon beam is unpolarized) and the electron total angular momentum (atoms are unpolarized) and sum over the spin of the final continuum electron (spin is not observed). Thenwe get
    \begin{align}
    W=\frac{2 \pi}{\hbar} \rho(E_f) \frac{1}{2 (2 j_n+1)} \sum_{\mu_n} \ \sum_{m_s=\pm 1/2} \ \sum_{\lambda=\pm 1} |H_{p,n}(m_s,\lambda,\mu_n)|^2
    \end{align}
    where [itex]\rho(E_f)[/itex] is the density of the final continuum electron states with energy [itex]E_f[/itex] and [itex]H_{p,n}(m_s,\lambda,\mu_n)[/itex] includes the initial and final wavefunctions and the operator for photoabsorption, which can, for example, be taken from Sakurai and reads
    \begin{align}
    c \sqrt{\frac{\hbar}{2 \omega V}} u_\lambda \exp(i \mathbf{kr}).
    \end{align}
    . For the density of states we get, if we neglect the motion of the nuclei,
    \begin{align}
    \rho(E_f) dE_f= \frac{V d^3 p_f}{(2 \pi \hbar)^3} = \frac{V p_f^2 dp_f d\Omega}{(2 \pi \hbar)^3} = \frac{V p_f E_f dE_f d\Omega}{(2 \pi \hbar)^3 c^2}.
    \end{align}
    This depends however on the energy and momentum of the final electron, which the stated formula from the reference does not (even if one would use an relativistic invariant version of fermis golden rule this would also be true). I do not see how the derived the formula in the beginning. Has anybody got an idea, how the have derived it?

    Thank you very much in advance,
    Syrius
     
  2. jcsd
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