Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Matrix element for direct reactions

  1. Mar 10, 2014 #1
    I apologise since I already have a question on this board, but I've been stuck for a good few hours understanding exactly how this has been done. The differential cross section for a direct reaction from [itex]\alpha[/itex] to [itex]\beta[/itex] is given by

    [itex] \frac{d\sigma}{d\Omega} = f(k,k')|T_{\beta \alpha}|^2 [/itex]

    f(k,k') is just a constant dependent on the initial and final momentums. For a one-nucleon transfer, e.g. A(B,C)D where B may be a deuteron and C a proton, [itex] T_{\alpha \beta} [/itex] is given by

    [itex] T_{\alpha \beta} \int e^{-i\mathbf{k}\cdot\mathbf{r_{\beta}}} \int \psi^*_C \psi^*_D V \psi_A \psi_B d\tau e^{i\mathbf{k}\cdot\mathbf{r_{\alpha}}} d^3\mathbf{r_{\alpha}} d^3\mathbf{r_{\beta}} [/itex]

    [itex] \alpha [/itex] represents the relative vector distance between incident particle A and target B
    [itex] \beta [/itex] ^^ projectile C and residual nucleus D

    V is the potential responsible for the direct reaction process.

    [itex] d\tau [/itex] is the integral over all internal coordinates.

    For reference Bertulani (nuclear physics in a nutshell) chapter 11.1 and Wong (intro to nuclear) chapter 8.3 focus on this.

    There are three approximations then used:
    (1) [itex] r_{\alpha} = r_{\beta} = r [/itex] because of the short range of [itex] V [/itex].
    (2) [itex] V = V_0\delta(r-R) [/itex] because the interaction is short ranged, but below a certain distance the formation of a compound nucleus is more favourable.
    (3) For a (d,p) reaction, which is what we're considering, the neutron is transferred to a single particle state (M=0) of B. i.e. [itex] \psi_D = \psi_B \phi_n = \psi_B f(r) Y_{L0}(\theta,\phi) [/itex]

    What I'm then confused with is that I'm told these approximations are used to obtain

    [itex] T_{\alpha \beta} = V_0\int e^{i (\mathbf{k - k'}) \cdot r} Y_{L0}(\theta,\phi)^*\delta(r-R) d^3\mathbf{r} [/itex]

    I think some factors were left out because we're interesting in really the angular dependence of the differential cross section. I've tried getting to this myself, but I'm a bit confused as to what is a function of what variables for integration, and I also have no idea how [itex] Y_{L0} [/itex] makes its way outside of the integral for [itex] d\tau [/itex]. This is as far as I can get subbing in and using the approximations above;

    [itex] T_{\alpha \beta} = V_0 \int e^{i (\mathbf{k - k'})\cdot\mathbf{r}} \int \psi^*_C \psi^*_D \delta(r-R) \psi_A \psi_B d\tau d^3\mathbf{r} [/itex]

    So I guess all I need is

    [itex] \int \psi^*_C \psi^*_D \psi_A \psi_B d\tau = Y_{L0}(\theta,\phi)[/itex]

    But all I can show is;

    [itex] \int \psi^*_C \psi^*_D \psi_A \psi_B = \int \psi^*_C \psi^*_B f(r')Y_{L0}(\theta',\phi') \psi_A \psi_B d\tau [/itex]

    Then that's really as far as I can get. I don't understand how this results in [itex] Y_{L0} [/itex] when we're integrating over [itex] d\tau [/itex] - all internal coordinates of those wavefunctions. Anyone think they know?
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Matrix element for direct reactions
Loading...