# Matrix element for direct reactions

by Silversonic
Tags: element, matrix, reactions
 P: 129 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 $\alpha$ to $\beta$ is given by $\frac{d\sigma}{d\Omega} = f(k,k')|T_{\beta \alpha}|^2$ 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, $T_{\alpha \beta}$ is given by $T_{\alpha \beta} \int e^{-i\mathbf{k}\cdot\mathbf{r_{\beta}}} \int \psi^*_C \psi^*_D V \psi_A \psi_B d\tau e^{i\mathbf{k}\cdot\mathbf{r_{\alpha}}} d^3\mathbf{r_{\alpha}} d^3\mathbf{r_{\beta}}$ $\alpha$ represents the relative vector distance between incident particle A and target B $\beta$ ^^ projectile C and residual nucleus D V is the potential responsible for the direct reaction process. $d\tau$ 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) $r_{\alpha} = r_{\beta} = r$ because of the short range of $V$. (2) $V = V_0\delta(r-R)$ 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. $\psi_D = \psi_B \phi_n = \psi_B f(r) Y_{L0}(\theta,\phi)$ What I'm then confused with is that I'm told these approximations are used to obtain $T_{\alpha \beta} = V_0\int e^{i (\mathbf{k - k'}) \cdot r} Y_{L0}(\theta,\phi)^*\delta(r-R) d^3\mathbf{r}$ 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 $Y_{L0}$ makes its way outside of the integral for $d\tau$. This is as far as I can get subbing in and using the approximations above; $T_{\alpha \beta} = V_0 \int e^{i (\mathbf{k - k'})\cdot\mathbf{r}} \int \psi^*_C \psi^*_D \delta(r-R) \psi_A \psi_B d\tau d^3\mathbf{r}$ So I guess all I need is $\int \psi^*_C \psi^*_D \psi_A \psi_B d\tau = Y_{L0}(\theta,\phi)$ But all I can show is; $\int \psi^*_C \psi^*_D \psi_A \psi_B = \int \psi^*_C \psi^*_B f(r')Y_{L0}(\theta',\phi') \psi_A \psi_B d\tau$ Then that's really as far as I can get. I don't understand how this results in $Y_{L0}$ when we're integrating over $d\tau$ - all internal coordinates of those wavefunctions. Anyone think they know?