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Matrix element for direct reactions 
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#1
Mar1014, 09:06 PM

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 [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 onenucleon 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(rR) [/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(rR) 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(rR) \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? 


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