Matrix element for direct reactions

[Silversonic]

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?

 

[LightHero]

In order to get the expression you provided, we need to make a few more assumptions. The first is that the wave functions for particles A and B are the same (this implies that they are in the same state). Secondly, we need to assume that the wave function for the particle C is the same as the wave function for particle D. With these two assumptions, we can then reduce the integral over dτ to just an integration over the coordinates of particle C (or equivalently, particle D). When doing this, we can replace the wave functions of particles A and B with the single wave function they both have, leading to the integral depending only on the wave function of C/D and the delta function. Finally, since we are integrating over the coordinates of C/D, we can pull the angular part of the wave function out, leaving us with the expression you provided. I hope this helps!