Zero Energy n-p scattering

It is easiest to think about the scattering of a n and a p in terms of the scattering of partial waves denoted by spin and isospin (we have set $l=0$) as we are at vanishing incident energy. The total wavefunction must be antisymmetric as we are dealing with fermions, and hence the scattering states must have either ${I=0, S=1}$ or ${I=1,S=0}$. ( We are, of course, writing it in such a way to easily extend to $nn$ and $pp$ scattering. These processes can only occur in the ${I=1,S=0}$ for S-wave scattering as they must be $I=1$ only.) Each of these scattering channels will have an associated scattering length, effective range and any other parameter defined to describe the energy dependence of the scattering amplitude. By conservation of angular momentum, the $L=0,S=0$ and $L=0,S=1$ states do not mix under the scattering interaction. An example of where the channels do mix would be at non-zero incident energy, we would have the possibility of a $L=2,S=1$ and a $L=0,S=1$ state, both with $J=1$, mixing with each other. This would require a coupled channels analysis. Let us ignore this type of problem to begin with.

For an unpolarized beam incident upon an unpolarized target we can simply determine the relative contribution of the $S=1$ channel to $S=0$ channel to be $3$ to $1$, simply by the number of substates. The spin-averaged cross section is given in terms of the $S=0$ (spin-singlet) and $S=1$ (spin triplet) scattering lengths $a_1$ and $a_0$, respectively, is then

 

$$\sigma (E=0) = 4\pi \left( {3\over 4} a_1^2 + {1\over 4} a_0^2 \right) \ \ \ .$$ (33)

The spin averaging gives the factor of ${1\over 4}$ from the four different spin configurations of 2 spin ${1\over 2}$ particles.

Let us go back on step and write the amplitude for a given process in terms of the triplet and singlet scattering amplitudes. The cross section in terms of the scattering length is

 

$$\sigma = 4\pi a^2 \ \ ,$$ (34)

which lead to eq. (33) after performing spin averaging over the two spin channels available to $np$ S-wave interactions. It is convenient to write the amplitude $a$ in terms of a spin operator that acts on the particles involved. This saves us from defining the amplitude for the $S=1$ channel and the amplitude for the $S=0$ channel separately. We know that for $S=1$ the amplitude is $a=a_1$ and for $S=0$ the amplitude is $a=a_0$. Now $S=s_n+s_p$ and $S^2 = (s_n+s_p)^2$ and hence

 

$$s_n\cdot s_p = {1\over 2}\left[ S(S+1) - s_n(s_n+1) - s_p(s_p+1)\right]$$

$$= {1\over 4}\ \ {\rm for }\ \ S=1$$

$$= -{3\over 4}\ \ {\rm for }\ \ S=0 \ \ \ .$$ (35)

It follows that the amplitude given in terms of the above spin operator is

 

$$a = {1\over 4}\left( 3 a_1 + a_0 \right) + \left( a_1-a_0\right) s_n\cdot s_p \ \ \ .$$ (36)

So far we haven't really gained anything by rewriting the amplitude in an operator language, but in fact we shall see that this allows us to extend the analysis to systems composed of multiple nucleons. In particular, consider scattering neutrons from molecular hydrogen, $H_2$. If the wavelengths of the incident neutrons are short compared to the inter-atomic spacing between the two protons in $H_2$, then molecule is likely to be blown apart when interacting with the incident neutron. However, if the incident energy and hence inverse wavelength is much less than the inter-atomic spacing ( $\ll 7.8 \times 10^{-11} m$, in fact it is required that $E^{-1} \ll 2\times 10^{-10}m$) then the neutron will only be able to elastically scatter from the $H_2$ molecule and hence the amplitudes for scattering from the two protons will add coherently. The operator giving rise to the amplitude (denoted by $a_H$) for scattering from $H_2$ is the sum of eq. (36) over the two protons, giving

 

$$a = {1\over 2}\left( 3 a_1 + a_0 \right) + \left( a_1-a_0\right) s_n\cdot s_H \ \ \ ,$$ (37)

where $S_H$ is now the spin of the $H_2$ molecule, either $S_H=0$ for para-hydrogen or $S_H=1$ for ortho-hydrogen.

Now to determine the scattering amplitudes. For para-hydrogen ($S=0$) we have that $s_n\cdot s_H=0$, which leads to $a_{\rm para} = {1\over 2}\left( 3 a_1 + a_0 \right)$ and hence a cross section of $\sigma_{\rm para} = \pi \left( 3 a_1 + a_0 \right)^2$, after spin-averaging and summing over the number of substates (in this case 2 corresponding to the two spin states of the neutron). It is a bit trickier for ortho-hydrogen, and most texts have entirely unsatisfactory descriptions in my opinion. When we have $s_H=1$, then the $n-H_2$ system can have total $S=3/2$ or $1/2$ and we must perform the appropriate spin averaging over the summed matrix element. There are a total of $2\times 3=6$ spin channels available $4$ in the $S=3/2$ channel and $2$ in the $S=1/2$ channel. Therefore the cross section for n scattering off ortho-hydrogen is

 

$$\sigma = 4\pi {1\over 6}\left[ 4 \vert a_{S=3/2}\vert^2 + 2 \vert a_{S=1/2}\vert^2\right] \ \ \ ,$$ (38)

where is is easy to show that

 

$$a_{S=3/2} = 2 a_1$$

$$a_{S=1/2} = {1\over 2} a_1 + {3\over 2} a_0 \ \ \ ,$$ (39)

where we have used $s_n\cdot s_H = {1\over 2}\left[ S^2 - s_n^2 - s_H^2\right]$. One finds that eq. (39) can be rewritten as

 

$$\sigma = \pi(3 a_1+a_0)^2 + 2 \pi (a_1-a_0)^2 \ \ \ .$$ (40)

The total cross section for zero-energy $np$ scattering and the coherent amplitude for scattering long-wavelength neutrons from molecular hydrogen have been measured very well and it is found that

 

$$\sigma(0) = 20.442\pm 0.23\ b$$

$${1\over 2}\left( 3 a_1 + a_0\right) = -3.707\pm 0.008\ fm \ \ \ ,$$ (41)

where the "barn" is defined by

 

$$1 b = 10^{-24} {\rm cm^2} = 100 {\rm fm^2} \ \ \ .$$ (42)

We have two unknowns and two independent measurements from which we can determine that

 

$$a_0 = -23.710\pm 0.030\ {\rm fm}$$

$$a_1 = 5.432\pm 0.005\ {\rm fm} \ \ \ .$$ (43)

There is a lot of physics in these two numbers. Firstly, notices that they are both large, larger than one would have guessed from the naive length scale associated with nuclear interactions, $\sim 1{\rm fm}$. We learned last time that the scattering length is not really a measure of the range of the potential, but tells us more about the behaviour of the wavefunction, it is where the $E=0$ wavefunction intercepts the $\psi=0$ axis. Further, we saw that it is not a bad thing to have a large, or even infinite scattering length, it merely means that the $E=0$ wavefunction is a straight horizontal line, resulting from the boundary condition of the wavefunction emerging from the region of the potential. The case at hand, where both the spin singlet and triplet channels have large scattering lengths indicates that the nuclear force is somewhat insensitive to spin-dependent interactions (we will return to this later). It is clear that there is some spin dependence to the interaction, but it is weak. Such scattering lengths can be understood if in the spin triplet channel there is a weakly bound state, so that the wavefunction just turns over at the boundary of the potential giving a large but $+ve$ scattering length. Similarly, if there is a barely, unbound state in the spin-singlet channel, so that their wavefunction just fails to turn over at the edge of the potential, giving a large but negative scattering length.

Further, we know that in the $S=1$ channel we have the deuteron. Neglecting, for the moment the $D-wave$ admixture in its wavefunction we can use the formula relating the binding energy to the scattering length and effective range to find that

 

$$r^{(1)}_{\rm eff} = 1.749\pm 0.008 \ {\rm fm} \ \ \ .$$ (44)

This information then allows us to use low energy scattering to determine the effective range in the $S=0$channel, we find that

 

$$r^{(0)}_{\rm eff} = 2.73\pm 0.03 \ {\rm fm} \ \ \ .$$ (45)