A Simple Minded Look at the Deuteron

The deuteron is the lightest, nontrivial nucleus (where I am viewing the proton as the lightest nucleus). It is a bound state of a neutron and proton, and one can determine its binding energy to be $E~=~2.224644\pm 0.000034 {\rm MeV}$. In addition to the binding energy one can measure electromagnetic quantities, such as the magnetic moment $\mu_d~=~0.857406\pm 0.000001 {\rm N.M.}$ and a quadrupole moment of $Q_d~=~0.2860\pm 0.0015 {\rm fm^2}$. Let me just make a comment in passing here, we will return for a closer look later. The non-zero quadrupole moment tells us directly that the deuteron is not just and n-p bound into an S-wave. An S-wave object has spherical symmetry and therefore cannot have a quadrupole deformation (component of the shape $\sim Y_{20}$) and hence electric quadrupole moment.

You will recall that the spatial extent of a wavefunction is determined by its binding energy and not by the radius of the potential. The small binding energy would indicate that the deuteron is an extended object and that the details of the potential causing its formation are not really that relevant to its properties. Further, as the deuteron is the only bound state of the NN system, it is going to be an S-state (neglecting the mixing of different L-states) as we know from quantum mechanics that the higher angular momentum states are at higher energy.

Noting that the deuteron has a non-zero magnetic moment which would indicate that it is $S=1$ and not $S=0$, and also that the scattering length in the $S~=~1$ channel is $+ve$ and not $-ve$ as shown in the figure, we will use the square well as given in eq. (1) (which reproduce the scattering lengths and effective rangein both channels)

 

$$S=1\ \ :\ \ V_{\rm square}^{(1)} = 38.5\ {\rm MeV} \ \ ,\ \ R_{\r... ...e}^{(1)} = 1.93 {\rm fm} \ \ ;\ \ (VR^3)^{(1)}_{\rm square} = 1.40 \ {\rm fm^2}$$

$$S=0\ \ :\ \ V_{\rm square}^{(0)} = 14.3\ {\rm MeV}\ \ , \ \ R_{\r... ... = 2.50 {\rm fm} \ \ ;\ \ (VR^3)^{(0)}_{\rm square} = 1.13 \ {\rm fm^2} \ \ \ .$$ (1)

to look for a bound state.

We look for a wavefunction of the form

 

$$\psi_d ({\bf r}) = {1\over r} u_0(r) Y_{00}(\Omega) \ \ ,$$ (2)

that is a solution to the schrodinger equation

 

$$\left[ -{\nabla^2\over 2\mu} + V(r)\right]\psi = -\epsilon\psi \ \ ,$$ (3)

where $\epsilon$ is a $+ve$ number. The $-ve$ sign indicates we are looking for a bound state, with energy less than a zero-energy incident scattering wave. The potential is a square well as given in eq. (1) and $\mu$ is the reduced mass of the system $\mu=M_N/2$. It is easy to show that eq. (3) leads to an equation for the radial part of the wavefunction of the form

 

$$\left[ -{1\over 2\mu} {d^2\over dr^2} + V+\epsilon \right] u_0(r) = 0 \ \ \ .$$ (4)

When $V(r)$ is a discontinuous step potential we can solve this analytically without having to go to Mathematica.

In the region inside the range of the potential, which I will call R, with $V=-V_0$ (I hope the notation is not getting out of hand) the wavefunction behaves sinusoidally and

 

$$u_<(r) = A\ \sin (K_d r) \ \ \,$$ (5)

where $K_d$ is the momentum vector inside the well

 

$$K_d = \sqrt{2\mu (V_0-\epsilon )} \ \ \ .$$ (6)

Outside the range of the potential the wavefunction $u_0(r)$ dies off exponentially (after excluding the unphysical exponentially growing solution)

 

$$u_>(r) = B e^{-\kappa_d r} \ \ \ ,$$ (7)

where

 

$$\kappa_d = \sqrt{2\mu\epsilon} \ \ \ .$$ (8)

At the boundary of the potential these solutions in the two regions must join up continuously and smoothly. It is useful to form the logarithmic derivative of the wavefunction so as to remove the normalization constants $A$ and $B$ from consideration, and the continuity requirement can be written as an equality between the logarithmic derivatives at the boundary. This relation provides a constraint on $\epsilon$,

 

$$\left({1\over u_{<}}{d u_{<}\over dr}\right)_{r=R} = \left({1\over u_{>}}{d u_{>}\over dr}\right)_{r=R}$$

$$K_d\cot (K_d R) = -\kappa_d$$

$$\cot (K_d R) = -{\kappa_d\over K_d} = -\sqrt{\epsilon\over V_0-\epsilon} \sim -\sqrt{2.22\over 36.2} \ \ \ .$$ (9)

It is clear that in order to have a small binding energy energy, the wave number must satisfy $K_d R\sim \pi/2$ and hence

 

$$2\mu V_0 R^2 = {\pi^2\over 4} \ \ \ ,$$ (10)

and that no bound state is possible unless $2\mu V_0 R^2 > {\pi^2\over 4}$. Numerically we find that $M_N (VR^2)_{\rm square}^{(1)} = 3.46$ which is larger than $\pi^2/4=2.47$

I have plotted the numerical solution for the bound states to the $S=1$ square well potential given in eq. () , along with the $E=0$ scattering solution in Fig. ().

  

Figure: Square well and bound state wavefunction in $S~=~1$ channel.

One sees that the the nucleons in fact "live" most of their time outside the range of the potential binding them together! Further, they live at distances that are large compared to the scattering length in this channel. It is clear how the scattering length and bound state are related, as they both satisfy the continuity condition at the boundary of the potential.