Quadrupole Moment of the Deuteron

The quadrupole moment of the deuteron is $Q_d = 0.2860\pm 0.0015 {\rm fm^2}$. Writing the wavefunction of the deuteron as we have before as

 

$$\Psi_d^m = \cos\omega\ {1\over r} U_0(r) {\cal Y}^m_{101} + \sin\omega\ {1\over r} U_2 (r) {\cal Y}^m_{121} \ \ \ ,$$ (49)

we have that

 

$$Q = \sqrt{16 \pi\over 5} \int\ d^3x\ \left({r\over 2}\right)^2 \ Y_2^0(\Omega)$$

$$= \sqrt{ \pi\over 5} \int\ dr\ r^2\ d\Omega\ r^2 Y_2^0(\Omega)\ {... ...ga\cos\omega {\cal Y}^{1*}_{121}{\cal Y}^1_{101} U_0 (r) U_2(r) \right] \ \ \ .$$ (50)

You will recall that

 

$${\cal Y}^1_{121} = \sqrt{3\over 5} Y_2^2\chi_1^{-1} - \sqrt{3\over 10} Y_2^1 \chi_1^0 + \sqrt{1\over 10} Y_2^0 \chi_1^1$$

$${\cal Y}^1_{101} = Y_0^0\chi_1^1 \ \ \ ,$$ (51)

which leads to

 

$$Q = {1\over 5\sqrt{2}} \int\ dr r^2 \left[\sin\omega\cos\omega\ U_0(r) U_2(r) - {1\over\sqrt{2}} \vert U_2(r)\vert^2 \sin^2\omega \right] \ \ \ .$$ (52)

We notice that a pure D-state has a $-ve$ quadrupole moment. It is clear that we must have a non-trivial amount of S-D mixing in order to obtain a $+ve$ quadrupole moment as is observed for the deuteron. We recall that the deuteron magnetic moment indicated $\sin^2\omega\sim 0.04$. Inserting this value into eq. (52) and using the measured value of $Q_d$ we find that $\int\ dr\ r^2 U_0 U_2\sim 10 {\rm fm^2}$, entirely consistent with the size of the deuteron we have determined from scattering.

We should also note at this stage that it might appear that the quadrupole moment is a better observable from which to extract $\sin\omega$. The magnetic moment contribution from the D state is small and hence yields a large uncertainty compared to $Q_d$. However, you will also recall that the magnetic moment was independent of the radial wavefunctions and hence the nucleon-nucleon potential. This is not true of the quadrupole moment that depends directly on both the S and D wave radial wavefunctions.