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A Careful Look at the Deuteron

The previous section has taught us that in general the orbital angular momentum of strongly interacting states is not a conserved quantum number due to the presence of non-central forces. These forces commute with the total spin of the system and hence we can classify the states by the $J$ and $S$ quantum numbers. For example, the deuteron is a $S~=~J~=~1$ object, with orbital angular momentum being a mixture of $L~=~0$ and $L~=~2$. It is therefore convenient to define states of good $S,J, L$ and $M_J$ which are linear combinations of the spatial and spin wavefunctions with different $M_S, M_L$. The wavefunction for a given system is then a sum over these states of different $L$. For the deuteron, we need $S~=~J~=~1$ and $L~=~0,2$ for each $M_J$. The ${\cal Y}^M_{JSL}$ by

    $\displaystyle {\cal Y}^{M_J}_{JSL} =
\sum_{M_S,M_L}\ \langle S,M_S,L,M_L\vert J,M_J\rangle \ \chi_S^{M_S}\ Y^{M_L}_L
\ \ \ ,$ (1)

where $\chi_S^{M_S}$ is the spin wavefunction for the state. For the deuteron in the $M_J=1$ states the relevant configurations (orthonormal) are
    $\displaystyle {\cal Y}^1_{110} = \langle 11,00\vert 11\rangle\ \chi^1_1 Y_0^0(\Omega)$  
    $\displaystyle = \chi^1_1 Y_0^0(\Omega)$  
    $\displaystyle {\cal Y}^1_{112} = \langle 11 20\vert 11\rangle \ \chi^1_1 Y_2^0(...
...hi_1^0 Y_2^1(\Omega)
+ \langle 1-1 22\vert 11\rangle\ \chi_1^{-1} Y_2^2(\Omega)$  
    $\displaystyle = \sqrt{1\over 10}\ \chi^1_1 Y_2^0(\Omega)
- \sqrt{3\over 10}\ \chi_1^0 Y_2^1(\Omega)
+ \sqrt{3\over 5}\ \chi_1^{-1} Y_2^2(\Omega)
\ \ \ .$ (2)

A formula that you have seen before, but will be very useful to us is the triple $Y_L^M$ integration formula
    $\displaystyle \int\ d\Omega\ Y_{L_1}^{M_1} \ Y_{L_2}^{M_2} \ Y_{L_3}^{M_3 *}
= ...
...L_2 M_{2}\vert L_3 M_{3}\rangle
\langle L_1 0 , L_2 0\vert L_3 0\rangle
\ \ \ .$ (3)

We can now compute some angular matrix elements relevant to the deuteron. In order to make progress we rewrite the $S_{12}$ operator in terms of the total spin operator, $S=s_1+s_2$,

    $\displaystyle S_{12}\ =\ 3\left(\sigma_1\cdot {\bf\hat r}\right)\left(\sigma_2\...
...t {\bf\hat r}\right)\left(s_2\cdot {\bf\hat r}\right)
\ -\ s_1\cdot s_2
    $\displaystyle = 2\left[ 3 \left(S\cdot {\bf\hat r}\right)^2 - S^2\right]$  
    $\displaystyle = 2\left[
3\left( {1\over 2} S_+ (\hat r_x - i \hat r_y)\ +\
{1\over 2} S_- (\hat r_x + i \hat r_y)\ +\ S_z \hat r_z\right)^2 - S^2\right]
\ \ \ ,$ (4)

where we have used the fact that $S_{12}$ is symmetric under interchange $1\leftrightarrow 2$, and $S_\pm$ are the spin raisong and lowering operators. Acting on $\vert{\cal Y}^1_{110}\rangle$ it is easy to show that
    $\displaystyle \langle {\cal Y}^1_{112}\vert S_{12}\vert{\cal Y}^1_{110}\rangle
3\sqrt{3\over 10}\sqrt{4\pi\over 15}\int d\Omega\ Y_2^{1*} Y_2^1 Y_0^0
    $\displaystyle \left.
\ +\
6\sqrt{3\over 5}\sqrt{2\pi\over 15}\int d\Omega\ Y_2^{2*} Y_2^2 Y_0^0
    $\displaystyle \ =\ \sqrt{8}
\ \ \ .$ (5)

We have used
    $\displaystyle S_\pm \vert 1 m\rangle = \sqrt{2} \vert 1, m\pm 1\rangle
\ \ \ ,$ (6)

and the standard representation of the $Y_l^m$'s.

Similar calculations lead to

    $\displaystyle \langle {\cal Y}^1_{110}\vert S_{12}\vert{\cal Y}^1_{110}\rangle \ =\ 0$  
    $\displaystyle \langle {\cal Y}^1_{112}\vert S_{12}\vert{\cal Y}^1_{110}\rangle \ =\ \sqrt{8}$  
    $\displaystyle \langle {\cal Y}^1_{112}\vert S_{12}\vert{\cal Y}^1_{112}\rangle \ =\ -2
\ \ \ .$ (7)

The deuteron wavefunction $\Psi^M$ (where $M$ is the magnetic substate) is a linear combination of $L~=~0$ and $L~=~2$ components

    $\displaystyle \Psi^M = \cos\omega\ {1\over r} U_0 (r)\ {\cal Y}_{110}^M
+ \sin\omega\ {1\over r} U_2 (r)\ {\cal Y}_{112}^M
\ \ \ ,$ (8)

where this relation defines the mixing angle $\omega$ between the orthonormal states. The wavefunction satisfies the schrodinger equation for the nucleons moving in the nuclear potential $V(r)~=~ V_{\rm cen} + V_{\rm ten} S_{12}$
    $\displaystyle \left[-{1\over 2\mu}\nabla^2 + V(r) \right]\Psi = -\epsilon\Psi
\ \ \ ,$ (9)

where $\mu$ is the reduced mass of the N-N system and writing this form out explicitly
    $\displaystyle \left[ -{1\over 2\mu}{d^2\over dr^2} + V_{\rm cen} + V_{\rm ten} ...
...en} + V_{\rm ten} S_{12} +
{6\over r^2} \right]\sin\omega\ U_2 {\cal
    $\displaystyle = -\epsilon \left[\cos\omega\ U_0 {\cal Y}_{110}^M + \sin\omega\
U_2 {\cal Y}_{112}^M \right]
\ \ \ .$ (10)

This represents a set of couple equations, which we can obtain by projecting with $\int\ d\Omega\ {\cal Y}_{112}^M$ and $\int\ d\Omega\ {\cal Y}_{112}^M$ to obtain
    $\displaystyle \left[-{1\over 2\mu}{d^2\over dr^2} + V_{\rm cen} + V_{\rm ten}
...ght] U_0 =
-V_{\rm ten} \langle 110\vert S_{12}\vert 112\rangle \tan\omega\ U_2$  
    $\displaystyle \left[-{1\over 2\mu}{d^2\over dr^2} + V_{\rm cen} +{6\over 2\mu r...
... ten} \langle 112\vert S_{12}\vert 110\rangle {1\over \tan\omega} \ U_0
\ \ \ .$ (11)

These equations have exactly the form we described above. If there are non-zero matrix elements between states of different $L$, as induced by the tensor force, then the $S$ and $D$ states mix with each other as expected.

We recall that from our naive guestimates made previously, the deuteron is a very extended object compared to the range of the nuclear interaction. In the region outside the potential, the coupled equations become

    $\displaystyle \left[-{1\over 2\mu}{d\over dr^2} + \epsilon\right] U_0 =0$  
    $\displaystyle \left[-{1\over 2\mu}{d\over dr^2} +{6\over r^2} + \epsilon\right] U_2 =
\ \ \ ,$ (12)

which have solutions of the form
    $\displaystyle U_{0,2} \sim \ \left( a + {b\over r} + {c\over r^2} \right) e^{-\gamma r}
\ \ \ ,$ (13)

and by direct substitution we find that
    $\displaystyle U_0(r) \rightarrow N_0 e^{-\gamma r}$  
    $\displaystyle U_2(r) \rightarrow N_2 \left( 1 + {3\over\gamma r}
+ {3\over\gamma^2 r^2}\right) e^{-\gamma r}$  
    $\displaystyle \gamma = \sqrt{2 \mu \epsilon}
\ \ \ ,$ (14)

where the $N_{0,2}$ are normalization constants determined solving the equation directly including the potentials.

It is clear that solving the coupled equations is not going to be so easy. In particular, we see that there is the mixing angle $\omega$ that is apriori undetermined. We see that for $\omega~=~0$, the Swave wavefunction is uncoupled from the D-wave admixture, but the D-wave is infinitely coupled to the S-wave component. We see that in order to solve these equations, we must compare with data, so that the solution reproduce the binding energy, magnetic moment and quadrupole moment of the deuteron.

In order for us to proceed, we therefore need to know how to relate these wavefunctions to the magnetic and quadrupole moments that are measured.

next up previous
Next: Electromagnetic Interactions. Up: PHYS 560: Lectures During Previous: PHYS 560: Lectures During
Martin Savage