I was calculating deuterium's magnetic dipole moment in the $l=0$ state, given the value of its quantum number $j=1$ and $j=l+s=l+s_p+s_n$. The state that we want to calculate this in is the $\left| S=1, S_z=1 \right\rangle$ state and using the Clebsch-Gordan coefficients, I got to:

$$ \langle\mu\rangle =\left\langle S=1, S_z=1 \left| \sum_i \frac{\hat{S_z}^{(i)}}{\hbar} \mu_{_N} g_i \right| S=1, S_z=1 \right\rangle $$

$$\langle\mu\rangle =\left\langle S=1, S_z=1 \left| \sum_i \frac{\hat{S_z}^{(i)}}{\hbar} \mu_{_N} g_i \right| m_p= \frac{1}{2} , m_n= \frac{1}{2} \right\rangle $$

$$\left\langle \mu \right\rangle = 0.88\mu_{_N}$$

where $g_i$ represent the respective gyromagnetic moments $g_p=5.538$ and $g_n=-3.826$ and $\mu_{_N}$ is the nuclear magneton. I know that the observed value for $\left\langle \mu \right\rangle$ is $0.86 \mu_{_N}$. How do we explain this discrepancy? Can we conclude anything about the symmetry of the nuclear potential?


1 Answer 1


You have assumed an $s$-wave deuteron, and you’ve come within 2% of the correct magnetic moment. You’re probably aware that the deuteron has nonzero electric quadrupole moment, and therefore has a non-negligible $d$-wave part of its wavefunction. Most sources say that the deuteron is about 4% $d$-wave.

That’s an interesting place to look next (and it was interesting and new fifty years ago), but I don’t know that I would say we can immediately draw conclusions about the nuclear potential. My first observation is that you’ve undergone a precision loss, starting with four significant figures for $g_n$ and $g_p$ and ending with only two significant figures. If you do this calculation the dumb way, by looking up the neutron and proton moments and adding them, the cancellation gives you some precision loss, but it’s one digit instead of two. Furthermore your Clebsch-Gordan computation seems to be reproducing the result of doing the computation the dumb way. That’s probably correct, and it’s worth it for you to do the half-page of algebra to see how.

If I were expecting $0.86\mu_N$ and I computed $0.88\mu_N$, as you have, I would stop there and call it a success: the least significant digit is never very trustworthy. But if I were expecting $0.857\mu_N$ and I computed $0.880\mu_N$, that’s enough of a discrepancy to warrant some extra effort.


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