Electromagnetic multipole comes from the charge and current distribution of the nucleons.

Magnetic multipole in nucleus has 2 origins, one is the spin of the nucleons, another is the relative orbital motion of the nucleons.  the magnetic charge or monopoles either not exist or very small. the next one is the magnetic dipole, which cause by the current loop of protons.

Electric multipole is solely by the proton charge.

From electromagnetism, we knew that the multipole has  different radial properties, from the potential of the fields:

$\displaystyle \Psi(r) = \frac{1}{4\pi\epsilon_0} \int\frac{\rho(r')}{|r-r'|}d^3r'$

$\displaystyle A(r) = \frac{\mu_0}{4\pi}\int\frac{J(r')}{|r-r'|}d^3r'$

and expand them into spherical harmonic by using:

$\displaystyle \frac{1}{|r-r'|} = 4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l} \frac{1}{2l+1}\frac{r_{<}^l}{r_>^{l+1}} Y_{lm}^*(\theta',\phi')Y_{lm}(\theta,\phi)$

we have

$\displaystyle \Psi(r) = \frac{1}{\epsilon_0} \sum_{l,m}\frac{1}{2l+1}\int Y_{lm}^*(\theta',\phi') r'^l\rho(r')d^3r' \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}$

$\displaystyle A(r)=\mu_0 \sum_{l,m}\frac{1}{2l+1}\int Y_{lm}^*(\theta',\phi') r'^l J(r') d^3r' \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}$

we can see the integral give us the required multipole moment. the magnetic and electric are just different by the charge density and the current density. we summarize in this way :

$q_{lm} = \int Y^*_{lm}(\theta',\phi') r'^l \O(r') d^3 r'$

where O can be either charge or current density. The l determine the order of multipole. and the potential will be simplified :

$M(r)=\sum_{l,m}\frac{1}{2l+1} q_{lm} \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}$

were M can be either electric or magnetic potential, and i dropped the constant. since the field is given by 1st derivative, thus we have:

1. monopole has $1/r^2$ dependence
2. dipole has $1/r^3$
3. quadrapole has $1/r^4$
4. and so on

The above radial dependences are same for electric or magnetic. for easy name of the multipole, we use L-pole, which L can be 0 for monopole, 1 for dipole, 2 for quadrapole, etc.. and we use E0 for electric monopole, M0 for magnetic monopole.

Since the nucleus must preserver parity, and the parity for electric and magnetic moment are diffident.the different come from the charge density and current density has different parity. The parity for charge density is even, but for the current density is odd. and $1/r^2$ has even parity, $1/r^3$ has odd parity. therefore

• electric L-pole — $(-1)^{L}$
• magnetic L-pole — $(-1)^{L+1}$

for easy compare:

• E0, E2, E4… and M1,M3, M5 … are even
• E1,E3,E5…. and M0, M2, M4…. are odd

The expectation value for L-pole, we have to calculate :

$\int \psi^* Q_{lm} \psi dx$

where $Q_{lm}$ is multipole operator ( which is NOT $q_{lm}$), and its parity is follow the same rule. the parity of the wave function will be canceled out due to the square of itself. thus, only even parity are non-Zero. those are:

• E0, E2, E4…
• M1,M3, M5 …

that make sense, think about a proton orbits in a circular loop, which is the case for E1, in time-average, the dipole momentum should be zero.