## Electromagnetic multi-pole moment

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.

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## Deuteron

The deuteron is the nucleus that contains 1 proton and 1 neutron. The spin and isospin of proton and neutron are the same, both are equal to half.  It is the only stable state for 2 nucleons. Deuteron provides an unique aspect to study the inter nuclear force. The strong force are believed to be charge independent. Thus, the strong force can be more easily to study on deuteron due to the absent of other force or eliminate from the Coulomb force, which is understood very much.

The mass of deuteron is 1876.1244MeV. The binding energy is then 2.2245MeV. It was determined by the slow neutron capture of a proton. The emitted gamma ray is approximately equal to the binding energy and the deuteron mass was calculated.

Deuteron has no excited state. It is because any excitation will easily to make the system break apart.

The parity is positive from experiment. If we separate the deuteron wavefunction into 3 parts. The proton wavefunction, neutron wavefunction and the orbital wavefunction. Under the only force, the strong force in this system, proton and neutron are the same nucleon with different state. Thus, the parity are the same for proton and neutron. So, the product of these 2 wavefunction always has positive parity. The total parity then is solely given by the angular orbital.

Any orbital wave function can be represented by the spherical harmonic, $Y(l,m)$.

The parity transform is changing it to

$Y(l,m) \rightarrow (-1)^l Y(l,m)$

So, the experimental face of positive parity fixed the angular momentum must be even.

Ok, we just predicted the possible angular momentum from parity.

The experimental fact on spin is 1. Since J = L + S, and the value of J can take every integer from |L-S| to L + S. and L must be even. The spin of proton and neutron is 1/2. Thus the possible S is 0 or 1 ( we are using L-S coupling scheme ). J = 1 = L + S , that tell us S must be odd to give out 1 for an even L. Thus S=1. So, the only possible L is 0 and 1. Thus, the possible state of deuteron is (L,S) = (0,1) or (2,1). Therefore, a deuteron is a mixed state, if without any further argument.

Now, 2 out of 3 parts of the wave function symmetry were determined by symmetry argument. The isospin can now be fixed by the 2 fermions state must be antisymmetry. The spatial state symmetry is even by L = 0 or 2. And for the state (L , S) = ( 0, 1 ), the spin state is symmetric. Thus, the isospin must be antisymmetric. Since the algebra for isospin and spin are the same. We use T = 0 for the isospin. Thus a complete wavefunction is ( L , S , T ) = ( 0 , 1, 0 ). For the other possible state (L , S) = ( 2 , 1 ) , we can use same argument for isospin state. And for the degenerated state with Ms = +1, 0, -1. By the symmetry of the raising and lowering ladder operator, they all preserved the symmetry. Thus, the Ms = 0 state can only be the + state.

So, we now have 2 possible states of deuteron. If the hamiltonian is commute with L^2 and  S^2, both L and L is a good quantum number and those states are eigen state. And the deuteron ground state must be one of them.

## Parity

It is Pa-ri-ty, not Par-ty.

( it needs to clean up)

Parity is just a reflection on every space dimension.

$\begin {pmatrix} x \\ y \\ z \end {pmatrix} \rightarrow \begin{pmatrix} - x \\ - y \\-z \end {pmatrix}$

## General

This is just a mirror reflection, although mirror reflection only reflects on 1 dimension, the dimension that perpendicular to the mirror surface.

May be we start on 2-D space instead of 3-D, draw a F and flips it upside down, and left-side right. then, you have a F just rotated 180 degree, not a reflection. however, in 3-D, then there is something different.

The parity transform is taking everything reverted. For example, when you stand up, your arms place horizontal and you left arm points forward and your right arm points right. After a parity transform. You right arm point left. Your left arm point backward, and you are standing on the ceiling, upside down. The result is a mirror image of your self. If we rotate the reverted-self from the ceiling to the ground.

Thus, parity also related as mirror reflection. In physics, we like to call the right-hand system (RHS) or left-hand system (LHS).

A simple RHS and LHS are on your hands! Although our left hand and right hand has some minor different, in general, they are the mirror image of each other. And the great interesting thing is, your left hand cannot overlap the right hand. They are equal but not the same.

Another thing is spring, when a wire is rolled clockwise and going upward, it form a left-hand spring and vice aver. Thus 2 springs are not the same.

For those which keep function as before parity transform, we called it parity positive, for those who are not, we called it parity negative.

Be reminded that the chiral material that interact circularly polarized light different still the parity positive. For example, a material which only let right hand light passes through, but not let the left hand pass. After parity transform, it lets left hand light pass through but not right hand .Thus, the left hand and right hand are work equally well!

We also cannot say our left hand is more weak then our right hand, then we called it parity negative. It is because, if we reflected ourself, our left hand is as good and right hand and the right hand is as weak as left hand.

A more physical example is the polarization of light, there are lefthand rotating light and right hand rotating light, called circular polarization. And material which interact differenty with different circular polarization are called chiral material. We should stop talking about examples in here. Because in nature, there are so many things has chiral property. Never the less, potenient and drug also has chirality. One book I recommend on general science for the chirality is “right-hand, left-hand” by chris McManus.

## Physics

Physics encounters parity is because we believe if the whole world is reverted, every thing just work fine and the same. For example, if our orgasms are all reflected, left goes to right, right go to left. We still alive. In fact, there are some real cases, that some peole do have reverted orgasm. Because there should be symmetric in the world.

In normal day, parity positive never break. It is seem impossible to break. How coome some thing work differently under parity transform?

For position, linear momentum, parity just make them change

However, in mathematics, there are many parity negative things. One example is the spherical harmonic. It is can be parity positive and negative depends on the parameter.

Lets take a imaginary example in parity negative. If we use photon to hit a target, all photons are going left. Now, we reflet the whole system. But now, the photons are still going left.

The first discovery of parity negative is on beta- decay from Co-60. Whe. Applied an external magnetic field from down to up, the beta particle come out at left. When we change the magnetic field, now is from up to down, the beta particle should come out at right, if parity is positive. But it is not, it still keep coming out left!

The reason of it is beyond my understanding… Sorry.