## symmetry breaking, mass and higg field.

Above curie temperature, the spin of iron is isotropic. The spin can be rotated to any direction without any resistance, they like massless. Below curie temp, the iron has phase transition and all spin now point to a particular direction. And we need some force to rotate the spin direction. The spin has mass now. This is what symmetry breaking in simple manner.

When matter becomes superconduct, the magnetic field inside is decay exponentially,which is similar as Yukawa force. And we said the force carrier particle is massive. The magnetic field decay is due to the copper pair, Which respond to the magnetic field and tend to cancel it. Thus, the direction of copper pair is not isotropic and this is another symmetry breaking due to the external field and low temperature.

At very high temperature, the weak force carriers are massless. And we assign an isotropic field (scaler field) for the force carrier and call it Higg field. The Higg field quata is called Higg boson. It act like the copper pair, which respond with the force carrier. When there is a force carrier, a Higg boson will be induced. And the symmetric breaking in Higg field, the symmetry breaking makes the force carrier has mass. That we need to apply a force to change the motion.

## 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.

## Optical Model II

Last post on optical model, we did not include the spin. to introduced the spin, we just have to modify the wave function. For spin-½ case.

$\begin {pmatrix} \psi_i \\ \psi_2 \end {pmatrix} \rightarrow Exp( i k r ) \begin {pmatrix} a_1 \\ a_2 \end {pmatrix} + \frac { Exp ( i k r) }{r} M \begin {pmatrix} a_1 \\ a_2 \end {pmatrix}$

where the M is a matrix:

$M = f + g \vec{ \sigma } \cdot \vec{n}$

the f is for the spin-Independence part of the wave function. For the incident wave and the scattered plane wave.

$\begin {pmatrix} a_1 \\ a_2 \end {pmatrix} = \begin {pmatrix} Exp( - i \phi_s /2 ) cos ( \theta_s /2 ) \\ Exp( i \phi_s /2 ) sin ( \theta_s /2 ) \end {pmatrix}$

where $\theta_s$ and $\phi_s$ are the angle of spin . not the detector angle.

after calculation by routine algebra, we have the scattered spherical wave.

$\chi = M \cdot \begin {pmatrix} a_1 \\ a_2 \end {pmatrix} =$ $\begin {pmatrix} (f+g)Exp( - i \phi_s /2 ) cos ( \theta_s /2 ) \\ (f-g)Exp( i \phi_s /2 ) sin ( \theta_s /2 ) \end {pmatrix}$

The expectation of the wavefunction, or the intensity of the spherical part will be:

$I(\theta_s) = \chi^{\dagger} \chi = |f|^2 + |g|^2 + 2 Re( f^* g ) cos( \theta_s)$

the beam polarization should be equal the intensity and normalized polarization.

$I P_z = \chi^{\dagger} \sigma_z \chi = ( |f|^2+ |g|^2 ) cos ( \theta_s) + 2 Re(f^* g)$

Thus, we have the induced polarization when incident beam is unpolarized:

$P_z ( \theta_s = \pi /2 ) = \frac { 2 Re ( f^* g ) }{ |f|^2 + |g|^2 }$

for a beam of many particle and formed an ensemble, the $\theta_s$ is the average.

and Analyzing power, which is a short term for Polarization Analyzing Power , or the spin asymmetry, is given by

$A_y=\frac { I(\theta_s = 0 ) - I( \theta_s = \pi ) } { I ( \theta_s = 0 ) + I ( \theta_s = \pi ) } = \frac { 2 Re( f^*g) }{ |f|^2 + |g|^2 } =P_z$

Therefore, in order to get the spin asymmetry, we have to use 2 polarized beams, one is up-polarized, and another is down-polarized, to see the different between the scattering result.

However, to have 100% polarized beam is a luxury. in most cases, we only have certain polarization. thus, the spin-asymmetry is not equal to the analyzing power. the spin-asymmetry $\epsilon$ is from the yield measurement.

$\epsilon = \frac { I(\theta_s) - I(\theta_s) }{ I(\theta_s ) + I(\theta_s) }$

since f and g only depend on the detector angle. and we can assume they are symmetry. Thus

$\epsilon = \frac {2 Re( f^* g ) }{|f|^2 +|g|^2 } cos ( \theta_s) = A_y P$

the P is the polarization of the target.