## Group table of Mirror, 180 degree Rotation and Parity

first, we use Cartesian coordinate to define the matrix representation of the Mirror, 180 degree Rotation and Parity.

$M_x= \left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$

$R_x= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right)$

$P= \left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right)$

i guess the reader can think the other matrix.

the group table is:

$\left( \begin{array}{cccccccc} i & M_x & M_y & M_z & R_x & R_x & R_z & P \\ M_x & i & R_z & R_x & P & M_z & M_y & R_x \\ M_y & R_z & i & R_x & M_z & P & M_x & R_x \\ M_z & R_x & R_x & i & M_y & M_x & P & R_z \\ R_x & P & M_z & M_y & i & R_z & R_x & M_x \\ R_x & M_z & P & M_x & R_z & i & R_x & M_y \\ R_z & M_y & M_x & P & R_x & R_x & i & M_z \\ P & R_x & R_x & R_x & M_x & M_y & M_z & i \end{array} \right)$

where i is identity matrix. we can see it is a group of order 8. there is a sub-group of $(i, R_x, R_y, R_z)$. since the group table are diagonal symmetry, each element from its own class.

since a rotational group can be represented by 2X2 matrix instead of 3X3 by introducing complex number. those matrix is Pauli’s spinor.

$\sigma_x= \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$

the transform from 3X3 matrix to 2X2 matrix is that…… then we can map the mirror matrix and parity into 2X2 matrix.

## Impedance matching and Smith Chart

the circles with center on x-axis of the Smith chart represent the real part of impedance.

the circles with center on the x = 1 are the imaginary part of the impedance.

and a circle with radius 1 and centered at (0,0) represent a coaxial line with 50Ω, the (-1,0) is zero length and (1,0) is ¼ wavelength from the load.

thus, we can construct the input impedance by Smith Chart very easily.

AND i always think, the normalized input impedance should be 1 in order to give minimum reflection. However, when the source impedance is not 50Ω, it is not the case. the impedance matching for normalized source impedance is

$z_{in} = z_s^*$

this is come form the Maximum Power Theory.

anyway, to construct the input impedance, we just need to  remember when circuit element is in series, we use normal Smith chart, if the circuit element is in parallel, we use a 180 degree rotated Smith chart, since adding by reactance is more easy.

there is an iphone app for this: rfCalc

it is very easy to use!

the question is how do i know the source impedance!???

## Lorentz Force and Stress tensor

$F_{ij} =\begin {pmatrix} 0 & D_1 & D_2 & D_ 3\\ -D_1 & 0& H_3 & -H_2 \\ -D_2 & - H_3 & 0 & H_1 \\ -D_3 & H_2 & -H_1 & 0 \end {pmatrix}$

$G_{ij} =\begin {pmatrix} 0 & B_1 & B_2 & B_ 3\\ -B_1 & 0& -E_3 & E_2 \\ -B_2 & E_3 & 0 & -E_1 \\ -B_3 & -E_2 & E_1 & 0 \end {pmatrix}$

The field equation are:

$\partial /\partial x_i F_{i j } = - J_i$

$\partial /\partial x_i G_{i j } =0$

That is the result from last time.

the conservation of charge is:

$\partial/\partial x_i J_i = 0$

thus the 4-Laplacian of the F-field is :

$\partial^2/\partial x_i^2 F_{ij} = 0$

The physical meaning of the simplification of the field equation by the field tensor is, a gradient in the tensor field is equation to the minus of 4-current, or zero. recall that the gradient in 3-D vector space, the conservation of charge density is :

$\nabla \cdot \vec{J} = - \partial \rho/\partial t$

we have the same form in the 4-D tensor space. the creation of field is conservation of the 4-charge displacement, if we integrate the 4-current. i dun know what physical meaning of the G-field. personally, i believe that the F-field and G-field can be related by some transform.

we have another interesting things. we can write the Lorentz force into the field tensor:

$\vec{f}_j = \frac{d\vec{P}_j}{d\tau}=q \frac{d\vec{X}_j}{d\tau}F_{ij}$

the reason why we can write this, i don’t know. any physical meaning? i don’t know. may be we can think in this way, the force depends on the motion of the 4-vector and the field and the charge. thus, it is natural to multiple them together to get the force. But why not the G field? never the less, the field tensor reduce the number of Field qualities into 2.

the Lorents Force can be more simple

$\vec{f}_j = J_i F_{ij}$

that the force is created by the field and the current.

The Electromagnetic stress tensor can also be related with the 4-force by :

$\vec{f}_j = - \partial/\partial x_i T_{ij}$

Thus, combined with the Field tensor:

$\partial/\partial x_iT_{ij}+ J_iF_{ij} = 0$

## Q-value and Biding energy

I already talked on binding energy.

And the Q-value is the mass different between nuclear reaction. In same case binding energy is same as the Q-value.

Q-value is :

$Q= \sum{m_i} - \sum{m_f}$

By energy conservation, it can be rewritten by kinetic energy.

$Q= \sum{T_f} - \sum{T_i}$

But using energy is bit troublesome due to the frame transform. Using mass is simple, since it is a Lorentzian invariance.

However, during scattering experiment. The kinetic energy is much more easy to measure. Thus the kinetic form is used more frequently in experiment.

I will give the expression of K.E. in lab frame later.

## Clebsch – Gordan Coefficient

i am kind of stupid, so, for most text book with algebra example, i am easy to lost in the middle.

Thus, now, i am going to present a detail calculation based on Recursion Relations.

we just need equation and few observations to calculate all. i like to use the J- relation:

$K(j,-m-1) C_{m_1 m_2}^{j m}= K(j_1,m_1) C_{m_1+1 m_2}^{j m+1}+ K(j_2,m_2) C_{m_1 m_2+1}^ {j m+1}$

$K(j,m) = \sqrt{ j(j+1) - m(m+1)}$

$C_{m_1 m_2}^{j m}$ is the coefficient.

Notice that the relation is only on fixed j, thus, we will have our $m_1 m_2$ plane with fixed j, so, we have many planes from $j = j_1+j_2$ down to $j = |j_1-j_2|$.

We have 2 observations:

1. $C_{j_1, j_2}^{j_1+j_2 , j_1+j_2} = 1$ which is the maximum state. the minimum state also equal 1.
2. For $m \ne m_1+m_2$ the coefficient is ZERO.

Thus, on the $j = j_1 + j_2$ plane. the right-upper corner is 1. then using the relation, we can have all element down and left. and then, we can have all element on the plane.

the problem comes when we consider $j = j_1 + j_2 -1$ plane. no relation is working! and no book tells us how to find it!

Lets take an example, a super easy one, $j_1 = 1/2 , j_2 = 1/2$. possible $j = 0, 1$, so we have 2 planes.

The j = 1 plane is no big deal.

but the j = 0 plane, there are only 2 coefficient. and we can just related them and know they are different only a sign. and we have to use the orthonormal condition to find out the value.

See? i really doubt is there somebody really do the actually calculation. J.J.Sakuarai just skip the j = l-1/2 case. he cheats!

when going to higher j1+j2 case, we have w=to use the J- relation to evaluate all coefficient. the way is start from the lower left corner, and use the J- relation to find out the relationship between each lower diagonal coefficients. then, since all lower diagonal coefficients have same m value, thus, the sum of them should be normalized. then, we have our base line and use the J+ to find the rest.

i will add graph and another example. say, j1 = 3, j2 = 1.

## Frequency Modulation

Frequency Modulation is encode a message signal on a carrier signal by changing the frequency of it.

the carrier signal has a basis frequency and the form is like:

$S_C(t) = A_C cos( \omega_C t )$

the message signal can take any form. after the Modulation, the output signal is:

$S_O (t) = A_C cos( \omega_C t + \Delta f \int{S_M dt } )$

the integrated message signal should be normalized to 1. and the Δf is the range of frequency change.

______________________________________

The frequency produced by the Gunn Oscillator can be modulated, in order to matching the resonance frequency of the microwave chamber.

there is a adjusting knob on the Gunn Oscillator, which determine the basic frequency of the microwave. and a modulation signal is from the power supply. the frequency of the Gunn Oscillator is ranging from 8.6 GHz to 9.6 GHz.

the modulation signal is a simple linear function with frequency 23ms.

since our modulation frequency is a linear function. thus, the change of  the amplitude will change the output frequency of the microwave.

but the actually frequency modulation is by a Varicap or Varactor, which the capacitance can be changed by applied voltage. by changing the capacitor of the Gunn Oscillator, the output frequency changed.

However, the principle of frequency modulation unchanged.

In microwave engineering,  the Gunn Oscillator with the modulator will be called Voltage-Controlled Oscillator or VCO.

I still not fully understand the mathematic of the Gunn Oscialltor, and how it reacts with modulation signal. since i don’t have the internal structure of the oscillator.

## 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 place to study the inter nuclear force. The strong force is 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 1876.1244 MeV. 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 extracted.

Deuteron has no excited state. It is because any excitation will easily to make the system break apart. When think deuteron as one of the family of NN system. Because of tensor force, which favor T=0 pn pair, thus only T=0, S=1 pn pair, which is deuteron is bounded. Any excitation will change the isospin from T=0 to T=1, which is unbound.

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.