## Lorentz Transform (a derivation )

we notate the 4 – coordinate as x in S frame and y is S’ frame.

let the Lorentz transform be $y_i = L_{ij} x_j$

notice that when writing like this, it is equivalent as a matrix equation on column vector: $\begin {pmatrix} y_0 \\ y_1 \\ y_2 \\ y_3 \end {pmatrix} = L \cdot x$

For row vector, the Einstein notation read: $y_i = x_j L_{ji}$

be careful the order of the subscript of L.

to derive the Lorentz transform, we only need to know 3 things,

1. the wave equation does not change ( i.e. the speed of light does not change )
2. the Lorentz transform has inverse
3. the relative motion of 2 frame is given by a speed $\beta$

we need Jacobian when relate the derivative between 2 frames. $\frac { \partial }{\partial x_i} = \frac { \partial } { \partial y_j} \frac { d y_j }{ d x_i }$

Thus, $\frac { \partial }{\partial x_i} = \frac { \partial } { \partial y_j} L_{ji}$

(this is a row vector equation )

The wave equation is : $\left( \frac{\partial^2}{\partial x_0^2} - \nabla^2 \right) \Psi = 0$

this wave equation can be viewed by using this: $a^2-b^2 = (a+b) (a - b) = ( a, b ) \cdot ( a, -b) = (a,b) \cdot \begin {pmatrix} 1 & 0 \\ 0 & -1 \end {pmatrix} \cdot ( a, b)$

There, we can write the wave equation as : $\frac { \partial } {\partial x_i } g_{ij}\frac {\partial}{\partial x_j} \Psi = 0$

where the g is a matrix: $g = \begin {pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {pmatrix}$

Subsitute the Jacobian, we have $\frac {\partial }{ \partial y_{\alpha} }L_{\alpha i}g_{ij}L_{j \beta}\frac {\partial}{\partial y_\beta} \Psi = 0$

the wave equation does not change with frame. there fore, this must be hold: $L_{\alpha i} g_{ij} L_{j \beta} = g_{\alpha \beta }$

ntoics that the left Lorentz transform is transpose, and the right one is normal one. For simplicity, for a 1-D spatial dimension. we can let L be: $\begin {pmatrix} a & b \\ c & d \end {pmatrix}$

by solving, we have 3 indenpent equations: $a^2 - c^2 = 1$ $ab-cd = 0$ $b^2 - d^2 = -1$

Using the inverse of Lorentz transform, we have another 3 equations: $a^2 - bc = 1$ $b(a - d) = 0$ $d^2 - bc = 1$

Solving this 6 equations for non-trivial solution. $L = \begin {pmatrix} a & b \\ b & a \end {pmatrix}$ $a^2 - b^2 = 1$

Now, using the original of frame S ( $x_1 = 0$ ) is moving with speed $\beta$ in the S’-frame. $y_1' / y_0' = b/a = \beta$

then we have the Lorentz Transform !

## Optical Pumping

DOI : 10.1119/1.1935926

This paper is easy to read and give a very easy understanding on optical pumping. it also demonstrated an experimental setup on detecting this effect. worth to read.

i am not going to reveal the detail and the experiment on the paper this time, but i just share my understanding on this subject.

The electron’s spin is polarized by optical pumping. Photon has intrinsic spin 1, and, it can be liner polarized or circular polarized. A linear polarized photon carries 0 magnetic angular momentum, and a circular polarized photon carries ±1 magnetic angular momentum. Therefore, an absorption of a circular polarized photon will make an electron undergoes ΔJ=±1and Δm=±1 transition.

In atomic energy level, a ground state is 3S½ and the next level is 3P½ and 3P3/2, the splitting of the P state is due to spin-orbital coupling between electron spin and the magnetic field generated by the orbital motion of the electron. Due to hyperfine coupling with the nuclear spin, the sub-shell will further split up. The good quantum numbers for hyper structure are |I-J|≤ F≤I+J and -F ≤ m_F≤F. The following graph demonstrated the I=3/2. The selection rule is ΔF=±1 and Δm_F=±1.

The red line is the transition by absorbed a circular polarized photon, and the black lines are possible de-excitation. Since there are 2/3 chance for the lowest state to the higher states, then, the population will move and become unbalanced. If we set up a set of rate equations, we can find out the population of each state.

A magnetic field is not necessary for optical pumping, as the graph demonstrated. The magnetic field for optical pumping is for measuring the effect. When a magnetic field is applied, the population will unbiased to, say, m_F=+1 state. Since there is no more state for absorption of the circular polarized light, then, the medium will become transparent. If we switch the magnetic field, the population will be inverted absorption will occur, and the medium become no-transparent. By measuring the intensity of the light pass through the medium, we can observe the effect of the pumping.

we can see, if the number higher energy state is more than the number of lower energy state, the maximum state has an escape channel. Look at the graph, if we have transition from F=1 to F=2 state, the m_F=+1 can absorb a photon and go to the m_F=+2 state (red dotted line). Due to relaxation, the m_F=+2 can transit to the F=2, at 3S½ state ( the black dotted lines).

Due to the lamb shift, the S-state is always little higher than the P-state in same shell. therefore, is it not possible to polarize the electron for Zero nuclear spin.

## [Pol. p target] Looking for a signal

the NMR signal was calibrated by water, since it has contains proton. the pule frequency is 12.78 MHz. this frequency is also equal to the Larmor frequency under a static B field. $2 \pi f = \mu_p B /2$

the time of the pule is 5 μs. this should be the π/2 pulse. thus $\mu_p B_R \tau_{\pi/2} /2 = \pi / 2$

we did a parameter search for any possible NMR signal. the parameters are the microwave power and the delay time.

(the following is a rough understanding, many places are unclear. )

the dynamic nuclear polarization requires a magnetic field and an microwave. the microwave is function as a power source or photo source. when nuclear spin flips, it requires a photo to provide the necessary energy. since our microwave frequency is fixed by the microwave cavity, so, in order to matching the correct frequency of absorption, we use a Sweeping magnetic field, when the magnetic field is at right magnitude, than the energy gap between spin up and spin down will equal to the photo energy and then absorbed.

after the electrons are all polarized by optical pumping, which means the electron polarization is saturated. the applied sweeping B-field and microwave will able to transfer the polarization to the proton.

thus, when the laser stopped by a chopper, the sweeping field and microwave will be turned on. the time delay between the laser and the sweeping field is called delay time. the upper line is the chopper signal ( channel 4) . the next one is the sweeping field ( channel 2 ). the 3rd one is the reflected microwave, we can see a very small change on it, the base line are off screen, since the signal is very small compare to the overall signal ( channel 1). the last one is the RF Amp trigger signal, which control the microwave on-off ( channel 3 ).

## Mass measurement

there are 2 ways to measure the mass of nucleus. one is by mass spectroscopy, the other one is from nuclear reaction.

## Mass Spectroscopy

the mass spectrometer contains 2 parts. the particles first enter a constant E-field. they will be separated into different radius  according to their kinetic energy by formular: $r_E = \frac {m }{q} \frac{ v^2}{E}$

after, they will enter a region of constant B field, and the radius will further separated according to their momentum. $r_M = \frac {m}{q} \frac { v }{B}$

another way is, Particles first passed a velocity filter by perpendicular electric field and magnetic filed. then use either the electric field and magnetic field to separate them. but this method will reduced the intensity of the beam, due to the velocity filter.

## Nuclear reaction

the basic idea of this method is the energy conservation and binding energy will converted to kinetic energy. there are many different way to do so, but the principle is that.

one example is neutron capture by proton : $n + p \rightarrow D + \gamma$

the binding energy is: $B.E. = m_n + m_p - m_D$

## Scattered amplitude

the scattered amplitude is: $f(\theta) = \frac {m } { 2 \pi \hbar^2 } \int { V(r') Exp ( i q \cdot r' )dr'^3}$

where $q = k - k'$ . using this: $Exp( i q \cdot r ) dr^3 = \frac { sin(q r) } {q r } 4 \pi r^2 dr$

then the amplitude reduce to $f(\theta ) = \frac { 2 m } { \hbar^2 q } \int { r V(r) sin(q r ) dr }$

For Coulomb force: $V(r) = Z e^2 / r$ $f(\theta ) = 2 m Z e^2 / q^2$

which is Rutherford Scattering equation!

For electron scattering, the potential from finite size of the nucleus is $\Delta V(r) =\frac { Ze^2 } { |r- r'| } \rho(r') dr'$ $f(\theta) = \frac { m Z e^2 } { 2 \pi \hbar ^2 } \int \int \frac {\rho(r') } {|r-r'| } Exp( i q \cdot ( r' + |r - r'| ) ) d r' d |r-r'|$

rearrange $f(\theta) = \int \rho(r') Exp( i q \cdot r' ) dr' \times f_{Ruth} (\theta) = f_{Ruth}(\theta ) \times F(\theta)$

the $F(\theta)$ is called the Form factor, which is due to the distribution of the charge. $F(\theta ) = \int \rho(r) Exp( i q \cdot r) dr$

the Form-factor is Fourier transform of the charge distribution alone. If reader familiarized with optics, the Form-factor is also the diffraction pattern. so, we can see, when the incoming particle wave has wavelength similar to the size of the charge distribution, then a diffraction occurred. here are some form of the Form factor.

for point charge: $\rho(r) = \delta(r)/4 \pi$ $F(\theta) = 1$

for exponential : $\rho(r) = (a^3/8 \pi) Exp( - a r)$ $F(\theta ) = ( 1+ q^2/a^2 \hbar^2)^{-2}$

Solid sphere : $\rho(r) = 3/4 \pi R^3 , r < R$ $F(\theta) = 3 \alpha^{-3} ( sin ( \alpha) - \alpha cos(\alpha) ) , \alpha = |q| R / \hbar$

## Lippmann-Schwinger equation

For elastic scattering, the scattered state’s energy should equal the incident state’s energy. However, the equation for them are different. $H_0 \left| \phi \right> = E \left| \phi \right>$ $H_0 + V \left| \psi \right> = E \left| \psi \right>$

the Lippmann-Schwinger equation proposed an equation ( or solution ) $\left| \psi ^{\pm} \right> = \frac{1} { E - H_0 \pm i \epsilon } V \left|\psi^{\pm} \right> + \left| \phi \right>$

the complex energy is the key, without it, the operator will be singular. to compute the wave function, we apply the position bra on the right: $\left< x | \psi ^{\pm} \right> = \int {dx'^3} \left< x| \frac {1}{E-H_0 \pm \epsilon } | x' \right> \left< x' | V | \psi^{\pm} \right> + \left< x | \phi \right>$

to compute the $< x | \frac {1} {E-H_0 \pm \epsilon } |x'>$

we have to use the momentum ket, since it is commute with the Hamiltonian. then we have $< x | \frac {1} {E-H_0 \pm \epsilon } |x'> = - \frac { 2 m } { 4 \pi \hbar^2 } \frac {1} {|x-x'| } Exp ( \pm i k |x-x'| )$

which is the Green function. We can also see this by compare the equation with the scatter equation.thus , the solution is $\left< x | \psi ^{\pm} \right> =- \frac { 2 m } { \hbar^2 } \int {dx'^3} \frac{1}{4 \pi |x-x'| }Exp( \pm i k |x-x'|) \left< x' | V | \psi^{\pm} \right> + \left< x | \phi \right>$

for the potential is local. which means it is confined in a small space. or $\left < x' | V | x'' \right> = V(x') \delta(x'-x'')$

and for the detector is far, plus using plane wave as incident wave the solution is $\psi^{pm}(x) = \frac{1}{\sqrt {2 \pi \hbar}^3 } \left( Exp( \pm i k z ) + \frac { Exp( \pm i k \cdot r ) } {r} f(k, k') \right)$ $f(k,k') = - \frac { (2 \pi )^3 } { 4 \pi } \frac { 2 m } {\hbar ^2 } < k' |V | \psi ^{pm} >$ $k = p / \hbar$

since the scattered wave can be a composition of  different wave-number k. thus, the matrix $T = < k'| V | \psi^ {\pm} >$

also called the transition matrix, which mean, it is the transition through potential V to a wave-number k.

we can see that the equation is very similar to the scatter equation, since basically, they are the solution of the same Schrödinger equation.

if we further apply the Born approximation, we will get the same result that the scattered wave amplitude is a Fourier transform of the nuclear potential.

## Born first approximation

in treating a scattering problem, the incident particles are treated as a plane matter wave in Schrödinger equation. a plane wave is: $\phi(r') = \frac {1}{\sqrt{2 \hbar}^3} Exp( i k' \cdot r')$

and the scattered wave is combined a spherical wave with un-scattered plane wave : $Exp( \pm i k z ) \rightarrow Exp( i k z ) + f ( \theta ) \frac { Exp ( i k r ) } {r}$

Recall from the scatter equation: $\psi(r) = \phi(r) +\int {G(r,r') U(r') \psi(r') dr'}$

where the integral is taking inside the nucleus or the effective field from the nucleus. the Born first approximation is that, when the nuclear interaction on the incident wave is weak, that the wave within the potential does not distored, the wave function inside the integral is equal to the plane wave. $\psi(r') \simeq \phi(r') = \frac {1}{ \sqrt{ 2 \hbar } ^3 } Exp ( i k' \cdot r' )$

since the distance is far away, the Green function can be approximated by $G(r,r') \simeq - \frac{1}{4 \pi} \frac {1}{r} Exp( i k \cdot r - i k' \cdot r')$

combined everything, $f(\theta) = \frac {m } { 2 \pi \hbar^2 } \int { V(r') Exp ( i (k-k') \cdot r' )dr'}$

we can see that it is a Fourier transform of the potential. and since the amplitude of the spherical wave is related with the differential cross section by: $\frac { d \sigma }{ d \theta } = | f(\theta) |^2$

thus we can see why the differential cross section can reveal the nuclear potential.