## 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.

## Optical Model

In nuclear physics, the Optical Model means, we are treating the scattering problem is like optical wave problem. due to the incident beam can be treated as a wave-function. and this wave will be scattered by the target.

when the beam is far away from the target, the wave function of the incident beam should satisfy the Schrödinger equation in free space :

$\left( \frac {\hbar^2 } {2m} \nabla^2 + V(r) \right) \psi( \vec{r} ) = E \psi ( \vec{r} )$

and the plane wave solution is

$\psi ( \vec{r} ) \sim Exp ( \pm i \vec{k} \cdot \vec {r} )$

after the scattering, there will be some spherical wave come out. the spherical wave should also satisfy the free-space Schrödinger equation.

$\psi( \vec{r} ) \sim Y(\theta, \phi) \frac {Exp( \pm i \vec{k} \cdot \vec{r} ) }{r}$

Thus, the process of scattering can be think in this way:

$Exp( \pm i k z ) \rightarrow Exp( i k z ) + f ( \theta ) \frac { Exp ( i k r ) } {r}$

where f(θ) is a combination of spherical wave.

one consequence of using Optical Model is, we use complex potential to describe the nuclear potential terms in quantum mechanics.

when using a complex potential, the flux of the incident beam wave function can be non-zero. meanings that the particles in the beam are being absorbed or emitted. This corresponding to the inelastic scattering.

The reason for the “OPTICAL” is come form the permittivity and permeability of the EM field. for metallic matter, their permittivity or permeability may have a imaginary part. and this imaginary part corresponding to the absorption of the light. so, nuclear physics borrow the same idea.

the flux is defined as:

$J = \frac { \hbar }{ 2 i m} ( \psi^*(r) \nabla \psi(r) - \psi(r) \nabla \psi^* (r) )$

and the gradient of the flux, which is the absorption (sink) or emission ( source ) is:

$\nabla J = \frac {\hbar }{ 2 i m }( \psi^* \nabla^2 \psi - \psi \nabla^2 \psi^* )$

The Schrödinger equation gives the equation for the wave function:

$\nabla^2 \psi(r) = \frac { 2m} {\hbar^2} ( E - V(r)) \psi(r)$

when sub the Schrödinger equation in to the gradient of flux, we have:

$\nabla J = \frac {1} {i \hbar } ( V(r) - V^*(r) ) | \psi |^2 = \frac { 2} {\hbar } Im ( V) | \psi |^2$

we can see, if the source and the sink depend on the complex part of the potential. if the imaginary part is zero, the gradient of the flux is zero, and the wave function of the beam is conserved.

## Projection theorem

The simplest way to say is:

a operator can be projected on another one, for example, The orbital angular momentum cab be projected on the total angular momentum.

$L = L\cdot J \frac {J}{j(j+1)}$

a simple application is on the Zeeman effect on spin-orbital coupling. the Hamiltonian is:

$H_B = - \mu \cdot B = - ( \mu_l L + \mu_s S ) \cdot B$

by the Wigner-Eckart theorem:

$L = L\cdot J \frac {J}{j(j+1)}$

$S = S\cdot J \frac {J}{j(j+1)}$

then the Hamiltonian becomes:

$H_B = - \frac{1}{j(j+1)} ( \mu_l (L \cdot J) + \mu_s (S \cdot J) ) J\cdot B$

and introduce the Bohr Magneton and g-factor:

$H_B = - g \mu_B J \cdot B$

$g = - \frac{1}{j(j+1)} ( g_l (L \cdot J) + g_s (S \cdot J) )$

## a review on Hydrogen’s atomic structure

I found that most of the book only talk part of it or present it separately. Now, I am going to treat it at 1 place. And I will give numerical value as well. the following context is on SI unit.

a very central idea when writing down the state quantum number is, is it a good quantum number? a good quantum number means that its operator commute with the Hamiltonian. and the eigenstate states are stationary or the invariant of motion. the prove on the commutation relation will be on some post later. i don’t want to make this post too long, and with hyperlink, it is more reader-friendly. since somebody may like to go deeper, down to the cornerstone.  but some may like to have a general review.

the Hamiltonian of a isolated hydrogen atom is given by fews terms, deceasing by their strength.

$H = H_{Coul} + H_{K.E.} + H_{Rel} + H_{Darwin} + H_{s-0} + H_{i-j} + H_{lamb} + H_{vol} + O$

the Hamiltonian can be separated into 3 classes.

__________________________________________________________

## Bohr model

$H_{Coul} = - \left(\frac {e^2}{4 \pi \epsilon_0} \right) \frac {1}{r}$

is the Coulomb potential, which dominate the energy. recalled that the ground state energy is -13.6 eV. and it is equal to half of the Coulomb potential energy, thus, the energy is about 27.2 eV, for ground state.

$H_{K.E.} = \frac {P^2}{ 2 m}$

is the non-relativistic kinetic energy, it magnitude is half of the Coulomb potential, so, it is 13.6 eV, for ground state.

comment on this level

this 2 terms are consider in the Bohr model, the quantum number, which describe the state of the quantum state, are

$n$ = principle number. the energy level.

$l$ = orbital angular momentum. this give the degeneracy of each energy level.

$m_l$ = magnetic angular momentum.

it is reasonable to have 3 parameters to describe a state of electron. each parameter gives 1 degree of freedom. and a electron in space have 3. thus, change of basis will not change the degree of freedom. The mathematic for these are good quantum number and the eigenstate $\left| n, l, m_l \right>$ is invariant of motion, will be explain in later post. But it is very easy to understand why the angular momentum is invariant, since the electron is under a central force, no torque on it. and the magnetic angular momentum is an invariant can also been understood by there is no magnetic field.

the principle quantum number $n$ is an invariance. because it is the eigenstate state of the principle Hamiltonian( the total Hamiltonian )!

the center of mass also introduced to make more correct result prediction on energy level. but it is just minor and not much new physics in it.

## Fine structure

$H_{Rel} = - \frac{1}{8} \frac{P^4}{m^3 c^2}$

is the 1st order correction of the relativistic kinetic energy. from $K.E. = E - mc^2 = \sqrt { p^2 c^2 + m^2c^4} - mc^2$, the zero-order term is the non-relativistic kinetic energy. the 1st order therm is the in here. the magnitude is about $1.8 \times 10^{-4} eV$. ( the order has to be recalculate, i think i am wrong. )

$H_{Darwin} = \frac{\hbar^{2}}{8m_{e}^{2}c^{2}}4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\delta^{3}\left(\vec r\right)$

is the Darwin-term. this term is result from the zitterbewegung, or rapid quantum oscillations of the electron. it is interesting that this term only affect the S-orbit. To understand it require Quantization of electromagnetic field, which i don’t know. the magnitude of this term is about $10^{-3} eV$

$H_{s-o} = \left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\left(\frac{1}{2m_{e}^{2}c^{2}}\right)\frac{1}{r^3} L \cdot S$

is the Spin-Orbital coupling term. this express the magnetic field generated by the proton while it orbiting around the electron when taking electron’s moving frame. the magnitude of this term is about $10^{-4} eV$

comment on this level

this fine structure was explained by P.M.Dirac on the Dirac equation. The Dirac equation found that the spin was automatically come out due to special relativistic effect. the quantum number in this stage are

$n$ = principle quantum number does not affected.

$l$ = orbital angular momentum.

$m_l$ = magnetic total angular momentum.

$s$ = spin angular momentum. since s is always half for electron, we usually omit it. since it does not give any degree of freedom.

$m_s$ = magnetic total angular momentum.

at this stage, the state can be stated by $\left| n, l, m_l, m_s \right>$, which shown all the degree of freedom an electron can possible have.

However, $L_z$ is no longer a good quantum number. it does not commute with the Hamiltonian. so, $m_l$ does not be the eigenstate anymore. the total angular momentum was introduced $J = L + S$ . and $J^2$ and $J_z$ commute with the Hamiltonian.  therefore,

$j$ = total angular momentum.

$m_j$ = magnetic total angular momentum.

an eigenstate can be stated as $\left| n, l, s, j, m_j \right>$. in spectroscopy, we denote it as $^{2 s+1} L _j$, where $L$ is the spectroscopy notation for $l$.

there are 5 degrees of freedom, but in fact, s always half, so, there are only 4 real degree of freedom, which is imposed by the spin ( can up and down).  the reason for stating the s in the eigenstate is for general discussion. when there are 2 electrons, s can be different and this is 1 degree of freedom.

## Hyperfine Structure

$H_{i-j} = \alpha I \cdot J$

is the nuclear spin- electron total angular momentum coupling. the coefficient of this term, i don’t know. Sorry. the nuclear has spin, and this spin react with the magnetic field generate by the electron. the magnitude is $10^{-5}$

$H_{lamb}$

is the lamb shift, which also only affect the S-orbit.the magnitude is $10^{-6}$

comment on this level

the hyperfine structure always makes alot questions in my mind. the immediate question is why not separate the orbital angular momentum and the electron spin angular momentum? why they first combined together, then interact with the nuclear spin?

may be i open another post to talk about.

The quantum number are:

$n$ = principle quantum number

$l$ = orbital angular momentum

$s$ = electron spin angular momentum.

$j$ = spin-orbital angular momentum of electron.

$i$ = nuclear spin. for hydrogen, it is half.

$f$ = total angular momentum

$m_f$ = total magnetic angular momentum

a quantum state is $\left| n, l, s, j,i, f , m_f \right>$. but since the s and i are always a half. so, the total degree of freedom will be 5. the nuclear spin added 1 on it.

## Smaller Structure

$H_{vol}$

this term is for the volume shift. the magnitude is $10^{-10}$.

in diagram:

## terminology

this part is for people who confess and poor in english :P

• Nucleus : (noun) is the core of element, which contain protons and neutrons
• Nuclei : (noun) many of nucleus
• Nuclear : (adj) to relate something to nucleus. e.g. nuclear energy, nuclear plant, nuclear phsyics
• Nucleon: (noun) is the thing make the nucleus, which is a single word. a nucleon can be a proton or neutron
• Nuclide: ( noun) [from google translate : A distinct kind of atom or nucleus characterized by a specific number of protons and neutrons ] (what??)
• Nuke: (noun?) (adj?) it is generally used for bomb.

## WKB approximation

I was scared by this term once before. ( the approach an explanation from J.J. Sakurai’s book is not so good)  in fact, don’t panic, it is easy. Let me explain.

i just copy what written in Introduction to Quantum Mechanics by David Griffiths (1995) Chapter 8.

The approx. can be applied when the potential is varies slowly compare the wavelength of the wave function. when it expressed in $Exp( i k x)$, wavelength = 2 π / k, when it expressed in $Exp( - \kappa x )$, wavelength = 1/κ.

in general, the wavefunction can be expressed as amplitude and phase:

$\Psi(x) = A(x)Exp(i \phi(x))$

where $A(x)$ and $\phi(x)$ are real function

sub this into the time-independent Schrödinger equation (TISE)

$\Psi '' (x) = - \frac {2 m} {\hbar^2 } ( E - V(x) ) \Psi (x)$

$\Psi ''(x) = ( A''(x)- A(x) \phi'(x)^2 + 2 i A'(x) \phi'(x)+ i A(x)\phi''(x) ) Exp(i \phi (x) )$

and separate the imaginary part and real part.

The imaginary part is can be simplified as:

$2 A'(x) \phi '(x) + A(x) \phi ''(x) = 0 = \frac {d}{dx} ( A^2(x) \phi '(x)$

$A(x) = \frac {const.} {\sqrt {\phi '(x)}}$

The real part is

$A''(x) = \left ( \phi ''(x) - \frac {2m}{\hbar^2 } ( E - V(x) ) \right) A(x)$

we use the approx. that $A''(x) = 0$ ,  since it varies slowly.

Thus,

$\phi '(x) = \sqrt { \frac {2m}{\hbar^2} (E - V(x) ) }$

$\Rightarrow \phi(x) = \int \sqrt { \frac {2m}{\hbar ^2} ( E - V(x ) )} dx$

if we set,

$p(x) = \sqrt { \frac {2m}{ \hbar^2 } ( E - V(x) )}$

for clear display and $p(x)$ is the energy different between energy and the potential. the solution is :

$\Psi(x) = \frac {const.}{\sqrt {p(x)}} Exp \left( i \int p(x) dx \right)$

Simple! but one thing should keep in mind that, the WKB approx is not OK when Energy = potential.

This tell you, the phase part of the wave function is equal the square of the area of the different of Energy and the Potential.

when the energy is smaller then the potential, than, the wavefunction is under decay.

one direct application of WKB approxi is on the Tunneling effect.

if the potential is large enough, so, the transmittance is dominated by the decay, Thus, the probability of the tunneling is equal to

$Exp \left( - 2 \sqrt { \frac {2m}{\hbar ^2 } A_{area} ( V(x) - E )} \right)$

Therefore, when we have an ugly potential, we can approx it by a rectangular potential with same area to give the similar estimation.

## Scattering phase shift

for a central potential, the angular momentum is a conserved quantity. Thus, we can expand the wave function by the angular momentum wave function:

$\sum a_l Y_{l , m=0} R_l(k, r)$

the m=0 is because the spherical symmetry. the R is the radial part of the wave function. and a is a constant. k is the linear momentum and r is the radial distance.

$R_l(k,r) \rightarrow J_{Bessel} (l, kr )$

which is reasonable when r is infinite and the nuclear potential is very short distance. when r goes to infinity,

$J_{Bessel} (l,kr) \rightarrow \frac {1}{kr} sin( k r - \frac{1}{2} l \pi )$

for elastic scattering, the probability of the current density is conserved in each angular wave function, thus,

the effect of the nuclear potential can only change the phase inside the sin function:

$\frac{1}{kr} sin( k r - \frac {1}{2} l \pi +\delta_l )$

with further treatment, the total cross section is proportional to $sin^2(\delta_l)$.

thus, by knowing the scattering phase shift, we can know the properties of the nuclear potential.

for more detail : check this website

## decay time constant and line width

the spectrum of energy always has a peak and a line width.

the reason for the line width is, this is decay.

i give 2 explanations, once is from classical point of view and i skipped the explanation for the imaginary part. so, i am not fully understand. the 2nd explanation is look better, but it is from QM. however, there is one hide question for that explanation is, why the imaginary energy is negative?

the simplest understanding of the relation is using fourier transform. (i think)

Fourier transform is changing the time-frame into the frequency frame. i.e, i have a wave, propagating with frequency w. we can see a wave shape when plot with time. and we only see a line, when we plot with frequency, since there is only 1 single frequency. however, for a general wave, it is composite of many different frequencies, using fourier transform can tell us which frequency are involved. And energy is proportional to frequency.

when the particle or state under decay. the function is like

$f(t) = Exp(-R t) Exp ( i \omega_0 t)$

where the R is decay constant, and ω0 is the wave frequency.

after fourier transform, assume there is nothing for t < 0

$F(t) = \frac {1} { R + i ( \omega_0 - \omega )}$

the real part is

$Re(F(t)) = \frac {R} { R^2 + ( \omega_0 - \omega )^2}$

which is a Lorentzian shape and have Full-Width-Half-Maximum (FWHM) is 2R. it comes from the cosine part of the fourier transform. thus, the real part.

and the imaginary part is

$Im(F(t)) = \frac {\omega_0 - \omega}{R^2 + ( \omega_0 -\omega )^2 }$

the imaginary part is corresponding to the since part, so, we can neglect it. (how exactly why we can neglect it? )

Thus, we can see, if there is no decay, R → 0, thus, there is no line width.

therefore, we can see the line width in atomic transition, say, 2p to 1s. but there are many other mechanism to the line width, like Doppler broadening, or power broadening. So, Decay will product line width, but not every line width is from decay.

**********************************

another view of this relation is from the quantum mechanics.

the solution of Schroedinger equation is

$\Psi (x,t) = \phi(x) Exp \left( - i \frac {E}{\hbar} t \right)$

so, the probability conserved with time, i.e.:

$|\Psi(x,t)|^2 = |\Psi (x,0)|^2$

if we assume the energy has small imaginary part

$E = E_0 - \frac {i} {2} R \hbar$

( why the imaginary energy is nagative?)

$|\Psi(x,t)|^2 = |\Psi (x,0)|^2 Exp ( - R t)$

that make the wavefunction be :

$\Psi (x,t) = \phi (x) Exp( - i \frac {E}{\hbar} t ) Exp( - \frac {R}{2} t )$

what is the meaning of the imaginary energy?

the wave function is on time-domain, but what is “physical”, or observable is in Energy -domain. so, we want Psi[x,E] rather then Psi[x,t], the way to do the transform is by fourier transform.

and after the transform, the probability of finding particle at energy E is given by

$|\Psi(x,E)|^2 = \frac {Const.}{R^2 +(\omega_0 - \omega )^2}$

which give out the line width in energy.

and the relation between the FWHM(line width) and the decay time is

mean life time ≥ hbar / FWHM

which once again verify the uncertainty principle.