## Complete derivation from Schrodinger equation to Laguerre equation for Coulomb potential

The Hamiltonian is

$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$

Separate the radial and angular part. The radial equation is

$\displaystyle \left( -\frac{\hbar^2}{2m}\frac{1}{r^2}\left(\frac{d}{dr} r^2 \frac{d}{dr} \right) - \frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} \right) R(r) = E R(r)$

rearrange, using Atomic unit

$\displaystyle \left( -\frac{1}{2}\frac{1}{r^2}\left(\frac{d}{dr} r^2 \frac{d}{dr} \right) - E -\frac{Z}{ r} + \frac{1}{2} \frac{l(l+1)}{r^2} \right) R(r) = 0$

Since the normalization condition of $R(r)$ is $\int_0^\infty R(r)R(r) r^2 dr = 1$, it is natural to define $u(r) = r R(r)$

$\displaystyle R = \frac{u}{r}, \frac{d}{dr}\left(r^2\frac{dR}{dr}\right) = r \frac{d^2u}{dr^2}$

The radial equation becomes

$\displaystyle -\frac{1}{2}\frac{d^2u}{dr^2} - \left( E + \frac{Z}{ r} - \frac{l(l+1)}{2r^2} \right) u(r)= 0$

Substitute the eigen energy $E_n = - \frac{Z^2}{2n^2}$

$\displaystyle \frac{d^2u}{dr^2} + \left( - \frac{Z^2}{n^2} + \frac{2Z}{ r} - \frac{l(l+1)}{r^2} \right) u(r)= 0$

Set

$\displaystyle x = \frac{2Z}{n} r, \frac{d}{dr} = \frac{d}{dx} \frac{2Z}{n}$

The radial equation becomes

$\displaystyle \left(\frac{2Z}{n}\right)^2\frac{d^2u(x)}{dx^2} - \left( - \frac{Z^2}{n^2} + \frac{4Z^2}{n x} - \frac{4Z^2}{n^2}\frac{l(l+1)}{x^2} \right) u(x)= 0$

Take out the $4Z^2/n^2$

$\displaystyle \frac{d^2u(x)}{dx^2} - \left(\frac{n}{ x} - \frac{1}{4} - \frac{l(l+1)}{x^2} \right) u(x)= 0$

Now, we know the short and long-range behaviour of $u(x)$, assume it to be

$u(x) = \exp\left(-\frac{x}{2} \right) x^{l+1} y(x)$

Then the equation of $y(x)$ is

$\displaystyle x \frac{d^2y}{dx^2} + \left( 2(l+1) - x\right) \frac{dy}{dx} +\left( n - l- 1 \right) y(x) = 0$

This is the Laguerre equation

$\displaystyle x \frac{d^2y}{dx^2} + \left( \alpha+1 - x\right) \frac{dy}{dx} + m y(x) = 0$

where $\alpha = 2l+1$ and $m = n-l -1$. Therefore, the solution of the radial equation is,

$R(r) = \frac{1}{r} A \exp\left(-\frac{x}{2} \right) x^{l+1} L_{n-l-1}^{2l+1}(x)$

where $A$ is normalization factor.

Notice that the Laguerre polynomial is only defined for $m \geq 0$, thus, $n > l$.

## Cross seciton of Coulomb Scattering

The coulomb scattering looks very easy, the formula of the differential cross section in CM frame is,

$\frac{d\sigma}{d\Omega} = (\frac{Z1 Z2 e}{4 E_{cm}})^2 \frac{1}{sin^4(\theta_{cm}/2)}$,

where $e = 1.44 MeV fm$ and $1 fm^2 = 10 mb$. The tricky point is, in most experiment, we are working in Laboratory frame that require frame transformation.

The relationship of the energy in CM frame and the energy in the Lab frame can be found by Lorentz transform, and use the total kinematic energy (both particle 1 and particle 2). In the CM frame, we can image we have a fixed virtual target on the center of mass, and there is only 1 object moving at energy of the total kinematic energy.

For example, we have a target of mass $m_2$, a projectile with mass $m_1$ and energy $T_1$, a classical energy in CM frame is

$E_{cm} = \frac{m_2}{m_1+m_2} T_1$

In fact the $E_{cm}$ has only 5% difference between relativistic and non-relativistic even up to 500 MeV

When calculating the integrated cross section, we can do it in the CM frame, but it is more intuitive to do it in the Lab frame. In this case, we need to transform the differential cross section from the CM frame to the Lab frame, mathematically, the transformation is done by a factor called Jacobian.

We can compare the result using the kinematic calculator in LISE++.

In the above plot, the blue line is the d.s.c. in CM frame, and the red line is d.s.c. of the 9Be in Lab frame. Jacobian was added, therefore, the zero degree d.s.c. of 9Be is larger than the 180 degree d.s.c. in the CM frame.

The grazing angle of the scattering, can be calculated by the shorted distance between the target and the projectile. In the Lab frame, the target is not fixed, so it is not easy to know the shortest distance. But in the CM frame, the virtual target is fixed, and we can calculate the distance using the $E_{cm}$ and $\theta_{cm}$.

## Energies in a nucleus

There are many kinds of energies, such as single particle energy, potential energy, kinetic energy, separation energy, and Fermi energy. How these energies are related?

I summarized in the following picture that the occupation number as a function of kinetic energy.

Since the nucleus is a highly interactive system, although the temperature of the ground state of a nucleus should be absolute Zero, but the Fermi surface is not sharp but diffusive.

The Fermi energy of a nucleon , which is the maximum kinetic energy of a nucleon, is approximately ~35 MeV.

The potential energy is approximately ~ 50 MeV per nucleon.

There is an additional energy for proton due to Coulomb force, which is a Coulomb barrier.

The separation energy  is the different between the potential energy and Fermi energy.

The single particle energy is the energy for each single particle orbit.

The binding energy for a nucleon is the energy requires to set that nucleon to be free, i.e. the energy difference between the single particle energy and the potential energy.

There is a minimum kinetic energy, using the 3D spherical well as an approximation. The n-th root of the spherical Bessel function can give the energy of n-orbit with angular momentum $l$, so that.

$j_{l}(kR) = 0, k^2=\frac{2 m E}{\hbar^2}$

For $l=0$, $j_0(x) = \frac{sin(x)}{x}$, the 1-st root is $x = \pi/2$, then

$k=\frac{\pi}{2R}$

$\frac{2 m E}{\hbar^2} = \frac{\pi^2}{4 R^2}$

use $R = 1.25 A^{1/3}$,

$E = \frac{\pi^2 \hbar^2}{8 m 1.25^2 A^{2/3}}$

use $\hbar c = 200 MeV \cdot fm$, $mc^2 = 940 MeV$

$E = \frac{33.6}{A^{2/3}} MeV$

we can see, for $^{16}O$, the minimum KE, which is the 1s-orbit, is about 5 MeV.

## Stability of a nucleus ( Liquid Drop Model )

when look at the table of the nuclear world, why there are some nucleus more stable then the other? which mean, why some will decay while some are not?

OK, this basically the ultimate question that nuclear physics want to answer.

so, the very fundamental reason, no one know.

but in the elementary level, or by experimental fact and some assumption. we have Binding Energy to estimate or predict the stability of a nucleus. when the Binding Energy is larger then Zero, it must be unstable and will decay under conservation laws. if it is less then zero, it may be stable or not, it depends on whether it reach the bottom of energy level.

Binding Energy can also be though as the energy required to break the nucleus.

In liquid drop model, we imagine the nucleus is like a liquid. and nucleons inside just like liquid molecules. experiments show that nucleus is a spherical object. and it density is a constant. and the interaction range of nuclear force is short, few fm. thus, it likes a incompressible liquid drop. the radius of it is related to the mass number:

$R^3 = A$

the Binding Energy ( = $\Delta M(A,Z,N)$ = mass deficit) is given by theoretical assumption and experimental fact.

$\Delta M(A,Z,N) = - a_1 A + a_2 A^{\frac {2}{3} } + a_3 Z^2 / A^{ \frac{1}{3}} + a_4 (N-Z)^2 /A \pm a_5 A^{- \frac{3}{4} }$

the first 3 terms are theoretical assumption and the lat 2 terms are from experimental fact. All coefficients are given by experimental measurement.

The first term is the “volume energy” by the nuclear force, which is proportional to the number of nucleons.

the 2nd term is the “surface tension” from the “liquid”. we can see its dimension is area.  (why this term is + ? ) it  explained why smaller nucleus has less Binding energy.

the 3rd term is the Coulomb potential term.

the 4th term is the balance term.  if the number of neutron and proton is no balance,

the 5th term is the “Symmetry term“. for even-even of neutron and proton number, the nucleus is more stable, thus, we choose minus sign for it. for odd-odd combination, nucleus are more unstable, thus, plus sign for it. for other, like ood – even or even-odd combination, this term is zero.

the value of the coefficients are:

$a_1 \simeq 15.6 MeV$

$a_2 \simeq 16.8 MeV$

$a_3 \simeq 0.72 MeV$

$a_4 \simeq 23.3 MeV$

$a_5 \simeq 34 MeV$

The below plot is the Binding Energy per nucleon in  Z against N.

Lets use the liquid drop model and Binding Energy to look the β-decay. the β-decay conserved the mass number A. there are 2 β-decays.

$\beta_- : n \rightarrow p + e^- + \bar{\nu_e}$

$\beta_+ : p \rightarrow n + e^+ + \nu_e$

so, the β+ decay decrease the number of proton while β– decay increase the number of proton.

The below diagram show the β-decay for A = 22. we can see the 22Ne is stable, since no more β-decay can help to reach a lower energy level.

## 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:

## Mass of particles and nucleus

in Nuclear physics, the particle we deal with are so small and so light, if we use standard unit, then there will be many zero and we will lost in the zeros. for example, the electron has mass:

Mass( electron ) = 9.11 × 10-31 kilograms
Mass( proton ) = 1.67 × 10-27 kilograms

see? as the special relativity give us a translation tool – E = m c^2, thus, we can use MeV to talk about mass.

Mass ( electron ) = 0.511 MeV
Mass ( proton ) = 938.3 MeV

thus, we can see, Proton is roughtly 2000 times heavier then electron ( 1000 : 0.5 ).

Mass( neutron ) = 939.6 MeV

neutron is just 1.3 MeV heavier then proton.

The nucleus is formed by proton and neutron. so, in simple thought, an nucleus with Z proton and ( A-Z ) neutron should have mass

Z x Mass( proton ) + ( A – Z ) x Mass ( neutron ) = Mass ( A, Z )

where A is the atomic mass number, which is equal the number of nucleons in the nucleus, and Z is the proton number.

However, scientists found that it is not true.

Z x Mass( proton ) + ( A – Z ) x Mass ( neutron ) > Mass ( A, Z )

Some of the mass is missing! But that is explained why nucleus will not break down automatically. since it need extra energy to break it down.

we called the mass different is Mass Deficit. or Binding energy.

Mass Deficit = Mass( A, Z) – Mass ( proton + neutron )

some one may think that the binding energy is the energy for holding the nucleus together. in order to hold the nucleus, some mass was converted into the energy to holding it. this is INCORRECT. the correct argument is, the binding energy is th energy require to break it down.

think about a simple 2 bodies system, like sun and earth. at far far away, when both of them are at rest, the total energy is Mass( sun) + Mass ( earth ) + Potential energy.

when the earth moves toward to sun, the potential energy converted to the Kinetic energy, so the earth moving faster and faster. but, in order to stay in the orbit, some K.E. must be lost so that it does not have enough ( or the same) energy to run away. Thus, the total energy of the system is lesser then the total mass.

another analogy is electron orbit. when an electron was captured by an atom, it radiate energy in order to stay in some energy level. thus, the total energy of the system again less then the total mass.

any any case, the mass of the sun and earth and electron does not change, but the potential changes to negative, thus it makes to total energy lesser.

similar idea hold for nucleus, but the potential of it are great different, because there are a Coulomb Barrier. Thus, in order to make a nucleus. we have to put so many K.E. to again this barrier, then the resultant nucleus release the Mass Deficit energy and also the input K.E..

 a scratch on the nuclear potential. there are a Coulomb Barrie. ( by wolframalpha.com)
When the nucleus is radioactive and undergoes decay. this mean, it Mass deficit is positive. thus, it will automatically break down to another nucleus until it mass deficit is negative again. during this process, the emitted particle carry K.E. which is from the potential. Not the mass for one nucleons.
Remember, Mass( nucleus ) = Mass ( protons + neutrons ) + Potential

## alpha decay

different decay cause by deferent mechanism, we first start on alpha decay.

i assume we know what is alpha decay, which is a process that bring excited nucleus to lower energy state by emitting an alpha particle.

The force govern this process is the strong force, due to the force is so strong, the interaction time is very short, base on the uncertainty principle that large change in energy leads to short time interval. however, the observed alpha decay constant is about 1.3 × 1010 year, which is about the age of our universe. That’s why we still able to find it at the beginning of nuclear physics : discovery of radioactive matter.

The reason for such a long decay time is due to the Coulomb barrier of the nuclear potential. since the proton carry positive charge, thus. it creates a positive potential wall in the nucleus. that potential not only repulse proton from outside but also the proton from inside which try to get out. thus, the inside protons are bounded back and forth inside the nucleus. due to the momentum carried by the protons, it has frequency 6  × 1021 per sec.

Due to the Quantum tunneling effect, the probability of tunneling is 4 × 10-40. which is a very small chance. But , don’t forget there are  6  × 1021 trails per sec. Thus, the chance per sec is 2.4 × 10-18 . and the mean life time is inverse of the probability, thus it is approx 1.3 × 1010 year.