## 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.

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## 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 $10^{-4} eV$

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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

## 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