The origin of spin orbit coupling

[amjad-sh]

I have read in the internet that "One naturally derive the dirac equation when starting from the relativistic expression of kinetic energy:

$\mathbf H^2=c^2\mathbf P^2 +m^2c^4$ where $\mathbf P$ is the canonical momentum.
Inclusion of electric and magnetic potentials $\phi$ and $A$ by substituting $\mathbf P -εA$in $\mathbf P$ and $\mathbf H -ε\phi$ in $\mathbf H$ we get $(\mathbf H -ε\phi)^2=(c\mathbf P -εA)^2+m^2c^4$

1-My first question is: Why potential and electric potentials are included in that way, I mean $\mathbf P -εA$in $\mathbf P$ and $\mathbf H -ε\phi$ in $\mathbf H$?

Then the text I am reading continued to the part where he stated the dirac equation which is:

$(\mathbf H-c\sum_{\mu}p_{\mu}-\beta mc^2)\psi=0$

Now with the fact that$[\mathbf H -ε\phi -cα.(\mathbf p -ε/c\mathbf A) -\beta mc^2][\mathbf H -ε\phi +cα.(\mathbf p -ε/c\mathbf A) +\beta mc^2]\psi=0$

Using the approximation that the kinetic and the potential energies are small compared to mc^2, two components of the spin function can be neglected and the equation above take the form :

$[1/2m(\mathbf p -ε/c\mathbf A)^2 +ε\phi -(ε\hbar/2mc)\sigma \cdot \mathbf B -(ε\hbar/4m^2c^2)\mathbf E \cdot \mathbf p -(ε\hbar/4m^2c^2)σ \cdot (\mathbf E \times \mathbf p)]\psi=W\psi$
Where W +mc^2 is the total energy.

2-My second question is how we can reach the second formula, If somebody can give me some hints?

 

[BvU]

What internet was that? Mine says

No results found for "naturally derive the dirac equation when starting from the relativistic expression of kinetic".​

 

[amjad-sh]

I downloaded this chapter.It is in the first page.

 

[BvU]

It says "One naturally arrives at the Dirac equation when starting from the relativistic expression for the kinetic energy" . Hard to find an author, though.

Anyway, taking $e\phi$ and $e\vec A$ into account as shown is standard (naturally ), see e.g here or here

 

[amjad-sh]

It says "One naturally arrives at the Dirac equation when starting from the relativistic expression for the kinetic energy" . Hard to find an author, though.

Anyway, taking $e\phi$ and $e\vec A$ into account as shown is standard (naturally ), see e.g here or here

Thanks,this was helpful.
I got the answer of my first question, can you give me some hints to reach the last equation?

 

[DrClaude]

I got the answer of my first question, can you give me some hints to reach the last equation?

Do you know what $\alpha$ and $\beta$ are? You are going to need that, along with the identity
$$
(\mathbf{\sigma} \cdot \mathbf{A})(\mathbf{\sigma} \cdot \mathbf{B}) = \mathbf{A} \cdot \mathbf{B} + i \mathbf{\sigma} \cdot (\mathbf{A} \times \mathbf{B})
$$

By the way, you can find much better derivations of the spin-orbit coupling than the document you posted. Check out instead one of
Sakurai and Napolitano, Modern Quantum Mechanics
Bransden and Joachain, Physics of Atoms and Molecules
Friedrich, Theoretical Atomic Physics