How Does the Dirac Equation Utilize Component Functions in Its Derivation?

[exponent137]

In
https://quantummechanics.ucsd.edu/ph130a/130_notes/node45.html
after
"Instead of an equation which is second order in the time derivative, we can make a first order equation, like the Schrödinger equation, by extending this equation to four components."

it is evident that the solution is obtained with help of $a^2-b^2=(a+b)(a-b)$

I cannot follow in this derivation, how rows $\phi^{(L)}=...$ and $\phi^{(R)}=...$ are used. Maybe more steps instead of these two rows will help.

Although I think that Feynman once described this more clearly in his book about QED.

Can someone, please, gives a link or more clearly explains this type of derivation of Dirac equation?

 

[stevendaryl]

I'm not sure what your point of confusion is. Do you understand that the following equations are true, by definition of $\phi^{(R)}$ and $\phi^{(L)}$ (plus the fact that $\phi$ obeys the second-order equation)?

$(i \hbar \frac{\partial}{\partial t} - i \hbar \sigma \cdot \nabla) \phi^{(R)} = mc \phi^{(L)}$

$(i \hbar \frac{\partial}{\partial t} + i \hbar \sigma \cdot \nabla)\phi^{(L)} = mc \phi^{(R)}$

 

[exponent137]

I'm not sure what your point of confusion is. Do you understand that the following equations are true, by definition of $\phi^{(R)}$ and $\phi^{(L)}$ (plus the fact that $\phi$ obeys the second-order equation)?

$(i \hbar \frac{\partial}{\partial t} - i \hbar \sigma \cdot \nabla) \phi^{(R)} = mc \phi^{(L)}$

$(i \hbar \frac{\partial}{\partial t} + i \hbar \sigma \cdot \nabla)\phi^{(L)} = mc \phi^{(R)}$

If I put the second equation in the first one, I obtain:

$(i \hbar \frac{\partial}{\partial t} - i \hbar \sigma \cdot \nabla)(i \hbar \frac{\partial}{\partial t}+ i \hbar \sigma \cdot \nabla)\phi^{(L)} = (mc)^2 \phi^{(L)}$

One problem is solved, I think that now I understand this derivation of these two rows, as I mentioned. Thanks.

I think that Feynman used this type of calculation, as you wrote. But I think that he continued I little bit more cleary that in my link?