Dirac equation, one question

  • #1
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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 Feynmann 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?
 
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  • #2
stevendaryl
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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)}##
 
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  • #3
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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 Feynmann used this type of calculation, as you wrote. But I think that he continued I little bit more cleary that in my link?
 

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