Physics Asked by test00alone on February 22, 2021
So in the Wikipedia Page of Dirac equation we are presented with this equation
$$nabla^2 – frac{1}{c^2}frac{partial^2}{partial t^2} = left(A partial_x + B partial_y + C partial_z + frac{i}{c}D partial_tright)left(A partial_x + B partial_y + C partial_z + frac{i}{c}D partial_tright).$$
My question is: Is the multiplication of two first-order equal to second-order as presented above?
When in doubt always use a test function to find out what differential operators might do, as an example (assuming $A,B,C$ and $D$ do not depend on position and do not commute): $$begin{align} left(Apartial_x + Bright)left(Cpartial_x + Dright)f(x) &= left(Apartial_x + Bright)left(Cpartial_x f(x) +D f(x)right) &= Apartial_x left(Cpartial_x f(x) +D f(x)right) + B left(Cpartial_x f(x) +D f(x)right) &= A Cpartial_x partial_x f(x) + A Dpartial_x f(x) + B Cpartial_x f(x) + BDf(x) &= A Cpartial^2_x f(x) + A Dpartial_x f(x) + B Cpartial_x f(x) + BDf(x) end{align}$$
as you can see operating with a differential operator of order 1 on another differential operator of order 1 produces a differential operator of order 2. Perhaps the notation is what is misleading to you: $$partial_x^2 = left(frac{partial}{partial x}right)^2 = frac{partial}{partial x}frac{partial}{partial x}$$ is a second order derivative, while $$left(frac{partial f}{partial x}right)^2 = (partial_x f) (partial_x f)$$ is a square of a first derivative.
Correct answer by ohneVal on February 22, 2021
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