# Expansion coefficients in the solution of the Dirac equation for a free particle

So my question is why do we need to write the coefficients $b$ (that after the second quantization are going to be promoted as the antiparticle creation operators) as complex conjugate? I mean, why not just write $b$, without the complex conjugation sign in the solution
$$\psi(x) =\int \frac{d^{3}p}{(2\pi)^{3}}\sum_{s=1,2}(a_{p,s}u_s(p)e^{-ipx} + b^{*}_{p,s}v_s(p)e^{ipx})$$ I hope my question is clear enough. It bothers me that in a lot of textbooks you just find $b^{*}$ without a clear argument why it isn't just $b$.

It is just notation. If you schematically write $$\psi\sim A\mathrm e^{-ipx}+B\mathrm e^{+ipx}$$ then you can check that \begin{aligned} {}[H,A]=-\omega A\\ [H,B]=+\omega B \end{aligned} so that $A$ behaves like an annihilation operator (it lowers the energy) and $B$ behaves like a creation operator (it increases the energy). Therefore, it makes sense to write $B\equiv b^\dagger$ - it is just convenient notation. You could omit this relabelling if you wanted to, but why would you.