# How does the Lorenz gauge eliminate the scalar component of the vector field?

Wikipedia states that by using the Lorenz gauge, $$\partial_\mu A^\mu=0$$, we eliminate the scalar part (spin-0) of the vector potential that previously had spin-1 and spin-0 components$${}^1$$.

However, this excellent Phys.SE answer by @AccidentalFourierTransform states that for a gauge fixing term in the Lagrangian $$\delta\mathcal L = -\frac{1}{2\xi}(\partial_\mu A^\mu)^2$$, the propagator for $$A^\mu$$ is given by$${}^2$$ $$\tilde\Delta_{\mu\nu}(p)=\underbrace{\frac{\eta_{\mu\nu}-\frac{p_\mu p_\nu}{\color{red}{m^2}}}{p^2-m^2+\text i\epsilon}}_{j=1} + \underbrace{\frac{\frac{p_\mu p_\nu}{m^2}}{p^2-\xi \,m^2+\text i\epsilon}}_{j=0}\tag{1}$$ where we can clearly identify the spin-1 and spin-0 parts.

The Lorenz gauge corresponds to setting $$\xi=0$$, which yields the following propagator: $$\tilde\Delta_{\mu\nu}(p)\stackrel{\xi=0}=\dfrac{\eta_{\mu\nu}-\frac{p_\mu p_\nu}{\color{red}{p^2}}}{p^2-m^2+\text i\epsilon}$$

Due to the statement on Wikipedia I would have assumed that $$\xi=0$$, i.e. choosing the Lorenz gauge, just eliminates the second (scalar) term in (1). Apparently, it is not that simple. Where lies my misunderstanding?

$${}^1$$ Because $$A^\mu$$ belongs to the Lorentz representation $$\big(\tfrac{1}{2},\tfrac{1}{2}\big)$$.

$${}^2$$ The textbook source for this is Itzykson & Zuber's QFT book.

$$p_\mu A^\mu = 0$$
$$\bar p_\mu=(M,0,0,0) \implies \bar p_\mu A^\mu =M A^0 = 0 \implies A^0 = 0$$