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.


1 Answer 1


The easiest way to see it is to write it in momentum space

$$p_\mu A^\mu = 0$$

In the center of mass frame, where the spin components are well defined, you have

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


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