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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.