4
$\begingroup$

The full gauge propagator in the $R_\xi$ gauge is $$D_{\mu\nu} = \frac{i}{k^2+i\epsilon}\left(-g_{\mu\nu}+\frac{1-\xi}{k^2}k_\mu k_\nu\right).\tag{1}$$ Now if we take $\xi=0$, we get the Lorenz gauge, and we see that $$k^\mu D_{\mu\nu}=0.\tag{2}$$ However if we take $\xi = 1$ (Feynman gauge) or $\xi\rightarrow\infty$ (unitary gauge), we get $k^\mu D_{\mu\nu}\neq0.$

Why does this equation (2) only hold in the Lorenz gauge? I of course see it mathematically but what property of the Lorenz gauge ensures this equality, so could we have guess this property a priori?

I suspect it has something to do with the ward identities and the propagator being transverse in the Lorenz gauge (perhaps ghost and longitudinal degrees of freedom canceling in this gauge), but I'm not sure.

$\endgroup$

2 Answers 2

3
$\begingroup$

OP's eq. (1) is the free/bare propagator, which is not a physical observable and does not have to obey OP's eq. (2). The pertinent Ward identity involves instead the self-energy/vacuum-polarization $$ k_{\mu}\Pi^{\mu\nu}(k)~=~0,\tag{62.6}$$ cf. e.g. Ref. 1.

References:

  1. M. Srednicki, QFT, 2007; chapters 62 p. 377. A prepublication draft PDF file is available here.
$\endgroup$
2
$\begingroup$

Defining the propagator is a little tricky in gauge theory, but schematically at least, the propagator is essentially $$D^{\mu\nu}(k)=\int d^{4}x\,e^{ik\cdot(x-y)}\langle 0|\mathcal{T}A^{\mu}(x)A^{\nu}(y)|0\rangle.$$ If this is contracted with the four-momentum $k$, the $k_{\mu}$ in $k_{\mu}D^{\mu\nu}$ can be moved inside the integral and changed to a derivative acting on the exponential, to give $$k_{\mu}D^{\mu\nu}(k)=\int d^{4}x\left[-i\partial_{\mu}e^{ik\cdot(x-y)}\right]\langle 0|\mathcal{T}A^{\mu}(x)A^{\nu}(y)|0\rangle.$$ The integration by parts moves the derivative to the matrix element, and we have $$k_{\mu}D^{\mu\nu}(k)=i\int d^{4}x\,e^{ik\cdot(x-y)}\langle 0|\mathcal{T}\left[\partial_{\mu}A^{\mu}(x)\right]A^{\nu}(y)|0\rangle,$$ which is manifestly vanishing in the Lorenz gauge,* $\partial_{\mu}A^{\mu}=0$.

In a different gauge, there is no reason for the expression $k_{\mu}D^{\mu\nu}$ to vanish on its own. Instead, what happens is that interacting matrix elements with external photon Lorentz indices will necessarily vanish. How this works can be seen by looking at the simplest version, in which the photon propagator is attached to a source current $J_{\mu}$. The source $J_{\mu}$ may be the Dirac current $\bar{\psi}\gamma_{\mu}\psi$, or an external current, for example—but the key thing is that it is conserved, $\partial^{\mu}J_{\mu}=0$. Calculating an expression such as $k_{\mu}D^{\mu\nu}(k)J_{\mu}$ in Fourier space, the $k_{\mu}$ can once again be pulled inside an integral as a derivative and moved via integration by parts, which results in an expression with $\partial^{\mu}J_{\mu}=0$.

*Actually, when the theory is described in terms of propagators, this is normally known as the "Landau gauge."

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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