Derivation of propagator for Proca action in QFT book by A.Zee

Question

Without considering gauge invariance, A.Zee derives Green function of electromagnetic field in his famous book, Quantum Field Theory in Nutshell. In chapter I.5, the Proca action would be, $$S(A) = \int{d^{4}x\,\left\{\frac{1}{2}A_{\mu}\left[(\partial^2+m^2)g^{\mu\nu}-\partial^{\mu}\partial^{\nu}\right]A_{\nu}+A_{\mu}J^{\mu}\right\}},\tag{1}$$ And the inverse of the differential operator in the square bracket would statisfy, $$\left[(\partial^2+m^2)g^{\mu\nu}-\partial^{\mu}\partial^{\nu}\right]D_{\nu\lambda}(x) = \delta^{\mu}_{\lambda}\delta^{(4)}(x)\tag{2}$$ Using Fourier transform $D_{\nu\lambda}(x) = \int{d^4k/(2\pi)^4D_{\nu\lambda}(k)e^{-ikx}}$ to solve the differential equation, we got the following equation, $$\left[-\left(k^2-m^2\right)g^{\mu\nu}+k^{\mu}k^{\nu}\right]D_{\nu\lambda}(k) = \delta^{\mu}_{\lambda}$$ Then I expect the solution would be, $$D_{\nu\lambda}(k) = \frac{g_{\nu\lambda}}{-\left(k^2-m^2\right)g^{\mu\nu}g_{\mu\nu}+k^{\mu}k^{\nu}g_{\mu\nu}}$$ However, A.Zee claims that it should be, $$D_{\nu\lambda}(k) = \frac{-g_{\nu\lambda}+k_{\nu}k_{\lambda}/m^2}{k^2-m^2}\tag{3}$$ I don't understand why the numerator would have second term, I suppose it should belong to the denominator. Please tell me how does he get (3)?

Answer 1

Your question can be formulated as the more general problem of finding the inverse $(T^{-1})^{\nu\rho}$ of the (momentum space) tensor $$T_{\mu \nu}=a(k^2) g_{\mu \nu}+b(k^2)k_\mu k_\nu \tag{1} \label{eq1}$$ satisfying the relation $$T_{\mu\nu}(T^{-1})^{\nu \lambda}=\delta_\mu^{\;\lambda}. \tag{2} \label{eq2}$$ Inserting the ansatz $$(T^{-1})^{\nu \lambda}=A(k^2) g^{\nu \lambda}+B(k^2) k^\nu k^\lambda \tag{3} \label{eq3}$$ into \eqref{eq2}, one finds $$a(k^2)A(k^2)\delta_\mu^{\;\lambda}+ \left[a(k^2)B(k^2) +b(k^2)A(k^2) +b(k^2)B(k^2) k^2\right] k_\mu k^\lambda=\delta_\mu^{\;\lambda}. \tag{4} \label{eq4} $$ Assuming $a(k^2)\ne 0$, $A(k^2)$ and $B(k^2)$ are obtained by comparing the coefficients of the linear independent terms $\delta_\mu^{\;\lambda}$ and $k_\mu k^\lambda$ in \eqref{eq4}, $$ A(k^2) = \frac{1}{a(k^2)}, \quad B(k^2)=-\frac{b(k^2)}{a(k^2) \left[a(k^2)+b(k^2)k^2\right]}, \tag{5} \label{eq5}$$ leading to the final result $$(T^{-1})_{\mu\nu}= \frac{1}{a(k^2)} \left(g_{\mu \nu} - \frac{k_\mu k_\nu \, b(k^2)}{a(k^2)+b(k^2)k^2}\right). \tag{6} \label{eq6}$$ It is now an easy task to find the solution for the special case $a(k^2)=m^2-k^2$, $b(k^2)=1$.

Answer 2

Firstly, consider that $g^{\mu\nu}g_{\nu\lambda}=\delta^\mu_\lambda$

So, if the equation was $$-(k^2-m^2)g^{\mu\nu}\tilde D_{\nu\lambda}=\delta^\mu_\lambda$$ then the solution would be $$\tilde D_{\nu\lambda}=\frac{-g_{\nu\lambda}}{k^2-m^2}$$ but that is not the equation we have. We can try and see what happens if we try this solution on the extra term, namely, $$k^\mu k^\nu\frac{-g_{\nu\lambda}}{k^2-m^2}=\frac{-k^\mu k_\lambda}{k^2-m^2}$$ Let us check whether Zee's solution works: $$-(k^2-m^2)g^{\mu\nu}\frac{k_\nu k_\lambda/m^2}{k^2-m^2}=\frac{-k^\mu k_\lambda}{m^2} \qquad\bigwedge\qquad k^\mu k^\nu\frac{k_\nu k_\lambda/m^2}{k^2-m^2}=\frac{k^\mu k_\lambda k^2/m^2}{k^2-m^2}$$ The combination of these last 3 terms is an exact cancellation, so Zee's solution is the correct one.

That is, I am not trying to tell you how to get Zee's solution, but that it should be easy to simply substitute into the equation and check that Zee's solution is the one that works. Solutions to DE are difficult to explain; oftentimes it is a matter of "we tried this and found that it works". In particular, the first extra term that we found, gave us a hint as to what kind of entity we should try to guess for, and then we could play with the terms that appear, until we either get the answer, or we find that such games do not work. It is not a straight-forward process.