Tensor structure of the one-loop vacuum polarization in scalar QED I'm working on the book by Schwartz to study QFT. This question concerns the evaluation of the vacuum polarization loop in scalar QED. Some more details of the calculation may be found in Schwartz chap. 16.2 p.304-308. The $e^2$ corrections to the photon propagator are the loop diagram and the seagull in scalar QED. Adding them one gets the following expression:
\begin{equation}
i\Pi_2^{\mu\nu}=-e^2\int\frac{d^4k}{(2\pi)^4}\frac{-4k^{\mu}k^{\nu}+2p^\mu k^\nu+2k^\mu p^\nu-p^\mu p^\nu+2g^{\mu\nu}[(p-k)^2-m^2]}{[(p-k)^2-m^2+i0^+][k^2-m^2+i0^+]}\tag{16.24}
\end{equation}
and to simplify things, Schwartz argues about the general structure of this correction. By Lorentz invariance, we know what kind of tensor structure to expect:
$$\Pi_2^{\mu\nu}=\Delta_1(p^2,m^2)p^2g^{\mu\nu}+\Delta_2(p^2,m^2)p^\mu p^\nu\tag{16.25}$$
for some $\Delta_{1,2}$. Now, working in Feynman gauge, he argues that the full Green's function will have the form
$$\begin{align}iG^{\mu\nu}(p)=&\frac{-ig^{\mu\nu}}{p^2+i0^+}+\frac{-ig^{\mu\alpha}}{p^2+i0^+}i\Pi^2_{\alpha\beta}\frac{-ig^{\beta\mu}}{p^2+i0^+}+...=...\cr =&-i\frac{(1+\Delta_1)g^{\mu\nu}+\Delta_2\frac{p^\mu p^\nu}{p^2}}{p^2+i0^+}+\mathcal{O}(e^4)\end{align}\tag{16.28}$$
and thus the $\Delta_2$ term corresponds to a change of gauge, which implies $\Delta_2$ is non-physical and not observable. This also implies we only need to care about the $\propto g^{\mu\nu}$ term in the integral, so only $\Delta_1$. So far so ok. The calculations are not the issue, one can use dimensional regularization as described in detail in the book. Explicitly one gets:
\begin{equation}
\Delta_1 = -\frac{\alpha}{2\pi}\int_0^1dx\;x(2x-1)\left[\frac{2}{\epsilon}+\ln\left(\frac{4\pi e^{-\gamma_E}\mu^2}{m^2-p^2x(1-x)}\right)+\mathcal{O}(\epsilon)\right]
\end{equation}
where the loop momentum was shifted $k^\mu\to k^\mu+p^\mu(1-x)$ and we use dimensional regularization. In the book, it is not done, but this shift gives the $\Delta_2$ term as 
\begin{equation}
\Delta_2 = \frac{\alpha}{2\pi}\int_0^1dx\;\left(2x^2-2x+\frac{1}{2}\right)\left[\frac{2}{\epsilon}+\ln\left(\frac{4\pi e^{-\gamma_E}\mu^2}{m^2-p^2x(1-x)}\right)+\mathcal{O}(\epsilon)\right]
\end{equation}
where one only looks at the $\propto p^\mu p^\nu$ part. Also, any odd function of $k$ in the numerator gives zero by symmetry.My question concerns the $\propto p^\mu p^\nu$ part, i.e. $\Delta_2$. Schwartz argues, that in order to fulfil the Ward identity we get $\Delta_1=-\Delta_2$. But I'm wondering why the identity should be fulfilled? We're not talking about some amplitude which may stem from a physical process. We're merely addressing a correction to the 2-point Green's function. When I calculate $\Delta_2$ I get some (divergent) integral as shown above, yet not $-\Delta_1$. I'm not surprised in as far as I don't expect the Ward identity to be forced to hold just for any photon n-point function, but only in actual amplitudes. I'm mainly surprised as in the book he states "it is the unique result [...] that satisfies the Ward identity" which is used as an argument to deduce the form of $\Delta_2$ without calculating it and thus simply claiming $\Delta_2=-\Delta_1$ because of Ward. I'm fairly sure that I didn't simply mess up the calculation because after all, it's the same procedure as the one for $\Delta_1$. But now I'm confused: Is Ward supposed to hold here? Do I need to fix a specific gauge so I can have $\Delta_2=-\Delta_1$? Why should some order $e^2$ photon propagator correction satisfy the Ward identity, especially as the leading order term itself does not?
I'm guessing that Schwartz means exactly that we may use the unphysical nature of $\Delta_2$ to "ignore" most of it and use it to get some "nice" property such as the Ward identity. It's simply confusing that he suggests to brute-force calculate $\Delta_2$ from the diagram in Feynman gauge and then he gives some quite different result.
EDIT:
To calculate the $\Delta_2$ term we have to look at the transformation of the numerator:
$N^{\mu\nu}:=-4k^{\mu}k^{\nu}+2p^\mu k^\nu+2k^\mu p^\nu-p^\mu p^\nu+2g^{\mu\nu}[(p-k)^2-m^2]$ under $k^\mu\to k^\mu+p^\mu(1-x)$. This gives
\begin{align}
N^{\mu\nu}=&-4k^{\mu}k^{\nu}+2p^\mu k^\nu+2k^\mu p^\nu-p^\mu p^\nu+2g^{\mu\nu}[(p-k)^2-m^2]\\
\to&-4(k^\mu+p^\mu(1-x))(k^\nu+p^\nu(1-x))+2p^\mu(k^\nu+p^\nu(1-x))+2(k^\mu+p^\mu(1-x))p^\nu-p^\mu p^\nu+2g^{\mu\nu}[(p-(k+p(1-x)))^2-m^2]\\
=&p^\mu p^\nu[-4(1-x)^2+4(1-x)-1]-4k^\mu k^\nu+2g^{\mu\nu}(k^2+x^2p^2-m^2)+\mathcal{O}(p\cdot k, k^\mu, k^\nu)
\end{align}
where the last terms are zero by symmetry of the integral. Doing the dimensional regularization for the second terms gives exactly $\Delta_1$. The term $\propto p^\mu p^\nu$ gives the $\Delta_2$ term and it turns out the only difference is the polynomial in $x$.
EDIT:
To see that indeed $\Delta_1=-\Delta_2$, first note that in the integral in $\Delta_2$ we can rewrite $2x^2-2x+\frac{1}{2}=x(2x-1)-(x-\frac{1}{2})$ and thus it suffices to see that the second term vanishes in the integral. For this, we can substitute $y=(x-\frac{1}{2})$, which gives $m^2-p^2x(1-x)\to m^2-p^2(\frac{1}{4}-y^2)$ and thus the second part vanishes by symmetry of the logarithm in $y$.
 A: Properly speaking @iDslash is right: Ward identity concern physically possible scattering processes and thus have all their external particles on-shell. But it could be generalized to the Ward–Takahashi identity which holds for every correlation function.
Hint: without speaking about Ward identity, you can explicitly check that $\Pi^{\mu \nu} p_{\nu} = 0$
Solution for "proper" QED: (hold the mouse over the grey box)

 $$\Pi^{\mu \nu} k_{\nu} = i \int \frac{d^p}{(2\pi)^4} Tr \left( \gamma^{\mu} \frac{ {k\!\!\!/} + {p\!\!\!/} + m}{(k+p)^2 - m^2} {k\!\!\!/} \frac{{p\!\!\!/} + m}{p^2 - m^2} \right) \, .$$ Now, $$ ({k\!\!\!/} + {p\!\!\!/} +m){k\!\!\!/}({p\!\!\!/}+m) = ({k\!\!\!/} + {p\!\!\!/} +m) \left[ ({k\!\!\!/} + {p\!\!\!/} -m) - ({p\!\!\!/} -m) \right] ({p\!\!\!/} +m) \, ,$$ so that $$ \Pi^{\mu \nu} k_{\nu} = i \int\frac{d^4}{(2\pi^4)} Tr \left( \gamma^{\mu} \left[ \frac{{p\!\!\!/} +m}{p^2 - m^2} - \frac{({k\!\!\!/} + {p\!\!\!/} +m)}{(k+p)^2 - m^2} \right] \right) \, .$$ This is zero after a change of variable in the second integral.

Edit:
In the case of scalar QED the computation are longer but not really different. You can rewrite the numerator of $\Pi^{\mu \nu} p_{\nu}$ as
$$ \left[ (p-k)^2 - m^2 \right] (2k^{\mu} + p^{\mu}) - (k^2 - m^2)(2k^{\mu} - p^{\mu}) $$
simplifying with the denominator and substituting $(k-p)^2 = q^2$ in one of the two integrals ends the game.
