Frame-independent probability amplitude for Compton scattering in scalar QED In scalar QED, the matrix element for compton scattering is the sum of three diagrams: the s-channel, t-channel and the seagull diagram. After simplifcation, making use of the orthogonality between photon polarization vectors and photon momenta, we get the following expression for the matrix element.
$$ \begin{align}
i\mathcal{M} &= -ie^{2} \left[ \frac{(2p+k)_\mu (2p'+k')_\nu}{(p+k)^2-m^2}+\frac{(2p'-k)_\mu (2p-k')_\nu}{(p-k')^2-m^2} -2\eta_{\mu\nu} \right]\epsilon^\mu \epsilon'^{\nu *} \\
&=-i2e^{2} \left[ \frac{p_\mu p'_\nu}{p.k}-\frac{p'_\mu p_\nu}{p.k'} -\eta_{\mu\nu} \right]\epsilon^\mu \epsilon'^{\nu *}
\end{align}
$$
where $p$ and $p'$ are the four momenta of the incoming and outgoing (scalar) electrons, while $k$ and $k'$ are the incoming and outgoing photons four momenta, which correspond with the polarization vectors $\epsilon $ and $ \epsilon'$. $\eta_{\mu \nu}$ is the metric tensor.
I want to find an expression for the probability amplitude, corresponding with this matrix element for unpolarized photons and which is frame independent. So I want to calculate 
$$ \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4}\left[ \frac{p_\mu p'_\nu}{p.k} - \frac{p'_\mu p_\nu}{p.k'} -\eta_{\mu \nu} \right]\left[ \frac{p_\alpha p'_\beta}{p.k}-\frac{p'_\alpha p_\beta}{p.k'} -\eta_{\alpha\beta} \right]\sum_{\epsilon \epsilon'}\epsilon^\mu \epsilon^{\alpha *} \epsilon'^{\nu *} \epsilon'^{\beta}
$$
The polarization vector sum
$
\sum_{\epsilon \epsilon}\epsilon^\mu \epsilon^{\alpha *}$ is equal to $-\eta^{\mu \alpha}$ (up to a term that does not contribute to the amplitude due to the Ward identity). Working out this expression and using
$$ \eta_{\mu \nu}\eta^{\mu \nu}=4
$$ and eliminating $p'$ using $$p'=p+k-k'$$ I get the following result ($m$ is the electron mass). 
$$ \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ m^4 \left(\frac{1}{p.k}-\frac{1}{p.k'} \right)^2 +2m^2\left(\frac{1}{p.k}-\frac{1}{p.k'} \right)+4 \right]
$$
However, the last term of this expression, i.e. the 4, does not seem right to me. If I  apply this formula  in the LAB frame, I get 
$$  \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ 3 +  \cos^2 \theta \right] $$
with $ \theta $ the deviation angle of the photon. However, there is another way to calculate this probability amplitude, i.e. by calculating  $\vert \mathcal M \vert ^2 $ in the LAB frame for each polarization state separately and taking the sum afterwards. This leads to
$$  \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ 1 +  \cos^2 \theta \right] $$
which is more credible, as it is proportional to the (classical) result for Thomson scattering. So, it seems that the factor between brackets should be $1+ \cos^2 \theta$ and that the term 4 in the frame independent expression should be 2. 
Can somebody help me out and tell me where I made a mistake?
Details of the calculation below.

 A: In the first expression that you write for $i\mathcal{M}$, you have simplified to the second line using a gauge condition (Lorentz gauge)  $\epsilon_\mu k^\mu=0$. This is a legal choice but it is incompatible with replacing both polarizations sums with $-\eta_{\mu\nu}$, since you have partially fixed the gauge in this way. It's simple to check that squaring the expression in the first line for $i\mathcal{M}$ and replacing in there the polarization sums with the metric tensors yields the correct answer. Alternatively, you get the correct answer squaring the second line after having used $\epsilon_\mu k^\mu=0$ but then including consistently the general expression for (at least one of the) two polarizations sums. You can find a discussion on this point, as well as the explicit calculation of the squared amplitude in the Lorentz gauge, in some hand written notes here, from page 11 to the last page. 
A: Just to fill in a few of the blanks of TwoBs answer:
Using
\begin{align*}
\sum_{s} \epsilon^s(k)^{\mu}\epsilon^s(k)^{\nu *} = -\eta^{\mu\nu} + \frac{1}{2E^2}(k^{\mu}\bar{k}^{\nu}+k^{\nu}\bar{k}^{\mu}),
\end{align*}
from TwoBs notes, we would like to calculate 
\begin{align*}
\sum_{s,s'} |\mathcal{M}|^2 = 4e^4 \left( \frac{p_{\mu}p'_{\nu}}{p\cdot k}-\frac{p'_{\mu}p_{\nu}}{p\cdot k'}-\eta_{\mu\nu}\right) \left( \frac{p_{\alpha}p'_{\beta}}{p\cdot k}-\frac{p'_{\alpha}p_{\beta}}{p\cdot k'}-\eta_{\alpha\beta}\right) \epsilon^{s}(k)^{\mu}\epsilon^{s}(k)^{\alpha *} \epsilon^{s'}(k')^{\beta } \epsilon^{s'}(k')^{\nu *}.
\end{align*}
We can see that the term in brackets in the spin sum involving $k$ always vanish. You can just do it once at a time: under  $\epsilon^{s}(k)^{\mu} \rightarrow k^\mu$ the expression is invariant, and similarly under $\epsilon^{s}(k)^{\nu *} \rightarrow k^\nu$. Both of these are thanks to $\epsilon(k)\cdot k=0$. So now we know that we can replace one of the spin sums with just the metric, as the momentum terms vanish. If we do the next spin sum, we don't have the luxury of dotting into the polarization vectors anymore - they're all gone. You can show that under $\epsilon(k') \rightarrow k'$, you are left with a term that is now $k' \cdot \bar{k}'$, which is not 0.
Doing one of these, say $\epsilon(k)^{\mu} \rightarrow k^{\mu}$, we have 
\begin{align*}
(\frac{p\cdot k}{p\cdot k}p'_{\nu} - \frac{p'\cdot k}{p\cdot k'}p_{\nu}-k_{\nu})\epsilon(k')^{\nu*}... &= -k'\cdot \epsilon(k')^* = 0.
\end{align*}
A: Another reasoning leading to the conclusion that one cannot just simplify the expression for $\mathcal{M}$ (first equation) using the Lorentz gauge condition $\epsilon . k=0$ and subsequently replace in the simplified expression (second equation) $\sum \epsilon_\mu \epsilon_\nu^*$ by $ -\eta_{\mu \nu}$ is the following.
The matrix element of the first equation can be written as follows
$$\mathcal{M}= \mathcal{M}^{\mu \nu} \epsilon_\mu \epsilon_{'\nu}^*=\left( \mathcal{M}^{\mu \nu }_{(1)} + \mathcal{M}^{\mu \nu}_{(2)}\right) \epsilon_\mu \epsilon^{'*}_\nu $$ where $\mathcal{M}^{\mu \nu}_{(2)}$ represents the terms for which in the Lorentz gauge $\mathcal{M}^{\mu \nu}_{(2)}\epsilon_\mu \epsilon^{'*}_\nu=0$, i.e. the terms that we dropped in simplifying the first equation. The Ward identity states that 
$$\mathcal{M}^{\mu \nu} k_\mu =0$$ and that $$\mathcal{M}^{\mu \nu}k^{'}_\nu=0$$
If we calculate the probability amplitude, we get
$$\frac{1}{2}\sum \limits_{\epsilon \epsilon'} \vert \mathcal{M}\vert^2= \left( \mathcal{M}^{\mu \nu }_{(1)} + \mathcal{M}^{\mu \nu}_{(2)}\right)\left( \mathcal{M}^{\alpha \beta }_{(1)} + \mathcal{M}^{\alpha \beta}_{(2)}\right) \sum \limits_{\epsilon \epsilon'}\epsilon_\mu \epsilon_\alpha^* \epsilon^{'*}_\nu\epsilon^{'}_\beta $$
We know that the polarization sum can be written as follows
$$\sum \limits_{\epsilon}\epsilon_\mu \epsilon_\alpha^*=\left( -\eta_{\mu \alpha}+W_{\mu \alpha}(k) \right) 
$$where $W_{\mu \alpha}(k)$ is a term that is linear in $k_\mu$ and $k_\alpha$.
Applying now the previous expression we get
$$\frac{1}{2}\sum \limits_{\epsilon \epsilon'} \vert \mathcal{M}\vert^2= \left( \mathcal{M}^{\mu \nu }_{(1)} + \mathcal{M}^{\mu \nu}_{(2)}\right)\left( \mathcal{M}^{\alpha \beta }_{(1)} + \mathcal{M}^{\alpha \beta}_{(2)}\right) \left( -\eta_{\mu \alpha}+W_{\mu \alpha}(k) \right)\left( -\eta_{\nu \beta}+W_{\nu \beta}(k') \right)  $$ The Ward indentity allows us to rewrite this as follows.$$\frac{1}{2}\sum \limits_{\epsilon \epsilon'} \vert \mathcal{M}\vert^2= \left( \mathcal{M}^{\mu \nu }_{(1)} + \mathcal{M}^{\mu \nu}_{(2)}\right)\left( \mathcal{M}^{\alpha \beta }_{(1)} + \mathcal{M}^{\alpha \beta}_{(2)}\right) \left( -\eta_{\mu \alpha} \right)\left( -\eta_{\nu \beta} \right)  $$
However, if we first simplify the expression for the matrix element using the Lorentz gauge condition we get
$$\frac{1}{2}\sum \limits_{\epsilon \epsilon'} \vert \mathcal{M}\vert^2=  \mathcal{M}^{\mu \nu }_{(1)}  \mathcal{M}^{\alpha \beta }_{(1)} \left( -\eta_{\mu \alpha }+W_{\mu \alpha}(k) \right)\left( -\eta_{\nu \beta} +W_{\nu \beta}(k')\right)  $$ which is an expression that cannot further be simplified using the Ward identity.
Conclusion of all this is that when first using the Lorentz gauge condition to simplify the expression for the matrix element, we cannot anymore do the replacement $$\sum \limits_{\epsilon}\epsilon_\mu \epsilon_\alpha^* \rightarrow -\eta_{\mu \alpha} $$
Correction: as explained in the answers from TwoBs and gmarocco, you can do the replacement for one of the two spin sums, but not for the second.
So, when doing the calculation of the probability amplitude using the very first expression for the matrix element in the initial question and applying the aforementioned replacement, I get after quite some algebra
$$ \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ m^4 \left(\frac{1}{p.k}-\frac{1}{p.k'} \right)^2 +2m^2\left(\frac{1}{p.k}-\frac{1}{p.k'} \right)+2 \right]
$$ which is the expected result.
