How to derive the Lagrangian of "Heavy Scalar Effective Theory"? For a fermion field (spinor QED), taking $m\rightarrow\infty$ limit, we have 
$$\mathcal{L}=\bar Q(iD\!\!\!/-m)Q\rightarrow \bar Q_v(iv\cdot D)Q_v+\mathcal{O}(\frac{1}{m})$$
It's commonly known as heavy quark effective theory. 
But for scalar QED
$$\mathcal{L}=(D_{\mu}\phi)^{\dagger}D^{\mu}\phi-m^2\phi^{\dagger}\phi$$
clearly, we can't deploy the same technique of using $v\!\!\!/$ as a projection operator as in HQET, so how should I take the limit and integrate the anti-particle out? 
In Schwartz's QFT (Chap. 35) he mentioned a choice of $\chi_v$ and $\tilde \chi_v$:
$$\phi(x)=e^{imv\cdot x}\frac{1}{\sqrt{2m}}(\chi_v(x)+\tilde\chi_v(x))$$
$$\chi_v(x)=e^{imv\cdot x}\frac{1}{\sqrt{2m}}(iv\cdot D+m)\phi(x),\;
\tilde\chi_v(x)=e^{imv\cdot x}\frac{1}{\sqrt{2m}}(-iv\cdot D+m)\phi(x)$$
but I can't find the projection operator in this scenerio. My guess is, the projection operator is the linear combination of $v\cdot D$ and m, and supposely when it multiplies with $(iv\cdot D+m)$ or $(-iv\cdot D+m)$ the result should at least also be a linear combination of $v\cdot D$ and m, but I failed to prove it.
Schwartz also gives an answer:
$$\mathcal{L}=\chi^{\dagger}iv\cdot D\chi-\tilde \chi^{\dagger}(iv\cdot D+2m)\tilde\chi+\mathcal{O}(\frac{1}{m})$$
but I don't know how to derive the $\tilde \chi$ part. 
For now, I only have this result
$$\mathcal{L}=(\chi_v(x)+\tilde\chi_v(x))^{\dagger}(iv\cdot D)(\chi_v(x)+\tilde\chi_v(x))+\mathcal{O}(\frac{1}{m})$$
but I can't get rid of the anti-particle field. Can anyone help with this?  
 A: OK, now I have the answer to my own question. 
Given the scalar QED Lagrangian:
$$\mathcal{L}_{SQED}=(D_{\mu}\phi)^{\dagger}D^{\mu}\phi-m^2\phi^{\dagger}\phi$$
first we define
$$\phi(x)=e^{imv\cdot x}\frac{1}{\sqrt{2m}}(\chi_v(x)+\tilde\chi_v(x))$$
$$\chi_v(x)=e^{imv\cdot x}\frac{1}{\sqrt{2m}}(iv\cdot D+m)\phi(x),\;
\tilde\chi_v(x)=e^{imv\cdot x}\frac{1}{\sqrt{2m}}(-iv\cdot D+m)\phi(x)$$
as a start, put the expansion of $\phi$ (the first equation) into the next line, a simple relation is derived:
$$(-iv\cdot D)\chi(x)=(2m+iv\cdot D)\tilde\chi(x)$$
Use the definition of $\chi_v(x)$ and $\tilde\chi_v(x)$, we can transform $\mathcal{L}_{SQED}$ into
$$\mathcal{L}_1=(\chi_v(x)+\tilde\chi_v(x))^{\dagger}(iv\cdot D)(\chi_v(x)+\tilde\chi_v(x))+\mathcal{O}(\frac{1}{m})$$
put the relation between $\chi_v(x)$ and $\tilde\chi_v(x)$ we derived earlier into $\mathcal{L}_1$, we can have the final form
$$\mathcal{L}=\chi^{\dagger}iv\cdot D\chi-\tilde \chi^{\dagger}(iv\cdot D+2m)\tilde\chi+\mathcal{O}(\frac{1}{m})$$
(I'm not sure about the coefficient of anti-particle mass, Schwartz gives 2 but my result is 4.)
To derive the $1/m$ correction or even higher order correction, the equation-of-motion must also be considered. 
