1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

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

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