By user168146

2017-12-14 14:27:49 8 Comments

I know that derivative couplings in a Lagrangian interaction, such as

$$\mathcal{L}_{int} = \lambda \phi (\partial_{\mu}\phi)(\partial^{\mu}\phi)$$

bring down two momentum factors into the matrix element $\mathcal{M}$.

Why doesn't this extend to a typical kinetic term

$$\mathcal{L}_{kin} = (\partial_{\mu}\phi)(\partial^{\mu}\phi)\quad ?$$

It seems odd that derivatives in $\mathcal{L}_{int}$ contribute a factor of momentum$^2$ to $\mathcal{M}$, but the derivatives in $\mathcal{L}_{kin}$ contribute a factor of the propagator $$\frac{1}{\mbox{momentum}^2}$$

to $\mathcal{M}$.


@coconut 2017-12-14 14:57:53

Informally speaking (I don't know if this can be stated more rigorously) we can treat the kinetic terms as having Feynman rule $\sim p^2$, in some sense.

Consider the massive case: $\mathcal{L}=|\partial\phi|^2 - m^2\phi^2$. We could say that we have a Feynman rule giving $-p^2$ for vertices with two fields. Then we'd have $1/m^2$ as the leading order propagator and the sum of all the contributions would be:

$$ \frac{1}{m^2} - \frac{1}{m^2} p^2 \frac{1}{m^2} + \frac{1}{m^2} p^2 \frac{1}{m^2} p^2 \frac{1}{m^2} + \cdots = \frac{1}{p^2 + m^2}. $$

What is unavoidable to do perturbation theory is to separate a quadratic part so you can compute

$$ \left<\mathcal{O}\right> = \int D\phi \, \mathcal{O}(\phi) \,e^{\int\phi D \phi + iS_{\text{int}}(\phi)} $$

by calculating the inverse of some differential operator $D$, which gives propagator, and expanding $e^{S_{\text{int}}}$ in powers of $\phi$. So in the end we get whatever is inside $D$ (including $p^2$) in the denominator and the things that appear in $S_{int}$ in the numerator.

@knzhou 2017-12-14 14:56:14

This has to do with how we split the Lagrangian into 'free' and 'interacting' pieces, which get treated differently. As a very simple example consider the Gaussian integral $$I = \int_{-\infty}^\infty dx \, e^{-a^2 x^2 - \epsilon^2 x^2}.$$ The exact value of this integral is $$I = \sqrt{\frac{\pi}{a^2 + \epsilon^2}}$$ where both $a$ and $\epsilon$ end up in the denominator. On the other hand, we can think of $e^{-a^2 x^2}$ as the 'free' part and $e^{-\epsilon^2 x^2}$ as the 'perturbation'. Then to lowest order in $\epsilon$, $$I \approx \int_{-\infty}^\infty dx \, e^{-a^2 x^2} (1 - \epsilon^2 x^2) = \frac{\sqrt{\pi}}{a} - \frac{\sqrt{\pi}}{2a^3} \epsilon^2.$$ So while $\epsilon$ was originally in the denominator, by treating it as a perturbation it ends up in the numerator; this is clear by just Taylor expanding our exact expression for $I$. By summing up all the contributions, $\epsilon$ ends up back in the denominator as expected.

Essentially the same thing is happening when you derive the Feynman rules with the path integral formulation. This is most visible in renormalized perturbation theory where the very same kinetic term appears in both the 'free' and 'perturbation' parts.

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