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 $\mathcal{M}$.

### Related Questions

#### Sponsored Content

#### 1 Answered Questions

### Feynman rules from Lagrangian

**2018-07-09 09:56:33****MeMeansMe****943**View**3**Score**1**Answer- Tags: quantum-field-theory lagrangian-formalism symmetry feynman-diagrams propagator

#### 1 Answered Questions

### [SOLVED] Interacting lagrangian with multiple terms

**2019-11-21 16:29:47****redhood****70**View**0**Score**1**Answer- Tags: quantum-field-theory lagrangian-formalism feynman-diagrams interactions

#### 2 Answered Questions

### [SOLVED] Why my 4-divergence term added to a Lagrangian modifies the equation of motion?

**2017-03-31 20:29:26****StarBucK****1327**View**5**Score**2**Answer- Tags: lagrangian-formalism field-theory variational-principle boundary-conditions classical-field-theory

#### 1 Answered Questions

### [SOLVED] Feynman rule for propagator with derivatives

**2014-11-20 20:46:03****Your Majesty****850**View**1**Score**1**Answer- Tags: quantum-field-theory feynman-diagrams

#### 2 Answered Questions

### [SOLVED] What´s the importance of the normalization of the Kinetic term?

**2012-08-29 17:44:28****Forever_a_Newcomer****2977**View**16**Score**2**Answer- Tags: quantum-field-theory

## 2 comments

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

This has to do with how we set up perturbation theory, treating part of the Lagrangian as "free" and hence treated exactly, and the rest as the "perturbation". 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.

## @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 partso 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.