导数相互作用及单圈图顶点因子推导的相关疑问

Let's walk through your derivation step by step and clarify the key points about derivative interactions in Srednicki's QFT conventions.

First, Recap Srednicki's Core Rules for Vertex Factors #

In Srednicki's text, the path integral is defined as $\exp\left(i\int \mathcal{L}d^4x\right)$, where $\mathcal{L} = \mathcal{L}0 + \mathcal{L}{\text{int}}$ (with $\mathcal{L}_0$ the free Lagrangian). For vertex factors:

• The base factor for any interaction term is $-i$ multiplied by the coefficient of the term in $\mathcal{L}_{\text{int}}$.
• Each derivative $\partial_\mu$ acting on a scalar field $\phi$ contributes a momentum factor: $-ik_\mu$ for an incoming particle (corresponding to the annihilation operator term $e^{-ikx}$) and $ik_\mu$ for an outgoing particle (corresponding to the creation operator term $e^{ikx}$).
• Derivative contractions (like $\partial_\mu\phi\partial^\mu\phi$) translate to contracting the Lorentz indices of the corresponding momentum factors in the vertex.

Analyzing Your 3-Point Vertex ($V_a$) #

Your interaction term is $\mathcal{L}{\text{int}}^{(3)} = a\phi\partial\mu\phi\partial^\mu\phi$, a 3-point vertex with one plain $\phi$ and two derivative-coupled $\phi$s. Here are the key issues with your current derivation:

1. Momentum Labeling: You use the same $k^\mu$ for both derivative fields, but each external line has its own distinct momentum. Let’s call the incoming momenta of the two derivative fields $k_1^\mu$ and $k_2^\mu$, and the incoming momentum of the plain $\phi$ $p^\mu$. Momentum conservation requires $p + k_1 + k_2 = 0$.
2. Factor Calculation: Applying the rules:
   • The base coefficient factor is $-i \times a$.
   • Each incoming derivative field contributes $-ik_1^\mu$ and $-ik_2^\mu$ respectively.
   • Contract the $\mu$ indices (from $\partial_\mu\partial^\mu$).

Putting it all together, the correct 3-point vertex factor (for all incoming particles) is:
$$V_a = -i a \left(-ik_1^{\mu\right)\left(-ik_2}\mu\right) = i a k_1^\mu k_2^\mu$$

If one of the derivative fields is outgoing (say momentum $k_2^\mu$ leaving the vertex), its factor becomes $ik_2^\mu$, so the vertex adjusts to:
$$V_a = -i a \left(-ik_1^{\mu\right)\left(ik_2}\mu\right) = -i a k_1^\mu k_2^\mu$$

Analyzing Your 4-Point Vertex ($V_b$) #

For the 4-point term $\mathcal{L}_{\text{int}}^{(4)} = b\phi^{2\partial_\mu\phi\partial}\mu\phi$ (two plain $\phi$s, two derivative $\phi$s), the same logic applies:

• You can’t use a single $k^\mu$ for both derivative fields—they need distinct momenta $k_1^\mu$ and $k_2^\mu$, with the two plain $\phi$s having momenta $p_1^\mu$ and $p_2^\mu$ (momentum conservation: $p_1 + p_2 + k_1 + k_2 = 0$).
• The correct vertex factor (for all incoming particles) is:
  $$V_b = -i b \left(-ik_1^{\mu\right)\left(-ik_2}\mu\right) = i b k_1^\mu k_2^\mu$$

Key Takeaways for Derivative Interactions #

• Always assign unique momenta to each external line—never reuse the same momentum label unless explicitly dealing with identical particles in a symmetric configuration.
• The sign of the derivative momentum factor depends strictly on whether the particle is incoming ($-ik_\mu$) or outgoing ($ik_\mu$).
• Lorentz index contractions in the Lagrangian must be preserved in the vertex factor (e.g., $\partial_\mu\partial^\mu$ becomes $k_1^\mu k_2^\mu$).

内容的提问来源于stack exchange，提问作者Melvin