Scalar derivative couplings: Effects on S-matrix and Feynman Rules

Question

In Schwartz's field theory book ch. 7.4.2 he claims that interaction Lagrangians like

$${\cal L}_{\rm int} = \lambda \phi_1(\partial_{\mu}\phi_2)(\partial_{\mu}\phi_3)\tag{7.101}$$

lead to the Feynman rules that incoming momenta yield a factor $-ip_{\mu}$ and an outgoing one produces $+ip_{\mu}$. According to Schwartz, internal lines still yield a Feynman propagator.

However, I have two issues with this claim:

1. The $S$-matrix is defined by $$S=\exp\left(-i\int H_{\rm int} d^4x\right),$$ i.e. the Hamiltonian $H_{\rm int}$ appears, and not the Lagrangian. In the case of derivative couplings, there is no longer $H=-L$, but actually $$H= \sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q_i}}- L = 2 \partial_t \phi_2 \partial_t \phi_3 \phi_1 - L.$$ Thus, the expansion of the S-matrix has an extra factor, which Schwartz seems to omit: $$S=\exp\left(-i\int H d^4x\right) = \exp\left(-i\int (2 \partial_t \phi_2 \partial_t \phi_3 \phi_1 - L) d^4x\right).$$

2. The commutation relation between creation and annihilation operator $a$ and $a^{\dagger}$ will change since the canonical momentum $\pi$ will also change: $$\pi_i = \frac{\partial L}{\partial \dot{\phi_i}} = \dot{\phi_i}-\lambda \dot{\phi_i} \phi_i.$$ Therefore, since $[\pi(x), \phi(y)]=i \delta(x-y)$ must hold due to causality, it holds that $[a(p),a^{\dagger}(p')] \neq \delta(p-p')$. And this leads to a different propagator when performing the contractions in the expansion of the S-matrix, since contracting two scalar fields $\phi(x)$ and $\phi(y)$ is according to Wick's theorem proportional to $[a(p),a^{\dagger}(p')]$, which is no longer just a delta function, but contains additional terms.

Answer 1

That's a good question. The topic of derivative couplings is full of subtleties. First and foremost it is important to distinguish between the usual time ordering $T$ and the covariant time ordering $T_{\rm cov}$, cf. e.g. this Phys.SE post.

OP is correct that the time evolution operator $\hat{U}_I$ in the interaction picture is given by the interaction Hamiltonian $H_{\rm int}$. However, it can formally be rewritten into the interaction Lagrangian $L_{\rm int}$ if we also change$^1$ the time-ordering $T\longrightarrow T_{\rm cov} $:

$$\begin{align} \hat{U}_I~=~& T \exp\left\{ -\frac{i}{\hbar}\int\!dt~H_{\rm int}(\hat{q},\hat{p})\right\}\cr ~=~& T \exp\left\{ \frac{i}{\hbar}\int\!dt\left(\frac{1}{2}\hat{p}^2 -H(\hat{q},\hat{p})\right)\right\}\cr ~=~&\int \! {\cal D}v~ T \exp\left\{ \frac{i}{\hbar}\int\!dt\left(\frac{1}{2}\hat{p}^2 -\hat{p}_iv^i+ L(\hat{q},v)\right)\right\}\cr ~\stackrel{\hat{p}=G\dot{\hat{q}}}{=}&\int \! {\cal D}v~ T \exp\left\{ \frac{i}{\hbar}\int\!dt\left(\frac{1}{2}(\dot{\hat{q}}-v)^2+ L_{\rm int}(\hat{q},v)\right)\right\} \cr ~=~&\int \! {\cal D}v~{\cal D}p~ T \exp\left\{ \frac{i}{\hbar}\int\!dt\left(p_i(\dot{\hat{q}}^i-v^i)-\frac{1}{2}p^2+ L_{\rm int}(\hat{q},v)\right)\right\}| \cr ~\stackrel{(10)}{=}~&\int \! {\cal D}v~{\cal D}p~ T_{\rm cov} \exp\left\{ \frac{i}{\hbar}\int\!dt\left(p_i(\dot{\hat{q}}^i-v^i)+ L_{\rm int}(\hat{q},v)\right)\right\}\cr ~=~& T_{\rm cov} \exp\left\{ \frac{i}{\hbar}\int\!dt~ L_{\rm int}(\hat{q},\dot{\hat{q}})\right\} ,\end{align}$$ where we used eq. (10) and notation from my Phys.SE answer here.

Ref. 1 is in subsection 7.4.2 implicitly referring to the last formula for the time evolution operator $\hat{U}_I$ in terms of the interaction Lagrangian $L_{\rm int}$ and the covariant time ordering $T_{\rm cov}$.

References:

1. M.D. Schwartz, QFT & the standard model, 2014; sections 7.2-7.4.

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$^1$ Ref. 1 fails to mention that the time-ordering in the Lagrangian interaction picture, eqs. (7.63) & (7.64), is the covariant time ordering $T_{\rm cov}$.