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.


1 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 $T_{\rm cov}$.


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


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


Your Answer

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

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