Consider the theory consisting of a complex scalar field $\phi$ and a real scalar field $\psi$: \begin{align} \mathcal{L}&=\partial _{\mu } \phi \partial ^{\mu }\phi ^*+\tfrac{1}{2} \partial _{\mu } \psi \partial ^{\mu }\psi -U[\phi,\psi] \end{align} with $U(1)$-symmetric scalar potential \begin{align} U[\phi,\psi]&=m_{\phi}^2|\phi|^2+\tfrac{1}{2}m_{\psi}^2\psi ^2+b |\phi| ^4+c |\phi| ^2 \psi +d |\phi| ^2\psi ^2+e \psi ^3+a \psi ^4. \end{align} Here, we can recognize $m_\phi$ and $m_\psi$ as being the masses of the respective fields, and $a,b,c,d,e$ are positive real constants.

Suppose we want to derive the equations of motion for this theory, for which we can use the Euler-Lagrange equations: \begin{align} \partial_\mu \partial^\mu \phi +\frac{\partial U}{\partial \phi^*} =0\,, && \partial_\mu \partial^\mu \psi +\frac{\partial U}{\partial \psi} =0\,. \label{eq:eoms} \end{align}

Now assume $m_\psi\gg m_\phi$. Apparently, under this assumption, one can neglect the kinetic term in the second Euler-Lagrange equation (so it just becomes $\partial U/\partial \psi=0$).

Why is this statement true? I am having trouble seeing why a large mass might allow you to neglect the dynamical kinetic term. Heuristically, I can somewhat accept that a massive field might be less dynamical, but I can't see how to argue this mathematically.

EDIT: see Sec. VI of this paper for a reference


1 Answer 1


The standard way to understand this is to do some power counting analysis. In this power counting, derivatives scale like the energy of the system, and their relevance depends on the regime at which we are doing our approximations. If we are considering the $\psi$ field to have a giantic mass, this means that we are considering to analyze the system at energies $E \ll m_\psi$. In this regime, $\partial_\mu \sim E \ll m_\psi$ and can be neglected compared to $U$, which contains terms scaling like $m_\psi^2$.

You can even push this analysis further and solve the equations of motion order by order in $m_\psi$. This is one way of integrating out the heavy fields to obtain an effective field theory describing the system at low energies (relative to the mass of $\psi$).


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.