In looking at the Lagrangian of a (free for simplicity) complex scalar field $\phi$, we have a kinetic term that goes like:


Given instead, a kinetic term of the form:


Is this not equivalent to the first term? Just by simply varying each term with respect to $\phi^{*}$ and it's derivative we arrive at the same equation of motion (up to some arbitary multiplacative factor) namely:


I'm sure this can be expressed in terms of formal adjoints more precisely, but if they are equivalent, why do we always write scalar fields in terms of first order lagrangians (boundary conditions maybe)?


1 Answer 1


Almost, you're off by a minus sign, but otherwise yes. You're supposed to use (higher-dimensional) integration by parts on the action $$ S = - \int_{\Omega} d^4 x\ \big( \partial^\mu \phi^\ast(x) \big) \big( \partial_{\mu} \phi(x) \big) = - \int_{\partial \Omega} d^3 \Sigma\ \phi^\ast(x) n^{\mu} \partial_{\mu} \phi(x) \big) + \int_{\Omega} d^4 x\ \phi^\ast(x) \partial^\mu \partial_{\mu} \phi(x) $$ where $\Omega \to \mathbb{R}^{4}$ and $n^{\mu}$ is the outward unit normal to the boundary $\partial \Omega$ (and $d \Sigma$ is the volume form on the boundary). The standard thing to do is assume that the boundary term vanishes and you find that $\partial^\mu \phi^\ast \partial_\mu \phi$ and $-\phi^\ast\partial^\mu \partial_{\mu} \phi$ can be used interchangebly.

  • 2
    $\begingroup$ I mostly agree, but here are a few comments. 1. It is not a "standard thing" to assume that the boundary term vanishes, in fact, this boundary term does not vanish if $\Omega$ is a finite region of spacetime. The vanishing occurs for a very specific calculation, for example, when we are calculating a QFT propagator in Euclidean signature and the field is assumed to decay at infinity. 2. OP asked why the first form is preferred. Probably the best reason to prefer it is that it admits a description in terms if Hamiltonian mechanics which is a prelude to canonical quantization. $\endgroup$ Oct 5, 2020 at 1:28
  • $\begingroup$ I definitely agree and should have clarified more about that - I meant that the boundary term vanishes when $\Omega \to \mathbb{R}^4$. There are certainly situations where the boundary term can be important. And I think you must be right about (2), with the preferred form of kinetic term the Hamiltonian contains the square of the conjugate momentum field $\pi := \partial_t \phi$ explicitly $\endgroup$ Oct 5, 2020 at 1:35
  • $\begingroup$ @QuantumEyedea suppose we now generalize this to the gauge covariant (or just covariant) derivatvie: $$\Phi^{\dagger}\nabla^{\mu}\nabla_{\mu}\Phi$$ $$=\Phi^{\dagger}\left(\partial^{\mu}-A^{\mu,i}T_{i}\right)\left(\partial_{\mu}+A_{\mu}^{i}T_{i}\right)\Phi$$ $$=\Phi^{\dagger}\partial^{\mu}\partial_{\mu}\Phi-\Phi^{\dagger}A^{\mu,i}T_{i}\partial_{\mu}\Phi+\Phi^{\dagger}\partial^{\mu}\left(A_{\mu}^{i}T_{i}\Phi\right)-\Phi^{\dagger}A^{\mu,i}T_{i}A_{\mu}^{i}T_{i}\Phi$$ we still have a term equivalent up to boundary, but now we have "extra" terms correct? a mass term for the connection field? $\endgroup$
    – R. Rankin
    Oct 13, 2020 at 17:22
  • $\begingroup$ @R.Rankin I'm pretty sure you can generalize this notation to covariant derivatives by considering Stoke's Theorem on manifolds. Off the top of my head I'm not sure of the details of that though $\endgroup$ Oct 13, 2020 at 17:30

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