I have a doubt about the problem 3.5 in Schwartz's "QFT and the Standard Model". The problem states:

Spontaneous symmetry breaking is an important subject, to be discussed in depth in Chapter 28. A simple classical example that demonstrates spontaneous symmetry breaking is described by the Lagrangian for a scalar with a negative mass term: $$ \mathcal{L}=-\frac{1}{2}\phi\Box\phi + \frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4 $$ How many constants $c$ can you find for which $\phi(x) = c$ is a solution to the equations of motion? Which solution has the lowest energy (the ground state)?

My doubt is whether the Lagrangian is correct. This looks like the $\lambda\phi^4$ Lagrangian, which indeed exhibits a symmetry break for a negative mass term. However, in $\lambda\phi^4$ theories, we do not have a d'Alambertian, but $(\partial_\mu\phi)^2$, and I have also never seen Lagrangians depending on second order derivatives (I seem to remember that they are problematic in classical physics, but in QFT maybe they are valid Lagrangians?). This last form of the Lagrangian is also the one used by other books to introduce symmetry break (e.g. Modern Particle Physics, Thomson). Also Schwartz himself states

If kinetic terms have too many derivatives, for example $\frac{1}{2}\phi\Box\phi$, there will generally be disastrous consequences.

Therefore, I am wondering if there is a typo in Schwartz, or if the Lagrangian he uses is indeed a known Lagrangian in QFT.

  • 1
    $\begingroup$ There is no mistake. Did you bother to integrate the kinetic term by parts? The author is just helping you to contrast this to the conventional K-G equation. $\endgroup$ Feb 19, 2022 at 22:06
  • $\begingroup$ Do integration by parts and remember in physics, boundary terms are always zero :) $\endgroup$
    – paul230_x
    Feb 19, 2022 at 22:12
  • $\begingroup$ Are you sure he really gave $\frac{1}{2} \phi \square \phi$ as an example of a kinetic term with two many derivatives? It is generically true that equations of motion higher than second order in time have instabilities, but the equations of motion for $\phi \square \phi$ are second order in time. $\endgroup$
    – Andrew
    Feb 20, 2022 at 2:44
  • 2
    $\begingroup$ Your second quote is a misquote. The example Schwartz uses is $\phi \Box^2 \phi$ (on page 32). $\endgroup$
    – d_b
    Feb 20, 2022 at 2:56
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/389421/2451 and links therein. $\endgroup$
    – Qmechanic
    Feb 20, 2022 at 3:04

1 Answer 1


Note that \begin{equation} \phi \square \phi = - (\partial \phi)^2 + \partial_\mu (\phi \partial^\mu \phi) \end{equation} The second term is a total derivative. In the action, when we integrate this term, it becomes an integral over the boundary of the spacetime region we're interested in. Typically we take this region to have infinite spatial extent, and we assume the fields die off asymptotically at a fast enough rate that such boundary terms vanish.

Under this (extremely common) assumption, we can consider two Lagrangians equivalent if they differ by a total derivative. Therefore, the Lagrangian in Schwartz is equivalent to the normal $(\partial \phi)^2$ Lagrangian you are used to.


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