When we write down the lagrangian of a general $U(1)$ scalar field theory we generally write

$$\mathcal{L} = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi^* - \frac{m^2}{2}\phi \phi^* - V(|\phi|^2)$$

I.e., we don't write terms linear in $| \phi |$. Why is this? $|\phi|$ is just as $U(1)$-invariant as $|\phi|^2$. Can any $U(1)$-invariant potential $V$ be expressed in powers only of $|\phi|^2$?

  • 1
    $\begingroup$ @AccidentalFourierTransform Do you know of a one-sentence reason for why such an assumption is made, or where I could find an explanation if it is lengthy? Thanks! $\endgroup$
    – Diffycue
    Oct 7, 2018 at 21:51
  • $\begingroup$ A possibly simplistic, but still practical, reason is that it is not obvious how to make sense perturbatively of a vertex like $c |\phi |$. The basic reason is that while a polynomial action in $\phi$ has a simple expression in terms of the Fourier transform of $\phi$, a non-polynomial term does not. $\endgroup$
    – pppqqq
    Oct 7, 2018 at 22:29
  • $\begingroup$ In addition to the comment by @AccidentalFourierTransform, in this framework, you can redefine the field according to $\phi\to \phi+c$, which allows you to eliminate the linear term. There are situations where this does not apply, like e.g. the Polonyi model in supergravity, where the shift changes the Kähler potential. The reason why we demand that the Lagrangean be a real analytic of the fields is that we do not want to deal with square roots of operators: $|\hat{\phi}|=\sqrt{\hat{\phi}^\dagger\hat{\phi}}$ where I introduced a hat to make clear that I am referring to field operators here. $\endgroup$
    – user178876
    Oct 7, 2018 at 22:54

2 Answers 2


The whole philosophy of perturbative QFT is that you assume the field's fluctuations about its energy- or action-minimizing value are small, so that you can Taylor expand the potential energy (and other quantities) about zero field. A Taylor expansion can't result in a $|\phi|$ term because it isn't complex-analytic, so such a term doesn't fit naturally into the framework of perturbative QFT. You could in principle image a non-perturbative action with such a term, but it would probably need to be studied numerically rather than analytically.


Unless you make assumptions on $V$, any expression in $|\phi|$ is of the form $V(|\phi|^2)$. If you do make assumptions on the shape of $V$, it is probably because you want it to be well-behaved. Since $|\phi|$ itself is not that well-behaved (e.g. it is not differentiable at 0, unlike $|\phi|^2$), restricting $V$ to have a special form, e.g. to be analytic, but allowing the argument to be something like $|\phi|$, somewhat defeats the purpose of requiring a well-behaved potential.


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