3
$\begingroup$

The Lagrangian for a complex scalar is

$$\mathcal{L}=-\left(\partial_\mu\phi\right)^*\partial^\mu\phi-V(\phi,\phi^*)$$

where $V$ is the potential, $*$ stands for complex conjugate and I use the metric $\eta_{\mu\nu}={\rm diag}(-1,1,1,1)$.

The most general potential leading to a renormalizable theory is:

$$V(\phi,\phi^*)=\lambda_0+\lambda_{11}\phi+\lambda_{12}\phi^*+\lambda_{21}\phi^2+\lambda_{22}\phi\phi^*+\lambda_{23}\phi^{*2}+\lambda_{31}\phi^3+\lambda_{32}\phi^2\phi^*+\lambda_{33}\phi\phi^{*2}+\lambda_{34}\phi^{*3}+\lambda_{41}\phi^4+\lambda_{42}\phi^3\phi^*+\lambda_{43}\phi^2\phi^{*2}+\lambda_{44}\phi\phi^{*3}+\lambda_{45}\phi^{*4}$$

where the $\lambda$'s are constants.

Why are we dropping always almost all terms and use

$$V(\phi,\phi^*)=\lambda_{22}|\phi|^2+\lambda_{43}|\phi|^4~~?$$

I get that the Lagrangian has to be bounded from below to find meaningful solutions. This probably eliminates the solutions with odd powers of $\phi$. So one remains with

$$V(\phi,\phi^*)=\lambda_0+\lambda_{21}\phi^2+\lambda_{22}\phi\phi^*+\lambda_{23}\phi^{*2}+\lambda_{41}\phi^4+\lambda_{42}\phi^3\phi^*+\lambda_{43}\phi^2\phi^{*2}+\lambda_{44}\phi\phi^{*3}+\lambda_{45}\phi^{*4}$$

I also understand that the Lagrangian has to be real, but this does not eliminate all the terms. The general potential is a real valued function if I impose certain relations between the $\lambda$'s. For example I can set $\lambda_{21}=\lambda_{23}^*$ and then $\lambda_{21}\phi^2+\lambda_{23}\phi^{*2}$ is real...

$\endgroup$

1 Answer 1

2
$\begingroup$

Because of "charge".

First of all, you obtain the Lagrangian $$V\left(\phi, \phi^{*}\right)=\lambda_{22}|\phi|^{2}+\lambda_{43}|\phi|^{4},$$ by requiring that your original Lagrangian is invariant under multiplication by a global phase $$ \phi \to \phi e^{i\alpha}.\tag{1} $$

Why do we require this? This is a bit handwavy, but a possible justification goes along the lines: In general, you want the Lagrangian to be invariant under this transformation as in classical quantization these fields will get promoted to operators in the Hilbert space that create particle states. Since multiplying those states by a global phase does not change their properties, then applying the same transformation on the fields should not change the Lagrangian.

A slightly better argument can be made if one considers the coupling of the scalar field to the electromagnetic field. In that case, the minimal gauging procedure requires the fields to transform as $$ \psi \to \psi e^{iq{_\psi}\alpha(x)}, $$ where $q_\psi$ is the charge of the field $\psi$. From this, you can see that the charge of $\phi$ is 1 and the charge of $\phi^*$ is -1. If you look at the Lagrangian from this perspective, you will see that it won't preserve the electric charge. However, the minimal gauge procedure explicitly takes into consideration that the fields are invariant under Eq. (1), such that this is also not a good answer.

The actual answer is that it is a choice. Invariance under Eq.(1) (global U(1) symmetry) is an extra condition that we impose. Don't get me wrong, it is a well-motivated choice that allows you to describe many interesting stuff (e.g. defining the notion of electric charge) but still a choice.

$\endgroup$
3
  • 1
    $\begingroup$ Ultimately, observable Nature tells us which extra-mathematical choices to make: choose what describes reality. $\endgroup$
    – J.G.
    Feb 23, 2022 at 21:48
  • $\begingroup$ Is this U(1) symmetry the one that the Higgs field spontaneously breaks? $\endgroup$ Feb 24, 2022 at 8:01
  • $\begingroup$ In the strict sense no, since the "usual" Higgs mechanism refers to the breaking of $SU(2)_L \times U(1) _Y\to U(1)$. But again, in the spirit of the answer, there is nothing to stop you from considering other groups and other symmetry breaking patterns, except their usefulness in describing reality. The Higgs mechanism is associated with the symmetry breaking pattern $SU(2)_L \times U(1)_Y \to U(1)$ simply because we want three massive ($W^\pm$, Z) and a massless ($\gamma$) gauge fields. $\endgroup$
    – JGBM
    Feb 24, 2022 at 10:09

Your Answer

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