I will use the example of the Abelian Higgs model to explain my problem. Consider the Lagrangian:

$ \mathcal{L} = - \frac{1}{4} F^{\mu \nu}F_{\mu \nu} + \left(D^\mu \phi\right)^\dagger \left( D_\mu \phi\right) - \mu^2 \phi^\dagger \phi - \lambda \left( \phi^\dagger \phi \right)^2 $

where $\phi(x)$ is a complex bosonic scalar field and $F^{\mu \nu} =\partial^\mu A^\nu - \partial^\nu A^\mu$ where $A^\mu (x)$ is a real bosonic vector field. $D^\mu$ is the usual covariant derivative.

This Lagrangian shows symmetry breaking and in my lecture notes and everywhere in the literature I find that they do the so called unitary gauge as follows:

$\phi(x) = \frac{v+\sigma(x)}{\sqrt{2}}$

where both $v$ and $\sigma(x)$ are real.

(In a non-Abelian model they would write similarly $\phi(x) = \frac{v+\sigma(x)}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \end{pmatrix}$, but as far as I can see this has the same problem as the example above that I am asking about).

I think I understand why all that is. My question is why we choose the $\frac{1}{\sqrt{2}}$ to be there. As far as I can see it is just convention, some people seem to called it a normalisation. I cannot see what the merit of this convention is and neither in what sense it is a normalisation. In fact there is one issue that makes the convention seem particularly inconvenient to me. When I expand out the Lagrangian in the $\sigma(x)$ field choosing v such that the potential is minimized as usual I get the following kinetic term in the Lagrangian:

$\mathcal{L} \supset \frac{1}{2} \left(\partial^\mu \sigma\right)^\dagger \left( \partial_\mu \sigma\right) - \lambda v^2 \sigma^2(x)$

where the factor $\frac{1}{2}$ in front essentially comes straight from the $\frac{1}{\sqrt{2}}$ mentioned above. So the factor seems to have changed the usual form of our kinetic term, e.g. unless I am missing something one would have to write down the Feynman propagator now as $ \frac{1}{k^2/2 - m^2} $ instead of $ \frac{1}{k^2 - m^2} $ as we are used to.

Long story short: Why on earth do we have that $\frac{1}{\sqrt{2}}$ in there?

  • 1
    $\begingroup$ $\sigma$ is a real field so the factor of $1/2$ in front of the kinetic term gives you the propagator normalized in the usual way (ie, the residue of the pole of the propagator is 1) $\endgroup$ – Andrew Mar 27 '16 at 22:25
  • $\begingroup$ @Andrew I see, so that's what I missed. Can I ask how come that the 1/2 isn't required for complex scalar field kinetic terms then? $\endgroup$ – Wolpertinger Mar 27 '16 at 22:37
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    $\begingroup$ Yes there's different ways of looking at it. Basically it comes down to, a complex field is really 2 fields (you can vary the action wrt $\Phi$ and $\Phi^\dagger$ independently). So varying $\partial_\mu \Phi^\dagger \partial^\mu \Phi$ wrt $\Phi^\dagger$ gives you $\square \Phi$, whereas varying $(\partial_\mu \sigma)^2$ with respect to $\sigma$ gives you $2 \square \sigma$. A more physical way to look at it is in terms of identical particles--for a real scalar a particle is its own antiparticle, so you need the 1/2 to avoid double counting. $\endgroup$ – Andrew Mar 27 '16 at 22:39
  • $\begingroup$ @Andrew thank you, that answers the question! (also I feel slightly stupid now xD) $\endgroup$ – Wolpertinger Mar 27 '16 at 22:42

@Andrew answered the question in a comment. Here a summary: $\sigma$ is a real field while the initial field was complex. Therefore to obtain the canonical propagator a factor of 1/2 is in fact necessary in the kintetic term.


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