# How do we get to the classical Higgs mass equation?

If I got it right that's $$m_H=\sqrt{2\lambda}\nu,$$ where $\nu=246GeV$ is the non-zero VEV of the Higgs field, and $\lambda$ is the coefficient of the $\phi^4$ interaction. But how can we get to that equation?

• And if I have got it right (which I may not have, or you may have read it already, sorry) its here en.wikipedia.org/wiki/Higgs_boson – user108787 Nov 22 '16 at 15:15
• By "that equation", do you mean $m_H = \sqrt {2\lambda} \nu$ or $\nu = 246 \,\text {GeV}$? – pppqqq Nov 22 '16 at 17:56
• You need to ask How do we get from $X$ to $Y$? If we do not know your starting point, how do we know where to begin an explanation? You also need to show effort in trying to answer the question yourself. – sammy gerbil Nov 23 '16 at 0:24

Let's take the following Lagrangian, which is quite similar to that of the Higgs : \begin{equation} \mathcal{L} = \dfrac{1}{2}\partial_\mu \phi \partial^\mu \phi^{\dagger} + \dfrac{1}{2}\mu^2 \phi^\dagger \phi - \dfrac{1}{4}\lambda (\phi^\dagger \phi)^2 = \dfrac{1}{2}\partial_\mu \phi \partial^\mu \phi^{\dagger} - V(\phi) \end{equation} First, we are looking for the vacuum expectation value of the field. Clearly, this Lagrangian is symmetric under a rotation. Therefore, we can choose any direction $e^{i\theta}$ for the minimum, and we decide to take $\phi$ to be real and positive, $\phi = \phi_0 \in \mathbb{R}^{+}$. Then : \begin{equation} V(\phi_0) = -\mu^2 \phi_0^2/2+\lambda \phi_0^4/4 \end{equation} The minimum is then reached for $\phi_0 = v$. This gives $v = \sqrt{\mu^2/\lambda}$.
Now, we can expand the field $\phi$ around its vacuum expectation value : $\phi = (v+a, b)$. This means we redefine $\phi$ in terms of two fields $a$ and $b$. Injecting this into the Lagrangian, and expanding, we should get terms of various orders in $a$ and $b$. The mass term of a scalar field $S$ of mass $m$ has the form $-\frac{1}{2}m^2 S^2$. If we calculate the terms in $a^2$, we find : \begin{equation} \mathcal{L} \supset \dfrac{1}{2}\mu^2a^2 - \dfrac{1}{4}\lambda \left(v^2+2va+a^2\right)^2 = \left [\dfrac{1}{2}\lambda v^2 -\dfrac{6}{4} \lambda v^2\right]a^2 = -\lambda v^2a^2 \equiv -\dfrac{1}{2}m^2 a^2 \end{equation} And finally : $m = \sqrt{2\lambda}v$. Note that the field $b$ is massless.