# What is the role of the vacuum expectation value in symmetry breaking and the generation of mass?

Consider a theory of one complex scalar field with the following Lagrangian. $$\mathcal{L}=\partial _\mu \phi ^*\partial ^\mu \phi +\mu ^2\phi ^*\phi -\frac{\lambda}{2}(\phi ^*\phi )^2.$$ The potential is $$V(\phi )=-\mu ^2|\phi |^2+\frac{\lambda}{2}|\phi|^4.$$ The classic stable minimum of this potential is given by $\phi =\frac{\mu}{\sqrt{\lambda}}e^{i\theta}=:v$ for any $\theta \in \mathbb{R}$.

We then define a new field $\psi: =\phi -|v|$, rewrite the Lagrangian in terms of the new field $\psi$, et voilà: out pops the mass term $$2|v|^2\lambda \psi ^*\psi .$$ People usually explain that this shows the existence of a field whose quanta are particles of mass $|v|\sqrt{2\lambda}$. In this sense, the original field has $\phi$ acquired a mass.

I don't buy it. I don't see where this argument has actually used the fact that $v$ is the vacuum expectation value of the field. Yes, it is natural to expand about the classical minima, but why not do something stupid instead and define a new field $\psi :=\phi -7$. Once again, after rewriting the Lagrangian in terms of the new field, you should find that it has indeed acquired has a mass term. Playing this trick over and over, by picking a number different than $7$, you should be able to find a mass term with any mass you like. Obviously, this doesn't make any physical sense.

There is something special about the substitution $\psi :=\phi -|v|$. There must be more to it than a tedious algebraic manipulation, but what is it? Why does nature possess particles of the mass that this substitution yields, as opposed to any other substitution?

On another note, what exactly does this have to do with the global $U(1)$ symmetry present. It seems that the only thing that has played a role so far is that the vacuum expectation value of the field is non-zero, but yet I've always seen this mass generation presented alongside symmetry breaking. What precisely is the relationship between the two?

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As is easily checked, fields linear in creation and annihilation operators (and hence amenable to a particle interpretation) have zero vacuum expectation value. Thus the $\phi$ field with its nonvanishing vacuum expectation value cannot be given a particle interpretation. But the field $\psi=\phi-v$ has such an interpretation as its vacuum expectation value is zero. This works only if $v$ is the vacuum expectation of $\phi$.

Note that the field $\phi$ is and remains massless; it is the field $\psi$ that had acquired a mass term.

The 1-loop approximation to a quantum field theory is given by the saddle-point approximation of the functional integral. For that you have to expand around a stationary point, and for stability reasons this stationary point has to be a local minimizer. If the local minimum is not global, the vacuum state is metastable only; so one usually expands around the global minimizer.

A mass term breaks the scaling symmetry of a previously scale-invariant theory. It may or may not break other symmetries. In the above case, the symmetry $\phi\to-\phi~~~$ of the action is broken in the stable vacuum.

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Firstly, note that 'particles' are quantised small oscillations of a field.

Now, to the potential. You would find a 'mass' term expanding around any point, but you'll also generically find a linear term. If you tried to perturbatively quantise the 'small oscillations' around such a point, you would find non-zero Feynman diagrams where particles are created from nothing. This represents an instability: the theory is telling you that you are doing the wrong thing.

Edit: I forgot to address the symmetry-breaking aspect. If you consider just the scalar field model, then the broken global symmetry means that you will have an exactly massless particle: the Goldstone boson. This corresponds to the angular oscillations (you only discussed the radial mode). In a gauge theory, a gauge boson is massless if and only if the vacuum preserves the corresponding symmetry.

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 I don't quite see how the requirement that you not have a linear term requires us to make the substitution $\psi :=\phi -|v|$. For example, the trivial substitution $\psi := \phi$ doesn't have a linear term, but this can't be correct, because we don't see a massless particle and antiparticle pair, we see a Goldstone boson and a massive real scalar particle. – Jonathan Gleason Nov 8 '12 at 15:25 If you expand around the origin, $\phi = 0$, you find a negative mass-squared term. This also indicates an instability: the classical solutions at small wavenumber are now either decaying or growing exponentially. – Rhys Nov 8 '12 at 16:44