A symmetry is spontaneously broken by an operator $\hat{O}$ which acquires a non-zero vacuum expectation value. This is expressed as

$$ \langle 0 | \hat{O} | 0 \rangle = v$$

The parameter $v$ is said to have dimensions of mass. Can someone demonstrate why this is the case? It makes intuitive sense, considering the general aspects of spontaneous symmetry breaking. However, I have not yet found a general proof of the fact that $v$ is a mass parameter. Thanks in advance.


1 Answer 1


We set $c = \hbar = 1$. This means that length and time have the same dimension, energy and mass have the same dimension, and action is dimensionless (since $\hbar$ has units of action). Further the dimension of length is the inverse of that of mass.

Now consider the Lagrangian density for a free scalar field $\phi$ $$\mathcal L = \frac{1}{2}(\partial_\mu \phi) (\partial^\mu \phi) - m^2\phi^2.$$ Since $\partial_\mu$ has the dimension of inverse length which is the same as the dimension of mass, this is dimensionally correct. Since action is dimensionless, and $$S = \int d^4 x \, \mathcal L$$ and $d^4 x$ has units of $L^4 = [m]^{-4}$, $\mathcal L$ must have the dimensions of $m^4$. This means that the field $\phi$ must have the dimension of mass. Then naturally, its expectation values also have this dimension.

In QCD there is also symmetry breaking from vacuum expectation values of the form $$v_q = \langle \overline{q} \gamma^5 q \rangle$$ where $q$ is a quark field, that is, a fermionic spinor field. The Lagrangian density for a free Dirac spinor field $\psi$ is $$\mathcal L = \overline{\psi}(i\gamma^\mu \partial_\mu - m)\psi.$$ By the same dimensional analysis argument you see that $[\psi] = [m]^{3/2}$, and so the VEV $v_q$ has dimensions of mass cubed.

  • $\begingroup$ This is very clear, much appreciated! $\endgroup$
    – user46837
    May 19, 2014 at 13:04

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