I have read in many places that one-point functions, like the one below:

$$\langle \Omega|\phi(x) |\Omega \rangle$$

are equal to zero ($|\Omega \rangle$ is the vacuum of some interacting theory, $\phi$ is the field operator - scalar, spinorial, etc...)

Peskin's book, for instance, says (page 212) this is USUALLY zero by symmetry in the case of a scalar field ("usually" probably means $\lambda \phi^4$ theory) and by Lorentz invariance for higher spins . How can I see that?

In a more general question: can someone point out a counterexample? A case when this functions are not zero?


1 Answer 1


First of all, if $\phi(x)$ were not a scalar field but a field with spin (1/2 or 1), then a nonzero expectation value would break the Lorentz symmetry because the expectation value would be a preferred spinor or vector in spacetime (similarly for other more complicated representations of the Lorentz group).

So only vevs of scalar fields may be compatible with a relativistic theory.

For scalar fields $\phi(x)$, it is a matter of field redefinition what value is called "zero" and what value is called something else. For example, in the electroweak theory, the Higgs field is a doublet $h=(h_1,h_2)$ and its vacuum expectation value is normally $(v,0)$ i.e. nonzero. However, one may rewrite the first component as $h_1=v+H$ and the the expectation value of the new field $H$ is zero again.

Classically, the vacuum is a stationary configuration so the scalar fields must be stationary points of the potential, $$ \frac{\partial}{\partial \phi_i} V = 0 |_{\phi_i =\text{vacuum values}}.$$ So when the vev of $\phi_i$ is zero in the vacuum, it means that there are no linaer terms of the type $\phi_i$ in the (Taylor-expanded) potential $V$. That's often the case; if there were such linear terms, the point $\phi=0$ would be unstable and we would probably reparameterize $\phi\to \phi'(\phi)$ in such a way that $\phi'=0$ would correspond to a stationary point.

As I hinted above, the Higgs field is a natural field for which the most natural parameterization implies $\langle h(x)\rangle \neq 0$. However, string theory and supersymmetric quantum field theories are full of additional counterexamples, the so-called moduli. The potential for the moduli is, by definition, zero (or universal constant) so any point is a stationary point. The physical properties of the vacuum and particles upon it depend on the moduli. For example, the coupling constant is often written as $$ g = \exp(\phi)$$ where $\phi$ is the so-called dilaton. It would be counterproductive to shift the value of $\phi$ in such cases because $\phi$ stores some important physical information. So moduli are other examples whose one-point function is nonzero in the most natural field redefinition.

Once again, it's always possible to redefine the fields so all of them have vanishing one-point functions in the vacuum.


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