Constant Solutions of Real Scalar Field

[hilbert2]

Suppose I have a self interacting real scalar field $\phi$ with equation of motion

$\partial^i \partial_i \phi + m^2 \phi = -A \phi^2 - B\phi^3$,

and I attempt to find constant solutions $\phi (x,t) = C$ for it. The trivial solution is the zero solution $\phi (x,t) = 0$, but there can also be two more constant solutions depending on the values of $A$, $B$ and $m$. Obviously $B$ needs to be positive for the system's energy to be bounded from below, but $A$ seems to be arbitrary. Higher order interaction terms would probably make this non-renormalizable.

Do the two nonzero constant solutions have any physical significance? Does this equation allow situations where the static solutions are unstable, becoming something else very quickly if perturbed even a little?

 

[TeethWhitener]

Does it matter that the operators $\phi(\mathbf{x},t)=C$ don't go to zero as $\mathbf{x} \rightarrow \infty$? I know that's a requirement for wavefunctions in QM, but I'm not sure what the constraints on the operators in QFT are.

 

[hilbert2]

I'm interpreting the field as a classical field here, similar to an air pressure field or an electromagnetic vector potential. I don't think this kind of fields have to be normalizable.

 

[A. Neumaier]

1. The quadratic term on the right just changes the mass, and of course $m^2+A$ must be nonnegative for a physical theory.

2. Without an additional $\phi^4$ term, the action is classically unbounded, hence one wouldn't expect a meaningful quantum solution.

3. With the additional $\phi^4$ term, the zero solution may be unstable. Then the physical vacuum state is (in the classical limit) in a global minimizer of the action, corresponding to a nonzero constant solution. This is just a slightly generaized version of the familiar process of mass generation through broken symmetry.

 

[Avodyne]

A. Neumaier, you are misreading the equation. The terms on the right come from cubic and quartic terms in the potential.

The complete potential is $V(\phi)=\frac12 m^2 \phi^2 + \frac13 A \phi^3 + \frac14 B\phi^4$. This potential has a local minimum at $\phi=0$, but for positive $B$ and large enough $A$, another minimum at a lower energy, with a maximum in between. For parameters in this range, the minimum at $\phi=0$ is a "false vacuum" in QFT.

 

[A. Neumaier]

A. Neumaier, you are misreading the equation. The terms on the right come from cubic and quartic terms in the potential.

Oh, sorry, yes, of course. I was thinking of the right hand side in terms of the potential - was too tired yesterday, after a day of travel...

This potential has a local minimum at ϕ=0

In general only a stationary point. If $A$ is large enough then this stationary point is an unstable local maximizer.