A Constant Solutions of Real Scalar Field

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1. Jul 12, 2017

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?

2. Jul 12, 2017

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.

3. Jul 12, 2017

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.

4. Jul 28, 2017

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.

5. Jul 28, 2017

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.

6. Jul 29, 2017

A. Neumaier

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....

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