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A Constant Solutions of Real Scalar Field

  1. Jul 12, 2017 #1

    hilbert2

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    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?
     
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  3. Jul 12, 2017 #2

    TeethWhitener

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    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.
     
  4. Jul 12, 2017 #3

    hilbert2

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    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.
     
  5. Jul 28, 2017 #4

    A. Neumaier

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    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.
     
  6. Jul 28, 2017 #5

    Avodyne

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    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.
     
  7. Jul 29, 2017 #6

    A. Neumaier

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    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.
     
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