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Consider a real scalar field $\Phi(x)$ in one spatial dimension which asymptotically goes to its vacuum values $\Phi_{+}$ or $\Phi_{-}$. Given the requirement of finiteness of energy, we deduce that no continuous transformation can change a field configuration to one with different asymptotic values. Thus the field values at $\pm\infty$ classify all possible field configurations into (four) homotopy sectors.

Now consider the configuration space. I am not so sure about this part, but to visualize this space, what I do is imagine an infinite-dimensional vector space with an orthogonal (or orthonormal, if necessary) basis (like Euclidean space, but infinite-dimensional), where every basis vector is labeled $\Phi_x$ and $\Phi_x=\Phi(x)$, with $x$ varying continuously from -$\infty$ to +$\infty$ .

Now, every point in this space corresponds to a field configuration and every path in this space is made up of series of such points.

My questions are:

  1. What is wrong with the image I have of the configuration space of a scalar field?

If my image is acceptable:

  1. What do we mean by a path in this space, in more physical terms?

  2. If I can draw a path from one vacuum to the other in this space, how is that possible, given the homotopy argument mentioned above? Does this imply that configuration space is somehow larger than merely the space of all configurations?

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  • $\begingroup$ 1. How is a field with non-zero asymptotical values supposed to obey the very finiteness of energy you invoke after that? Note that the functions that are not asymptotically zero but have finite energy usually simply don't have an asymptotic value at all. 2. How would classifying functions by their asymptotic values imply "four homotopy sectors"? What are these four sectors, and what do they have to do with homotopy? $\endgroup$
    – ACuriousMind ♦
    Aug 19, 2017 at 11:30
  • $\begingroup$ @ACuriousMind: Do sections 2.3 and 2.4 of Rajaraman elsevier.com/books/solitons-and-instantons/rajaraman/… not answer both your questions? Or perhaps I've misunderstood you. $\endgroup$
    – Optimus Prime
    Aug 19, 2017 at 11:56
  • $\begingroup$ I do not have access to that particular book, and understanding your question should not require the reader to read another resource to be able to understand it. Please at least write down the energy functional you are considering here when referring to "finiteness of energy" and what the four homotopy sectors are. $\endgroup$
    – ACuriousMind ♦
    Aug 19, 2017 at 12:04
  • $\begingroup$ For instance, if you are talking about the model discussed also here, that should be evident from your question. $\endgroup$
    – ACuriousMind ♦
    Aug 19, 2017 at 12:21
  • $\begingroup$ @ACuriousMind: Oh my apologies. Consider the kink solution of the $\lambda\phi^4$ theory. The field tends to nonzero values at spatial infinity while the energy is finite (Q1) and the four topological sectors would be the kink, antikink, and the two sectors with trivial topology, i.e, where the field approaches the same value in either direction asymptotically (Q2). Though I have a feeling you are already aware of this and I've made some mistake along the way which is still unknown to me. Perhaps mentioning a particular model in the question helps? $\endgroup$
    – Optimus Prime
    Aug 19, 2017 at 12:32

1 Answer 1

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Your image of the space of finite-energy solutions is indeed wrong. It's not a vector space. Take, for example, a model with an energy functional $$ \int \left(\frac{1}{2}(\partial_t\phi)^2 + \frac{1}{2}(\partial_x \phi)^2 + \lambda\phi^4 - \frac{1}{4}\phi^2 + \frac{1}{64\lambda}\right)\mathrm{d}x\mathrm{d}t.$$ In this model, there are only two possible asymptotic values for a stationary finite-energy solution $\phi(x,t)$: $\lim_{x\to\pm\infty}\phi(x,t)$ has to be either $\frac{1}{\sqrt{8\lambda}}$ or $-\frac{1}{\sqrt{8\lambda}}$, otherwise the energy cannot be finite. This solution space is not a vector space: $\phi_+(x,t) = \frac{1}{\sqrt{8\lambda}}$ is a finite-energy solution, but $2\phi_+$ is not.

Rather, we observe indeed that the space of finite-energy solutions has four homotopy components: There are solutions which take the positive value at both spatial infinities, those that take the negative value, and those that take the positive at one infinity and the negative at the other. There are no continuous deformations which would deform one such class of solutions into the other while keeping the function as a finite-energy solution all along the way. Of course you could deform one solution into any other if you didn't care whether it remained a solution along the way, but that's not what we're talking about here.

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  • $\begingroup$ Obviously, the space of finite-energy solutions would fail to be a vector space, based on the sound argument you made. But wouldn't such a space be a subset of the vector space that was described in the question? In other words, would the addition of the impossible configurations(due to finiteness of energy) to the finite-energy space constitute a vector space? Also, isn't the last paragraph in your answer merely a restatement of the first paragraph of the original question and what was discussed in its comments? $\endgroup$
    – Optimus Prime
    Aug 19, 2017 at 13:56
  • $\begingroup$ @OptimusPrime Yes, the space of finite-energy solutions is a subset of many function vector spaces. and yes, I'm not saying much more than you did in the second paragraph except applying it to this specific model (there are certainly models with more or less than four homotopy sectors in general!). $\endgroup$
    – ACuriousMind ♦
    Aug 19, 2017 at 14:13

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