content_image/319884.jpg content_image/319883.jpg content_image/319882.jpg content_image/319881.jpg content_image/319880.jpg
The equations of motion for φ and χ are identical, so we write the generic field as Φ, and its equation of motion is 

<img src='content_image/319884.jpg'>

We assume that the mesons will be non-interacting and plane-wave, so we make the ans¨ atz Φ = f ( ρ ) e $^{ik.x}$, with k 2 = − M 2 ( M being the mass of the meson). We then look for normalisable solutions to (3.13): these correspond to consistent field theory solutions. 

(3.13) can be solved exactly by: 

$_{2}$F$_{1}$ ( a, b ; c ; z ) is the regular hypergeometric function, defined as 

<img src='content_image/319883.jpg'>

<img src='content_image/319882.jpg'>

where ( a )$_{n}$ is the rising factorial. α is as yet undetermined. We can de- termine it by considering our requirements on f ( ρ ): the solution must be normalizable and real. To stop ( ρ $^{2}$+ L $^{2}$) − α $_{2}$F$_{1}$ ( − α, − α +1; 2; − ρ $^{2}$/L$^{2}$) run- ning off to ±∞ as ρ → ∞ we need to terminate the series at order ρ − 2 $^{α}$. A little maths shows that this can be done by setting 

<img src='content_image/319881.jpg'>

in which case the hypergeometric function terminates at order ( ρ $^{2}$/L$^{2}$) $^{n}$. Hence φ ∼ ρ − 2 as ρ → ∞ . So the final solution to (3.13) is 

<img src='content_image/319880.jpg'>

