Is it possible to factorize $$(-\partial^2+\phi^2(r))^2-\left(\frac{\partial\phi(r)}{\partial r}\right)^2,$$ where $\phi(r)$ is a function of $r\equiv\sqrt{x^2+y^2+z^2+\xi^2}$ and $\partial^2$ is the 4-dimensional Laplacian? Actually since we have the $SO(4)$ symmetry, we can separate the angular part and focus on the radial part. That is, we can take $$\partial^2\rightarrow \frac{1}{r^3}\frac{\partial}{\partial r}r^3\frac{\partial }{\partial r}\equiv \hat{L}(r)$$. Note that we can not simply factorize it as $$\left(-\hat{L}(r)+\phi^2(r)+\left(\frac{\partial\phi(r)}{\partial r}\right)\right)\cdot\left(-\hat{L}(r)+\phi^2(r)-\left(\frac{\partial\phi(r)}{\partial r}\right)\right)$$ because of the non-commutativity of derivative operators. Here I just wonder whether there is a subject that show us how to factorize operator polynomials systematically?

If it can not be factorized, can anyone give a proof?

  • $\begingroup$ What does the term $-\partial^2$ stand for? Is it second order derivative operator? With respect to what do we take derivative then? $\endgroup$ – Vlad Mar 10 '16 at 11:47
  • $\begingroup$ @Vlad $\partial\equiv\partial_{x^i}$, where $x^i=x, y, z, \xi$. $\endgroup$ – Wein Eld Mar 10 '16 at 16:11
  • $\begingroup$ Please go more into the definition of terms in that equation for those of us who might know how to factorize in general but not know enough about differential operators. For example, is $\partial_{x^i} = $ the partial derivative operator, and why can you add the two terms above, like what space do they belong to? Thank you! $\endgroup$ – BananaCats Category Theory App Mar 11 '16 at 23:25
  • $\begingroup$ Is there any more information about $\phi$ other than you can take a partial with respect to $r$? $\endgroup$ – BananaCats Category Theory App Mar 11 '16 at 23:35
  • $\begingroup$ What do you mean by $\phi^2(r)$, power or composition? $\endgroup$ – BananaCats Category Theory App Mar 12 '16 at 0:01

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