0
$\begingroup$

I am currently reading a physics paper in which the authors have complexified an ordinary differential equation (ODE). They mention the following statement in the paper:

"These branch points reflect the fact that two linearly independent solutions of the wave equation will mix when they are transported around a non-trivial loop in the complex radial plane. "

I believe this is a straightforward complex analysis problem, but I have not been able to find an appropriate reference to help me fully grasp the concept. If anyone could provide a suitable reference to better understand this problem, I would greatly appreciate it.

I could refine my question a bit. Suppose we have two independent solutions given by $(x-1)^{i\alpha}$ and $(x-1)^{-i\alpha}$. When I transport them around $x=1$, I don't observe any mixing.

However, if I choose these independent solutions as $a_1 (x-1)^{i \alpha}+a_2 (x-1)^{-i \alpha}$ and $b_1 (x-1)^{i \alpha}+b_2 (x-1)^{-i \alpha}$, the first basis transforms into $a_1 e^{2\pi \alpha} (x-1)^{i\alpha}+a_2 e^{-2\pi \alpha}(x-1)^{-i\alpha}$ when I transport it around $x=1$. In this case, I can argue that the two bases mix together.

Is this a valid interpretation of the problem, and if so, can you provide any references to help me understand the underlying concepts?

$\endgroup$
3

1 Answer 1

0
$\begingroup$

It just means that when you have a complex-valued function $f(z)$ with a branch point (and therefore a branch cut) somewhere, the function is single-valued only as long as you do not have a contour that traverses a loop around it. As soon as you complete a full loop of $f(z)$ around a branch point, the function becomes discontinuous (or multi-valued). That's why we define branch cuts which provide artificial barriers that shouldn't be crossed. In the context you are referring to, a complex-valued solution to a differential equation will no longer remain the same if you traverse a loop around a branch point in the complex plane.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.