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  Axion strings and spontaneously broken symmetry

+ 2 like - 0 dislike
2171 views

I have two question about axion strings:

  1. Why their appearance is connected with spontaneously broken symmetry? How to demonstrate that?
  2. Why they are stable topological configurations (look to the "Addition" text below)?
  3. Why when we choose string located along $z-$axis and set solution for Peccei-Quinn scalar field $\varphi $ in a string-like form $\varphi = ve^{i\theta}$, where $v$ is VEV of $\varphi$, $\theta$ is axion, then we have $$ [\partial_{x}, \partial_{y}]\theta = 2\pi \delta (x) \delta (y)? $$ How to demonstrate that?

Addition

Let's assume axion "bare" lagrangian $$ \tag 1 L = \frac{1}{2}|\partial_{\mu}\varphi |^{2} - \frac{\lambda}{4} (|\varphi |^{2} - v^{2})^{2} $$ One of solution of corresponding e.o.m. is axion string - stable topological configuration. If string is located along z-axis and if it is static, then corresponding solution is simply ($\rho$ is polar radius, $\varphi$ corresponds to polar angle and, in fact, to axion) $$ \varphi (x) = f(\rho ) e^{i n \varphi}, \quad f(0) = 0, \quad f(\infty ) = v, $$ where $n$ is winding number.

Statement that configurations with different winding numbers are stable means that they are separated by infinite potential barriers. But I don't understand how $(1)$ creates barriers for different $n$.

Addition 2 Thank to the Meng Cheng comment. The first and the third questions are closed. Explicit proof of the statement of the third question: $$ [\partial_{x}, \partial_{y}]e^{iarctg\left[\frac{y}{x}\right]} = i\partial_{x}\left[ \frac{x}{x^{2} + y^{2} + a^{2}}\right]_{\lim a \to 0} + i\partial_{y}\left[ \frac{y}{x^{2} + y^{2} + a^{2}}\right]_{\lim a \to 0} = $$ $$ =i\left[\frac{2a^{2}}{(x^{2} + y^{2} + a^{2})^{2}}\right]_{\lim a \to 0} = 2 \pi i \left[\frac{a^{2}}{\pi}\frac{1}{(r^{2} + a^{2})} \right]_{\lim a =0} = 2 \pi i \delta_{a}(\mathbf r) $$

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
asked Jun 29, 2015 in Theoretical Physics by NAME_XXX (1,060 points) [ no revision ]
What's an "axion string"? Googling gives me only "axions in string theory" stuff.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user ACuriousMind
@ACuriousMind : string is topologically stable solution of equation of motion for case of one space dimension in a case of spontaneously broken continuous symmetry. As for axion string, some outlook is here: arxiv.org/pdf/hep-ph/9807374v2.pdf . However, I don't understand explicitly how spontaneously broken symmetry causes existence of string as topologically stable solution.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
Sounds like what you are talking about is just vortex lines in a superfluid.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Meng Cheng
@MengCheng : maybe you're right, and this case is very similar to superconductors. But I still have problem with understanding its topological nature.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
@ACuriousMind : section II in an article.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
The topological nature is the winding number of the order parameter phase $\theta$. Since the order parameter (your scalar $\varphi$ ) is $ve^{i\theta}$ and the fluctuations of $v$ is ignored, the order parameter lives in $\mathrm{U}(1)\simeq S^1$. The topological nature of vortices is classified by $\pi_1(S^1)=\mathbb{Z}$. For your last question, assume the string is aligned along $z$ axis, then $\theta(x,y,z)\propto\arctan \frac{y}{x}$.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Meng Cheng
@MengCheng : thank you! Also, can you help me to understand why the solutions with different winding numbers are topologically stable (i.e. are separated by barriers)?

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
The point I think is that we keep $v$ fixed so that the order parameter manifold is $S^1$. This is actually important, because for vortices the expression for $\theta$ I wrote in the previous comment breaks down near the origin, which in reality means that $v$ actually has to go to zero there. But otherwise, especially far away from the origin where the homotopy argument works, $v$ stays a constant. The only way to change the winding number is to let $v$ go to zero at some other places, which means there are multiple vortices.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Meng Cheng
@MengCheng : excuse me, I don't understand why the only way to change winding number is to set $v$ to zero.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
Here is a more formal way to put it: whenever $\theta$ is well-defined, you can define the winding number as $\oint \nabla\theta\cdot d\mathbf{l}$ along a closed path. You can prove that this is not going to change under smooth deformation of $\theta$. When $\theta$ is singular -- which can only happen when $v=0$, the value of this topological invariant can change.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Meng Cheng
@MengCheng : thank you! And the last question, if you please: we add boundary conditions for $\varphi = f(r)e^{i\theta}$ such that $f(0) = 0, f(\infty ) = v$ to avoid bad behaviour of $\varphi$ at $r = 0$ and to satisfy the minimum condition of potential?

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
@NameYYY That is exactly what happens for vortices: the amplitude of the order parameter is suppressed at the core (so there is an energy associated), and far away from the core the amplitude is just $v$. I think in my previous comments I was abusing $v$ for $f$, so all I mean by $v=0$ is that the amplitude has to vanish. Sorry for the confusion.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Meng Cheng

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