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?


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: $$ [\partial_{x}, \partial_{y}]arctg\left[\frac{y}{x}\right] = \partial_{x}\left[ \frac{x}{x^{2} + y^{2} + a^{2}}\right]_{\lim a \to 0} + \partial_{y}\left[ \frac{y}{x^{2} + y^{2} + a^{2}}\right]_{\lim a \to 0} = $$ $$ =\left[\frac{2a^{2}}{(x^{2} + y^{2} + a^{2})^{2}}\right]_{\lim a \to 0} = 2 \pi \left[\frac{a^{2}}{\pi}\frac{1}{(r^{2} + a^{2})} \right]_{\lim a =0} = 2 \pi \delta_{a}(\mathbf r) $$

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    $\begingroup$ Sounds like what you are talking about is just vortex lines in a superfluid. $\endgroup$
    – Meng Cheng
    Jun 29 '15 at 15:21
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    $\begingroup$ 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 $\theta(x,y,z)\propto\arctan \frac{y}{x}$. $\endgroup$
    – Meng Cheng
    Jun 29 '15 at 15:48
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    $\begingroup$ 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. $\endgroup$
    – Meng Cheng
    Jun 29 '15 at 16:46
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    $\begingroup$ 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. $\endgroup$
    – Meng Cheng
    Jun 29 '15 at 18:22
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    $\begingroup$ @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. $\endgroup$
    – Meng Cheng
    Jun 29 '15 at 20:22

The topological stability of global $U(1)$ vortices or strings has been discussed in several introductory books on the topic (see, for instance, Advanced Topics in Quantum Field Theory by M.Shifman).

In cylindrical coordinates $(\rho,\theta, z)$ of physical space $\mathbb R^3,$ a static, straight vortex or string of core size $\delta$, aligned along the $z$-axis, is given by the map

$$ (\rho,\theta, z) \in \mathbb R^3\to \mathfrak M \ni \varphi(\rho,\theta) \simeq \begin{cases} 0 &\text{ for } \rho=0\,,\\ v e^{in\theta} &\text{ for } \rho \gg\delta \,, \end{cases} $$

where $\mathfrak M$ denotes the vacuum manifold and $n\in\mathbb Z$ is called the winding number. Let us emphasize that $\theta \in S^1$ is only a spatial coordinate, not the vortex itself. The vortex is a continuous assignment of vacua in $\mathfrak M$ to every point in physical space. In our case, $\mathfrak M$ is itself an abstract topological circle $S^1$ (because the vacua are defined by $|\varphi|=v$). Note that if we traverse a circular contour in physical space around the vortex going from $\theta = 0$ to $\theta = 2\pi$, then the phase of the vortex $\varphi$ winds in the vacuum manifold $\mathfrak M$ from $0$ to $2\pi n$. This is why we say that (at $\rho \gg \delta$) the map from physical space $S^1\to \mathfrak M$ has winding number $n$.

The claim is that $n$ characterises different homotopy classes of non-contractible loops in $\mathfrak M$. In other words, $n$ is a homotopy-invariant, that is there exists no homotopy which can continuously deform $\varphi$ to another $\varphi'$ with a different winding number. For example, for the two vector fields shown below (the left is a string with $n=2$ and the right is a string with $n=3$), the claim is that there is no homotopy that morphs one into the other.

winding number = 2winding number = 3

This claim is readily established by the observation that $\pi_1(\mathfrak M) = \mathbb Z$. Of course, it is obvious that if you loop a rope say twice around the rim of a wheel and tie the ends, it is impossible to unwrap the rope with local shenanigans into a loop that winds only once, unless you do something discontinuous such as unwrap it outside the wheel. It is a consequence of the global topological structure of the vacuum manifold.

The above argument is one of purely topological origin and has nothing to do with any model-specific tunnelling amplitudes being suppressed by an infinite potential barrier in this case. In fact, a quantum field theory does not allow tunnelling between degenerate vacua at all, regardless of the specifics of the model. The Hilbert space of each degenerate vacuum is completely separate from the Hilbert spaces of other degenerate vacua. This is the very reason why symmetries can spontaneously break in quantum field theory (but not in quantum mechanics). You see, granted that a global $U(1)$ symmetry completely breaks outside a vortex of a certain core size and all points in spacetime pick a certain vacuum during this phase transition, the question is not whether this vacuum configuration can dynamically tunnel into another vacuum configuration with a different winding number (because it simply cannot, due to the general fact that degenerate vacua cannot tunnel at all in quantum field theory). The question is whether the winding numbers are stable under arbitrary local perturbations of the Hamiltonian of our theory (the perturbations being respectful of the original symmetry). If they are stable (and they are, because of topological protection), then it either takes an infinite amount of energy to change the winding numbers or you have to do something nasty (discontinuous) with your order parameter, namely the vortex.


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