Let's have simple scalar $\Phi$ action involves spontaneously symmetry breaking in a form $$ \tag 1 S = \int d^{4}x\left( |\partial_{\mu}\psi|^{2} + \psi^{2}|\partial_{\mu}\theta |^{2} - \frac{\lambda}{4} (\psi^{2} - v^{2})^{2}\right) $$ It is usual higgs-like action where complex scalar field is represented in a form $\Phi = \psi e^{i\theta}$. E.o.m. generated by $(1)$ implies string-like solutions $$ \tag 2 \Phi = vf(r)e^{in\varphi}, \quad f(0) = 0, \quad f(\infty ) = 1 $$ ($r, \varphi$ denote polar coordinates). From substituting $(2)$ in $(1)$ and calculating of stress-energy tensor follows that strings have nonzero tension, thus they have to radiate.

One says (see, for example, this article, p.5, section "Dual representation...") that by using duality relation $$ \tag 3 \psi^{2}\partial_{\mu}\theta = \frac{1}{2}f_{a}\epsilon_{\mu \nu \alpha \beta}\partial^{\nu}B^{\lambda \rho} $$ one provides analytical description of radiation of $\theta $ bosons by string-like solutions $(2)$.

The question: I completely don't understand how action generated by $(3)$ (it is called Kalb-Ramond action) describes radiation of string-like solution $(3)$. Can you explain this?


This is a standard particle-vortex duality transformation. The idea is there exists a "dual" description of a superfluid, in which the bosons (to be precise, the Goldstone mode) is described by a 2-form(tensor) gauge field, and the vortices are now string-like "matter" charged under the gauge field. The effective action for phase fluctuations (i.e. Goldstone mode) is just

$ L=\frac{1}{2}\rho^2 (\partial_\mu\theta)^2$

Here $\rho$ is the superfluid density, roughly your $f$ at infinity. $\theta$ can be divided into two parts: first the smooth fluctuations, and the singular ones (vortices). We will write $\theta=\eta + \theta_v$. $\eta$ satisfies $[\partial_x,\partial_y]\eta=0$.

The next thing is to rewrite the Lagrangian using Hubbard-Stratonovich transformation, by introducing an auxiliary vector field $\zeta_\mu$:

$L=\partial_\mu \theta \zeta^\mu -\frac{1}{\rho^2} \zeta^2=\partial_\mu \eta\zeta^\mu+\partial_\mu \theta_v\zeta^\mu -\frac{1}{\rho^2} \zeta^2$.

Since $\eta$ is a smooth function, the functional integration over $\eta$ can be easily performed, yielding a constraint $\partial_\mu \zeta^\mu=0$ which can be resolved by writing $\zeta^\mu=\varepsilon^{\mu\nu\lambda\rho}\partial_\nu B_{\lambda\rho}$. Notice that $B$ is a 2-form gauge field, since this expression has exactly the gauge redundancy. Then the Lagrangian can be casted into a gauge theory of 2-form tensor gauge field coupled to string-like sources (i.e. the $\theta_v$).

  • $\begingroup$ Thank you! If I understand correctly, you've introduced $\zeta$ field by formal transformation $$ \tag 1 \int D \rho D \theta e^{i\int d^{4}x\left[... + \frac{1}{2}\rho^{2}(\partial_{\mu}\theta )^{2}\right]} \sim \int D \theta D \rho D \zeta e^{i \int d^{4}x\left[... + \partial_{\mu}\theta \zeta^{\mu} - \frac{1}{\rho^{2}}\zeta^{2}\right]}? $$ If yes, how to integrate over $\eta$ here? Action $(1)$ depends on $\eta$ only linear, so corresponding path integral isn't gaussian. $\endgroup$ – Name YYY Jul 1 '15 at 12:52
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    $\begingroup$ Integrating over some linear action is even easier than Gaussian: it's just $\int dx e^{ikx}=\delta(k)$. $\endgroup$ – Meng Cheng Jul 1 '15 at 14:52

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