4
$\begingroup$

The Hamiltonian for 3+1D compact $U(1)$ gauge theory on a cubic lattice is of the form \begin{equation} H = J\sum_{\text{links}\ l}E^2_{l}+ g \sum_{\text{plaquettes}\ P}\cos(\Phi_P), \end{equation} where the operator $E_l$ is the integer valued electric field, canonically conjugate to the vector potential $A_l$: $[A_l,E_l]=i$. I can use either the $2\pi$-periodic vector potential or the electric field to label the links of the cubic lattice and they correspond to two different bases of the Hilbert space of the theory.

The charges in this theory is very easy to represent. In particular, they can be defined in the electric field line basis by the Gauss' law: \begin{equation} q_i = \sum_{<ij>}E_{ij} \end{equation} Then I can add a $\sum_{\text{sites} \ i}q_i^2$ term to the above Hamiltonian, and consider a charge to be an excitation of the theory living on the lattice sites.

We also know that a 3+1D compact U(1) gauge theory hosts monopoles.

My question is: How should we represent a monopole in the lattice description?

Naively, one can mimic the definition of an electric charge and use the following Gauss's law to define a monopole: \begin{equation} m_{\tilde{i}\ \text{on the dual lattice}} = \sum_{\text{plaquetttes}\ P \ containing\ \tilde{i}} \Phi_P \end{equation} Then I can add a $\sum_{\text{sites on the dual lattice} \ \tilde{i}}m_{\tilde i}^2$ term to the above Hamiltonian, and consider a monopole to be an excitation of the theory living on the dual lattice sites. But there seems to be something wrong with this guess, because the following identity always holds: \begin{equation} \sum_{\text{boundary plaquettes}\ P \ \text{of a cube}} \Phi_P = 0. \end{equation} This can be verified by applying a discrete version of the Stokes' theorem ($\int_D B = \int_{\partial D} A$), $\Phi_P = \sum_{\text{links}\ l \ \text{of}\ P} a_l$, and the sum of fluxes over the boundary plaquettes of a cube reduces to the sum of gauge fields over all links the cube, with each gauge field summed over twice, once with each sign, and the sum vanishes. So it seems that the monopoles are always absent in the theory.

$\endgroup$
  • $\begingroup$ One way to do it is the following: instead of defining F=dA (the lattice version), write F = dA + S, where *dS=m and m is the (integer valued) monopole current. This is kind of expected since in continuum monopole means A is not globally well-defined. $\endgroup$ – Meng Cheng Jan 9 '17 at 2:06
  • $\begingroup$ Hi Meng, Thanks! I also find this reference journals.aps.org/prd/pdf/10.1103/PhysRevD.22.2478. In section III (eqn 3.2), it defines $F = dA - 2\pi n$, where $n$ is the number of Dirac strings through the plaquette. I think this is essentially what you wrote down above. $\endgroup$ – Zitao Wang Jan 9 '17 at 2:26
  • $\begingroup$ But I'm still a little bit confused because even though the vector potential cannot be defined globally, it can still be defined locally (I'm having the Wu-Yang monopole in mind). So the presence of monopole should still modify the vector potential somehow, but this cannot be reflected at the lattice level, because no matter how I redefine the vector potential $a_{ij}^{\prime} = a_{ij}+\delta_{ij}$ and calculate the modified $F$ for $a_{ij}^{\prime}$, the result would always be zero. $\endgroup$ – Zitao Wang Jan 9 '17 at 2:27
  • $\begingroup$ For example, in eqn 11 of journals.aps.org/prd/pdf/10.1103/PhysRevD.12.3845. The vector potential would give you the magnetic field of a monopole. Is it possible to write down this vector potential at the lattice level? Then we can still use $F^{\prime} = dA^{\prime}$. $\endgroup$ – Zitao Wang Jan 9 '17 at 2:39
  • $\begingroup$ I'm not sure what your "n" means exactly, in the expression I wrote to actually find S I need to invert the kernel, and probably end up with some highly non-local-looking expression for S. I somehow think this is the best one can do on lattice... $\endgroup$ – Meng Cheng Jan 9 '17 at 4:00
1
$\begingroup$

But there seems to be something wrong with this guess, because the following identity always holds: \begin{equation} \sum_{\text{boundary plaquettes}\ P \ \text{of a cube}} \Phi_P = 0. \end{equation} This can be verified by applying a discrete version of the Stokes' theorem ($\int_D B = \int_{\partial D} A$)$\Phi_P = \sum_{\text{links}\ l \ \text{of}\ P} a_l$, and the sum of fluxes over the boundary plaquettes of a cube reduces to the sum of gauge fields over all links the cube, with each gauge field summed over twice, once with each sign, and the sum vanishes. So it seems that the monopoles are always absent in the theory.

Recall that in the continuum the monople is precisely an object that violates Stokes' theorem. It does so because $A$ is compact and non-single-valued. On a lattice it violates the discrete version of the Stokes' theorem by the same token. So by definition you cannot use discrete Stokes' theorem to argue for its absence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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