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Gauging away the constant gauge field

+ 3 like - 0 dislike
153 views

In few papers (see, for example, here, the bottom of the left column on the page 6, or here, the upper part of the page 5) I've met the strange calculations using the constant gauge field

$$
A_{\mu}(x) = (0,0,0,A_{3} = \text{const}), \quad\text{or}\quad A_{\mu}(x) = (A_{0} = \text{const},0,0,0)
$$
By using these fields authors obtain observable effects like chiral and vector currents.

One might think that these constant gauge fields can be gauged away, but the authors of the first linked article say that (at least about constant $A_{3}$)

One might think that a constant gauge field could be gauged away, but this is not possible by a gauge transformation satisfying the periodic boundary condition.

I don't understand this statement. Could You clarify it? Also I don't understand what is the problem with $A_{0} = \text{const}$.

asked Nov 25, 2016 in Theoretical Physics by NAME_XXX (1,010 points) [ revision history ]
recategorized Nov 25, 2016 by Dilaton

Apparently it is not a usual (or purely) gauge field since it must satisfy some boundary conditions.

Imagine there is no equation for $A_{\mu}(x)$, but there are boundary conditions $A_{\mu}(0)=a,\;A_{\mu}(L)=a$. ($\mu$ is fixed.) At least $A_{\mu}(x)=a$ satisfies the boundary conditions. It is a momentum contribution, hence current contribution too.

@VladimirKalitvianski : thank You. But what about constant $A_{0}$?

It was just my guess. It remains valid for $A_0$. Some energy contribution, maybe an energy gap somewhere.

@VladimirKalitvianski : it seems that this is entirely true for the case of there are temporary boundary conditions, say
$$
A_{\mu}(\mathbf r,  t) = A_{\mu}(\mathbf r, t + T)
​$$
​However, it seems that in the linked article there are no such conditions. The author just writes that this field is

if it's periodic you can put the field on a circle and compute holonomy (Wilson loop), which is gauge invariant.

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