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In Yang-Mills theory with periodic boundary conditions in time, is the temporale gauge, i.e. $A_0 = 0$, well defined? Periodic boundary conditions would be $$A_\mu(T_2,x) = A_\mu(T_1,x).$$

Naively I would say that the gauge is not well defined because enforcing the periodicity of gauge tranformations would make a solution of the condition $$U A_0 U^\dagger+ i U \partial_0 U^\dagger = 0$$ impossible, since: $$U(t,x) = P \exp\left(i\int_{T_1}^t A_0dx_0\right) U(T_1,x)$$ then enforcing $U(T_2,x) = U(T_1,x)$ is not possible.

Am I getting this wrong, is there some subtlety that I am missing? Maybe gauge transformations don't need to be periodic? But if so why?

Qmechanic
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Fra
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1 Answers1

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The problem is that your definition of the $U$ operator is incorrect.

The $U$ operators satisfy in Yang-Mills theory:

$U(t,x) = e^{ig\theta(t,x)^aT^a}$

With $T^a$ the generators of the algebra in the correct representation.

For example take the U(1) EM theory: then $T^a = 1$ and $U(t,x) = e^{ie\theta(t,x)}$

Take the SU(2) theory in the fundamental representation: You have that $T^a = \sigma^a/2$ and then

$U(t,x) = e^{\big(i\frac{g}{2}(\theta(t,x)^1\sigma^1 + \theta(t,x)^2\sigma^2+\theta(t,x)^3\sigma^3)\big)}$

You see that in the simple $U(1)$ case that

$U(T_1,x) = e^{ie\theta(T_1,x)} = e^{ie\theta(T_2,x)} = U(T_2,x)$

The gauge equation then reads:

$U A_0 U^\dagger = 0$ since $A_0 = 0$

And

$\frac{i}{e}U \partial_0 U^\dagger = \frac{i}{e}e^{ie\theta(t,x)}(-ie)e^{-ie\theta(t,x)}\frac{\partial\theta(t,x)}{\partial t} = \frac{\partial\theta(t,x)}{\partial t} $

The gauge conditions then read:

$\frac{\partial\theta(t,x)}{\partial t} = 0$

Which implies that $\theta$ does not depend on time, so:

$\theta(T_1, x) = \theta(T_2,x) =\theta(t,x) \forall t$

You can do the same thing for each group, it just takes more derivative and more steps to solve the equation for the gauge condition.

Be also careful because for other groups you have that:

$A_\mu = A_{\mu}^a T^a$

So you have to set to zero everyone of the $N$ gauge fields $A_{0}^a$

LolloBoldo
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