Momentum of $A_0$-field in non-abelian gauge theory

Question

My question: why the momentum of the $A_0$-field in non abelian gauge theories is $0$?

In my class on gauge theories, we end up with this Lagrangian:

$$ \mathcal{L}=F_a^{\mu \nu}F^a_{\mu \nu} $$

Where:

$$F_a^{\mu \nu}=\partial^{\mu} A_a^{\nu}-\partial^\mu A_a^\mu-g f^{abc} A_b^\mu A_c^\nu$$

At order $0$ in $g$, we find the same term than the abelian case.

At order $1$ in $g$, we find :

$$ 2g f^{abc}(\partial_\mu A_\nu^a-\partial_\nu A_\mu^a)A^\mu_b A^\nu_b$$

Where $f^{abc}$ are totally antisymmetric : $[T_a,T_b]=if_{abc}T_c$ where $T_k$ are the generators of the Lie algebra.

My question is :

In my course, it is written that we have the same problem about canonical quantization here than in abelian theories : $\pi_0^a=0$ (the momentum associated to the fields is $0$).

For the order $0$ I totally agree bc it has the same form as in the Abelian theory.

But for the order $1$ it is not $0$ ?

Indeed :

$$B=f^{abc}(\partial_\mu A_\nu^a-\partial_\nu A_\mu^a)A^\mu_b A^\nu_b$$

$$ \frac{\partial B}{\partial \dot{A}^a_{\beta}}=f^{abc}A^0_b A^{\beta}_c-f^{abc}A_b^{\beta}A^0_c=f^{abc}A^0_b A^{\beta}_c-f^{acb}A_c^{\beta}A^0_b=A^0_b A^{\beta}_c(f^{abc}-f^{acb})\neq 0$$

It is not $0$ because $f^{abc}$ is totally antisymmetric...

Thus, why in non Abelian gauge theories we have the same problem about canonical quantization than in the Abelian theories ?

Answer 1

Not all of the canonical field momenta in the Abelian gauge field are zero. The electromagnetic Lagrangian (the prototypical example of an Abelian field) is \begin{align} \mathcal{L} & = -\frac{1}{4} F^{\mu\nu}F_{\mu\nu} \\ & = \frac{1}{2} \vec{E}\cdot\vec{E} - \frac{1}{2}\vec{B}\cdot\vec{B} \\ & = \frac{1}{2} \left(\partial_i A^0 + \partial_0 A^i\right)\left(\partial_i A^0 + \partial_0 A^i\right) - \frac{1}{2}\left(\epsilon_{ijk}\partial_j A^k\right)\left(\epsilon_{ilm}\partial_l A^m\right) \end{align} In Heaviside-Lorentz units with $c=1$ and metric signature $(+,-,-,-)$. The field momentum canonically conjugate to $A_\mu$ is given by \begin{align} \Pi^\mu & \equiv \eta^{\mu\nu}\frac{\partial \mathcal{L}}{\partial \left(\partial_0 A^\nu\right)} \Rightarrow\\ \Pi^i & = -\partial_i A^0 - \partial_0 A^i. \end{align} Note that $\Pi^0$ is identically $0$ because $\partial_0 A^0$ doesn't even appear in the Lagrangian. For this reason, $A^0$ can be regarded as a non-dynamical, or auxiliary, field. This is a manifestation of the Gauge symmetry - there is an entire field worth of freedom in $A^\mu$ that has no inertia, and can therefore take any value the problem needs it to have at any time. The form that these unconstrained degrees of freedom take is set by the gauge fixing condition used. This can be seen in gauges that are frequently fixed by adding a term to the Lagrangian that can add $\partial_0 A^0$ terms to the Lagrangian (e.g. $R_\xi$ gauges).

Non-Abelian fields act like multiple interacting Abelian fields, so this picture of the Lagrangian not depending on $\partial_0A^0$ extends to those fields, too.

All that said, you could imagine integrating the cross terms from expanding $\vec{E}^2$ by parts to swap the space and time derivatives to get $$\mathcal{L} = \frac{1}{2} \partial_iA^0 \partial_iA^0 + \partial_0A^0 \partial_iA^i + \frac{1}{2} \partial_0A^i \partial_0A^i + \vec{B}\cdot\vec{B} + (\mathrm{surface\ terms}).$$ I haven't actually looked at this last Lagrangian (I literally just derived it), but I don't think it would say anything particularly interesting since it differs from the Lagrangian without $\partial_0A^0$ by a couple of surface terms ($-\partial_0\left[A^0\partial_iA^i\right]+\partial_i\left[A^0\partial_0A^i\right]$), and is thus equivalent. Odd to note, though, that in this Lagrangian $\Pi^0=\nabla\cdot\vec{A}$ is exactly what is set to zero in the Coulomb gauge.