# Gauss' law from Hamiltonian density of electromagnetic field

I am going through David Tong's QFT course, for which lecture notes and exercises are available online at http://www.damtp.cam.ac.uk/user/tong/qft.html.

In Question 1.8 we have the Lagrangian (density) $$L = -\frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \frac{1}{2} m^2 C_\mu C^\mu,$$ $$F_{\mu \nu} = \partial_\mu C_\nu - \partial_\nu C_\mu,$$ which is like the standard electromagnetic field in the case $$m=0$$. I eventually derive the conjugate momenta $$\Pi_\mu$$ to $$C_\mu$$ and convert the Lagrangian to a Hamiltonian $$H = -\frac{1}{2} \Pi_i \Pi^i + \frac{1}{4}F^{ij}F_{ij} - \frac{1}{2} m^2 C^{\mu}C_{\mu} - \Pi_i \partial^i C_0,$$ answering the question.

However in a pdf of tutor's solutions I came across online (which I maybe shouldn't link), the tutor comments and interprets further: they rearrange the last term, $$H = -\frac{1}{2} \Pi_i \Pi^i + \frac{1}{4}F^{ij}F_{ij} - \frac{1}{2} m^2 C^{\mu}C_{\mu} - C^0(\partial_i \Pi^i) - \partial_i (\Pi^i C^0),$$ and comment

[the term] involves an irrelevant three-divergence term. Since the remainder of the Hamiltonian contains no derivatives in $$C^0$$, $$C^0$$ may be regarded as a multiplier that, in the $$m=0$$ theory, imposes the constraint $$\nabla \cdot \Pi = m^2 C^0 = 0$$, which is precisely Gauss' law.

Since we are back to examining the $$m=0$$ case, this is a statement about the standard electromagnetic field.

I don't understand either statement here.

How is $$\partial_i (\Pi^i C^0)$$ "irrelevant"? Can we just ignore this divergence, which as far as I can see has a nonzero value?

$$- C^0(\partial_i \Pi^i)$$ could be a (Lagrange) multiplier, how is it rearranged to include the $$m^2$$ term and (together) constrain to $$\nabla \cdot \Pi=0$$?

What you found is actually the Hamiltonian Density $$\mathcal{H}$$. The Hamiltonian is the spatial integral of Hamiltonian Density

$$\begin{equation} H=\int_\mathcal{M} d^Nx\,\mathcal{H}(x). \end{equation}$$

If you have an spatial divergence in your Hamiltonian Density, its contribution to the Hamiltonian can be converted into a hypersurface integral using divergence theorem

$$\begin{equation} \int_\mathcal{M} d^Nx\,\partial_i\Pi^i=\int_{\partial\mathcal{M}}d\sigma\,n_i \Pi^i, \end{equation}$$

where $$\partial\mathcal{M}$$ is the boundary hypersurface, and $$n_i$$ is the normal vector to it. As that boundary is in infinite, where $$\Pi^i$$ must vanish, the contribution of the $$\partial_i\Pi^i$$ term to the Hamiltonian Density is nule.

The Lagrange multiplier would be just $$C_0$$ in the $$m^2=0$$ case, enforcing $$\partial_i \Pi^i=0$$. When $$m^2\neq0$$ that constraint is not true.

Main points:

• A total spacetime divergence in the Lagrangian (or Hamiltonian) does not change the EOMs, cf. e.g. this Phys.SE post.

• If we know that the fields vanishes on the boundary, e.g. by imposing pertinent boundary conditions, we can use the divergence theorem to argue that a divergence term cannot contribute to, say the EM energy.

• The EOM for $$C_0$$ reads $$\nabla \cdot \Pi = m^2 C^0$$.

• In the massless limit $$m=0$$, this EOM becomes Gauss's law in vacuum.